Mix Examples - Triangles Questions in English

(A) In rectangle $ABCD$,we know that opposite sides are equal,so $CD = AB$ and $DA = BC$.
Also,$\angle B = 90^{\circ}$.
Given the equation: $AB^{2} + BC^{2} + CD^{2} + DA^{2} = 338$.
Substituting the equal sides: $AB^{2} + BC^{2} + AB^{2} + BC^{2} = 338$.
$2(AB^{2} + BC^{2}) = 338$.
$AB^{2} + BC^{2} = 169$.
In $\triangle ABC$,by Pythagoras theorem: $AC^{2} = AB^{2} + BC^{2}$.
Therefore,$AC^{2} = 169$.
$AC = \sqrt{169} = 13$.

(C) In $\Delta AMB$,$m \angle M = 90^{\circ}$ and $m \angle B = 45^{\circ}$.
Since the sum of angles in a triangle is $180^{\circ}$,$m \angle MAB = 180^{\circ} - 90^{\circ} - 45^{\circ} = 45^{\circ}$.
Since $m \angle MAB = m \angle B = 45^{\circ}$,$\Delta AMB$ is an isosceles right triangle.
Therefore,$AM = BM = 4$.
Given $M \in \overline{BC}$,we have $BC = BM + MC$.
$7 = 4 + MC$,which implies $MC = 7 - 4 = 3$.
In right-angled $\Delta AMC$,by the Pythagorean theorem:
$AC^{2} = AM^{2} + MC^{2}$
$AC^{2} = 4^{2} + 3^{2} = 16 + 9 = 25$
$AC = \sqrt{25} = 5$.

(C) In parallelogram $ABCD$,let $\overline{AC} \cap \overline{BD} = \{O\}$.
Since the diagonals of a parallelogram bisect each other,$O$ is the midpoint of $\overline{AC}$ and $\overline{BD}$.
Therefore,$BO = \frac{1}{2} BD = \frac{1}{2}(12) = 6$.
In $\Delta ABD$,$\overline{AO}$ is the median to side $\overline{BD}$.
By Apollonius' theorem,$AB^{2} + AD^{2} = 2(AO^{2} + BO^{2})$.
Substituting the given values: $200 = 2(AO^{2} + 6^{2})$.
$100 = AO^{2} + 36$.
$AO^{2} = 100 - 36 = 64$.
$AO = \sqrt{64} = 8$.
Since $O$ is the midpoint of $\overline{AC}$,$AC = 2 AO = 2 \times 8 = 16$.

(B) In a parallelogram $ABCD$,the sum of the squares of the diagonals is equal to the sum of the squares of its four sides.
This is known as the Parallelogram Law: $AC^{2} + BD^{2} = 2(AB^{2} + BC^{2})$.
Given $AB^{2} + BC^{2} = 260$ and $AC = 18$.
Substituting the values into the formula: $18^{2} + BD^{2} = 2(260)$.
$324 + BD^{2} = 520$.
$BD^{2} = 520 - 324$.
$BD^{2} = 196$.
$BD = \sqrt{196} = 14$.

(D) The perimeter of an equilateral triangle is given by the formula $P = 3 \times \text{side}$.
Given $P = 18$,we have $18 = 3 \times \text{side}$,which implies $\text{side} = 6$.
The altitude $(h)$ of an equilateral triangle is given by the formula $h = \frac{\sqrt{3}}{2} \times \text{side}$.
Substituting the value of the side,we get $h = \frac{\sqrt{3}}{2} \times 6 = 3 \sqrt{3}$.

(B) $1.$ In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse. Thus,$BM = \frac{1}{2} AC$ $(1-d)$.
$2.$ In a right-angled triangle $\Delta ABC$ with altitude $\overline{AD}$ to the hypotenuse,by Apollonius theorem or properties of similar triangles,the relation $AC^2 = CD \cdot BC$ is derived from $\Delta ACD \sim \Delta BCA$ $(2-c)$.
$3.$ In a $30^\circ-60^\circ-90^\circ$ triangle,the side opposite to $30^\circ$ is half the hypotenuse. Here,$BC$ is opposite to $30^\circ$,so $BC = \frac{1}{2} AB$ $(3-b)$.
$4.$ Apollonius theorem states that for any triangle $\Delta ABC$ with median $\overline{BD}$,$AB^2 + BC^2 = 2(BD^2 + CD^2)$ $(4-a)$.
Therefore,the correct match is $(1-d), (2-c), (3-b), (4-a)$.

(D) In $\Delta PQR$,$\angle Q = 90^{\circ}$.
Therefore,$\overline{PR}$ is the hypotenuse.
By Pythagoras theorem,$PR^2 = PQ^2 + QR^2$.
$PR^2 = 8^2 + 15^2 = 64 + 225 = 289$.
$PR = \sqrt{289} = 17$.
The radius $(r)$ of the incircle of a right-angled triangle is given by the formula $r = \frac{PQ + QR - PR}{2}$.
$r = \frac{8 + 15 - 17}{2} = \frac{23 - 17}{2} = \frac{6}{2} = 3$.

(B) In a right-angled triangle,the radius $r$ of the incircle is given by the formula $r = \frac{AB + BC - AC}{2}$.
First,we find the hypotenuse $AC$ using the Pythagorean theorem: $AC = \sqrt{AB^2 + BC^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
Now,substitute the values into the formula: $r = \frac{4 + 3 - 5}{2} = \frac{7 - 5}{2} = \frac{2}{2} = 1$.
Therefore,the radius of the incircle is $1$.

(D) For a right-angled triangle with sides $a, b$ and hypotenuse $c$,the in-radius $r$ is given by the formula $r = \frac{a+b-c}{2}$.
$1.$ For $\Delta ABC$ with sides $3, 4, 5$: $r = \frac{3+4-5}{2} = \frac{2}{2} = 1$. Matches $(b)$.
$2.$ For $\Delta PQR$ with sides $5, 12, 13$: $r = \frac{5+12-13}{2} = \frac{4}{2} = 2$. Matches $(a)$.
$3.$ For $\Delta XYZ$ with sides $8, 15, 17$: $r = \frac{8+15-17}{2} = \frac{6}{2} = 3$. Matches $(d)$.
$4.$ For $\Delta MNP$ with sides $20, 21, 29$: $r = \frac{20+21-29}{2} = \frac{12}{2} = 6$. Matches $(c)$.
Thus,the correct matching is $(1-b), (2-a), (3-d), (4-c)$.