Similarity of Triangles and Basic Proportionality Theorem Questions in English

(A) In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.
Let the diagonals $AC$ and $BD$ intersect at point $O$.
Then,$AO = OC = \frac{AC}{2} = \frac{16}{2} = 8$ and $BO = OD = \frac{BD}{2} = \frac{30}{2} = 15$.
In the right-angled triangle $\triangle AOB$,by the Pythagoras theorem:
$AB^2 = AO^2 + BO^2$
$AB^2 = 8^2 + 15^2$
$AB^2 = 64 + 225 = 289$
$AB = \sqrt{289} = 17$.
The perimeter of a rhombus is $4 \times \text{side}$.
Perimeter $= 4 \times 17 = 68$.

(B) The total height of the bamboo tree is $8 \, m$.
It breaks at a height of $3 \, m$ from the ground.
Let the total height be $H = 8 \, m$ and the height at which it breaks be $h = 3 \, m$.
The length of the broken part (hypotenuse of the right-angled triangle formed) is $L = H - h = 8 - 3 = 5 \, m$.
The height of the remaining trunk is $h = 3 \, m$.
Let $x$ be the distance between the top of the tree on the ground and the base of the tree.
According to the Pythagorean theorem in the right-angled triangle formed:
$x^2 + h^2 = L^2$
$x^2 + 3^2 = 5^2$
$x^2 + 9 = 25$
$x^2 = 25 - 9 = 16$
$x = \sqrt{16} = 4 \, m$.
Thus,the distance is $4 \, m$.

(C) Given that $\Delta ABC$ is a right-angled triangle with $m \angle C = 90^\circ$.
According to the Pythagorean 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,$AB$ is the hypotenuse,and $AC$ and $BC$ are the other two sides.
Therefore,$AB^2 = AC^2 + BC^2$.
Substituting the given values: $(12.5)^2 = AC^2 + (12)^2$.
$156.25 = AC^2 + 144$.
$AC^2 = 156.25 - 144$.
$AC^2 = 12.25$.
Taking the square root of both sides,$AC = \sqrt{12.25} = 3.5$.

(D) Since $\Delta ABC$ is a right-angled triangle with $m \angle B = 90^{\circ}$,we can apply the Pythagorean theorem:
$AC^2 = AB^2 + BC^2$
Given $AB = 4$ and $BC = 7.5$,we substitute these values into the equation:
$AC^2 = 4^2 + (7.5)^2$
$AC^2 = 16 + 56.25$
$AC^2 = 72.25$
Taking the square root of both sides:
$AC = \sqrt{72.25} = 8.5$
Therefore,the length of $AC$ is $8.5$.

(D) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$,the hypotenuse $AC$ can be calculated using the Pythagorean theorem:
$AC^2 = AB^2 + BC^2$
$AC^2 = 15^2 + 20^2 = 225 + 400 = 625$
$AC = \sqrt{625} = 25$.
In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.
Therefore,$BM = \frac{1}{2} \times AC = \frac{1}{2} \times 25 = 12.5$.

(A) In a right-angled triangle $\Delta XYZ$,where $m\angle Y = 90^{\circ}$,the side $XZ$ is the hypotenuse.
According to the Pythagorean theorem,$(XZ)^{2} = (XY)^{2} + (YZ)^{2}$.
Substituting the given values: $(XZ)^{2} = (a^{2} - b^{2})^{2} + (2ab)^{2}$.
Expanding the terms: $(XZ)^{2} = (a^{4} - 2a^{2}b^{2} + b^{4}) + 4a^{2}b^{2}$.
Simplifying the expression: $(XZ)^{2} = a^{4} + 2a^{2}b^{2} + b^{4}$.
This is a perfect square: $(XZ)^{2} = (a^{2} + b^{2})^{2}$.
Taking the square root of both sides,we get $XZ = a^{2} + b^{2}$.

(C) In a rectangle $ABCD$,the angle $\angle ABC = 90^{\circ}$.
Therefore,$\triangle ABC$ is a right-angled triangle with $AC$ as the hypotenuse.
According to the Pythagorean theorem: $AB^2 + BC^2 = AC^2$.
Given $AB = 7.5$ and $AC = 19.5$.
Substituting the values: $(7.5)^2 + BC^2 = (19.5)^2$.
$56.25 + BC^2 = 380.25$.
$BC^2 = 380.25 - 56.25$.
$BC^2 = 324$.
$BC = \sqrt{324} = 18$.

