Mix Examples - Triangles Questions in English

(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 $PQ : PR$.
$PQ : PR = x : x\sqrt{2} = 1 : \sqrt{2}$.

(D) Given that $\Delta XYZ$ is a right-angled triangle with $m\angle X = 90^{\circ}$.
Here,$YZ$ is the hypotenuse and $XY$ is one of the legs.
Using the Pythagorean theorem: $XY^2 + XZ^2 = YZ^2$.
Substituting the given values: $8^2 + XZ^2 = 17^2$.
$64 + XZ^2 = 289$.
$XZ^2 = 289 - 64 = 225$.
$XZ = \sqrt{225} = 15$.
The area of a right-angled triangle is given by $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
$\text{Area} = \frac{1}{2} \times XY \times XZ$.
$\text{Area} = \frac{1}{2} \times 8 \times 15 = 4 \times 15 = 60$.

(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 $NP = 2 \times NX$.
$NP = 2 \times 4 = 8$.

(B) In $\Delta ABC$,$\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$.
Consider $\Delta ABM$ and $\Delta ABC$.
In $\Delta ABM$,$\angle AMB = 90^{\circ}$. Thus,$\angle BAM + \angle ABM = 90^{\circ}$.
In $\Delta ABC$,$\angle B = 90^{\circ}$. Thus,$\angle BAM + \angle C = 90^{\circ}$.
Comparing these two equations,we get $\angle ABM = \angle C$ (or $\angle MCB$).
Also,in $\Delta BMC$,$\angle BMC = 90^{\circ}$,so $\angle MBC + \angle C = 90^{\circ}$.
Since $\angle BAM + \angle C = 90^{\circ}$ and $\angle MBC + \angle C = 90^{\circ}$,it follows that $\angle BAM = \angle MBC$.

(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.

(D) In $\Delta PQR$,$\angle Q = 90^{\circ}$ and $\overline{QD} \perp \overline{PR}$.
By the property of geometric mean in a right-angled triangle,we have $QD^2 = PD \times DR$.
Also,in $\Delta PQD$ and $\Delta PQR$,by similarity criteria,we have $PQ^2 = PD \times PR$.
Given $PD = 9 DR$,we can write $PR = PD + DR = 9 DR + DR = 10 DR$.
Substituting these into the expression for $PQ^2$:
$PQ^2 = (9 DR) \times (10 DR) = 90 DR^2$.
Similarly,in $\Delta QDR$ and $\Delta PQR$,we have $QR^2 = DR \times PR$.
$QR^2 = DR \times (10 DR) = 10 DR^2$.
Now,find the ratio $\frac{PQ^2}{QR^2} = \frac{90 DR^2}{10 DR^2} = 9$.
Taking the square root on both sides,we get $\frac{PQ}{QR} = \sqrt{9} = 3$.
Therefore,$PQ = 3 \times QR$.

(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 $CM$.

(B) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$,the triangle is divided into two triangles $\Delta AMB$ and $\Delta BMC$ which are similar to the original triangle $\Delta ABC$ and to each other.
By the property of similarity,$\Delta AMB \sim \Delta ABC$.
This implies that the ratio of corresponding sides is equal: $\frac{AM}{AB} = \frac{AB}{AC}$.
Cross-multiplying gives $AB^2 = AM \cdot AC$.
Therefore,$AB$ is the geometric mean of $AM$ and $AC$.

(C) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$,if $\overline{BM}$ is the altitude to the hypotenuse $AC$,then by the properties of similar triangles $(\Delta BMC \sim \Delta BCA)$:
$\frac{BC}{AC} = \frac{MC}{BC}$
Cross-multiplying gives:
$BC^2 = MC \cdot AC$
This implies that $BC$ is the geometric mean of $CM$ and $AC$.

(D) 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,we have the ratio of corresponding sides:
$\frac{AM}{BM} = \frac{BM}{CM}$.
This leads to the geometric mean theorem: $BM^2 = AM \times CM$.
Given $AM = 8$ and $CM = 2$,we substitute these values into the equation:
$BM^2 = 8 \times 2 = 16$.
Taking the square root of both sides,we get $BM = \sqrt{16} = 4$.
Therefore,the length of $BM$ is $4$.

