Similarity of Triangles and Basic Proportionality Theorem Questions in English

(A) In $\Delta ABC$,since $m \angle B = 90^{\circ}$,the triangle is a right-angled triangle.
Using the Pythagorean theorem,the hypotenuse $AC = \sqrt{AB^2 + BC^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$.
The area of $\Delta ABC$ can be calculated in two ways:
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 7 \times 24 = 84$.
Also,using $AC$ as the base and $BM$ as the altitude,Area $= \frac{1}{2} \times AC \times BM = \frac{1}{2} \times 25 \times BM$.
Equating the two areas: $\frac{1}{2} \times 25 \times BM = 84$.
$25 \times BM = 168$.
$BM = \frac{168}{25} = 6.72$.

(N/A) Given: In $\Delta ABC$,$\angle B = 90^{\circ}$. $N$ is a point on $AB$ and $M$ is a point on $BC$.
To prove: $AM^{2} + CN^{2} = AC^{2} + MN^{2}$.
Proof:
$1$. In right-angled $\Delta ABM$,by Pythagoras theorem: $AM^{2} = AB^{2} + BM^{2}$.
$2$. In right-angled $\Delta CBN$,by Pythagoras theorem: $CN^{2} = CB^{2} + BN^{2}$.
$3$. Adding these two equations: $AM^{2} + CN^{2} = AB^{2} + BM^{2} + CB^{2} + BN^{2}$.
$4$. Rearranging the terms: $AM^{2} + CN^{2} = (AB^{2} + BC^{2}) + (BM^{2} + BN^{2})$.
$5$. In $\Delta ABC$,by Pythagoras theorem: $AC^{2} = AB^{2} + BC^{2}$.
$6$. In right-angled $\Delta MBN$,by Pythagoras theorem: $MN^{2} = BM^{2} + BN^{2}$.
$7$. Substituting these into the equation from step $4$: $AM^{2} + CN^{2} = AC^{2} + MN^{2}$.
Hence,the statement is proved.

(N/A) Let the diagonals $XZ$ and $YW$ of the quadrilateral $\square XYZW$ intersect at point $O$ at right angles $(90^{\circ})$.
In $\triangle XOY$,$\triangle YOZ$,$\triangle ZOW$,and $\triangle WOX$,by the Pythagorean theorem:
$XY^{2} = XO^{2} + YO^{2}$ $(1)$
$YZ^{2} = YO^{2} + ZO^{2}$ $(2)$
$ZW^{2} = ZO^{2} + WO^{2}$ $(3)$
$XW^{2} = WO^{2} + XO^{2}$ $(4)$
Adding equations $(1)$ and $(3)$:
$XY^{2} + ZW^{2} = (XO^{2} + YO^{2}) + (ZO^{2} + WO^{2})$
Rearranging the terms:
$XY^{2} + ZW^{2} = (YO^{2} + ZO^{2}) + (WO^{2} + XO^{2})$
Substituting equations $(2)$ and $(4)$ into the expression:
$XY^{2} + ZW^{2} = YZ^{2} + XW^{2}$
Hence,it is proved.

(N/A) Let $ABCD$ be a rectangle with sides $AB = CD = l$ and $BC = DA = b$. Let the diagonals be $AC$ and $BD$.
In a rectangle,all interior angles are $90^{\circ}$.
Consider the right-angled triangle $\triangle ABC$. By the Pythagoras theorem:
$AC^2 = AB^2 + BC^2 = l^2 + b^2$
Similarly,in the right-angled triangle $\triangle BCD$:
$BD^2 = BC^2 + CD^2 = b^2 + l^2$
Sum of the squares of the diagonals:
$AC^2 + BD^2 = (l^2 + b^2) + (b^2 + l^2) = 2l^2 + 2b^2$
Sum of the squares of the sides:
$AB^2 + BC^2 + CD^2 + DA^2 = l^2 + b^2 + l^2 + b^2 = 2l^2 + 2b^2$
Thus,the sum of the squares of the sides is equal to the sum of the squares of the diagonals.

