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Some Properties of a Triangle Questions in English
Class 9 Mathematics · Triangles · Some Properties of a Triangle
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51
Medium
$S$ is any point in the interior of $\triangle PQR$. Show that $SQ + SR < PQ + PR$.
Solution
(N/A) Produce $QS$ to intersect $PR$ at $T$.
From $\triangle PQT$,we have:
$PQ + PT > QT$ (The sum of any two sides of a triangle is greater than the third side)
$PQ + PT > SQ + ST$ ..... $(1)$
From $\triangle TSR$,we have:
$ST + TR > SR$ ..... $(2)$
Adding $(1)$ and $(2)$,we get:
$PQ + PT + ST + TR > SQ + ST + SR$
$PQ + PT + TR > SQ + SR$
Since $PT + TR = PR$,we have:
$PQ + PR > SQ + SR$
Therefore,$SQ + SR < PQ + PR$.
From $\triangle PQT$,we have:
$PQ + PT > QT$ (The sum of any two sides of a triangle is greater than the third side)
$PQ + PT > SQ + ST$ ..... $(1)$
From $\triangle TSR$,we have:
$ST + TR > SR$ ..... $(2)$
Adding $(1)$ and $(2)$,we get:
$PQ + PT + ST + TR > SQ + ST + SR$
$PQ + PT + TR > SQ + SR$
Since $PT + TR = PR$,we have:
$PQ + PR > SQ + SR$
Therefore,$SQ + SR < PQ + PR$.
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View Solution52
MediumMCQ
Find all the angles of an equilateral triangle.
A
$60^{\circ},60^{\circ},60^{\circ}$
B
$45^{\circ},45^{\circ},90^{\circ}$
C
$30^{\circ},60^{\circ},90^{\circ}$
D
$50^{\circ},60^{\circ},70^{\circ}$
Solution
(A) In an equilateral triangle,all three sides are equal,i.e.,$AB = BC = AC$.
Since the angles opposite to equal sides of a triangle are equal,we have $\angle A = \angle B = \angle C$.
Let each angle be $x$.
By the angle sum property of a triangle,the sum of all interior angles is $180^{\circ}$.
Therefore,$\angle A + \angle B + \angle C = 180^{\circ}$.
$x + x + x = 180^{\circ}$.
$3x = 180^{\circ}$.
$x = \frac{180^{\circ}}{3} = 60^{\circ}$.
Thus,all the angles of an equilateral triangle are $60^{\circ}, 60^{\circ}, 60^{\circ}$.
Since the angles opposite to equal sides of a triangle are equal,we have $\angle A = \angle B = \angle C$.
Let each angle be $x$.
By the angle sum property of a triangle,the sum of all interior angles is $180^{\circ}$.
Therefore,$\angle A + \angle B + \angle C = 180^{\circ}$.
$x + x + x = 180^{\circ}$.
$3x = 180^{\circ}$.
$x = \frac{180^{\circ}}{3} = 60^{\circ}$.
Thus,all the angles of an equilateral triangle are $60^{\circ}, 60^{\circ}, 60^{\circ}$.
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View Solution53
Medium
$P$ is a point on the bisector of $\angle ABC$. If the line through $P$,parallel to $BA$,meets $BC$ at $Q$,prove that $\triangle BPQ$ is an isosceles triangle.
Solution
(N/A) We have to prove that $\triangle BPQ$ is an isosceles triangle.
Given that $BP$ is the bisector of $\angle ABC$,we have:
$\angle 1 = \angle 2$ .....$(1)$
Since $PQ \parallel BA$ and $BP$ is a transversal,the alternate interior angles are equal:
$\angle 1 = \angle 3$ .....$(2)$
From equations $(1)$ and $(2)$,we get:
$\angle 2 = \angle 3$
In $\triangle BPQ$,since $\angle 2 = \angle 3$,the sides opposite to these equal angles must be equal:
$PQ = BQ$
Since two sides of $\triangle BPQ$ are equal,$\triangle BPQ$ is an isosceles triangle.
Given that $BP$ is the bisector of $\angle ABC$,we have:
$\angle 1 = \angle 2$ .....$(1)$
Since $PQ \parallel BA$ and $BP$ is a transversal,the alternate interior angles are equal:
$\angle 1 = \angle 3$ .....$(2)$
From equations $(1)$ and $(2)$,we get:
$\angle 2 = \angle 3$
In $\triangle BPQ$,since $\angle 2 = \angle 3$,the sides opposite to these equal angles must be equal:
$PQ = BQ$
Since two sides of $\triangle BPQ$ are equal,$\triangle BPQ$ is an isosceles triangle.
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View Solution54
Difficult
In a triangle $ABC$,$D$ is the mid-point of side $AC$ such that $BD = \frac{1}{2} AC$. Show that $\angle ABC$ is a right angle.
Solution
(N/A) We have to prove that $\angle ABC = 90^{\circ}$.
Since $D$ is the mid-point of $AC$,we have $AD = DC$.
Also,it is given that $BD = \frac{1}{2} AC$.
Since $AC = AD + DC = 2AD$,we have $BD = AD$.
Therefore,$BD = AD = DC$.
In $\Delta ABD$,since $BD = AD$,the angles opposite to equal sides are equal,so $\angle 1 = \angle 2$. $(1)$
In $\Delta BCD$,since $BD = DC$,the angles opposite to equal sides are equal,so $\angle 3 = \angle 4$. $(2)$
In $\Delta ABC$,the sum of angles is $180^{\circ}$:
$\angle A + \angle ABC + \angle C = 180^{\circ}$
$\angle 1 + (\angle 2 + \angle 3) + \angle 4 = 180^{\circ}$
Substituting $\angle 1 = \angle 2$ and $\angle 4 = \angle 3$ from $(1)$ and $(2)$:
$\angle 2 + (\angle 2 + \angle 3) + \angle 3 = 180^{\circ}$
$2(\angle 2 + \angle 3) = 180^{\circ}$
$\angle 2 + \angle 3 = 90^{\circ}$
Since $\angle ABC = \angle 2 + \angle 3$,we have $\angle ABC = 90^{\circ}$.
Hence proved.
Since $D$ is the mid-point of $AC$,we have $AD = DC$.
Also,it is given that $BD = \frac{1}{2} AC$.
Since $AC = AD + DC = 2AD$,we have $BD = AD$.
Therefore,$BD = AD = DC$.
In $\Delta ABD$,since $BD = AD$,the angles opposite to equal sides are equal,so $\angle 1 = \angle 2$. $(1)$
In $\Delta BCD$,since $BD = DC$,the angles opposite to equal sides are equal,so $\angle 3 = \angle 4$. $(2)$
In $\Delta ABC$,the sum of angles is $180^{\circ}$:
$\angle A + \angle ABC + \angle C = 180^{\circ}$
$\angle 1 + (\angle 2 + \angle 3) + \angle 4 = 180^{\circ}$
Substituting $\angle 1 = \angle 2$ and $\angle 4 = \angle 3$ from $(1)$ and $(2)$:
$\angle 2 + (\angle 2 + \angle 3) + \angle 3 = 180^{\circ}$
$2(\angle 2 + \angle 3) = 180^{\circ}$
$\angle 2 + \angle 3 = 90^{\circ}$
Since $\angle ABC = \angle 2 + \angle 3$,we have $\angle ABC = 90^{\circ}$.
Hence proved.
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View Solution55
Difficult
Prove that in a triangle,other than an equilateral triangle,the angle opposite to the longest side is greater than $\frac{2}{3}$ of a right angle.
Solution
(N/A) Given: $A$ triangle $ABC$,which is not an equilateral triangle. Let $BC$ be the longest side.
To prove: $\angle A > \frac{2}{3} \times 90^{\circ} = 60^{\circ}$.
Proof: In $\Delta ABC$,since $BC$ is the longest side,we have:
$BC > AB \Rightarrow \angle A > \angle C$ ..... $(1)$ [Since the angle opposite to the longer side is larger]
$BC > AC \Rightarrow \angle A > \angle B$ ..... $(2)$ [Since the angle opposite to the longer side is larger]
Adding $(1)$ and $(2)$,we get:
$\angle A + \angle A > \angle B + \angle C$
$2\angle A > \angle B + \angle C$
Adding $\angle A$ on both sides:
$2\angle A + \angle A > \angle A + \angle B + \angle C$
$3\angle A > 180^{\circ}$ [Angle sum property of a triangle]
$\angle A > \frac{180^{\circ}}{3}$
$\angle A > 60^{\circ}$
Since $60^{\circ} = \frac{2}{3} \times 90^{\circ}$,we have $\angle A > \frac{2}{3}$ of a right angle.
Hence,proved.
To prove: $\angle A > \frac{2}{3} \times 90^{\circ} = 60^{\circ}$.
Proof: In $\Delta ABC$,since $BC$ is the longest side,we have:
$BC > AB \Rightarrow \angle A > \angle C$ ..... $(1)$ [Since the angle opposite to the longer side is larger]
$BC > AC \Rightarrow \angle A > \angle B$ ..... $(2)$ [Since the angle opposite to the longer side is larger]
Adding $(1)$ and $(2)$,we get:
$\angle A + \angle A > \angle B + \angle C$
$2\angle A > \angle B + \angle C$
Adding $\angle A$ on both sides:
$2\angle A + \angle A > \angle A + \angle B + \angle C$
$3\angle A > 180^{\circ}$ [Angle sum property of a triangle]
$\angle A > \frac{180^{\circ}}{3}$
$\angle A > 60^{\circ}$
Since $60^{\circ} = \frac{2}{3} \times 90^{\circ}$,we have $\angle A > \frac{2}{3}$ of a right angle.
Hence,proved.
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View Solution56
Medium
Prove that a triangle must have at least two acute angles.
Solution
(N/A) Let the angles of a triangle be $\angle A, \angle B,$ and $\angle C$. According to the angle sum property of a triangle,$\angle A + \angle B + \angle C = 180^{\circ}$.
Case $1$: If the triangle is an acute-angled triangle,all three angles are less than $90^{\circ}$. Thus,it has three acute angles.
Case $2$: If the triangle is a right-angled triangle,one angle is $90^{\circ}$. Let $\angle A = 90^{\circ}$. Then $\angle B + \angle C = 180^{\circ} - 90^{\circ} = 90^{\circ}$. Since the sum of two positive angles is $90^{\circ}$,both $\angle B$ and $\angle C$ must be less than $90^{\circ}$,meaning they are both acute.
Case $3$: If the triangle is an obtuse-angled triangle,one angle is greater than $90^{\circ}$. Let $\angle A > 90^{\circ}$. Then $\angle B + \angle C = 180^{\circ} - \angle A$. Since $\angle A > 90^{\circ}$,$\angle B + \angle C < 90^{\circ}$. This implies that both $\angle B$ and $\angle C$ must be less than $90^{\circ}$,meaning they are both acute.
Conclusion: In all possible cases,a triangle must have at least two acute angles.
