Similarity of Triangles and Basic Proportionality Theorem Questions in English

(D) Since $G$ is the centroid of $\Delta ABC$,it divides the median $\overline{AD}$ in the ratio $2:1$. Therefore,$\frac{GD}{AD} = \frac{1}{3}$.
In $\Delta ADE$,since $\overline{GK} \parallel \overline{DE}$,by the Basic Proportionality Theorem,we have $\frac{GD}{AD} = \frac{EK}{AE} = \frac{1}{3}$.
This implies $AE = 3 \times EK = 3 \times 1.8 = 5.4$.
Since $\overline{BE}$ is a median to $\overline{AC}$,$E$ is the midpoint of $\overline{AC}$.
Therefore,$AC = 2 \times AE = 2 \times 5.4 = 10.8$.

(A) According to the Angle Bisector Theorem,in $\Delta ABC$,if the bisector of $\angle B$ intersects $\overline{AC}$ at $D$,then the ratio of the segments of the opposite side is equal to the ratio of the other two sides of the triangle.
Therefore,$\frac{AB}{BC} = \frac{AD}{DC}$.
Given that $AB = 7.5$ and $\frac{AD}{DC} = \frac{3}{4}$,we substitute these values into the equation:
$\frac{7.5}{BC} = \frac{3}{4}$
By cross-multiplying,we get:
$3 \times BC = 7.5 \times 4$
$3 \times BC = 30$
$BC = \frac{30}{3}$
$BC = 10$

(B) According to the Angle Bisector Theorem,if a ray bisects an angle of a triangle,then it divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta XYZ$,$YM$ is the bisector of $\angle Y$,which intersects $\overline{XZ}$ at $M$.
Therefore,by the Angle Bisector Theorem,the ratio of the segments of the side $\overline{XZ}$ is equal to the ratio of the other two sides of the triangle:
$\frac{XM}{MZ} = \frac{XY}{YZ}$.
Thus,option $B$ is correct.

(D) According to the Angle Bisector Theorem,if a ray bisects an angle of a triangle,then it divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta ABC$,since $AD$ is the bisector of $\angle A$,it divides the side $BC$ such that:
$\frac{BD}{DC} = \frac{AB}{AC}$
By cross-multiplying the terms,we get:
$BD \times AC = DC \times AB$
Therefore,the correct option is $D$.

(B) In $\Delta ABC$,$\angle B = 90^{\circ}$.
According to Pythagoras' theorem,in a right-angled triangle,the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Here,the side opposite to $\angle B$ is $\overline{AC}$,which is the hypotenuse.
Therefore,$AC^{2} = AB^{2} + BC^{2}$.

(C) In a right-angled triangle $\Delta ABC$ where $m \angle A = 90^{\circ}$ and $\overline{AD} \perp \overline{BC}$,the triangle is divided into two triangles $\Delta ABD$ and $\Delta CAD$ which are similar to the original triangle $\Delta ABC$ and to each other.
By the property of similarity,$\Delta ABD \sim \Delta CAD$.
Therefore,the ratio of their corresponding sides is equal: $\frac{AD}{BD} = \frac{CD}{AD}$.
Cross-multiplying gives $AD^{2} = BD \cdot CD$ or $AD^{2} = BD \cdot DC$.

(A) Apollonius' theorem states that for any triangle $\Delta ABC$,if $\overline{AD}$ is the median to the side $\overline{BC}$,then the sum of the squares of the two sides is equal to twice the sum of the square of the median and the square of half the side bisected by the median.
Mathematically,this is expressed as $AB^{2} + AC^{2} = 2(AD^{2} + BD^{2})$,where $BD = DC = \frac{1}{2}BC$.
Comparing this with the given equation $AB^{2} + AC^{2} = 2(AD^{2} + BD^{2})$,it is clear that $\overline{AD}$ must be the median to side $\overline{BC}$.

(A) In $\Delta ABC$,$m \angle B = 90$ and $D$ is the midpoint of the hypotenuse $\overline{AC}$.
According to the theorem of the median to the hypotenuse in a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.
Therefore,$BD = \frac{1}{2} AC$.