(N/A) Given: In $\Delta PQR$,$\overline{QM} \perp \overline{PR}$. Let $PM = x$. Then $PQ = 2x$ and $RM = 3x$.
In right-angled $\Delta QMP$,by Pythagoras theorem: $PQ^2 = PM^2 + QM^2 \implies (2x)^2 = x^2 + QM^2 \implies 4x^2 = x^2 + QM^2 \implies QM^2 = 3x^2$.
In right-angled $\Delta QMR$,by Pythagoras theorem: $QR^2 = QM^2 + RM^2 \implies QR^2 = 3x^2 + (3x)^2 \implies QR^2 = 3x^2 + 9x^2 = 12x^2$.
Now,consider the sides of $\Delta PQR$: $PR = PM + RM = x + 3x = 4x$. So,$PR^2 = (4x)^2 = 16x^2$.
Check if $PQ^2 + QR^2 = PR^2$: $PQ^2 + QR^2 = (2x)^2 + 12x^2 = 4x^2 + 12x^2 = 16x^2$.
Since $PQ^2 + QR^2 = PR^2$,by the converse of Pythagoras theorem,$\Delta PQR$ is a right-angled triangle with $\angle PQR = 90^\circ$.

(N/A) In $\Delta ABC$,$\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$.
By the property of geometric mean in a right-angled triangle,we have $BM^2 = AM \cdot CM$.
Given that $BM = 2 CM$,substitute this into the equation:
$(2 CM)^2 = AM \cdot CM$
$4 CM^2 = AM \cdot CM$
Since $CM \neq 0$,we can divide both sides by $CM$ to get $AM = 4 CM$.
Now,the hypotenuse $AC$ is the sum of segments $AM$ and $CM$:
$AC = AM + CM$
$AC = 4 CM + CM$
$AC = 5 CM$.
Hence,the result is proved.

(N/A) $1$. In $\Delta PQR$,$\angle Q = 90^{\circ}$. By Pythagoras theorem,$PR^{2} = PQ^{2} + QR^{2}$.
$2$. Since $\overline{PM}$ is a median to side $\overline{QR}$,$M$ is the midpoint of $\overline{QR}$. Therefore,$QM = RM = \frac{1}{2} QR$,which implies $QR = 2RM$.
$3$. In right-angled $\Delta PQM$,by Pythagoras theorem,$PM^{2} = PQ^{2} + QM^{2}$. Thus,$PQ^{2} = PM^{2} - QM^{2}$.
$4$. Substitute $PQ^{2} = PM^{2} - QM^{2}$ and $QR = 2RM$ into the first equation: $PR^{2} = (PM^{2} - QM^{2}) + (2RM)^{2}$.
$5$. Since $QM = RM$,substitute $QM$ with $RM$: $PR^{2} = PM^{2} - RM^{2} + 4RM^{2}$.
$6$. Simplifying the expression,we get $PR^{2} = PM^{2} + 3RM^{2}$. Hence proved.

(N/A) Let $BC$ be divided into three equal parts by points $D$ and $E$. Let $BD = DE = EC = x$. Then $BE = 2x$ and $BC = 3x$.
In right-angled $\Delta ABE$,by Pythagoras theorem: $AE^{2} = AB^{2} + BE^{2} = AB^{2} + (2x)^{2} = AB^{2} + 4x^{2}$.
Multiplying by $8$: $8AE^{2} = 8AB^{2} + 32x^{2}$.
In right-angled $\Delta ABC$,by Pythagoras theorem: $AC^{2} = AB^{2} + BC^{2} = AB^{2} + (3x)^{2} = AB^{2} + 9x^{2}$.
Multiplying by $3$: $3AC^{2} = 3AB^{2} + 27x^{2}$.
Subtracting the two equations: $8AE^{2} - 3AC^{2} = (8AB^{2} + 32x^{2}) - (3AB^{2} + 27x^{2}) = 5AB^{2} + 5x^{2}$.
Since $DE = x$,$DE^{2} = x^{2}$. Thus,$8AE^{2} - 3AC^{2} = 5AB^{2} + 5DE^{2}$.

(N/A) Let the diagonals $AC$ and $BD$ of the rhombus $ABCD$ intersect at point $O$.
Since the diagonals of a rhombus bisect each other at right angles $(90^{\circ})$,we have $AO = OC = \frac{AC}{2}$ and $BO = OD = \frac{BD}{2}$.
In the right-angled triangle $\triangle AOB$,by the Pythagoras theorem:
$AB^{2} = AO^{2} + BO^{2}$
Substitute the values of $AO$ and $BO$:
$AB^{2} = (\frac{AC}{2})^{2} + (\frac{BD}{2})^{2}$
$AB^{2} = \frac{AC^{2}}{4} + \frac{BD^{2}}{4}$
$AB^{2} = \frac{AC^{2} + BD^{2}}{4}$
Multiplying both sides by $4$,we get:
$4AB^{2} = AC^{2} + BD^{2}$
Hence,it is proved that $AC^{2} + BD^{2} = 4AB^{2}$.