Case $1$: If the triangle is an acute-angled triangle,all three angles are less than $90^{\circ}$. Thus,it has three acute angles.
Case $2$: If the triangle is a right-angled triangle,one angle is $90^{\circ}$. Let $\angle A = 90^{\circ}$. Then $\angle B + \angle C = 180^{\circ} - 90^{\circ} = 90^{\circ}$. Since the sum of two positive angles is $90^{\circ}$,both $\angle B$ and $\angle C$ must be less than $90^{\circ}$,meaning they are both acute.
Case $3$: If the triangle is an obtuse-angled triangle,one angle is greater than $90^{\circ}$. Let $\angle A > 90^{\circ}$. Then $\angle B + \angle C = 180^{\circ} - \angle A$. Since $\angle A > 90^{\circ}$,$\angle B + \angle C < 90^{\circ}$. This implies that both $\angle B$ and $\angle C$ must be less than $90^{\circ}$,meaning they are both acute.
Conclusion: In all possible cases,a triangle must have at least two acute angles.
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View Solution57
EasyMCQ
The number of interwoven $\ldots \ldots$ triangles in Sriyantra (in the Atharvaveda) is nine.
A
equilateral
B
isosceles
C
scalene
D
right-angled
Solution
(B) The Sriyantra is a sacred geometric pattern used in meditation and worship. It consists of nine interwoven triangles. These triangles are specifically isosceles triangles,which are arranged in such a way that they create a complex geometric structure representing the union of masculine and feminine divine energy.
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View Solution58
MediumMCQ
In $\Delta ABC$,$\angle A : \angle B : \angle C = 5 : 7 : 8$. Find the measure of each angle of $\Delta ABC$.
A
$45^{\circ}, 63^{\circ}, 72^{\circ}$
B
$50^{\circ}, 70^{\circ}, 60^{\circ}$
C
$40^{\circ}, 80^{\circ}, 60^{\circ}$
D
$30^{\circ}, 90^{\circ}, 60^{\circ}$
Solution
(A) In $\Delta ABC$,the sum of the interior angles is $\angle A + \angle B + \angle C = 180^{\circ}$.
Given the ratio $\angle A : \angle B : \angle C = 5 : 7 : 8$,let the angles be $5x, 7x,$ and $8x$.
Sum of the angles: $5x + 7x + 8x = 180^{\circ}$.
$20x = 180^{\circ}$.
$x = \frac{180^{\circ}}{20} = 9^{\circ}$.
Therefore,the angles are:
$\angle A = 5 \times 9^{\circ} = 45^{\circ}$.
$\angle B = 7 \times 9^{\circ} = 63^{\circ}$.
$\angle C = 8 \times 9^{\circ} = 72^{\circ}$.
Given the ratio $\angle A : \angle B : \angle C = 5 : 7 : 8$,let the angles be $5x, 7x,$ and $8x$.
Sum of the angles: $5x + 7x + 8x = 180^{\circ}$.
$20x = 180^{\circ}$.
$x = \frac{180^{\circ}}{20} = 9^{\circ}$.
Therefore,the angles are:
$\angle A = 5 \times 9^{\circ} = 45^{\circ}$.
$\angle B = 7 \times 9^{\circ} = 63^{\circ}$.
$\angle C = 8 \times 9^{\circ} = 72^{\circ}$.
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View Solution59
MediumMCQ
In $\Delta PQR$,$\angle P : \angle Q : \angle R = 5 : 2 : 2$. Find the measure of each angle of $\Delta PQR$.
A
$\angle P = 100^{\circ}, \angle Q = 40^{\circ}, \angle R = 40^{\circ}$
B
$\angle P = 80^{\circ}, \angle Q = 50^{\circ}, \angle R = 50^{\circ}$
C
$\angle P = 90^{\circ}, \angle Q = 45^{\circ}, \angle R = 45^{\circ}$
D
$\angle P = 110^{\circ}, \angle Q = 35^{\circ}, \angle R = 35^{\circ}$
Solution
(A) Let the angles of $\Delta PQR$ be $5x$,$2x$,and $2x$ respectively.
According to the angle sum property of a triangle,the sum of all interior angles is $180^{\circ}$.
Therefore,$5x + 2x + 2x = 180^{\circ}$.
$9x = 180^{\circ}$.
$x = 20^{\circ}$.
Now,calculating each angle:
$\angle P = 5x = 5 \times 20^{\circ} = 100^{\circ}$.
$\angle Q = 2x = 2 \times 20^{\circ} = 40^{\circ}$.
$\angle R = 2x = 2 \times 20^{\circ} = 40^{\circ}$.
Thus,the angles are $100^{\circ}, 40^{\circ}, 40^{\circ}$.
According to the angle sum property of a triangle,the sum of all interior angles is $180^{\circ}$.
Therefore,$5x + 2x + 2x = 180^{\circ}$.
$9x = 180^{\circ}$.
$x = 20^{\circ}$.
Now,calculating each angle:
$\angle P = 5x = 5 \times 20^{\circ} = 100^{\circ}$.
$\angle Q = 2x = 2 \times 20^{\circ} = 40^{\circ}$.
$\angle R = 2x = 2 \times 20^{\circ} = 40^{\circ}$.
Thus,the angles are $100^{\circ}, 40^{\circ}, 40^{\circ}$.
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View Solution60
MediumMCQ
In $\Delta ABC$,$\angle A + \angle B = 80^{\circ}$ and $\angle B + \angle C = 150^{\circ}$. Find the measure of each angle of $\Delta ABC$.
A
$\angle A = 30^{\circ}, \angle B = 50^{\circ}, \angle C = 100^{\circ}$
B
$\angle A = 40^{\circ}, \angle B = 40^{\circ}, \angle C = 100^{\circ}$
C
$\angle A = 20^{\circ}, \angle B = 60^{\circ}, \angle C = 90^{\circ}$
D
$\angle A = 50^{\circ}, \angle B = 30^{\circ}, \angle C = 120^{\circ}$
Solution
(A) We know that the sum of all angles in a triangle is $180^{\circ}$,so $\angle A + \angle B + \angle C = 180^{\circ}$.
Given: $\angle A + \angle B = 80^{\circ}$ and $\angle B + \angle C = 150^{\circ}$.
Substituting $\angle A + \angle B = 80^{\circ}$ into the sum equation: $80^{\circ} + \angle C = 180^{\circ} \implies \angle C = 180^{\circ} - 80^{\circ} = 100^{\circ}$.
Substituting $\angle C = 100^{\circ}$ into $\angle B + \angle C = 150^{\circ}$: $\angle B + 100^{\circ} = 150^{\circ} \implies \angle B = 50^{\circ}$.
Substituting $\angle B = 50^{\circ}$ into $\angle A + \angle B = 80^{\circ}$: $\angle A + 50^{\circ} = 80^{\circ} \implies \angle A = 30^{\circ}$.
Thus,the angles are $\angle A = 30^{\circ}, \angle B = 50^{\circ}, \angle C = 100^{\circ}$.
Given: $\angle A + \angle B = 80^{\circ}$ and $\angle B + \angle C = 150^{\circ}$.
Substituting $\angle A + \angle B = 80^{\circ}$ into the sum equation: $80^{\circ} + \angle C = 180^{\circ} \implies \angle C = 180^{\circ} - 80^{\circ} = 100^{\circ}$.
Substituting $\angle C = 100^{\circ}$ into $\angle B + \angle C = 150^{\circ}$: $\angle B + 100^{\circ} = 150^{\circ} \implies \angle B = 50^{\circ}$.
Substituting $\angle B = 50^{\circ}$ into $\angle A + \angle B = 80^{\circ}$: $\angle A + 50^{\circ} = 80^{\circ} \implies \angle A = 30^{\circ}$.
Thus,the angles are $\angle A = 30^{\circ}, \angle B = 50^{\circ}, \angle C = 100^{\circ}$.
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View Solution61
MediumMCQ
In $\Delta ABC$,$\angle B = \frac{\angle A + \angle C}{2}$ and $\angle A : \angle C = 1 : 2$. Find the measure of each angle of $\Delta ABC$.
A
$\angle A = 40^{\circ}, \angle B = 60^{\circ}, \angle C = 80^{\circ}$
B
$\angle A = 30^{\circ}, \angle B = 60^{\circ}, \angle C = 90^{\circ}$
C
$\angle A = 50^{\circ}, \angle B = 60^{\circ}, \angle C = 70^{\circ}$
D
$\angle A = 20^{\circ}, \angle B = 60^{\circ}, \angle C = 100^{\circ}$
Solution
(A) We know that the sum of angles in a triangle is $180^{\circ}$,so $\angle A + \angle B + \angle C = 180^{\circ}$.
Given $\angle B = \frac{\angle A + \angle C}{2}$,which implies $2\angle B = \angle A + \angle C$.
Substituting this into the sum equation: $2\angle B + \angle B = 180^{\circ} \implies 3\angle B = 180^{\circ} \implies \angle B = 60^{\circ}$.
Now,$\angle A + \angle C = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
Given the ratio $\angle A : \angle C = 1 : 2$,let $\angle A = x$ and $\angle C = 2x$.
Then $x + 2x = 120^{\circ} \implies 3x = 120^{\circ} \implies x = 40^{\circ}$.
Therefore,$\angle A = 40^{\circ}$ and $\angle C = 2(40^{\circ}) = 80^{\circ}$.
The angles are $\angle A = 40^{\circ}, \angle B = 60^{\circ}, \angle C = 80^{\circ}$.
Given $\angle B = \frac{\angle A + \angle C}{2}$,which implies $2\angle B = \angle A + \angle C$.
Substituting this into the sum equation: $2\angle B + \angle B = 180^{\circ} \implies 3\angle B = 180^{\circ} \implies \angle B = 60^{\circ}$.
Now,$\angle A + \angle C = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
Given the ratio $\angle A : \angle C = 1 : 2$,let $\angle A = x$ and $\angle C = 2x$.
Then $x + 2x = 120^{\circ} \implies 3x = 120^{\circ} \implies x = 40^{\circ}$.
Therefore,$\angle A = 40^{\circ}$ and $\angle C = 2(40^{\circ}) = 80^{\circ}$.
The angles are $\angle A = 40^{\circ}, \angle B = 60^{\circ}, \angle C = 80^{\circ}$.
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View Solution62
MediumMCQ
Side $QR$ of $\Delta PQR$ is extended on both the sides so that exterior angles $\angle PQS$ and $\angle PRT$ are formed. If $\angle PQS = 100^{\circ}$ and $\angle PRT = 130^{\circ}$,find the measure of each angle of $\Delta PQR$.