(D) In $\Delta ABC$,since $m \angle B = 90^{\circ}$,by the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.
Substituting the given values,$AC^2 = 5^2 + 12^2 = 25 + 144 = 169$.
Thus,$AC = \sqrt{169} = 13$.
In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.
Therefore,$BD = \frac{1}{2} \times AC = \frac{1}{2} \times 13 = 6.5$.

(B) In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.
Let the diagonals $AC$ and $BD$ intersect at point $O$.
Then,$AO = \frac{AC}{2} = \frac{18}{2} = 9$ and $BO = \frac{BD}{2} = \frac{24}{2} = 12$.
In the right-angled triangle $\triangle AOB$,by the Pythagorean theorem:
$AB^2 = AO^2 + BO^2$
$AB^2 = 9^2 + 12^2$
$AB^2 = 81 + 144 = 225$
$AB = \sqrt{225} = 15$.
The perimeter of a rhombus is given by $4 \times \text{side}$.
Perimeter $= 4 \times 15 = 60$.

(C) In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.
Let the diagonals $AC$ and $BD$ intersect at point $O$.
Then,$AO = \frac{AC}{2} = \frac{15}{2} = 7.5$ and $BO = \frac{BD}{2} = \frac{36}{2} = 18$.
In the right-angled triangle $\triangle AOB$,by the Pythagorean theorem:
$AB^2 = AO^2 + BO^2$
$AB^2 = (7.5)^2 + (18)^2$
$AB^2 = 56.25 + 324 = 380.25$
$AB = \sqrt{380.25} = 19.5$.
The perimeter of a rhombus is $4 \times \text{side}$.
Perimeter $= 4 \times 19.5 = 78$.

(C) Given that $\Delta ABC$ is an isosceles triangle with $\overline{AB} \cong \overline{AC}$.
In an isosceles triangle,the median to the base is also the altitude to the base.
Since $\overline{AD}$ is the median to the base $\overline{BC}$,$\overline{AD} \perp \overline{BC}$ and $D$ is the midpoint of $\overline{BC}$.
Therefore,$BD = DC = \frac{BC}{2} = \frac{12}{2} = 6$.
In the right-angled triangle $\Delta ABD$,by the Pythagorean theorem:
$AB^2 = AD^2 + BD^2$
$AB^2 = 8^2 + 6^2$
$AB^2 = 64 + 36$
$AB^2 = 100$
$AB = \sqrt{100} = 10$.
Thus,the length of $AB$ is $10$.

(A) In a right-angled triangle $\Delta ABC$,where $\angle B = 90^{\circ}$,the sides $AB$ and $BC$ are the legs,and $AC$ is the hypotenuse.
According to the Pythagorean theorem: $AC^2 = AB^2 + BC^2$.
Given $AB = 11$ and $BC = 60$,we have $AC^2 = 11^2 + 60^2$.
$AC^2 = 121 + 3600 = 3721$.
Taking the square root,$AC = \sqrt{3721} = 61$.
The perimeter of $\Delta ABC$ is the sum of all its sides: $P = AB + BC + AC$.
$P = 11 + 60 + 61 = 132$.

(A) In a right-angled triangle $\Delta PQR$,where $\angle Q = 90^{\circ}$,the side $PR$ is the hypotenuse.
According to the Pythagorean theorem: $(PR)^2 = (PQ)^2 + (QR)^2$.
Given values are $PQ = 33$ and $PR = 65$.
Substituting the values: $(65)^2 = (33)^2 + (QR)^2$.
$4225 = 1089 + (QR)^2$.
$(QR)^2 = 4225 - 1089$.
$(QR)^2 = 3136$.
Taking the square root of both sides: $QR = \sqrt{3136} = 56$.
Therefore,the length of $QR$ is $56$.

(A) Let the diagonals $PR$ and $QS$ of the quadrilateral $PQRS$ intersect at point $O$ at right angles.
Thus,$\angle POQ = \angle QOR = \angle ROS = \angle SOP = 90^{\circ}$.
In right-angled $\triangle POQ$,by Pythagoras theorem: $PQ^{2} = PO^{2} + OQ^{2}$.
In right-angled $\triangle ROS$,by Pythagoras theorem: $RS^{2} = RO^{2} + OS^{2}$.
Adding these two equations: $PQ^{2} + RS^{2} = PO^{2} + OQ^{2} + RO^{2} + OS^{2}$ --- (Equation $1$).
In right-angled $\triangle SOP$,by Pythagoras theorem: $PS^{2} = SO^{2} + OP^{2}$.
In right-angled $\triangle QOR$,by Pythagoras theorem: $QR^{2} = OQ^{2} + OR^{2}$.
Adding these two equations: $PS^{2} + QR^{2} = SO^{2} + OP^{2} + OQ^{2} + OR^{2}$ --- (Equation $2$).
Comparing Equation $1$ and Equation $2$,we see that the right-hand sides are identical.
Therefore,$PQ^{2} + RS^{2} = PS^{2} + QR^{2}$.