(A) $1$. In $\Delta ABC$,since $\angle B = 90^{\circ}$,by Pythagoras theorem,$AC^{2} = AB^{2} + BC^{2}$ (Equation $1$).
$2$. Since $\overline{AD}$ is a median to side $\overline{BC}$,$D$ is the midpoint of $\overline{BC}$. Thus,$BD = DC = \frac{1}{2} BC$,which implies $BC = 2BD$.
$3$. In $\Delta ABD$,since $\angle B = 90^{\circ}$,by Pythagoras theorem,$AD^{2} = AB^{2} + BD^{2}$.
$4$. Rearranging for $BD^{2}$,we get $BD^{2} = AD^{2} - AB^{2}$.
$5$. Substitute $BC = 2BD$ into Equation $1$: $AC^{2} = AB^{2} + (2BD)^{2} = AB^{2} + 4BD^{2}$.
$6$. Substitute $BD^{2} = AD^{2} - AB^{2}$ into the expression: $AC^{2} = AB^{2} + 4(AD^{2} - AB^{2})$.
$7$. Simplify: $AC^{2} = AB^{2} + 4AD^{2} - 4AB^{2} = 4AD^{2} - 3AB^{2}$.
$8$. Hence,$AC^{2} = 4AD^{2} - 3AB^{2}$ is proved.

(N/A) Given: In quadrilateral $ABCD$,$\angle B = 90^{\circ}$ and $AD^{2} = AB^{2} + BC^{2} + CD^{2}$.
To prove: $\angle ACD = 90^{\circ}$.
Proof: Join $AC$. In $\triangle ABC$,since $\angle B = 90^{\circ}$,by Pythagoras theorem,$AC^{2} = AB^{2} + BC^{2}$.
Substitute this into the given equation: $AD^{2} = (AB^{2} + BC^{2}) + CD^{2}$.
This becomes $AD^{2} = AC^{2} + CD^{2}$.
In $\triangle ACD$,we have $AD^{2} = AC^{2} + CD^{2}$.
By the converse of the Pythagoras theorem,if the square of one side is equal to the sum of the squares of the other two sides,then the angle opposite to the first side is a right angle.
Therefore,$\angle ACD = 90^{\circ}$.

(N/A) In $\Delta ABM$,by Pythagoras theorem: $AB^{2} = AM^{2} + BM^{2}$ $(1)$.
In $\Delta ACM$,by Pythagoras theorem: $AC^{2} = AM^{2} + MC^{2}$ $(2)$.
Subtracting $(2)$ from $(1)$: $AB^{2} - AC^{2} = BM^{2} - MC^{2} = (BM - MC)(BM + MC)$.
Since $D$ is the midpoint of $BC$,$BD = DC = \frac{BC}{2}$.
We can express $BM$ as $BD + DM$ and $MC$ as $DC - DM = BD - DM$.
Substituting these: $BM - MC = (BD + DM) - (BD - DM) = 2DM$.
Also,$BM + MC = BC$.
Therefore,$AB^{2} - AC^{2} = (2DM)(BC) = 2 \cdot BC \cdot DM$.

(A) Given that in $\Delta ABC$,$m\angle A + m\angle C = m\angle B$.
We know that the sum of all angles in a triangle is $180^{\circ}$,so $m\angle A + m\angle B + m\angle C = 180^{\circ}$.
Substituting $m\angle A + m\angle C = m\angle B$ into the equation,we get $m\angle B + m\angle B = 180^{\circ}$,which implies $2m\angle B = 180^{\circ}$,so $m\angle B = 90^{\circ}$.
Since $\angle B = 90^{\circ}$,$\Delta ABC$ is a right-angled triangle with $AC$ as the hypotenuse.
By the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.
Substituting the given values,$AC^2 = 7^2 + 24^2 = 49 + 576 = 625$.
Therefore,$AC = \sqrt{625} = 25$.

(D) In $\Delta PQR$,$\angle P = 90^{\circ}$ and $\overline{PM} \perp \overline{QR}$.
According to the geometric mean theorem (or property of altitude to the hypotenuse),$PQ^2 = QM \cdot QR$.
Given $PQ = \sqrt{20}$,so $PQ^2 = 20$.
Substituting the values: $20 = 4 \cdot QR$.
Therefore,$QR = \frac{20}{4} = 5$.
Since $QR = QM + RM$,we have $5 = 4 + RM$.
Thus,$RM = 5 - 4 = 1$.