(C) In $\triangle ABC$,since $m \angle A = 90^{\circ}$,the triangle is a right-angled triangle.
By the Pythagoras theorem,the square of the hypotenuse is equal to the sum of the squares of the other two sides.
$BC^{2} = AB^{2} + AC^{2}$
$BC^{2} = 8^{2} + 15^{2}$
$BC^{2} = 64 + 225$
$BC^{2} = 289$
Taking the square root on both sides,$BC = \sqrt{289} = 17$.

(A) In $\Delta PQR$,for $\angle R$ to be a right angle,the triangle must satisfy the Pythagorean theorem: $PR^2 + QR^2 = PQ^2$.
Given $PQ = 25$ and $QR = 24$.
Substituting the values: $PR^2 + 24^2 = 25^2$.
$PR^2 + 576 = 625$.
$PR^2 = 625 - 576 = 49$.
$PR = \sqrt{49} = 7$.

(A) In a right-angled triangle $\Delta PQR$,where $\angle Q = 90^{\circ}$ and $\overline{QM}$ is the altitude to the hypotenuse $\overline{PR}$,the geometric mean theorem states that the altitude to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and to each other.
According to the property of the altitude to the hypotenuse in a right triangle:
$QM^{2} = PM \cdot RM$
Given $PM = 8$ and $RM = 12$:
$QM^{2} = 8 \times 12$
$QM^{2} = 96$
Taking the square root of both sides:
$QM = \sqrt{96}$
$QM = \sqrt{16 \times 6}$
$QM = 4 \sqrt{6}$

(D) In $\Delta ABC$,$m \angle B = 90^\circ$ and $\overline{BM}$ is an altitude to the hypotenuse $AC$.
According to the geometric mean theorem for right-angled triangles,the altitude to the hypotenuse divides the triangle into two triangles similar to the original triangle and to each other.
Specifically,$BM^2 = AM \cdot CM$.
Given $AM = 8$ and $BM = 8$,we substitute these values into the equation:
$8^2 = 8 \cdot CM$
$64 = 8 \cdot CM$
$CM = \frac{64}{8} = 8$.
Now,the length of the hypotenuse $AC$ is the sum of the segments $AM$ and $CM$:
$AC = AM + CM = 8 + 8 = 16$.
Therefore,the correct option is $D$.

(B) In a right-angled triangle,the altitude to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and to each other.
According to the geometric mean theorem,$BD^2 = AD \cdot CD$.
Given $BD = 2 \sqrt{30}$ and $CD = 6$.
Substituting the values: $(2 \sqrt{30})^2 = AD \cdot 6$.
$4 \cdot 30 = 6 \cdot AD$.
$120 = 6 \cdot AD$.
$AD = 20$.
Since $AC = AD + CD$,we have $AC = 20 + 6 = 26$.

(D) In $\Delta PQR$,since $m \angle Q = 90^{\circ}$ and $\overline{QS} \perp \overline{PR}$,by the properties of right-angled triangles,we have $PQ^2 = PS \cdot PR$ and $QR^2 = SR \cdot PR$.
Subtracting these two equations: $PQ^2 - QR^2 = (PS - SR) \cdot PR$.
Given $PQ^2 - QR^2 = 260$ and $PS - SR = 10$.
Substituting these values: $260 = 10 \cdot PR$.
Therefore,$PR = \frac{260}{10} = 26$.

(A) Given sides of $\Delta ABC$ are $AB = 17$,$BC = 15$,and $AC = 8$.
First,check if the triangle is a right-angled triangle using the converse of the Pythagorean theorem.
$BC^2 + AC^2 = 15^2 + 8^2 = 225 + 64 = 289$.
$AB^2 = 17^2 = 289$.
Since $BC^2 + AC^2 = AB^2$,$\Delta ABC$ is a right-angled triangle with the hypotenuse $AB = 17$.
The longest side of a right-angled triangle is its hypotenuse.
According to the property of a right-angled triangle,the length of the median drawn to the hypotenuse is half the length of the hypotenuse.
Length of median = $\frac{1}{2} \times AB = \frac{1}{2} \times 17 = 8.5$.