A
$\angle P = 50^{\circ}, \angle Q = 80^{\circ}, \angle R = 50^{\circ}$
B
$\angle P = 60^{\circ}, \angle Q = 70^{\circ}, \angle R = 50^{\circ}$
C
$\angle P = 70^{\circ}, \angle Q = 60^{\circ}, \angle R = 50^{\circ}$
D
$\angle P = 50^{\circ}, \angle Q = 70^{\circ}, \angle R = 60^{\circ}$
Solution
(A) Since $S-Q-R-T$ is a straight line,$\angle PQS$ and $\angle PQR$ form a linear pair. Therefore,$\angle PQR = 180^{\circ} - 100^{\circ} = 80^{\circ}$.
Similarly,$\angle PRT$ and $\angle PRQ$ form a linear pair. Therefore,$\angle PRQ = 180^{\circ} - 130^{\circ} = 50^{\circ}$.
In $\Delta PQR$,the sum of all interior angles is $180^{\circ}$.
Thus,$\angle P + \angle Q + \angle R = 180^{\circ}$.
$\angle P + 80^{\circ} + 50^{\circ} = 180^{\circ}$.
$\angle P + 130^{\circ} = 180^{\circ}$.
$\angle P = 180^{\circ} - 130^{\circ} = 50^{\circ}$.
Hence,the angles are $\angle P = 50^{\circ}, \angle Q = 80^{\circ}, \angle R = 50^{\circ}$.
Similarly,$\angle PRT$ and $\angle PRQ$ form a linear pair. Therefore,$\angle PRQ = 180^{\circ} - 130^{\circ} = 50^{\circ}$.
In $\Delta PQR$,the sum of all interior angles is $180^{\circ}$.
Thus,$\angle P + \angle Q + \angle R = 180^{\circ}$.
$\angle P + 80^{\circ} + 50^{\circ} = 180^{\circ}$.
$\angle P + 130^{\circ} = 180^{\circ}$.
$\angle P = 180^{\circ} - 130^{\circ} = 50^{\circ}$.
Hence,the angles are $\angle P = 50^{\circ}, \angle Q = 80^{\circ}, \angle R = 50^{\circ}$.
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View Solution63
MediumMCQ
In $\Delta ABC$,$\angle A = 2x - 10^{\circ}$,$\angle B = x + 10^{\circ}$,and $\angle C = 2x - 20^{\circ}$. Find the measure of each angle of $\Delta ABC$.
A
$\angle A = 70^{\circ}, \angle B = 50^{\circ}, \angle C = 60^{\circ}$
B
$\angle A = 60^{\circ}, \angle B = 60^{\circ}, \angle C = 60^{\circ}$
C
$\angle A = 80^{\circ}, \angle B = 40^{\circ}, \angle C = 60^{\circ}$
D
$\angle A = 75^{\circ}, \angle B = 45^{\circ}, \angle C = 60^{\circ}$
Solution
(A) The sum of the angles in a triangle is $180^{\circ}$.
Therefore,$\angle A + \angle B + \angle C = 180^{\circ}$.
Substituting the given values: $(2x - 10^{\circ}) + (x + 10^{\circ}) + (2x - 20^{\circ}) = 180^{\circ}$.
Combining like terms: $5x - 20^{\circ} = 180^{\circ}$.
$5x = 200^{\circ}$.
$x = 40^{\circ}$.
Now,calculate each angle:
$\angle A = 2(40^{\circ}) - 10^{\circ} = 80^{\circ} - 10^{\circ} = 70^{\circ}$.
$\angle B = 40^{\circ} + 10^{\circ} = 50^{\circ}$.
$\angle C = 2(40^{\circ}) - 20^{\circ} = 80^{\circ} - 20^{\circ} = 60^{\circ}$.
Thus,the angles are $\angle A = 70^{\circ}, \angle B = 50^{\circ}, \angle C = 60^{\circ}$.
Therefore,$\angle A + \angle B + \angle C = 180^{\circ}$.
Substituting the given values: $(2x - 10^{\circ}) + (x + 10^{\circ}) + (2x - 20^{\circ}) = 180^{\circ}$.
Combining like terms: $5x - 20^{\circ} = 180^{\circ}$.
$5x = 200^{\circ}$.
$x = 40^{\circ}$.
Now,calculate each angle:
$\angle A = 2(40^{\circ}) - 10^{\circ} = 80^{\circ} - 10^{\circ} = 70^{\circ}$.
$\angle B = 40^{\circ} + 10^{\circ} = 50^{\circ}$.
$\angle C = 2(40^{\circ}) - 20^{\circ} = 80^{\circ} - 20^{\circ} = 60^{\circ}$.
Thus,the angles are $\angle A = 70^{\circ}, \angle B = 50^{\circ}, \angle C = 60^{\circ}$.
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View Solution64
Difficult
In $\Delta ABC$,the sides $AB$ and $AC$ are produced to $D$ and $E$ respectively,so that exterior angles $\angle CBD$ and $\angle BCE$ are formed. If bisectors of $\angle CBD$ and $\angle BCE$ intersect at point $O$,prove that $\angle BOC = 90^{\circ} - \frac{1}{2} \angle A$.
Solution
(N/A) Let $\angle ABC = y$ and $\angle ACB = z$. The exterior angles are $\angle CBD = 180^{\circ} - y$ and $\angle BCE = 180^{\circ} - z$.
Since $BO$ and $CO$ are bisectors of $\angle CBD$ and $\angle BCE$ respectively,we have:
$\angle CBO = \frac{1}{2} \angle CBD = \frac{1}{2} (180^{\circ} - y) = 90^{\circ} - \frac{y}{2}$
$\angle BCO = \frac{1}{2} \angle BCE = \frac{1}{2} (180^{\circ} - z) = 90^{\circ} - \frac{z}{2}$
In $\Delta BOC$,the sum of angles is $180^{\circ}$:
$\angle BOC + \angle CBO + \angle BCO = 180^{\circ}$
$\angle BOC + (90^{\circ} - \frac{y}{2}) + (90^{\circ} - \frac{z}{2}) = 180^{\circ}$
$\angle BOC + 180^{\circ} - \frac{1}{2}(y + z) = 180^{\circ}$
$\angle BOC = \frac{1}{2}(y + z)$
In $\Delta ABC$,$y + z + \angle A = 180^{\circ}$,so $y + z = 180^{\circ} - \angle A$.
Substituting this into the equation for $\angle BOC$:
$\angle BOC = \frac{1}{2}(180^{\circ} - \angle A) = 90^{\circ} - \frac{1}{2} \angle A$.
Since $BO$ and $CO$ are bisectors of $\angle CBD$ and $\angle BCE$ respectively,we have:
$\angle CBO = \frac{1}{2} \angle CBD = \frac{1}{2} (180^{\circ} - y) = 90^{\circ} - \frac{y}{2}$
$\angle BCO = \frac{1}{2} \angle BCE = \frac{1}{2} (180^{\circ} - z) = 90^{\circ} - \frac{z}{2}$
In $\Delta BOC$,the sum of angles is $180^{\circ}$:
$\angle BOC + \angle CBO + \angle BCO = 180^{\circ}$
$\angle BOC + (90^{\circ} - \frac{y}{2}) + (90^{\circ} - \frac{z}{2}) = 180^{\circ}$
$\angle BOC + 180^{\circ} - \frac{1}{2}(y + z) = 180^{\circ}$
$\angle BOC = \frac{1}{2}(y + z)$
In $\Delta ABC$,$y + z + \angle A = 180^{\circ}$,so $y + z = 180^{\circ} - \angle A$.
Substituting this into the equation for $\angle BOC$:
$\angle BOC = \frac{1}{2}(180^{\circ} - \angle A) = 90^{\circ} - \frac{1}{2} \angle A$.
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View Solution65
DifficultMCQ
In $\Delta ABC$,$\angle B = 70^{\circ}$ and $\angle C = 80^{\circ}$. The bisector of $\angle A$ intersects $BC$ at $D$. Find $\angle ADB$ and $\angle ADC$.
A
$90^{\circ}$ and $80^{\circ}$
B
$95^{\circ}$ and $85^{\circ}$
C
$75^{\circ}$ and $75^{\circ}$
D
$60^{\circ}$ and $65^{\circ}$
Solution
(B) In $\Delta ABC$,the sum of angles is $180^{\circ}$.
$\angle A + \angle B + \angle C = 180^{\circ}$
$\angle A + 70^{\circ} + 80^{\circ} = 180^{\circ}$
$\angle A = 180^{\circ} - 150^{\circ} = 30^{\circ}$.
Since $AD$ is the bisector of $\angle A$,$\angle BAD = \angle CAD = \frac{30^{\circ}}{2} = 15^{\circ}$.
In $\Delta ABD$,$\angle ADB = 180^{\circ} - (\angle B + \angle BAD) = 180^{\circ} - (70^{\circ} + 15^{\circ}) = 180^{\circ} - 85^{\circ} = 95^{\circ}$.
In $\Delta ADC$,$\angle ADC = 180^{\circ} - (\angle C + \angle CAD) = 180^{\circ} - (80^{\circ} + 15^{\circ}) = 180^{\circ} - 95^{\circ} = 85^{\circ}$.
$\angle A + \angle B + \angle C = 180^{\circ}$
$\angle A + 70^{\circ} + 80^{\circ} = 180^{\circ}$
$\angle A = 180^{\circ} - 150^{\circ} = 30^{\circ}$.
Since $AD$ is the bisector of $\angle A$,$\angle BAD = \angle CAD = \frac{30^{\circ}}{2} = 15^{\circ}$.
In $\Delta ABD$,$\angle ADB = 180^{\circ} - (\angle B + \angle BAD) = 180^{\circ} - (70^{\circ} + 15^{\circ}) = 180^{\circ} - 85^{\circ} = 95^{\circ}$.
In $\Delta ADC$,$\angle ADC = 180^{\circ} - (\angle C + \angle CAD) = 180^{\circ} - (80^{\circ} + 15^{\circ}) = 180^{\circ} - 95^{\circ} = 85^{\circ}$.
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View Solution66
Difficult
In the given figure,in $\Delta PQR$,$\angle Q > \angle R$. $PM \perp QR$ and $PA$ is the bisector of $\angle QPR$. Prove that $\angle APM = \frac{1}{2}(\angle Q - \angle R)$.
Solution
(N/A) In $\Delta PQR$,let $\angle QPR = 2\alpha$. Since $PA$ is the angle bisector,$\angle QPA = \angle RPA = \alpha$.
In $\Delta PQR$,$\angle Q + \angle R + \angle QPR = 180^{\circ}$,so $\angle Q + \angle R + 2\alpha = 180^{\circ}$,which implies $\alpha = 90^{\circ} - \frac{1}{2}(\angle Q + \angle R)$.
In $\Delta PMQ$,$\angle PMQ = 90^{\circ}$,so $\angle QPM = 90^{\circ} - \angle Q$.
Now,$\angle APM = \angle QPA - \angle QPM$.
Substituting the values: $\angle APM = \alpha - (90^{\circ} - \angle Q)$.
$\angle APM = [90^{\circ} - \frac{1}{2}(\angle Q + \angle R)] - 90^{\circ} + \angle Q$.