(A) Let the rhombus be $ABCD$ with diagonals $AC = 24$ and $BD = 10$.
In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.
Let the intersection point be $O$.
Then,$AO = AC / 2 = 24 / 2 = 12$ and $BO = BD / 2 = 10 / 2 = 5$.
In the right-angled triangle $\triangle AOB$,by the Pythagorean theorem:
$AB^2 = AO^2 + BO^2$
$AB^2 = 12^2 + 5^2$
$AB^2 = 144 + 25 = 169$
$AB = \sqrt{169} = 13$.
Since all sides of a rhombus are equal,the length of each side is $13$.

(B) In $\Delta XYZ$,since $m\angle Y = 90^{\circ}$,the triangle is a right-angled triangle.
Using the Pythagorean theorem,$XZ^2 = XY^2 + YZ^2$.
Substituting the given values,$XZ^2 = 3^2 + 4^2 = 9 + 16 = 25$.
Therefore,$XZ = \sqrt{25} = 5$.
In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.
Since $M$ is the midpoint of $\overline{XZ}$,$\overline{YM}$ is the median to the hypotenuse $\overline{XZ}$.
Thus,$YM = \frac{1}{2} \times XZ = \frac{1}{2} \times 5 = 2.5$.

(C) Let the length of the ladder be $L = 17 \, m$.
Let the distance of the base of the ladder from the wall be $b = 8 \, m$.
Let the height reached by the ladder on the wall be $h$.
The wall,the ground,and the ladder form a right-angled triangle where the ladder is the hypotenuse.
According to the Pythagorean theorem: $h^2 + b^2 = L^2$.
Substituting the values: $h^2 + 8^2 = 17^2$.
$h^2 + 64 = 289$.
$h^2 = 289 - 64 = 225$.
$h = \sqrt{225} = 15 \, m$.
Thus,the upper end of the ladder reaches a height of $15 \, m$ on the wall.

(A) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$,we use the geometric mean theorem (or properties of similar triangles).
According to the property of altitudes in a right triangle,the square of the leg is equal to the product of the hypotenuse segment adjacent to it and the entire hypotenuse.
Specifically,$BC^2 = CM \times AC$.
Given $AM = 7$ and $CM = 9$,the total length of the hypotenuse $AC = AM + CM = 7 + 9 = 16$.
Substituting the values into the formula: $BC^2 = 9 \times 16$.
$BC^2 = 144$.
Taking the square root of both sides,$BC = \sqrt{144} = 12$.

(B) In $\Delta ABC$,since $m \angle A = 90^\circ$,the triangle is a right-angled triangle with the hypotenuse $a$ opposite to $\angle A$.
According to the Pythagorean theorem,$a^2 = b^2 + c^2$.
Given $b = 12$ and $c = 35$.
$a^2 = 12^2 + 35^2$
$a^2 = 144 + 1225$
$a^2 = 1369$
$a = \sqrt{1369} = 37$.
Therefore,$a = 37$.

(C) Given that in $\Delta PQR$,$m \angle Q = 90^\circ$,the triangle is a right-angled triangle.
Let $PQ = x$ and $QR = x$ (since $PQ : QR = 1 : 1$).
According to the Pythagorean theorem,$PR^2 = PQ^2 + QR^2$.
Substituting the values,$PR^2 = x^2 + x^2 = 2x^2$.
Therefore,$PR = \sqrt{2x^2} = x\sqrt{2}$.
We need to find the ratio $PQ : PR$.
$PQ : PR = x : x\sqrt{2} = 1 : \sqrt{2}$.

(A) According to Apollonius's Theorem for $\Delta MNP$ where $\overline{MX}$ is the median to side $\overline{NP}$:
$MN^2 + MP^2 = 2(MX^2 + NX^2)$
Given $MN^2 + MP^2 = 50$ and $MX = 3$,substitute these values into the formula:
$50 = 2(3^2 + NX^2)$
$50 = 2(9 + NX^2)$
$25 = 9 + NX^2$
$NX^2 = 25 - 9 = 16$
$NX = \sqrt{16} = 4$
Since $\overline{MX}$ is a median,$X$ is the midpoint of $\overline{NP}$,so $NP = 2 \times NX$.
$NP = 2 \times 4 = 8$.