(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 $6$.

(C) 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 property of similarity,the ratio of corresponding sides is equal:
$\frac{AM}{BM} = \frac{BM}{CM}$.
This implies $BM^2 = AM \times CM$.
Given $AM = 4$ and $CM = 6$,we have:
$BM^2 = 4 \times 6 = 24$.
Taking the square root of both sides:
$BM = \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}$.

(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,$QR = \sqrt{225} = 15$.

(B) In a rectangle $ABCD$,all interior angles are $90^{\circ}$.
Consider the right-angled triangle $\triangle BCD$ (or $\triangle ABD$).
By the Pythagorean theorem,in $\triangle BCD$,we have $BD^2 = BC^2 + CD^2$.
Since $ABCD$ is a rectangle,$CD = AB = 9$.
Substituting the values,$BD^2 = 12^2 + 9^2$.
$BD^2 = 144 + 81 = 225$.
Taking the square root,$BD = \sqrt{225} = 15$.

(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 $3 \sqrt{2}$.

(D) 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 $d = 5 \sqrt{2}$.
Equating the two expressions: $a \sqrt{2} = 5 \sqrt{2}$.
Dividing both sides by $\sqrt{2}$,we get $a = 5$.
Therefore,the length of the side of the square is $5$.

(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 $XZ = 6 \times 2 = 12$.

(A) According to Apollonius's theorem for $\Delta ABC$ with median $\overline{AD}$:
$AB^2 + AC^2 = 2(AD^2 + BD^2)$
Given that $\overline{AD}$ is a median to side $\overline{BC}$,$D$ is the midpoint of $\overline{BC}$.
Therefore,$BD = DC = \frac{BC}{2} = \frac{18}{2} = 9$.
Substituting the given values into the theorem:
$8^2 + AC^2 = 2(7^2 + 9^2)$
$64 + AC^2 = 2(49 + 81)$
$64 + AC^2 = 2(130)$
$64 + AC^2 = 260$
$AC^2 = 260 - 64$
$AC^2 = 196$
$AC = \sqrt{196} = 14$.

(B) In $\Delta ABC$,the sum of angles is $180^{\circ}$.
Since $m \angle B = 90^{\circ}$ and $m \angle C = 30^{\circ}$,we have $m \angle A = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$.
This is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.
In a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle,the side opposite to the $30^{\circ}$ angle is half the hypotenuse.
The side opposite to $30^{\circ}$ is $AB$.
The hypotenuse is $AC = 10$.
Therefore,$AB = \frac{1}{2} \times AC = \frac{1}{2} \times 10 = 5$.

(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.

(A) In rectangle $HIJK$,$HJ$ is the diagonal and $HI$ is one of the sides.
Let the sides be $HI = l = 5$ and $IJ = w$. The diagonal $HJ$ forms a right-angled triangle $\triangle HIJ$ with sides $HI$ and $IJ$.
By the Pythagorean theorem,$HI^2 + IJ^2 = HJ^2$.
$5^2 + IJ^2 = 13^2$.
$25 + IJ^2 = 169$.
$IJ^2 = 169 - 25 = 144$.
$IJ = \sqrt{144} = 12$.
Now,the perimeter of the rectangle is given by $P = 2(l + w)$.
$P = 2(5 + 12) = 2(17) = 34$.

(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 $8$.

(C) First,identify the longest side of $\Delta ABC$. We have $AB = 4$,$BC = 2\sqrt{3} \approx 3.46$,and $AC = 2\sqrt{7} \approx 5.29$. Since $AC$ is the longest side,we need to find the length of the median $BD$ drawn to side $AC$.
Using Apollonius Theorem for $\Delta ABC$ with median $BD$ on side $AC$:
$AB^2 + BC^2 = 2(BD^2 + AD^2)$
Since $D$ is the midpoint of $AC$,$AD = AC/2 = (2\sqrt{7})/2 = \sqrt{7}$.
Substitute the values: $4^2 + (2\sqrt{3})^2 = 2(BD^2 + (\sqrt{7})^2)$
$16 + 12 = 2(BD^2 + 7)$
$28 = 2(BD^2 + 7)$
$14 = BD^2 + 7$
$BD^2 = 7$
$BD = \sqrt{7}$.