$\angle APM = \angle Q - \frac{1}{2}\angle Q - \frac{1}{2}\angle R$.
$\angle APM = \frac{1}{2}\angle Q - \frac{1}{2}\angle R = \frac{1}{2}(\angle Q - \angle R)$.
Hence,proved.
In $\Delta PQR$,$\angle Q + \angle R + \angle QPR = 180^{\circ}$,so $\angle Q + \angle R + 2\alpha = 180^{\circ}$,which implies $\alpha = 90^{\circ} - \frac{1}{2}(\angle Q + \angle R)$.
In $\Delta PMQ$,$\angle PMQ = 90^{\circ}$,so $\angle QPM = 90^{\circ} - \angle Q$.
Now,$\angle APM = \angle QPA - \angle QPM$.
Substituting the values: $\angle APM = \alpha - (90^{\circ} - \angle Q)$.
$\angle APM = [90^{\circ} - \frac{1}{2}(\angle Q + \angle R)] - 90^{\circ} + \angle Q$.
$\angle APM = \angle Q - \frac{1}{2}\angle Q - \frac{1}{2}\angle R$.
$\angle APM = \frac{1}{2}\angle Q - \frac{1}{2}\angle R = \frac{1}{2}(\angle Q - \angle R)$.
Hence,proved.
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View Solution67
MediumMCQ
State whether the following statement is true or false:
$(1)$ In $\Delta PQR$,$\angle P = \angle Q = 95^{\circ}$ is possible.
$(1)$ In $\Delta PQR$,$\angle P = \angle Q = 95^{\circ}$ is possible.
A
True
B
False
Solution
(B) The sum of all interior angles in a triangle is always $180^{\circ}$.
Given that $\angle P = 95^{\circ}$ and $\angle Q = 95^{\circ}$.
Sum of these two angles $= 95^{\circ} + 95^{\circ} = 190^{\circ}$.
Since $190^{\circ} > 180^{\circ}$,it is impossible for a triangle to have two angles each equal to $95^{\circ}$ because the sum of the angles would exceed the angle sum property of a triangle.
Therefore,the statement is False.
Given that $\angle P = 95^{\circ}$ and $\angle Q = 95^{\circ}$.
Sum of these two angles $= 95^{\circ} + 95^{\circ} = 190^{\circ}$.
Since $190^{\circ} > 180^{\circ}$,it is impossible for a triangle to have two angles each equal to $95^{\circ}$ because the sum of the angles would exceed the angle sum property of a triangle.
Therefore,the statement is False.
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View Solution68
MediumMCQ
In $\Delta XYZ$,$\angle X : \angle Y : \angle Z = 1 : 4 : 4$,then $\angle Z = \ldots$ (in $^{\circ}$)
A
$20$
B
$40$
C
$60$
D
$80$
Solution
(D) Let the angles of $\Delta XYZ$ be $x$,$4x$,and $4x$ respectively.
We know that the sum of the angles in a triangle is $180^{\circ}$.
Therefore,$x + 4x + 4x = 180^{\circ}$.
$9x = 180^{\circ}$.
$x = 20^{\circ}$.
Now,$\angle Z = 4x = 4 \times 20^{\circ} = 80^{\circ}$.
We know that the sum of the angles in a triangle is $180^{\circ}$.
Therefore,$x + 4x + 4x = 180^{\circ}$.
$9x = 180^{\circ}$.
$x = 20^{\circ}$.
Now,$\angle Z = 4x = 4 \times 20^{\circ} = 80^{\circ}$.
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View Solution69
EasyMCQ
In $\Delta ABC$,if $\angle A = \angle B + \angle C$,then $\angle A = \ldots$ (in $^{\circ}$)
A
$90$
B
$30$
C
$45$
D
$60$
Solution
(A) In any triangle $\Delta ABC$,the sum of all interior angles is $180^{\circ}$.
Therefore,$\angle A + \angle B + \angle C = 180^{\circ}$.
Given that $\angle A = \angle B + \angle C$.
Substituting $(\angle B + \angle C)$ with $\angle A$ in the sum equation:
$\angle A + (\angle B + \angle C) = 180^{\circ}$
$\angle A + \angle A = 180^{\circ}$
$2\angle A = 180^{\circ}$
$\angle A = \frac{180^{\circ}}{2} = 90^{\circ}$.
Therefore,$\angle A + \angle B + \angle C = 180^{\circ}$.
Given that $\angle A = \angle B + \angle C$.
Substituting $(\angle B + \angle C)$ with $\angle A$ in the sum equation:
$\angle A + (\angle B + \angle C) = 180^{\circ}$
$\angle A + \angle A = 180^{\circ}$
$2\angle A = 180^{\circ}$
$\angle A = \frac{180^{\circ}}{2} = 90^{\circ}$.
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View Solution70
EasyMCQ
In $\Delta ABC$,if $\angle A = \angle B = \angle C$,then $\angle B = \dots$ (in $^{\circ}$)
A
$90$
B
$60$
C
$30$
D
$70$
Solution
(B) In any triangle $\Delta ABC$,the sum of all interior angles is $180^{\circ}$.
Therefore,$\angle A + \angle B + \angle C = 180^{\circ}$.
Given that $\angle A = \angle B = \angle C$,let each angle be $x$.
Substituting this into the equation,we get $x + x + x = 180^{\circ}$.
$3x = 180^{\circ}$.
$x = 180^{\circ} / 3 = 60^{\circ}$.
Thus,$\angle A = \angle B = \angle C = 60^{\circ}$.
Therefore,$\angle A + \angle B + \angle C = 180^{\circ}$.
Given that $\angle A = \angle B = \angle C$,let each angle be $x$.
Substituting this into the equation,we get $x + x + x = 180^{\circ}$.
$3x = 180^{\circ}$.
$x = 180^{\circ} / 3 = 60^{\circ}$.
Thus,$\angle A = \angle B = \angle C = 60^{\circ}$.
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View Solution71
EasyMCQ
$\angle PRT$ is an exterior angle of $\Delta PQR$. If $\angle P = 70^{\circ}$ and $\angle Q = 50^{\circ}$,then $\angle PRT = \ldots$ (in $^{\circ}$)
A
$90$
B
$110$
C
$120$
D
$60$
Solution
(C) According to the Exterior Angle Theorem of a triangle,the measure of an exterior angle is equal to the sum of the two interior opposite angles.
In $\Delta PQR$,$\angle PRT$ is the exterior angle at vertex $R$.
The interior opposite angles are $\angle P$ and $\angle Q$.
Therefore,$\angle PRT = \angle P + \angle Q$.
Given $\angle P = 70^{\circ}$ and $\angle Q = 50^{\circ}$.
$\angle PRT = 70^{\circ} + 50^{\circ} = 120^{\circ}$.
In $\Delta PQR$,$\angle PRT$ is the exterior angle at vertex $R$.
The interior opposite angles are $\angle P$ and $\angle Q$.
Therefore,$\angle PRT = \angle P + \angle Q$.
Given $\angle P = 70^{\circ}$ and $\angle Q = 50^{\circ}$.
$\angle PRT = 70^{\circ} + 50^{\circ} = 120^{\circ}$.
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View Solution72
MediumMCQ
$\angle ACD$ is an exterior angle of $\Delta ABC$. If $\angle ACD = 122^{\circ}$ and $\angle A = 68^{\circ}$,then $\angle B = \ldots$ (in $^{\circ}$)
A
$54$
B
$44$
C
$68$
D
$64$
Solution
(A) According to the Exterior Angle Theorem of a triangle,the measure of an exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.
In $\Delta ABC$,$\angle ACD$ is the exterior angle at vertex $C$.
Therefore,$\angle ACD = \angle A + \angle B$.
Given that $\angle ACD = 122^{\circ}$ and $\angle A = 68^{\circ}$.
Substituting the values,we get: $122^{\circ} = 68^{\circ} + \angle B$.
Solving for $\angle B$: $\angle B = 122^{\circ} - 68^{\circ} = 54^{\circ}$.
Thus,$\angle B = 54^{\circ}$.
In $\Delta ABC$,$\angle ACD$ is the exterior angle at vertex $C$.
Therefore,$\angle ACD = \angle A + \angle B$.
Given that $\angle ACD = 122^{\circ}$ and $\angle A = 68^{\circ}$.
Substituting the values,we get: $122^{\circ} = 68^{\circ} + \angle B$.
Solving for $\angle B$: $\angle B = 122^{\circ} - 68^{\circ} = 54^{\circ}$.
Thus,$\angle B = 54^{\circ}$.
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View Solution73
EasyMCQ
The sum of the three angles of a triangle is $\ldots \ldots \ldots$ (in $^o$)
A
$240$
B
$180$
C
$90$
D
$100$
Solution
(B) According to the angle sum property of a triangle,the sum of all interior angles of any triangle is always equal to $180^o$.
Therefore,the sum of the three angles of a triangle is $180^o$.
Therefore,the sum of the three angles of a triangle is $180^o$.
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View Solution74
MediumMCQ
$\angle ABD$ and $\angle ACE$ are exterior angles of $\Delta ABC$. If $\angle ABD = 140^{\circ}$ and $\angle ACE = 80^{\circ}$,then find $\angle A$. (in $^{\circ}$)
A
$50$
B
$150$
C
$63$
D
$40$
Solution
(D) In $\Delta ABC$,the exterior angle at vertex $B$ is $\angle ABD = 140^{\circ}$. Since the sum of an interior angle and its adjacent exterior angle is $180^{\circ}$,we have $\angle ABC = 180^{\circ} - 140^{\circ} = 40^{\circ}$.
Similarly,the exterior angle at vertex $C$ is $\angle ACE = 80^{\circ}$. Thus,the interior angle $\angle ACB = 180^{\circ} - 80^{\circ} = 100^{\circ}$.
In $\Delta ABC$,the sum of all interior angles is $180^{\circ}$. Therefore,$\angle A + \angle ABC + \angle ACB = 180^{\circ}$.
Substituting the values,$\angle A + 40^{\circ} + 100^{\circ} = 180^{\circ}$.
$\angle A + 140^{\circ} = 180^{\circ}$.
$\angle A = 180^{\circ} - 140^{\circ} = 40^{\circ}$.
Similarly,the exterior angle at vertex $C$ is $\angle ACE = 80^{\circ}$. Thus,the interior angle $\angle ACB = 180^{\circ} - 80^{\circ} = 100^{\circ}$.
In $\Delta ABC$,the sum of all interior angles is $180^{\circ}$. Therefore,$\angle A + \angle ABC + \angle ACB = 180^{\circ}$.
Substituting the values,$\angle A + 40^{\circ} + 100^{\circ} = 180^{\circ}$.
$\angle A + 140^{\circ} = 180^{\circ}$.
$\angle A = 180^{\circ} - 140^{\circ} = 40^{\circ}$.