(C) In a right-angled triangle $\Delta PQR$ where $m \angle Q = 90^{\circ}$,$\overline{QD}$ is the altitude drawn to the hypotenuse $\overline{PR}$.
According to the property of similar triangles in a right triangle,the altitude to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and to each other.
Specifically,$\Delta PQD \sim \Delta QRD$.
In $\Delta PQD$,the angles are $\angle P$,$\angle PQD$,and $\angle PDQ = 90^{\circ}$.
In $\Delta QRD$,the angles are $\angle R$,$\angle RQD$,and $\angle RDQ = 90^{\circ}$.
Since $\angle P + \angle R = 90^{\circ}$ and $\angle P + \angle PQD = 90^{\circ}$,it follows that $\angle PQD = \angle R$.
Looking at the options,the angle $\angle DRQ$ (which is the same as $\angle R$) is the correct corresponding angle to $\angle PQD$ in the context of the geometric properties of the altitude of a right triangle.

(A) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$,if $\overline{BM}$ is the altitude to the hypotenuse $\overline{AC}$,then $\Delta AMB \sim \Delta BMC$.
From the property of similar triangles,the ratio of corresponding sides is equal:
$\frac{AM}{BM} = \frac{BM}{CM}$.
Cross-multiplying gives $BM^2 = AM \cdot CM$.
Therefore,$BM = \sqrt{AM \cdot CM}$,which means $BM$ is the geometric mean of $AM$ and $CM$.

(A) In a right-angled triangle $\Delta PQR$ with $\angle Q = 90^{\circ}$,$\overline{QM}$ is the altitude to the hypotenuse $PR$.
According to the geometric mean theorem for right triangles,$QM^2 = PM \times MR$.
Given $QM = 6$ and $MR = 9$,we have $6^2 = PM \times 9$.
$36 = PM \times 9$,which implies $PM = 36 / 9 = 4$.
The length of the hypotenuse $PR = PM + MR = 4 + 9 = 13$.

(B) In a right-angled triangle,the altitude drawn to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and to each other.
Specifically,$\Delta AMB \sim \Delta BMC$.
From the similarity of these triangles,the ratio of their corresponding sides is equal:
$\frac{AM}{BM} = \frac{BM}{CM}$.
Substituting the given values:
$\frac{12}{BM} = \frac{BM}{3}$.
Cross-multiplying gives:
$BM^2 = 12 \times 3$.
$BM^2 = 36$.
Taking the square root of both sides:
$BM = \sqrt{36} = 6$.
Therefore,the length of $BM$ is $6$.

(D) Since $\Delta ABC$ is a right-angled triangle with $m\angle B = 90^\circ$,we can apply the Pythagorean theorem.
According to the Pythagorean theorem: $(AC)^2 = (AB)^2 + (BC)^2$.
Given $AB = 12$ and $BC = 5$.
Substituting the values: $(AC)^2 = (12)^2 + (5)^2$.
$(AC)^2 = 144 + 25 = 169$.
Taking the square root on both sides: $AC = \sqrt{169} = 13$.
Thus,the length of the hypotenuse $AC$ is $13$.

(A) In $\Delta PQR$,since $m \angle Q = 90^{\circ}$,the triangle is a right-angled triangle.
According to the Pythagoras theorem,the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Here,$PR$ is the hypotenuse (the side opposite to the $90^{\circ}$ angle).
Thus,$PR^{2} = PQ^{2} + QR^{2}$.
Substituting the given values: $17^{2} = 8^{2} + QR^{2}$.
$289 = 64 + QR^{2}$.
$QR^{2} = 289 - 64$.
$QR^{2} = 225$.
Taking the square root on both sides,$QR = \sqrt{225} = 15$.

(C) In a right-angled triangle $\Delta ABC$ where $m\angle B = 90^{\circ}$,the side $b$ represents the hypotenuse.
According to the Pythagorean theorem,$b^2 = a^2 + c^2$.
Given $a = 16$ and $c = 12$.
Substituting the values: $b^2 = 16^2 + 12^2$.
$b^2 = 256 + 144$.
$b^2 = 400$.
Taking the square root on both sides,$b = \sqrt{400} = 20$.

(D) In $\Delta PQR$,since $m\angle Q = 90^\circ$,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,$PR$ is the hypotenuse.
So,$PR^2 = PQ^2 + QR^2$.
Given $PQ = 8$ and $QR = 15$.
$PR^2 = 8^2 + 15^2$.
$PR^2 = 64 + 225$.
$PR^2 = 289$.
Taking the square root on both sides,$PR = \sqrt{289} = 17$.