(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,$AC = AM + MC = 9 + 16 = 25$.

(A) In $\Delta PQR$,$\angle P = 90^{\circ}$ and $\overline{PX} \perp \overline{QR}$.
According to the geometric mean theorem (or property of altitude to the hypotenuse),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.
Specifically,$\Delta QXP \sim \Delta QPR$ and $\Delta PXR \sim \Delta QPR$.
From the similarity $\Delta QXP \sim \Delta QPR$,we have $\frac{QX}{PQ} = \frac{PQ}{QR}$.
Thus,$PQ^2 = QX \cdot QR$.
Given $QX = 4$ and $RX = 5$,we have $QR = QX + RX = 4 + 5 = 9$.
Substituting these values,$PQ^2 = 4 \cdot 9 = 36$.
Therefore,$PQ = \sqrt{36} = 6$.

(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 $6 \sqrt{2}$.

(C) Let the side of the square be $a$.
The length of the diagonal of a square is given by the formula $d = a \sqrt{2}$.
Given that the diagonal $d = 8 \sqrt{2}$.
Equating the two,we get $a \sqrt{2} = 8 \sqrt{2}$.
Dividing both sides by $\sqrt{2}$,we find the side length $a = 8$.
The area of a square is calculated as $\text{Area} = a^2$.
Substituting the value of $a$,we get $\text{Area} = 8^2 = 64$.
Alternatively,the area of a square can be calculated using the diagonal $d$ as $\text{Area} = \frac{1}{2} \times d^2$.
Substituting $d = 8 \sqrt{2}$,we get $\text{Area} = \frac{1}{2} \times (8 \sqrt{2})^2 = \frac{1}{2} \times (64 \times 2) = \frac{1}{2} \times 128 = 64$.

(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.

(A) In $\Delta XYZ$,the ratio of angles is $m \angle X : m \angle Y : m \angle Z = 2 : 3 : 5$.
Let the angles be $2k, 3k, 5k$.
Since the sum of angles in a triangle is $180^{\circ}$,we have $2k + 3k + 5k = 180^{\circ}$,which gives $10k = 180^{\circ}$,so $k = 18^{\circ}$.
Thus,$m \angle X = 2(18^{\circ}) = 36^{\circ}$,$m \angle Y = 3(18^{\circ}) = 54^{\circ}$,and $m \angle Z = 5(18^{\circ}) = 90^{\circ}$.
The correspondence $XYZ \leftrightarrow EFD$ is a similarity,which implies $\Delta XYZ \sim \Delta EFD$.
Therefore,the corresponding angles are equal: $m \angle X = m \angle E$,$m \angle Y = m \angle F$,and $m \angle Z = m \angle D$.
Since $m \angle Z = 90^{\circ}$,it follows that $m \angle D = 90^{\circ}$.
Thus,$\angle D$ is a right angle.

(C) Given that $\Delta ABC \sim \Delta XYZ$ under the correspondence $ABC \leftrightarrow ZXY$.
This implies the ratio of corresponding sides is equal:
$\frac{AB}{ZX} = \frac{BC}{XY} = \frac{CA}{YZ}$.
From the properties of ratios,we have $\frac{AB}{ZX} = \frac{BC + CA}{XY + YZ}$.
Substituting the given values: $AB = 12, BC = 8, CA = 10, ZX = 10$.
$\frac{12}{10} = \frac{8 + 10}{XY + YZ}$.
$\frac{6}{5} = \frac{18}{XY + YZ}$.
$XY + YZ = \frac{18 \times 5}{6} = 3 \times 5 = 15$.
Thus,$XY + YZ = 15$.

(B) For two similar triangles,the ratio of their areas is equal to the square of the ratio of their corresponding sides.
Let the areas be $A_1 = 25$ and $A_2 = 16$.
Let the corresponding sides be $s_1$ and $s_2$.
Then,$\frac{A_1}{A_2} = (\frac{s_1}{s_2})^2$.
$\frac{25}{16} = (\frac{s_1}{s_2})^2$.
Taking the square root on both sides,$\frac{s_1}{s_2} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.
The ratio of the perimeters of two similar triangles is equal to the ratio of their corresponding sides.
Therefore,the ratio of their perimeters is $5: 4$.