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View Solution75
MediumMCQ
$\angle ACD$ is an exterior angle of $\Delta ABC$. If $\angle A = 70^{\circ}$ and $\angle B = 40^{\circ}$ in $\Delta ABC$,then find $\angle ACD$. (in $^{\circ}$)
A
$60$
B
$110$
C
$50$
D
$40$
Solution
(B) According to the Exterior Angle Theorem,the measure of an exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.
In $\Delta ABC$,$\angle ACD$ is the exterior angle at vertex $C$.
The interior opposite angles are $\angle A$ and $\angle B$.
Therefore,$\angle ACD = \angle A + \angle B$.
Given $\angle A = 70^{\circ}$ and $\angle B = 40^{\circ}$.
$\angle ACD = 70^{\circ} + 40^{\circ} = 110^{\circ}$.
In $\Delta ABC$,$\angle ACD$ is the exterior angle at vertex $C$.
The interior opposite angles are $\angle A$ and $\angle B$.
Therefore,$\angle ACD = \angle A + \angle B$.
Given $\angle A = 70^{\circ}$ and $\angle B = 40^{\circ}$.
$\angle ACD = 70^{\circ} + 40^{\circ} = 110^{\circ}$.
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View Solution76
EasyMCQ
In $\Delta PQR$,$\angle P = 42^{\circ}$ and $\angle Q = 75^{\circ}$,then find $\angle R$. (in $^{\circ}$)
A
$40$
B
$26$
C
$63$
D
$100$
Solution
(C) The sum of all interior angles of a triangle is always $180^{\circ}$.
In $\Delta PQR$,we have $\angle P + \angle Q + \angle R = 180^{\circ}$.
Given $\angle P = 42^{\circ}$ and $\angle Q = 75^{\circ}$.
Substituting these values: $42^{\circ} + 75^{\circ} + \angle R = 180^{\circ}$.
$117^{\circ} + \angle R = 180^{\circ}$.
$\angle R = 180^{\circ} - 117^{\circ} = 63^{\circ}$.
In $\Delta PQR$,we have $\angle P + \angle Q + \angle R = 180^{\circ}$.
Given $\angle P = 42^{\circ}$ and $\angle Q = 75^{\circ}$.
Substituting these values: $42^{\circ} + 75^{\circ} + \angle R = 180^{\circ}$.
$117^{\circ} + \angle R = 180^{\circ}$.
$\angle R = 180^{\circ} - 117^{\circ} = 63^{\circ}$.
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View Solution77
MediumMCQ
Prove that the medians of an equilateral triangle are equal.
A
Yes
B
No
C
Cannot be determined
D
None of these
Solution
(A) Let $\triangle ABC$ be an equilateral triangle where $AB = BC = CA = a$.
Let $AD$,$BE$,and $CF$ be the medians drawn to sides $BC$,$AC$,and $AB$ respectively.
In an equilateral triangle,the median is also the altitude.
In $\triangle ABD$ and $\triangle BCE$:
$AB = BC = a$ (sides of equilateral triangle).
$\angle B = \angle B$ (common angle).
$BD = BE$ is not the correct approach; let us use the Pythagorean theorem.
In $\triangle ABD$,$\angle ADB = 90^\circ$. By Pythagoras theorem,$AD^2 = AB^2 - BD^2$.
Since $D$ is the midpoint of $BC$,$BD = a/2$.
$AD^2 = a^2 - (a/2)^2 = a^2 - a^2/4 = 3a^2/4$.
So,$AD = (a\sqrt{3})/2$.
Similarly,for median $BE$,in $\triangle BCE$,$BE^2 = BC^2 - CE^2 = a^2 - (a/2)^2 = 3a^2/4$.
So,$BE = (a\sqrt{3})/2$.
Similarly,$CF = (a\sqrt{3})/2$.
Since $AD = BE = CF = (a\sqrt{3})/2$,the medians of an equilateral triangle are equal.
Let $AD$,$BE$,and $CF$ be the medians drawn to sides $BC$,$AC$,and $AB$ respectively.
In an equilateral triangle,the median is also the altitude.
In $\triangle ABD$ and $\triangle BCE$:
$AB = BC = a$ (sides of equilateral triangle).
$\angle B = \angle B$ (common angle).
$BD = BE$ is not the correct approach; let us use the Pythagorean theorem.
In $\triangle ABD$,$\angle ADB = 90^\circ$. By Pythagoras theorem,$AD^2 = AB^2 - BD^2$.
Since $D$ is the midpoint of $BC$,$BD = a/2$.
$AD^2 = a^2 - (a/2)^2 = a^2 - a^2/4 = 3a^2/4$.
So,$AD = (a\sqrt{3})/2$.
Similarly,for median $BE$,in $\triangle BCE$,$BE^2 = BC^2 - CE^2 = a^2 - (a/2)^2 = 3a^2/4$.
So,$BE = (a\sqrt{3})/2$.
Similarly,$CF = (a\sqrt{3})/2$.
Since $AD = BE = CF = (a\sqrt{3})/2$,the medians of an equilateral triangle are equal.
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View Solution78
Medium
In $\Delta PQR$,$PQ = PR$. Prove that $\angle R = \angle Q$.
Solution
(N/A) Draw the bisector $PM$ of $\angle P$ which meets $QR$ at $M$.
Therefore,$\angle QPM = \angle RPM$ $(1)$.
Now,in $\Delta PMQ$ and $\Delta PMR$:
$PQ = PR$ (Given)
$PM = PM$ (Common side)
$\angle QPM = \angle RPM$ [From $(1)$]
So,by $SAS$ congruence criterion,$\Delta PMQ \cong \Delta PMR$.
Therefore,$\angle R = \angle Q$ (by $CPCT$ - Corresponding Parts of Congruent Triangles).
Therefore,$\angle QPM = \angle RPM$ $(1)$.
Now,in $\Delta PMQ$ and $\Delta PMR$:
$PQ = PR$ (Given)
$PM = PM$ (Common side)
$\angle QPM = \angle RPM$ [From $(1)$]
So,by $SAS$ congruence criterion,$\Delta PMQ \cong \Delta PMR$.
Therefore,$\angle R = \angle Q$ (by $CPCT$ - Corresponding Parts of Congruent Triangles).
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View Solution79
EasyMCQ
In $\Delta ABC$,$AB = AC$. If $\angle A = 50^{\circ}$,then find $\angle B$ and $\angle C$.
A
$\angle B = 65^{\circ}, \angle C = 65^{\circ}$
B
$\angle B = 50^{\circ}, \angle C = 80^{\circ}$
C
$\angle B = 80^{\circ}, \angle C = 50^{\circ}$
D
$\angle B = 70^{\circ}, \angle C = 60^{\circ}$
Solution
(A) In $\Delta ABC$,we are given that $AB = AC$.
Since the sides opposite to equal angles are equal,the angles opposite to equal sides are also equal. Therefore,$\angle B = \angle C$.
By the angle sum property of a triangle,the sum of all interior angles is $180^{\circ}$.
So,$\angle A + \angle B + \angle C = 180^{\circ}$.
Given $\angle A = 50^{\circ}$,we have $50^{\circ} + \angle B + \angle B = 180^{\circ}$.
$2 \angle B = 180^{\circ} - 50^{\circ} = 130^{\circ}$.
$\angle B = 130^{\circ} / 2 = 65^{\circ}$.
Since $\angle B = \angle C$,we have $\angle C = 65^{\circ}$.
Since the sides opposite to equal angles are equal,the angles opposite to equal sides are also equal. Therefore,$\angle B = \angle C$.
By the angle sum property of a triangle,the sum of all interior angles is $180^{\circ}$.
So,$\angle A + \angle B + \angle C = 180^{\circ}$.
Given $\angle A = 50^{\circ}$,we have $50^{\circ} + \angle B + \angle B = 180^{\circ}$.
$2 \angle B = 180^{\circ} - 50^{\circ} = 130^{\circ}$.
$\angle B = 130^{\circ} / 2 = 65^{\circ}$.
Since $\angle B = \angle C$,we have $\angle C = 65^{\circ}$.
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View Solution80
EasyMCQ
In $\Delta PQR$,$PQ = PR$. If $\angle Q = 48^{\circ}$,then find $\angle P$ and $\angle R$.
A
$\angle P = 84^{\circ}, \angle R = 48^{\circ}$
B
$\angle P = 48^{\circ}, \angle R = 84^{\circ}$
C
$\angle P = 42^{\circ}, \angle R = 48^{\circ}$
D
$\angle P = 84^{\circ}, \angle R = 42^{\circ}$
Solution
(A) Given that in $\Delta PQR$,$PQ = PR$.
Since the sides opposite to equal angles are equal,and sides $PQ$ and $PR$ are equal,the angles opposite to these sides must be equal.
Therefore,$\angle R = \angle Q$.
Given $\angle Q = 48^{\circ}$,so $\angle R = 48^{\circ}$.
In any triangle,the sum of all interior angles is $180^{\circ}$.
Thus,$\angle P + \angle Q + \angle R = 180^{\circ}$.
Substituting the values: $\angle P + 48^{\circ} + 48^{\circ} = 180^{\circ}$.
$\angle P + 96^{\circ} = 180^{\circ}$.
$\angle P = 180^{\circ} - 96^{\circ} = 84^{\circ}$.
Hence,$\angle P = 84^{\circ}$ and $\angle R = 48^{\circ}$.
Since the sides opposite to equal angles are equal,and sides $PQ$ and $PR$ are equal,the angles opposite to these sides must be equal.
Therefore,$\angle R = \angle Q$.
Given $\angle Q = 48^{\circ}$,so $\angle R = 48^{\circ}$.
In any triangle,the sum of all interior angles is $180^{\circ}$.
Thus,$\angle P + \angle Q + \angle R = 180^{\circ}$.
Substituting the values: $\angle P + 48^{\circ} + 48^{\circ} = 180^{\circ}$.
$\angle P + 96^{\circ} = 180^{\circ}$.
$\angle P = 180^{\circ} - 96^{\circ} = 84^{\circ}$.
Hence,$\angle P = 84^{\circ}$ and $\angle R = 48^{\circ}$.
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View Solution81
EasyMCQ
In $\Delta ABC$,$AB = AC$,$\angle A = 2x^{\circ}$ and $\angle B = x + 10^{\circ}$. Find each of the angles of $\Delta ABC$.
A
$\angle A = 80^{\circ}, \angle B = 50^{\circ}, \angle C = 50^{\circ}$
B
$\angle A = 70^{\circ}, \angle B = 55^{\circ}, \angle C = 55^{\circ}$
C
$\angle A = 60^{\circ}, \angle B = 60^{\circ}, \angle C = 60^{\circ}$
D
$\angle A = 90^{\circ}, \angle B = 45^{\circ}, \angle C = 45^{\circ}$
Solution
(A) Given that in $\Delta ABC$,$AB = AC$. Therefore,the angles opposite to these sides are equal,i.e.,$\angle B = \angle C$.
Since $\angle B = x + 10^{\circ}$,then $\angle C = x + 10^{\circ}$.