(A) In $\Delta DEF$,since $m \angle D = 90^{\circ}$,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,$EF$ is the hypotenuse because it is opposite to the right angle at $D$.
Therefore,$EF^2 = DE^2 + DF^2$.
Given $EF = 6$ and $DF = 4$,we substitute these values into the equation:
$6^2 = DE^2 + 4^2$
$36 = DE^2 + 16$
$DE^2 = 36 - 16$
$DE^2 = 20$
$DE = \sqrt{20} = \sqrt{4 \times 5} = 2 \sqrt{5}$.

(B) Given that $\Delta XYZ$ is a right-angled triangle with $m\angle Y = 90^{\circ}$.
Since $XY = YZ$,let $XY = YZ = x$.
According to the Pythagorean theorem,$XY^2 + YZ^2 = XZ^2$.
Substituting the given values,$x^2 + x^2 = 10^2$.
$2x^2 = 100$.
$x^2 = 50$.
$x = \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$.
Therefore,$XY = 5\sqrt{2}$.

(C) Let the side of the square be $a$.
In a square, the diagonal $d$ is related to the side $a$ by the formula $d = a \sqrt{2}$.
Given that the diagonal $d = 6$, we have:
$a \sqrt{2} = 6$
$a = \frac{6}{\sqrt{2}}$
To rationalize the denominator, multiply the numerator and denominator by $\sqrt{2}$:
$a = \frac{6 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{6 \sqrt{2}}{2} = 3 \sqrt{2}$.
Therefore, the length of the side of the square is $3 \sqrt{2}$.

(A) In a rectangle,each angle is a right angle $(90^{\circ})$.
Let the sides of the rectangle be $a = 12$ and $b = 35$.
The diagonal $d$ of the rectangle forms a right-angled triangle with the two adjacent sides.
According to the Pythagorean theorem,$d^2 = a^2 + b^2$.
Substituting the given values: $d^2 = 12^2 + 35^2$.
$d^2 = 144 + 1225$.
$d^2 = 1369$.
Taking the square root of both sides: $d = \sqrt{1369} = 37$.
Therefore,the length of the diagonal is $37$.

(B) In a rectangle $ABCD$,the angle $\angle B = 90^{\circ}$.
Therefore,$\triangle ABC$ is a right-angled triangle with $AB$ and $BC$ as the legs and $AC$ as the hypotenuse.
According to the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.
Given $AB = 2.4$ and $BC = 3.2$.
$AC^2 = (2.4)^2 + (3.2)^2$.
$AC^2 = 5.76 + 10.24$.
$AC^2 = 16$.
Taking the square root on both sides,$AC = \sqrt{16} = 4$.

(C) In a right-angled triangle,the length of the median to the hypotenuse is equal to half the length of the hypotenuse.
Given that $\Delta ABC$ is a right-angled triangle with $m\angle B = 90^{\circ}$,the side $\overline{AC}$ is the hypotenuse.
$\overline{BM}$ is the median to the hypotenuse $\overline{AC}$.
Therefore,$BM = \frac{1}{2} \times AC$.
Given $AC = 20$,we have $BM = \frac{1}{2} \times 20 = 10$.
Thus,the correct option is $C$.

(D) In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.
Given that $\Delta XYZ$ is a right-angled triangle with $m\angle Y = 90^{\circ}$,the side $XZ$ is the hypotenuse.
$\overline{YP}$ is the median to the hypotenuse $XZ$.
Therefore,$YP = \frac{1}{2} \times XZ$.
Given $YP = 6$,we have $6 = \frac{1}{2} \times XZ$.
Multiplying both sides by $2$,we get $XZ = 6 \times 2 = 12$.

(C) In $\triangle PQR$,we are given $PQ = 10$,$QR = 10$,and $PR = 10\sqrt{2}$.
Check if $\triangle PQR$ is a right-angled triangle using the converse of the Pythagoras theorem:
$PQ^2 + QR^2 = 10^2 + 10^2 = 100 + 100 = 200$.
$PR^2 = (10\sqrt{2})^2 = 100 \times 2 = 200$.
Since $PQ^2 + QR^2 = PR^2$,$\triangle PQR$ is a right-angled triangle with $\angle PQR = 90^\circ$.
In a quadrilateral,if two adjacent sides are equal and the angle between them is $90^\circ$,and no further information is provided about the other sides,the most specific classification based on the given properties of the triangle formed by the diagonal is that it contains a right angle. However,among the given options,if $PQ = QR = 10$,it satisfies the condition for a rhombus or a square. Given the diagonal $PR = 10\sqrt{2}$ confirms the right angle at $Q$. Without information about $PS$ and $RS$,but assuming it is a standard geometry problem where $PQRS$ is a rhombus with a right angle,it becomes a square.