(N/A) Given: In $\Delta ABC$,$\angle B = 90^{\circ}$. $N$ is a point on $AB$ and $M$ is a point on $BC$.
Step $1$: Apply the Pythagorean theorem in $\Delta ABM$ and $\Delta CBN$.
In $\Delta ABM$,since $\angle B = 90^{\circ}$,we have $AM^{2} = AB^{2} + BM^{2}$.
In $\Delta CBN$,since $\angle B = 90^{\circ}$,we have $CN^{2} = CB^{2} + BN^{2}$.
Step $2$: Add the two equations.
$AM^{2} + CN^{2} = (AB^{2} + BM^{2}) + (CB^{2} + BN^{2})$.
Step $3$: Rearrange the terms.
$AM^{2} + CN^{2} = (AB^{2} + CB^{2}) + (BM^{2} + BN^{2})$.
Step $4$: Use the Pythagorean theorem for $\Delta ABC$ and $\Delta MBN$.
In $\Delta ABC$,$AC^{2} = AB^{2} + CB^{2}$.
In $\Delta MBN$,$MN^{2} = BM^{2} + BN^{2}$.
Step $5$: Substitute these into the equation from Step $3$.
$AM^{2} + CN^{2} = AC^{2} + MN^{2}$.
Hence,the result 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) Let the sides of the triangle be $a = m^{2}-n^{2}$,$b = 2mn$,and $c = m^{2}+n^{2}$.
To prove that the triangle is a right-angled triangle,we need to check if the square of the longest side is equal to the sum of the squares of the other two sides (Pythagoras Theorem).
Here,$c = m^{2}+n^{2}$ is the longest side since $m > n > 0$.
Calculate $a^{2} + b^{2}$:
$a^{2} + b^{2} = (m^{2}-n^{2})^{2} + (2mn)^{2}$
$= (m^{4} - 2m^{2}n^{2} + n^{4}) + 4m^{2}n^{2}$
$= m^{4} + 2m^{2}n^{2} + n^{4}$
$= (m^{2}+n^{2})^{2}$
$= c^{2}$
Since $a^{2} + b^{2} = c^{2}$,the triangle satisfies the Pythagoras Theorem.
Therefore,the triangle is a right-angled triangle.

(N/A) Given: In $\Delta ABC$,$\angle B = 90^{\circ}$ (since $\overline{AC}$ is the hypotenuse),and $\overline{BE}$ is a median to the hypotenuse $\overline{AC}$.
Since $\overline{BE}$ is a median,$E$ is the midpoint of $\overline{AC}$. Thus,$AE = EC = \frac{1}{2} AC$,which implies $AC = 2AE$.
In right-angled $\Delta ABC$,by Pythagoras theorem: $AB^{2} + BC^{2} = AC^{2}$.
We need to prove: $AB^{2} + BC^{2} + AC^{2} = 8AE^{2}$.
Substitute $AB^{2} + BC^{2} = AC^{2}$ into the expression:
$AC^{2} + AC^{2} = 2AC^{2}$.
Since $AC = 2AE$,substitute this into the expression:
$2(2AE)^{2} = 2(4AE^{2}) = 8AE^{2}$.
Thus,$AB^{2} + BC^{2} + AC^{2} = 8AE^{2}$ is proved.

(N/A) In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$.
By the area of a right-angled triangle,the area of $\Delta ABC$ can be expressed in two ways:
Area $= \frac{1}{2} \times AB \times BC$ (using legs as base and height)
Area $= \frac{1}{2} \times AC \times BM$ (using hypotenuse as base and altitude as height)
Equating the two expressions: $\frac{1}{2} \times AB \times BC = \frac{1}{2} \times AC \times BM \implies AB \times BC = AC \times BM$.
Squaring both sides: $AB^2 \times BC^2 = AC^2 \times BM^2$.
Since $\Delta ABC$ is a right-angled triangle,by the Pythagorean theorem: $AC^2 = AB^2 + BC^2$.
Substituting $AC^2$ into the equation: $AB^2 \times BC^2 = (AB^2 + BC^2) \times BM^2$.
Rearranging for $BM^2$: $BM^2 = \frac{AB^2 \times BC^2}{AB^2 + BC^2}$.
Taking the reciprocal of both sides: $\frac{1}{BM^2} = \frac{AB^2 + BC^2}{AB^2 \times BC^2}$.
Splitting the fraction: $\frac{1}{BM^2} = \frac{AB^2}{AB^2 \times BC^2} + \frac{BC^2}{AB^2 \times BC^2}$.
Simplifying: $\frac{1}{BM^2} = \frac{1}{BC^2} + \frac{1}{AB^2}$.