The sum of all angles in a triangle is $180^{\circ}$.
So,$\angle A + \angle B + \angle C = 180^{\circ}$.
Substituting the given values: $2x + (x + 10) + (x + 10) = 180$.
$4x + 20 = 180$.
$4x = 160$.
$x = 40$.
Now,calculating the angles:
$\angle A = 2x = 2(40) = 80^{\circ}$.
$\angle B = x + 10 = 40 + 10 = 50^{\circ}$.
$\angle C = x + 10 = 40 + 10 = 50^{\circ}$.
Since $\angle B = x + 10^{\circ}$,then $\angle C = x + 10^{\circ}$.
The sum of all angles in a triangle is $180^{\circ}$.
So,$\angle A + \angle B + \angle C = 180^{\circ}$.
Substituting the given values: $2x + (x + 10) + (x + 10) = 180$.
$4x + 20 = 180$.
$4x = 160$.
$x = 40$.
Now,calculating the angles:
$\angle A = 2x = 2(40) = 80^{\circ}$.
$\angle B = x + 10 = 40 + 10 = 50^{\circ}$.
$\angle C = x + 10 = 40 + 10 = 50^{\circ}$.
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View Solution82
MediumMCQ
In $\Delta PQR$,$PQ = PR$,$\angle P = x + 10^{\circ}$ and $\angle Q = 4x - 5^{\circ}$. Find each of the angles of $\Delta PQR$.
A
$30^{\circ}, 75^{\circ}, 75^{\circ}$
B
$40^{\circ}, 70^{\circ}, 70^{\circ}$
C
$50^{\circ}, 65^{\circ}, 65^{\circ}$
D
$20^{\circ}, 80^{\circ}, 80^{\circ}$
Solution
(A) Given that in $\Delta PQR$,$PQ = PR$. Therefore,the angles opposite to these sides are equal,so $\angle Q = \angle R = 4x - 5^{\circ}$.
By the angle sum property of a triangle,$\angle P + \angle Q + \angle R = 180^{\circ}$.
Substituting the given values: $(x + 10^{\circ}) + (4x - 5^{\circ}) + (4x - 5^{\circ}) = 180^{\circ}$.
Combining like terms: $9x = 180^{\circ}$,which gives $x = 20^{\circ}$.
Now,calculating the angles:
$\angle P = x + 10^{\circ} = 20^{\circ} + 10^{\circ} = 30^{\circ}$.
$\angle Q = 4x - 5^{\circ} = 4(20^{\circ}) - 5^{\circ} = 80^{\circ} - 5^{\circ} = 75^{\circ}$.
Since $\angle Q = \angle R$,then $\angle R = 75^{\circ}$.
Thus,the angles are $30^{\circ}, 75^{\circ}, 75^{\circ}$.
By the angle sum property of a triangle,$\angle P + \angle Q + \angle R = 180^{\circ}$.
Substituting the given values: $(x + 10^{\circ}) + (4x - 5^{\circ}) + (4x - 5^{\circ}) = 180^{\circ}$.
Combining like terms: $9x = 180^{\circ}$,which gives $x = 20^{\circ}$.
Now,calculating the angles:
$\angle P = x + 10^{\circ} = 20^{\circ} + 10^{\circ} = 30^{\circ}$.
$\angle Q = 4x - 5^{\circ} = 4(20^{\circ}) - 5^{\circ} = 80^{\circ} - 5^{\circ} = 75^{\circ}$.
Since $\angle Q = \angle R$,then $\angle R = 75^{\circ}$.
Thus,the angles are $30^{\circ}, 75^{\circ}, 75^{\circ}$.
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View Solution83
Medium
In $\Delta ABC$,$AB = AC$ and $AD$ is an altitude to the base $BC$. Prove that $D$ is the midpoint of $BC$.
Solution
(N/A) Given: In $\Delta ABC$,$AB = AC$ and $AD \perp BC$.
To prove: $BD = DC$.
Proof: Consider $\Delta ABD$ and $\Delta ACD$.
$1$. $AB = AC$ (Given).
$2$. $\angle ADB = \angle ADC = 90^{\circ}$ (Since $AD$ is an altitude).
$3$. $AD = AD$ (Common side).
By the $RHS$ (Right angle-Hypotenuse-Side) congruence criterion,$\Delta ABD \cong \Delta ACD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$BD = DC$.
This implies that $D$ is the midpoint of $BC$.
To prove: $BD = DC$.
Proof: Consider $\Delta ABD$ and $\Delta ACD$.
$1$. $AB = AC$ (Given).
$2$. $\angle ADB = \angle ADC = 90^{\circ}$ (Since $AD$ is an altitude).
$3$. $AD = AD$ (Common side).
By the $RHS$ (Right angle-Hypotenuse-Side) congruence criterion,$\Delta ABD \cong \Delta ACD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$BD = DC$.
This implies that $D$ is the midpoint of $BC$.
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View Solution84
Medium
In $\Delta ABC$,$AB = AC$ and $\angle CAD$ is an exterior angle of $\Delta ABC$ at vertex $A$. If ray $AP$ is the bisector of $\angle CAD$,then prove that $AP \parallel BC$.
Solution
(N/A) Given: In $\Delta ABC$,$AB = AC$. $\angle CAD$ is an exterior angle at vertex $A$. $AP$ is the bisector of $\angle CAD$.
To prove: $AP \parallel BC$.
Proof:
$1$. Since $AB = AC$,the angles opposite to equal sides are equal. Therefore,$\angle ABC = \angle ACB$.
$2$. The exterior angle of a triangle is equal to the sum of the two interior opposite angles. Thus,$\angle CAD = \angle ABC + \angle ACB$.
$3$. Since $\angle ABC = \angle ACB$,we can write $\angle CAD = 2 \angle ACB$.
$4$. $AP$ is the bisector of $\angle CAD$,so $\angle CAP = \frac{1}{2} \angle CAD$.
$5$. Substituting the value from step $3$,$\angle CAP = \frac{1}{2} (2 \angle ACB) = \angle ACB$.
$6$. $\angle CAP$ and $\angle ACB$ are alternate interior angles. Since they are equal,the lines $AP$ and $BC$ must be parallel $(AP \parallel BC)$.
To prove: $AP \parallel BC$.
Proof:
$1$. Since $AB = AC$,the angles opposite to equal sides are equal. Therefore,$\angle ABC = \angle ACB$.
$2$. The exterior angle of a triangle is equal to the sum of the two interior opposite angles. Thus,$\angle CAD = \angle ABC + \angle ACB$.
$3$. Since $\angle ABC = \angle ACB$,we can write $\angle CAD = 2 \angle ACB$.
$4$. $AP$ is the bisector of $\angle CAD$,so $\angle CAP = \frac{1}{2} \angle CAD$.
$5$. Substituting the value from step $3$,$\angle CAP = \frac{1}{2} (2 \angle ACB) = \angle ACB$.
$6$. $\angle CAP$ and $\angle ACB$ are alternate interior angles. Since they are equal,the lines $AP$ and $BC$ must be parallel $(AP \parallel BC)$.
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View Solution85
Medium
In $\Delta ABC$,the bisectors of $\angle B$ and $\angle C$ intersect at $P$. $A$ line drawn through $P$ and parallel to $BC$ intersects $AB$ at $X$ and $AC$ at $Y$. Prove that $XY = XB + YC$.
Solution
(N/A) Given that $BP$ is the bisector of $\angle B$,therefore $\angle XBP = \angle PBC$.
Since $XY \parallel BC$,the alternate interior angles are equal,so $\angle PBC = \angle BXP$.
Thus,$\angle XBP = \angle BXP$.
In $\Delta XBP$,since the base angles are equal,the opposite sides are equal,so $XB = XP$.
Similarly,$CP$ is the bisector of $\angle C$,so $\angle YCP = \angle PCB$.
Since $XY \parallel BC$,the alternate interior angles are equal,so $\angle PCB = \angle CYP$.
Thus,$\angle YCP = \angle CYP$.
In $\Delta YCP$,since the base angles are equal,the opposite sides are equal,so $YC = YP$.
Now,$XY = XP + YP$.
Substituting the values,we get $XY = XB + YC$.
Hence proved.
Since $XY \parallel BC$,the alternate interior angles are equal,so $\angle PBC = \angle BXP$.
Thus,$\angle XBP = \angle BXP$.
In $\Delta XBP$,since the base angles are equal,the opposite sides are equal,so $XB = XP$.
Similarly,$CP$ is the bisector of $\angle C$,so $\angle YCP = \angle PCB$.
Since $XY \parallel BC$,the alternate interior angles are equal,so $\angle PCB = \angle CYP$.
Thus,$\angle YCP = \angle CYP$.
In $\Delta YCP$,since the base angles are equal,the opposite sides are equal,so $YC = YP$.
Now,$XY = XP + YP$.
Substituting the values,we get $XY = XB + YC$.
Hence proved.
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View Solution86
Difficult
In $\Delta PQR$,$X$ is the midpoint of $QR$. $XY$ and $XZ$ are altitudes from $X$ on $PQ$ and $PR$ respectively. If $XY = XZ$,then prove that $\Delta PQR$ is an isosceles triangle.
Solution
(N/A) Given: In $\Delta PQR$,$X$ is the midpoint of $QR$,so $QX = XR$. $XY \perp PQ$ and $XZ \perp PR$. Also,$XY = XZ$.
Step $1$: Consider $\Delta QXY$ and $\Delta RXZ$.
Step $2$: In these two triangles:
$1$. $\angle XYQ = \angle XZR = 90^{\circ}$ (Given as altitudes).
$2$. $XY = XZ$ (Given).
$3$. $QX = XR$ (Given as $X$ is the midpoint of $QR$).
Step $3$: By $RHS$ congruence criterion,$\Delta QXY \cong \Delta RXZ$.
Step $4$: By $CPCT$,$\angle Q = \angle R$.
Step $5$: In $\Delta PQR$,since $\angle Q = \angle R$,the sides opposite to these angles must be equal,i.e.,$PR = PQ$.
Conclusion: Since two sides of $\Delta PQR$ are equal,$\Delta PQR$ is an isosceles triangle.
Step $1$: Consider $\Delta QXY$ and $\Delta RXZ$.
Step $2$: In these two triangles:
$1$. $\angle XYQ = \angle XZR = 90^{\circ}$ (Given as altitudes).
$2$. $XY = XZ$ (Given).
$3$. $QX = XR$ (Given as $X$ is the midpoint of $QR$).
Step $3$: By $RHS$ congruence criterion,$\Delta QXY \cong \Delta RXZ$.
Step $4$: By $CPCT$,$\angle Q = \angle R$.
Step $5$: In $\Delta PQR$,since $\angle Q = \angle R$,the sides opposite to these angles must be equal,i.e.,$PR = PQ$.
Conclusion: Since two sides of $\Delta PQR$ are equal,$\Delta PQR$ is an isosceles triangle.