(D) Let the length of the bamboo be the hypotenuse $c = 15 \, m$.
Let the height reached on the wall be one leg $a = 12 \, m$.
Let the distance from the base of the wall be the other leg $b$.
According to the Pythagorean theorem,$a^2 + b^2 = c^2$.
Substituting the values,we get $(12)^2 + b^2 = (15)^2$.
$144 + b^2 = 225$.
$b^2 = 225 - 144 = 81$.
$b = \sqrt{81} = 9 \, m$.
Thus,the lower end of the bamboo is $9 \, m$ away from the base of the wall.

(B) In a rectangle $ABCD$,$\angle B = 90^{\circ}$.
By the Pythagorean theorem in $\triangle ABC$,we have $AC^{2} = AB^{2} + BC^{2}$.
Given that $AB^{2} + BC^{2} = 64$.
Therefore,$AC^{2} = 64$.
Taking the square root on both sides,$AC = \sqrt{64} = 8$.
Thus,the length of the diagonal $AC$ is $8$.

(D) In $\Delta ABC$,$\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$.
By the property of geometric mean in a right-angled triangle,$BM^2 = AM \cdot MC$.
Given $BM = 12$ and $AM = 9$,we have $12^2 = 9 \cdot MC$.
$144 = 9 \cdot MC \implies MC = \frac{144}{9} = 16$.
In $\Delta ABM$,by Pythagoras theorem,$AB^2 = AM^2 + BM^2 = 9^2 + 12^2 = 81 + 144 = 225$.
So,$AB = \sqrt{225} = 15$.
In $\Delta ABC$,by Pythagoras theorem,$AC^2 = AB^2 + BC^2$.
First,find $BC^2$ in $\Delta BMC$: $BC^2 = BM^2 + MC^2 = 12^2 + 16^2 = 144 + 256 = 400$.
Thus,$AC^2 = 225 + 400 = 625$.
$AC = \sqrt{625} = 25$.
Alternatively,$AC = AM + MC = 9 + 16 = 25$.

(B) Let the side of the square be $a$.
We know that the diagonal $d$ of a square is given by the formula $d = a \sqrt{2}$.
Given that the diagonal $d = 12$,we have:
$12 = a \sqrt{2}$
$a = \frac{12}{\sqrt{2}}$
To rationalize the denominator,multiply the numerator and denominator by $\sqrt{2}$:
$a = \frac{12 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}}$
$a = \frac{12 \sqrt{2}}{2}$
$a = 6 \sqrt{2}$
Therefore,the length of each side of the square is $6 \sqrt{2}$.

(D) Given that $\Delta PQR$ is a right-angled triangle with $m\angle Q = 90^{\circ}$.
Using the Pythagorean theorem,$PR^2 = PQ^2 + QR^2$.
Substituting the given values,$PR^2 = 6^2 + 8^2 = 36 + 64 = 100$.
Thus,$PR = \sqrt{100} = 10$.
In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.
Since $T$ is the midpoint of $\overline{PR}$,$\overline{QT}$ is the median to the hypotenuse $\overline{PR}$.
Therefore,$QT = \frac{1}{2} \times PR = \frac{1}{2} \times 10 = 5$.

(A) Let the length of the ladder be the hypotenuse $h = 2.6 \,m$.
Let the distance of the lower end from the wall be the base $b = 1 \,m$.
Let the height reached by the ladder on the wall be $a$.
According to the Pythagorean theorem,$a^2 + b^2 = h^2$.
Substituting the values: $a^2 + (1)^2 = (2.6)^2$.
$a^2 + 1 = 6.76$.
$a^2 = 6.76 - 1 = 5.76$.
$a = \sqrt{5.76} = 2.4 \,m$.
Thus,the upper end reaches a height of $2.4 \,m$ on the wall.

(B) triangle is a right-angled triangle if the square of the longest side is equal to the sum of the squares of the other two sides (Pythagoras theorem).
For option $A$: $6^2 + 8^2 = 36 + 64 = 100$,while $12^2 = 144$. Since $100 \neq 144$,it is not a right-angled triangle.
For option $B$: $7^2 + 24^2 = 49 + 576 = 625$,and $25^2 = 625$. Since $625 = 625$,it satisfies the Pythagoras theorem.
For option $C$: $7^2 + 15^2 = 49 + 225 = 274$,while $17^2 = 289$. Since $274 \neq 289$,it is not a right-angled triangle.
For option $D$: $3^2 + 7^2 = 9 + 49 = 58$,while $9^2 = 81$. Since $58 \neq 81$,it is not a right-angled triangle.
Therefore,the sides $7, 24, 25$ form a right-angled triangle.