(C) Given that $\Delta ABC \sim \Delta PQR$ under the correspondence $ABC \leftrightarrow PQR$.
According to the theorem of areas of similar triangles,the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Therefore,$\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta PQR)} = \left(\frac{AB}{PQ}\right)^2$.
Substituting the given values: $\frac{36}{64} = \left(\frac{12}{PQ}\right)^2$.
Taking the square root on both sides: $\sqrt{\frac{36}{64}} = \frac{12}{PQ}$.
$\frac{6}{8} = \frac{12}{PQ}$.
Simplifying the fraction: $\frac{3}{4} = \frac{12}{PQ}$.
Cross-multiplying: $3 \times PQ = 12 \times 4$.
$3 \times PQ = 48$.
$PQ = \frac{48}{3} = 16$.

(D) Given that $\Delta ABC \sim \Delta XYZ$ for the correspondence $ABC \leftrightarrow XYZ$.
According to the theorem of areas of similar triangles,the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Therefore,$\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta XYZ)} = \left(\frac{BC}{YZ}\right)^2$.
Substituting the given values: $\frac{72}{\text{Area}(\Delta XYZ)} = \left(\frac{6}{10}\right)^2$.
$\frac{72}{\text{Area}(\Delta XYZ)} = \frac{36}{100}$.
$\text{Area}(\Delta XYZ) = \frac{72 \times 100}{36}$.
$\text{Area}(\Delta XYZ) = 2 \times 100 = 200$.

(C) In $\Delta ABC$,$D$,$E$,and $F$ are the midpoints of $\overline{AB}$,$\overline{BC}$,and $\overline{CA}$ respectively.
By the Midpoint Theorem,$EF \parallel AB$ and $EF = \frac{1}{2} AB$,$DF \parallel BC$ and $DF = \frac{1}{2} BC$,and $DE \parallel AC$ and $DE = \frac{1}{2} AC$.
Therefore,$\frac{DE}{AC} = \frac{EF}{AB} = \frac{DF}{BC} = \frac{1}{2}$.
By the $SSS$ similarity criterion,$\Delta DEF \sim \Delta CAB$ (since $D$ corresponds to $C$,$E$ corresponds to $A$,and $F$ corresponds to $B$).
Thus,the correspondence $DEF \leftrightarrow CAB$ is a similarity.

(A) In $\Delta ADB$ and $\Delta BDC$:
$\angle ADB = \angle BDC = 90^{\circ}$ (since $\overline{BD} \perp \overline{AC}$).
In $\Delta ABC$,$\angle A + \angle C = 90^{\circ}$.
In $\Delta ADB$,$\angle A + \angle ABD = 90^{\circ}$.
Therefore,$\angle ABD = \angle C$.
By the $AA$ similarity criterion,the correspondence $A \leftrightarrow B$,$D \leftrightarrow D$,and $B \leftrightarrow C$ holds.
Thus,$\Delta ADB \sim \Delta BDC$ is incorrect; the correct correspondence is $\Delta ADB \sim \Delta BDC$ is not the standard notation,but based on the vertices,$\Delta ADB \sim \Delta BDC$ implies $A \leftrightarrow B, D \leftrightarrow D, B \leftrightarrow C$.
Looking at the options,the correspondence $ADB \leftrightarrow BDC$ is the correct similarity.

(A) Given that the correspondences $ABC \leftrightarrow BAC$ and $ABC \leftrightarrow ACB$ are similarities.
From $ABC \leftrightarrow BAC$,we have $\angle A = \angle B$,$\angle B = \angle A$,and $\angle C = \angle C$.
From $ABC \leftrightarrow ACB$,we have $\angle A = \angle A$,$\angle B = \angle C$,and $\angle C = \angle B$.
Combining these,we get $\angle A = \angle B$ and $\angle B = \angle C$.
Therefore,$\angle A = \angle B = \angle C$.
Since all three angles are equal,the triangle must be equiangular.
An equiangular triangle is always an equilateral triangle.
Thus,$\Delta ABC$ is an equilateral triangle.