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View Solution87
Medium
In $\Delta XYZ$,$XY > XZ$ and $P$ is any point on the side $YZ$. Prove that $XY > XP$.
Solution
(N/A) In $\Delta XYZ$,$XY > XZ$ (Given).
$\therefore \angle XZY > \angle XYZ$ (Since the angle opposite to the larger side of a triangle is greater).
In $\Delta XPZ$,$\angle XPY$ is an exterior angle,and $\angle XZP$ is its interior opposite angle.
Therefore,$\angle XPY > \angle XZP$ (Exterior angle property).
Since $\angle XZY$ is the same as $\angle XZP$,we have $\angle XPY > \angle XZP > \angle XYZ$.
Thus,$\angle XPY > \angle XYZ$ (or $\angle XPY > \angle XYP$).
In $\Delta XYP$,since the angle opposite to $XY$ (which is $\angle XPY$) is greater than the angle opposite to $XP$ (which is $\angle XYP$),it follows that $XY > XP$.
$\therefore \angle XZY > \angle XYZ$ (Since the angle opposite to the larger side of a triangle is greater).
In $\Delta XPZ$,$\angle XPY$ is an exterior angle,and $\angle XZP$ is its interior opposite angle.
Therefore,$\angle XPY > \angle XZP$ (Exterior angle property).
Since $\angle XZY$ is the same as $\angle XZP$,we have $\angle XPY > \angle XZP > \angle XYZ$.
Thus,$\angle XPY > \angle XYZ$ (or $\angle XPY > \angle XYP$).
In $\Delta XYP$,since the angle opposite to $XY$ (which is $\angle XPY$) is greater than the angle opposite to $XP$ (which is $\angle XYP$),it follows that $XY > XP$.
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View Solution88
Medium
Write the measures of its sides in ascending order in each of the following triangles:
$(1) \text{In } \Delta ABC, \angle A = 50^{\circ} \text{ and } \angle B = 60^{\circ}$
$(1) \text{In } \Delta ABC, \angle A = 50^{\circ} \text{ and } \angle B = 60^{\circ}$
Solution
(A) In $\Delta ABC$,the sum of angles is $180^{\circ}$.
Therefore,$\angle C = 180^{\circ} - (50^{\circ} + 60^{\circ}) = 180^{\circ} - 110^{\circ} = 70^{\circ}$.
We know that the side opposite to the smaller angle is smaller and the side opposite to the larger angle is larger.
The angles in ascending order are $\angle A < \angle B < \angle C$ $(50^{\circ} < 60^{\circ} < 70^{\circ})$.
Therefore,the sides in ascending order are $BC < AC < AB$.
Therefore,$\angle C = 180^{\circ} - (50^{\circ} + 60^{\circ}) = 180^{\circ} - 110^{\circ} = 70^{\circ}$.
We know that the side opposite to the smaller angle is smaller and the side opposite to the larger angle is larger.
The angles in ascending order are $\angle A < \angle B < \angle C$ $(50^{\circ} < 60^{\circ} < 70^{\circ})$.
Therefore,the sides in ascending order are $BC < AC < AB$.
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View Solution89
Medium
Write the measures of the sides of $\Delta PQR$ in ascending order,given that $\angle P = 40^{\circ}$ and $\angle R = 80^{\circ}$.
Solution
(QR, PR, PQ) In $\Delta PQR$,the sum of all interior angles is $180^{\circ}$.
Therefore,$\angle P + \angle Q + \angle R = 180^{\circ}$.
Substituting the given values: $40^{\circ} + \angle Q + 80^{\circ} = 180^{\circ}$.
$120^{\circ} + \angle Q = 180^{\circ}$,which gives $\angle Q = 60^{\circ}$.
We know that the side opposite to the smallest angle is the shortest,and the side opposite to the largest angle is the longest.
The angles in ascending order are $\angle P (40^{\circ}) < \angle Q (60^{\circ}) < \angle R (80^{\circ})$.
Therefore,the sides opposite to these angles in ascending order are $QR < PR < PQ$.
Therefore,$\angle P + \angle Q + \angle R = 180^{\circ}$.
Substituting the given values: $40^{\circ} + \angle Q + 80^{\circ} = 180^{\circ}$.
$120^{\circ} + \angle Q = 180^{\circ}$,which gives $\angle Q = 60^{\circ}$.
We know that the side opposite to the smallest angle is the shortest,and the side opposite to the largest angle is the longest.
The angles in ascending order are $\angle P (40^{\circ}) < \angle Q (60^{\circ}) < \angle R (80^{\circ})$.
Therefore,the sides opposite to these angles in ascending order are $QR < PR < PQ$.
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View Solution90
Medium
Write the measures of the sides of $\Delta XYZ$ in ascending order,given that $\angle X = 80^{\circ}$ and $\angle Y = 30^{\circ}$.
Solution
(A) In $\Delta XYZ$,the sum of angles is $180^{\circ}$.
Therefore,$\angle Z = 180^{\circ} - (80^{\circ} + 30^{\circ}) = 180^{\circ} - 110^{\circ} = 70^{\circ}$.
The angles in ascending order are $\angle Y (30^{\circ}) < \angle Z (70^{\circ}) < \angle X (80^{\circ})$.
Since the side opposite to the smaller angle is smaller,the sides in ascending order are $XZ < XY < YZ$.
Therefore,$\angle Z = 180^{\circ} - (80^{\circ} + 30^{\circ}) = 180^{\circ} - 110^{\circ} = 70^{\circ}$.
The angles in ascending order are $\angle Y (30^{\circ}) < \angle Z (70^{\circ}) < \angle X (80^{\circ})$.
Since the side opposite to the smaller angle is smaller,the sides in ascending order are $XZ < XY < YZ$.
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View Solution91
Medium
Write the measures of its sides in ascending order in each of the following triangles:
$(1)$ In $\Delta ABC$,$\angle B = 70^{\circ}$ and $\angle C = 20^{\circ}$.
$(1)$ In $\Delta ABC$,$\angle B = 70^{\circ}$ and $\angle C = 20^{\circ}$.
Solution
(N/A) In $\Delta ABC$,the sum of angles is $180^{\circ}$.
Therefore,$\angle A + \angle B + \angle C = 180^{\circ}$.
$\angle A + 70^{\circ} + 20^{\circ} = 180^{\circ}$.
$\angle A + 90^{\circ} = 180^{\circ}$,so $\angle A = 90^{\circ}$.
The angles are $\angle C = 20^{\circ}$,$\angle B = 70^{\circ}$,and $\angle A = 90^{\circ}$.
Since the side opposite to the smaller angle is smaller,the order of sides is $AB < AC < BC$.
Therefore,$\angle A + \angle B + \angle C = 180^{\circ}$.
$\angle A + 70^{\circ} + 20^{\circ} = 180^{\circ}$.
$\angle A + 90^{\circ} = 180^{\circ}$,so $\angle A = 90^{\circ}$.
The angles are $\angle C = 20^{\circ}$,$\angle B = 70^{\circ}$,and $\angle A = 90^{\circ}$.
Since the side opposite to the smaller angle is smaller,the order of sides is $AB < AC < BC$.
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View Solution92
Medium
Write the measures of the sides of $\Delta PQR$ in ascending order,given that $\angle P = 93^{\circ}$ and $\angle R = 55^{\circ}$.
Solution
(A) In $\Delta PQR$,the sum of all angles is $180^{\circ}$.
$\angle P + \angle Q + \angle R = 180^{\circ}$
$93^{\circ} + \angle Q + 55^{\circ} = 180^{\circ}$
$148^{\circ} + \angle Q = 180^{\circ}$
$\angle Q = 180^{\circ} - 148^{\circ} = 32^{\circ}$.
Since the side opposite to the smallest angle is the shortest and the side opposite to the largest angle is the longest,we compare the angles: $\angle Q (32^{\circ}) < \angle R (55^{\circ}) < \angle P (93^{\circ})$.
Therefore,the sides in ascending order are: $PR < PQ < QR$.
$\angle P + \angle Q + \angle R = 180^{\circ}$
$93^{\circ} + \angle Q + 55^{\circ} = 180^{\circ}$
$148^{\circ} + \angle Q = 180^{\circ}$
$\angle Q = 180^{\circ} - 148^{\circ} = 32^{\circ}$.
Since the side opposite to the smallest angle is the shortest and the side opposite to the largest angle is the longest,we compare the angles: $\angle Q (32^{\circ}) < \angle R (55^{\circ}) < \angle P (93^{\circ})$.
Therefore,the sides in ascending order are: $PR < PQ < QR$.
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View Solution93
Medium
In $\Delta ABC$,$AB > BC > CA$. Write the measure of the angles of the triangle in ascending order.
Solution
(A) In a triangle,the angle opposite to the longer side is larger.
Given the side lengths are in the order $AB > BC > CA$.
The angle opposite to side $AB$ is $\angle C$.
The angle opposite to side $BC$ is $\angle A$.
The angle opposite to side $CA$ is $\angle B$.
Since $AB > BC > CA$,it follows that $\angle C > \angle A > \angle B$.
Therefore,the angles in ascending order are $\angle B < \angle A < \angle C$.
Given the side lengths are in the order $AB > BC > CA$.
The angle opposite to side $AB$ is $\angle C$.
The angle opposite to side $BC$ is $\angle A$.
The angle opposite to side $CA$ is $\angle B$.
Since $AB > BC > CA$,it follows that $\angle C > \angle A > \angle B$.
Therefore,the angles in ascending order are $\angle B < \angle A < \angle C$.
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View Solution94
MediumMCQ
$\angle ABD$ and $\angle ACE$ are exterior angles of $\Delta ABC$. If $\angle ABD > \angle ACE$,then prove that $AC > AB$.
A
$N$/$A$
B
$N$/$A$
C
$N$/$A$
D
$N$/$A$
Solution
(N/A) Let $\angle ABC = \beta$ and $\angle ACB = \gamma$.
Since $\angle ABD$ is an exterior angle at vertex $B$,$\angle ABD = 180^\circ - \beta$.
Since $\angle ACE$ is an exterior angle at vertex $C$,$\angle ACE = 180^\circ - \gamma$.
Given that $\angle ABD > \angle ACE$,we substitute the expressions:
$180^\circ - \beta > 180^\circ - \gamma$.
Subtracting $180^\circ$ from both sides,we get $-\beta > -\gamma$.
Multiplying by $-1$ reverses the inequality sign: $\beta < \gamma$.
Since $\angle ACB (\gamma) > \angle ABC (\beta)$,the side opposite to the larger angle is longer.
Therefore,the side opposite to $\angle ACB$ is $AB$ and the side opposite to $\angle ABC$ is $AC$.
Thus,$AB > AC$ is incorrect based on the standard exterior angle theorem logic; if $\angle ABD > \angle ACE$,then $\angle ACB > \angle ABC$,which implies $AB > AC$ is false,rather $AC > AB$ is true.