(B) In an isosceles right-angled triangle,the two sides containing the right angle are equal in length.
Let the length of each of these sides be $x \, cm$.
By the 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.
Therefore,$(x)^2 + (x)^2 = (8)^2$.
$2x^2 = 64$.
$x^2 = 32$.
$x = \sqrt{32} = \sqrt{16 \times 2} = 4 \sqrt{2} \, cm$.
Since the length cannot be negative,the length of each of the other two sides is $4 \sqrt{2} \, cm$.

(C) According to the Angle Bisector Theorem,in $\Delta PQR$,if $PS$ is the bisector of $\angle P$,then $\frac{PQ}{PR} = \frac{QS}{SR}$.
Given that $\frac{PQ}{PR} = \frac{5}{4}$ and $SR = 5.6 \text{ cm}$.
Substituting the values,we get $\frac{5}{4} = \frac{QS}{5.6}$.
Therefore,$QS = \frac{5 \times 5.6}{4} = 5 \times 1.4 = 7 \text{ cm}$.
Now,$QR = QS + SR = 7 + 5.6 = 12.6 \text{ cm}$.

(D) Given that in $\Delta ABC$,$\overline{MN} \parallel \overline{BC}$.
According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Therefore,$\frac{AM}{AB} = \frac{AN}{AC}$.
Given $\frac{AM}{AB} = \frac{2}{3}$ and $AC = 15$.
Substituting the values,we get $\frac{2}{3} = \frac{AN}{15}$.
$AN = \frac{2 \times 15}{3} = 10$.
Since $A-N-C$,we have $AC = AN + NC$.
$15 = 10 + NC$.
$NC = 15 - 10 = 5$.

(C) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Since $\overline{DE} \parallel \overline{BC}$,we have $\frac{AD}{DB} = \frac{AE}{EC}$.
Given $AB:AC = 3:4$,let $AB = 3k$ and $AC = 4k$.
By the property of similar triangles,$\Delta ADE \sim \Delta ABC$.
Therefore,$\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}$.
From the given ratio,$\frac{AD}{AB} = \frac{AE}{AC} = \frac{3}{4}$.
This implies $AD = 3x$ and $AB = 4x$ (where $AB$ is the total length),but here $AB$ is given as $3k$. Thus,$AD = \frac{3}{4} AB$ and $AE = \frac{3}{4} AC$.
Then $BD = AB - AD = AB - \frac{3}{4} AB = \frac{1}{4} AB$.
Similarly,$EC = AC - AE = AC - \frac{3}{4} AC = \frac{1}{4} AC$.
Now,calculating the ratio $\frac{EC}{BD} = \frac{\frac{1}{4} AC}{\frac{1}{4} AB} = \frac{AC}{AB} = \frac{4}{3}$.
Thus,$EC:BD = 4:3$.

(C) According to the Angle Bisector Theorem,in $\Delta ABC$,if the bisector of $\angle A$ intersects $\overline{BC}$ at $D$,then $\frac{AB}{AC} = \frac{BD}{DC}$.
Given $\frac{AB}{AC} = \frac{3}{4}$ and $BD = 4.5$.
Substituting the values: $\frac{3}{4} = \frac{4.5}{DC}$.
Therefore,$DC = \frac{4.5 \times 4}{3} = 1.5 \times 4 = 6$.
Since $BC = BD + DC$,we have $BC = 4.5 + 6 = 10.5$.

(B) In $\Delta ABC$,$\overline{BE}$ is a median to side $\overline{AC}$.
Therefore,$E$ is the midpoint of $\overline{AC}$,which implies $AE = EC$,or $AC = 2EC$.
In $\Delta CEB$,$D$ is the midpoint of $\overline{BC}$ (since $\overline{AD}$ is a median) and line $m$ (passing through $D$) is parallel to $\overline{BE}$.
By the Converse of the Midpoint Theorem in $\Delta CEB$,since $D$ is the midpoint of $\overline{BC}$ and $\overline{DK} \parallel \overline{BE}$,$K$ must be the midpoint of $\overline{EC}$.
Therefore,$EK = KC = 2.5$.
Thus,$EC = EK + KC = 2.5 + 2.5 = 5$.
Finally,$AC = 2EC = 2 \times 5 = 10$.