(B) Let the rhombus be $ABCD$ with diagonals $d_1 = 10$ and $d_2 = 24$.
In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.
Let the diagonals intersect at point $O$. Then,the segments of the diagonals are $d_1/2 = 5$ and $d_2/2 = 12$.
These segments form a right-angled triangle with the side of the rhombus as the hypotenuse.
Using the Pythagorean theorem: $side^2 = (d_1/2)^2 + (d_2/2)^2$.
$side^2 = 5^2 + 12^2 = 25 + 144 = 169$.
$side = \sqrt{169} = 13$.
Therefore,the length of each side of the rhombus is $13$.

(B) In a right-angled triangle $\Delta PQR$,where $\angle P = 90^{\circ}$ and $\overline{PM}$ is the altitude to the hypotenuse $\overline{QR}$,the geometric mean theorem states that the altitude to the hypotenuse is the geometric mean of the segments of the hypotenuse.
Therefore,the relationship is given by $PM^{2} = QM \cdot RM$.
Substituting the given values:
$PM^{2} = (2x^{2}) \cdot (8x^{2})$
$PM^{2} = 16x^{4}$
Taking the square root of both sides:
$PM = \sqrt{16x^{4}}$
$PM = 4x^{2}$

(C) In $\Delta PQR$,we are given $m \angle P = m \angle Q + m \angle R$.
We know that the sum of all angles in a triangle is $180^{\circ}$,so $m \angle P + m \angle Q + m \angle R = 180^{\circ}$.
Substituting $m \angle Q + m \angle R$ with $m \angle P$,we get $m \angle P + m \angle P = 180^{\circ}$,which implies $2 \cdot m \angle P = 180^{\circ}$,so $m \angle P = 90^{\circ}$.
Since $\Delta PQR$ is a right-angled triangle with the right angle at $P$,$QR$ is the hypotenuse.
By the Pythagorean theorem,$QR^2 = PQ^2 + PR^2$.
Substituting the given values,$25^2 = 20^2 + PR^2$.
$625 = 400 + PR^2$,so $PR^2 = 625 - 400 = 225$.
Thus,$PR = \sqrt{225} = 15$.
The perimeter of $\Delta PQR = PQ + QR + PR = 20 + 25 + 15 = 60$.

(D) In $\Delta ABC$,since $m \angle B = 90^{\circ}$,the triangle is a right-angled triangle.
According to the Pythagorean theorem,the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Here,$b$ is the hypotenuse (side opposite to $\angle B$).
Therefore,$b^{2} = a^{2} + c^{2}$.
Substituting the given values: $b^{2} = 9^{2} + 12^{2}$.
$b^{2} = 81 + 144 = 225$.
Taking the square root on both sides,$b = \sqrt{225} = 15$.

(B) In a right-angled triangle $\Delta ABC$ where $m \angle A = 90^{\circ}$,the side $a$ is the hypotenuse,and $b$ and $c$ are the other two sides.
According to the Pythagorean theorem: $a^2 = b^2 + c^2$.
Given $a = 20$ and $b = 12$,we substitute these values into the equation:
$20^2 = 12^2 + c^2$
$400 = 144 + c^2$
$c^2 = 400 - 144$
$c^2 = 256$
Taking the square root of both sides,we get $c = \sqrt{256} = 16$.

(D) triangle is right-angled if the square of the longest side is equal to the sum of the squares of the other two sides (Pythagoras theorem).
For option $A$: $5^{2} + 12^{2} = 25 + 144 = 169 = 13^{2}$. This is a right-angled triangle.
For option $B$: $3^{2} + 4^{2} = 9 + 16 = 25 = 5^{2}$. This is a right-angled triangle.
For option $C$: $7^{2} + 24^{2} = 49 + 576 = 625 = 25^{2}$. This is a right-angled triangle.
For option $D$: $8^{2} + 24^{2} = 64 + 576 = 640$,while $26^{2} = 676$. Since $640 \neq 676$,these are not the sides of a right-angled triangle.

(C) Apollonius' theorem states that for any triangle $\Delta ABC$ with median $\overline{AD}$ to side $\overline{BC}$,the sum of the squares of the two sides is equal to twice the sum of the square of the median and the square of half the third side.
Mathematically,this is expressed as: $AB^2 + AC^2 = 2(AD^2 + BD^2)$.
Since $\overline{AD}$ is a median,$D$ is the midpoint of $\overline{BC}$,so $BD = DC$.
Substituting $DC$ for $BD$,we get $AB^2 + AC^2 = 2(AD^2 + DC^2)$.