(N/A) $1$. In $\Delta ABC$,$\angle B = 90^{\circ}$ and $\overline{BD} \perp \overline{AC}$.
$2$. By the property of geometric mean in a right-angled triangle,we have $BD^2 = AD \cdot CD$.
$3$. Given $BD = 2CD$,substitute this into the equation: $(2CD)^2 = AD \cdot CD$.
$4$. This simplifies to $4CD^2 = AD \cdot CD$.
$5$. Dividing both sides by $CD$ (since $CD \neq 0$),we get $AD = 4CD$.
$6$. Since $AC = AD + CD$,substitute $AD = 4CD$ into the equation: $AC = 4CD + CD$.
$7$. Therefore,$AC = 5CD$.

(N/A) In $\Delta ABC$,since $m \angle B = 90^{\circ}$,by the Pythagorean theorem,we have $AC^{2} = AB^{2} + BC^{2}$. (Equation $1$)
In $\Delta ABD$,since $m \angle B = 90^{\circ}$,by the Pythagorean theorem,we have $AD^{2} = AB^{2} + BD^{2}$. (Equation $2$)
From Equation $2$,we can write $AB^{2} = AD^{2} - BD^{2}$.
Substitute this value of $AB^{2}$ into Equation $1$:
$AC^{2} = (AD^{2} - BD^{2}) + BC^{2}$.
Rearranging the terms,we get $AC^{2} + BD^{2} = AD^{2} + BC^{2}$.
Hence,the statement is proved.

(A) Given: In $\Delta ABC$,$\angle B = 90^{\circ}$ and $\overline{AD}$ is a median to side $\overline{BC}$.
Since $\overline{AD}$ is a median,$D$ is the midpoint of $\overline{BC}$. Therefore,$BD = DC = \frac{1}{2} BC$,which implies $BC = 2BD$.
In right-angled $\Delta ABC$,by Pythagoras theorem: $AC^{2} = AB^{2} + BC^{2}$.
Substituting $BC = 2BD$,we get $AC^{2} = AB^{2} + (2BD)^{2} = AB^{2} + 4BD^{2}$. (Equation $1$)
In right-angled $\Delta ABD$,by Pythagoras theorem: $AD^{2} = AB^{2} + BD^{2}$,which implies $AB^{2} = AD^{2} - BD^{2}$. (Equation $2$)
Substitute Equation $2$ into Equation $1$:
$AC^{2} = (AD^{2} - BD^{2}) + 4BD^{2}$
$AC^{2} = AD^{2} + 3BD^{2}$.
Hence,the result is proved.

(B) According to Apollonius's Theorem,for any triangle $\Delta ABC$ with median $\overline{AD}$ to side $\overline{BC}$,the following relation holds:
$AB^2 + AC^2 = 2(AD^2 + BD^2)$
Given that $AD = 7$ and $AB^2 + AC^2 = 148$,we substitute these values into the formula:
$148 = 2(7^2 + BD^2)$
$148 = 2(49 + BD^2)$
Divide both sides by $2$:
$74 = 49 + BD^2$
$BD^2 = 74 - 49 = 25$
$BD = \sqrt{25} = 5$
Since $\overline{AD}$ is a median,$D$ is the midpoint of $\overline{BC}$,so $BC = 2 \times BD$.
$BC = 2 \times 5 = 10$.

(A) According to Apollonius's Theorem for $\Delta PQR$ with median $\overline{PS}$:
$PQ^2 + PR^2 = 2(PS^2 + QS^2)$
Given $PQ = 9$,$PR = 40$,and $PS = 20.5$:
$9^2 + 40^2 = 2(20.5^2 + QS^2)$
$81 + 1600 = 2(420.25 + QS^2)$
$1681 = 2(420.25 + QS^2)$
$840.5 = 420.25 + QS^2$
$QS^2 = 840.5 - 420.25 = 420.25$
$QS = \sqrt{420.25} = 20.5$
Since $\overline{PS}$ is a median,$S$ is the midpoint of $\overline{QR}$,so $QR = 2 \times QS = 2 \times 20.5 = 41$.