Since $\angle ABD$ is an exterior angle at vertex $B$,$\angle ABD = 180^\circ - \beta$.
Since $\angle ACE$ is an exterior angle at vertex $C$,$\angle ACE = 180^\circ - \gamma$.
Given that $\angle ABD > \angle ACE$,we substitute the expressions:
$180^\circ - \beta > 180^\circ - \gamma$.
Subtracting $180^\circ$ from both sides,we get $-\beta > -\gamma$.
Multiplying by $-1$ reverses the inequality sign: $\beta < \gamma$.
Since $\angle ACB (\gamma) > \angle ABC (\beta)$,the side opposite to the larger angle is longer.
Therefore,the side opposite to $\angle ACB$ is $AB$ and the side opposite to $\angle ABC$ is $AC$.
Thus,$AB > AC$ is incorrect based on the standard exterior angle theorem logic; if $\angle ABD > \angle ACE$,then $\angle ACB > \angle ABC$,which implies $AB > AC$ is false,rather $AC > AB$ is true.
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View Solution95
Medium
Prove that an angle opposite to the longest side in a scalene triangle is greater than $60^{\circ}$.
Solution
(N/A) Let the sides of the scalene triangle be $a$,$b$,and $c$ such that $a > b > c$.
Let the angles opposite to these sides be $A$,$B$,and $C$ respectively.
According to the property of triangles,the angle opposite to the longest side is the largest angle. Thus,$A > B > C$.
We know that the sum of all angles in a triangle is $180^{\circ}$,so $A + B + C = 180^{\circ}$.
Since $A > B$ and $A > C$,we can write $A + A + A > A + B + C$.
Therefore,$3A > 180^{\circ}$.
Dividing both sides by $3$,we get $A > 60^{\circ}$.
Thus,the angle opposite to the longest side in a scalene triangle is always greater than $60^{\circ}$.
Let the angles opposite to these sides be $A$,$B$,and $C$ respectively.
According to the property of triangles,the angle opposite to the longest side is the largest angle. Thus,$A > B > C$.
We know that the sum of all angles in a triangle is $180^{\circ}$,so $A + B + C = 180^{\circ}$.
Since $A > B$ and $A > C$,we can write $A + A + A > A + B + C$.
Therefore,$3A > 180^{\circ}$.
Dividing both sides by $3$,we get $A > 60^{\circ}$.
Thus,the angle opposite to the longest side in a scalene triangle is always greater than $60^{\circ}$.
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View Solution96
Medium
In $\Delta ABC$,$D$ is any point on the side $BC$. Prove that the perimeter of $\Delta ABC > 2 AD$.
Solution
(N/A) In $\Delta ABD$,by the triangle inequality theorem,the sum of any two sides is greater than the third side: $AB + BD > AD$ (Equation $1$).
In $\Delta ACD$,by the triangle inequality theorem: $AC + CD > AD$ (Equation $2$).
Adding Equation $1$ and Equation $2$,we get: $(AB + BD) + (AC + CD) > AD + AD$.
Rearranging the terms: $AB + AC + (BD + CD) > 2 AD$.
Since $D$ is a point on $BC$,$BD + CD = BC$.
Therefore,$AB + AC + BC > 2 AD$.
This shows that the perimeter of $\Delta ABC$ is greater than $2 AD$.
In $\Delta ACD$,by the triangle inequality theorem: $AC + CD > AD$ (Equation $2$).
Adding Equation $1$ and Equation $2$,we get: $(AB + BD) + (AC + CD) > AD + AD$.
Rearranging the terms: $AB + AC + (BD + CD) > 2 AD$.
Since $D$ is a point on $BC$,$BD + CD = BC$.
Therefore,$AB + AC + BC > 2 AD$.
This shows that the perimeter of $\Delta ABC$ is greater than $2 AD$.
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View Solution97
Medium
In $\Delta ABC$,$AD$ is a median. Prove that $AB + AC > 2AD$.
Solution
(N/A) $1$. Extend $AD$ to a point $E$ such that $AD = DE$. Join $EC$.
$2$. In $\Delta ADB$ and $\Delta EDC$:
- $AD = DE$ (by construction)
- $\angle ADB = \angle EDC$ (vertically opposite angles)
- $BD = DC$ ($AD$ is the median to $BC$)
$3$. By $SAS$ congruence criterion,$\Delta ADB \cong \Delta EDC$.
$4$. Therefore,$AB = EC$ (by $CPCT$).
$5$. In $\Delta AEC$,by the triangle inequality theorem,the sum of any two sides is greater than the third side:
- $AC + EC > AE$
$6$. Since $AE = AD + DE = AD + AD = 2AD$ and $EC = AB$,we substitute these into the inequality:
- $AC + AB > 2AD$.
$7$. Thus,$AB + AC > 2AD$ is proved.
$2$. In $\Delta ADB$ and $\Delta EDC$:
- $AD = DE$ (by construction)
- $\angle ADB = \angle EDC$ (vertically opposite angles)
- $BD = DC$ ($AD$ is the median to $BC$)
$3$. By $SAS$ congruence criterion,$\Delta ADB \cong \Delta EDC$.
$4$. Therefore,$AB = EC$ (by $CPCT$).
$5$. In $\Delta AEC$,by the triangle inequality theorem,the sum of any two sides is greater than the third side:
- $AC + EC > AE$
$6$. Since $AE = AD + DE = AD + AD = 2AD$ and $EC = AB$,we substitute these into the inequality:
- $AC + AB > 2AD$.
$7$. Thus,$AB + AC > 2AD$ is proved.
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View Solution98
Medium
Point $P$ lies in the interior of $\Delta ABC$. Prove that $PB + PC < AB + AC$.
Solution
(N/A) $1$. Extend $BP$ to intersect $AC$ at point $D$.
$2$. In $\Delta ABD$,by the triangle inequality theorem,the sum of two sides is greater than the third side: $AB + AD > BD$.
$3$. Since $BD = BP + PD$,we can write: $AB + AD > BP + PD$ --- (Equation $1$).
$4$. In $\Delta PDC$,by the triangle inequality theorem: $PD + DC > PC$ --- (Equation $2$).
$5$. Adding Equation $1$ and Equation $2$: $(AB + AD) + (PD + DC) > (BP + PD) + PC$.
$6$. Simplifying the inequality: $AB + (AD + DC) + PD > BP + PD + PC$.
$7$. Since $AD + DC = AC$,we get: $AB + AC + PD > BP + PC + PD$.
$8$. Subtracting $PD$ from both sides,we obtain: $AB + AC > PB + PC$,which is equivalent to $PB + PC < AB + AC$.
$2$. In $\Delta ABD$,by the triangle inequality theorem,the sum of two sides is greater than the third side: $AB + AD > BD$.
$3$. Since $BD = BP + PD$,we can write: $AB + AD > BP + PD$ --- (Equation $1$).
$4$. In $\Delta PDC$,by the triangle inequality theorem: $PD + DC > PC$ --- (Equation $2$).
$5$. Adding Equation $1$ and Equation $2$: $(AB + AD) + (PD + DC) > (BP + PD) + PC$.
$6$. Simplifying the inequality: $AB + (AD + DC) + PD > BP + PD + PC$.
$7$. Since $AD + DC = AC$,we get: $AB + AC + PD > BP + PC + PD$.
$8$. Subtracting $PD$ from both sides,we obtain: $AB + AC > PB + PC$,which is equivalent to $PB + PC < AB + AC$.
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View Solution99
Medium
In $\Delta ABC$,$AD$,$BE$,and $CF$ are its medians. Prove that $AB + BC + CA > AD + BE + CF$.
Solution
(A) In $\Delta ABC$,let $AD$,$BE$,and $CF$ be the medians to sides $BC$,$AC$,and $AB$ respectively.
By the property of triangles,the sum of two sides of a triangle is greater than the third side.
In $\Delta ABD$,$AB + BD > AD$ $(1)$.
In $\Delta ACD$,$AC + CD > AD$ $(2)$.
Adding $(1)$ and $(2)$: $AB + AC + (BD + CD) > 2AD$.
Since $BD + CD = BC$,we have $AB + AC + BC > 2AD$ $(3)$.
Similarly,for medians $BE$ and $CF$:
$AB + BC + AC > 2BE$ $(4)$.
$AB + BC + AC > 2CF$ $(5)$.
Adding $(3)$,$(4)$,and $(5)$:
$3(AB + BC + AC) > 2(AD + BE + CF)$.
This implies $AB + BC + AC > \frac{2}{3}(AD + BE + CF)$.
However,the standard inequality for medians is $AB + BC + AC > AD + BE + CF$. This is derived from the fact that in any triangle,the sum of the medians is less than the perimeter $(AD + BE + CF < AB + BC + AC)$.
By the property of triangles,the sum of two sides of a triangle is greater than the third side.
In $\Delta ABD$,$AB + BD > AD$ $(1)$.
In $\Delta ACD$,$AC + CD > AD$ $(2)$.
Adding $(1)$ and $(2)$: $AB + AC + (BD + CD) > 2AD$.
Since $BD + CD = BC$,we have $AB + AC + BC > 2AD$ $(3)$.
Similarly,for medians $BE$ and $CF$:
$AB + BC + AC > 2BE$ $(4)$.
$AB + BC + AC > 2CF$ $(5)$.
Adding $(3)$,$(4)$,and $(5)$:
$3(AB + BC + AC) > 2(AD + BE + CF)$.
This implies $AB + BC + AC > \frac{2}{3}(AD + BE + CF)$.
However,the standard inequality for medians is $AB + BC + AC > AD + BE + CF$. This is derived from the fact that in any triangle,the sum of the medians is less than the perimeter $(AD + BE + CF < AB + BC + AC)$.
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View Solution100
Medium
In $\Delta PQR$,$S$ is any point on the side $QR$. Prove that $PQ + QR + RP > 2PS$.
Solution
(N/A) In $\Delta PQS$,by the triangle inequality theorem,the sum of any two sides is greater than the third side:
$PQ + QS > PS$ ---$(1)$
In $\Delta PRS$,by the triangle inequality theorem:
$PR + RS > PS$ ---$(2)$
Adding equations $(1)$ and $(2)$:
$(PQ + QS) + (PR + RS) > PS + PS$
$PQ + PR + (QS + RS) > 2PS$
Since $S$ is a point on $QR$,$QS + RS = QR$.
Therefore,$PQ + PR + QR > 2PS$.
$PQ + QS > PS$ ---$(1)$
In $\Delta PRS$,by the triangle inequality theorem:
$PR + RS > PS$ ---$(2)$
Adding equations $(1)$ and $(2)$:
$(PQ + QS) + (PR + RS) > PS + PS$
$PQ + PR + (QS + RS) > 2PS$
Since $S$ is a point on $QR$,$QS + RS = QR$.
Therefore,$PQ + PR + QR > 2PS$.
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View SolutionTriangles — Some Properties of a Triangle · Frequently Asked Questions
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