(B) According to Apollonius's Theorem,for any triangle $ABC$ with median $AD$ to side $BC$:
$AB^2 + AC^2 = 2(AD^2 + BD^2)$
Given $AB = 10$,$AC = 24$,and $AD = 13$.
Substituting the values:
$10^2 + 24^2 = 2(13^2 + BD^2)$
$100 + 576 = 2(169 + BD^2)$
$676 = 2(169 + BD^2)$
$338 = 169 + BD^2$
$BD^2 = 338 - 169 = 169$
$BD = \sqrt{169} = 13$
Since $AD$ is a median,$D$ is the midpoint of $BC$,so $BC = 2 \times BD$.
$BC = 2 \times 13 = 26$.

(C) Given that $\overline{AD}$ is a median to side $\overline{BC}$,so $D$ is the midpoint of $\overline{BC}$.
Since $AD = BD = 8.5$,in $\Delta ABD$,we have $AD = BD = 8.5$. This implies $\angle DAB = \angle DBA$.
In $\Delta ABD$,by the Law of Cosines or using the property of isosceles triangles,we can find $\angle ADB$. Since $AD = BD = 8.5$,$D$ is the center of a circle passing through $A, B,$ and $C$ if $CD = 8.5$ as well.
Actually,since $AD$ is a median,$BD = DC = 8.5$. Thus,$AD = BD = DC = 8.5$.
This means $D$ is the circumcenter of $\Delta ABC$,and $BC$ is the diameter of the circumcircle.
Therefore,$\angle BAC = 90^\circ$ (angle in a semicircle).
In right-angled $\Delta ABC$,by Pythagoras theorem,$AC^2 + AB^2 = BC^2$.
We know $AB = 8$ and $BC = BD + DC = 8.5 + 8.5 = 17$.
$AC^2 + 8^2 = 17^2$
$AC^2 + 64 = 289$
$AC^2 = 225$
$AC = 15$.

(A) Given that $\overline{PS}$ is a median of $\Delta PQR,$ it divides the side $QR$ into two equal parts. Thus,$QS = SR = 18.5.$
Since $PS = QS = SR = 18.5,$ the point $S$ is the circumcenter of $\Delta PQR$ with respect to the side $QR,$ implying $\angle QPR = 90^\circ$ (by the property that if the median to a side is half the length of that side,the triangle is a right-angled triangle).
In the right-angled $\Delta PQR,$ by the Pythagorean theorem:
$PQ^2 + PR^2 = QR^2$
$QR = QS + SR = 18.5 + 18.5 = 37$
$12^2 + PR^2 = 37^2$
$144 + PR^2 = 1369$
$PR^2 = 1369 - 144 = 1225$
$PR = \sqrt{1225} = 35.$

(D) Given that $\Delta ABC$ is a right-angled triangle with $m \angle B = 90^{\circ}$.
Here,$AC$ is the hypotenuse and $AB$ is one of the legs.
Using the Pythagorean theorem: $AC^2 = AB^2 + BC^2$.
Substituting the given values: $10^2 = 6^2 + BC^2$.
$100 = 36 + BC^2$.
$BC^2 = 100 - 36 = 64$.
$BC = \sqrt{64} = 8$.
The area of a right-angled triangle is given by the formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
Here,base $BC = 8$ and height $AB = 6$.
$\text{Area} = \frac{1}{2} \times 8 \times 6 = 4 \times 6 = 24$.

(A) $1$. In $\Delta ABC$,since $m\angle B = 90^{\circ}$,we can find the length of $BC$ using the Pythagorean theorem: $AB^2 + BC^2 = AC^2$.
$2$. Substituting the given values: $3^2 + BC^2 = 5^2$,which gives $9 + BC^2 = 25$.
$3$. Therefore,$BC^2 = 25 - 9 = 16$,so $BC = 4$.
$4$. The area of $\Delta ABC$ can be calculated in two ways: $\text{Area} = \frac{1}{2} \times AB \times BC$ or $\text{Area} = \frac{1}{2} \times AC \times BM$.
$5$. Equating the two expressions: $\frac{1}{2} \times 3 \times 4 = \frac{1}{2} \times 5 \times BM$.
$6$. Simplifying: $12 = 5 \times BM$,which leads to $BM = \frac{12}{5} = 2.4$.

(B) Given that $\Delta ABC$ is a right-angled triangle where $m \angle B = 90^{\circ}$.
Since $\overline{AB} \cong \overline{BC}$,let $AB = BC = x$.
According to the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.
Substituting the values,$AC^2 = x^2 + x^2 = 2x^2$.
Therefore,$AC = \sqrt{2x^2} = x\sqrt{2}$.
We need to find the ratio $AB : AC$.
$AB : AC = x : x\sqrt{2} = 1 : \sqrt{2}$.

(C) In $\Delta ABC$,since $m \angle B = 90^{\circ}$,the triangle is a right-angled triangle.
By the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.
Given $AB = 20$ and $AC = 29$,we have $29^2 = 20^2 + BC^2$.
$841 = 400 + BC^2$.
$BC^2 = 841 - 400 = 441$.
$BC = \sqrt{441} = 21$.
The perimeter of $\Delta ABC = AB + BC + AC$.
Perimeter $= 20 + 21 + 29 = 70$.

(A) In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.
First,find the length of the hypotenuse $AC$ using the Pythagorean theorem: $AC^2 = AB^2 + BC^2$.
$AC^2 = (3.6)^2 + (4.8)^2 = 12.96 + 23.04 = 36$.
$AC = \sqrt{36} = 6$.
Since $\overline{BE}$ is the median to the hypotenuse $AC$,its length is $BE = \frac{1}{2} AC$.
$BE = \frac{1}{2} \times 6 = 3$.

(B) Given that in $\Delta ABC$,$m\angle B = 90^{\circ}$.
Let $AB = 3x$ and $BC = 4x$ for some constant $x > 0$.
By the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.
$AC^2 = (3x)^2 + (4x)^2 = 9x^2 + 16x^2 = 25x^2$.
Taking the square root,$AC = \sqrt{25x^2} = 5x$.
We need to find the ratio $AB : AC$.
$AB : AC = 3x : 5x = 3 : 5$.
Therefore,the correct option is $B$.

(C) In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.
Here,$\Delta ABC$ is a right-angled triangle with $m\angle B = 90^{\circ}$.
Since $\overline{BM}$ is the median to the hypotenuse $\overline{AC}$,we have $BM = \frac{1}{2} AC$.
Given $BM = 12.5$,so $12.5 = \frac{1}{2} AC$,which implies $AC = 25$.
Now,using the Pythagorean theorem in $\Delta ABC$:
$AB^2 + BC^2 = AC^2$
$15^2 + BC^2 = 25^2$
$225 + BC^2 = 625$
$BC^2 = 625 - 225 = 400$
$BC = \sqrt{400} = 20$.

(A) The perimeter of an equilateral triangle is given by $P = 3a$, where $a$ is the side length.
Given $P = 24$, we have $3a = 24$, which implies $a = 8$.
The altitude $h$ of an equilateral triangle is given by the formula $h = \frac{\sqrt{3}}{2} a$.
Substituting $a = 8$ into the formula, we get $h = \frac{\sqrt{3}}{2} \times 8 = 4 \sqrt{3}$.
Therefore, the length of the altitude is $4 \sqrt{3}$.

(D) Given that $\Delta XYZ$ is a right-angled triangle with $m \angle Y = 90^{\circ}$.
According to the Pythagorean theorem,$XY^2 + YZ^2 = XZ^2$.
Let $XY = 15k$ and $XZ = 17k$ for some constant $k > 0$.
Substituting the values into the theorem: $(15k)^2 + 4^2 = (17k)^2$.
$225k^2 + 16 = 289k^2$.
$16 = 289k^2 - 225k^2$.
$16 = 64k^2$.
$k^2 = \frac{16}{64} = \frac{1}{4}$.
Therefore,$k = \frac{1}{2} = 0.5$.
Now,calculate the side lengths:
$XY = 15 \times 0.5 = 7.5$.
$XZ = 17 \times 0.5 = 8.5$.
$YZ = 4$.
The perimeter of $\Delta XYZ = XY + YZ + XZ = 7.5 + 4 + 8.5 = 20$.

(A) Given that $\Delta ABC$ is a right-angled triangle with $m \angle B = 90^{\circ}$ and $AB = BC$.
Let $AB = BC = x$.
According to the Pythagorean theorem,$AB^2 + BC^2 = AC^2$.
Substituting the values,we get $x^2 + x^2 = (14\sqrt{2})^2$.
$2x^2 = 196 \times 2$.
$2x^2 = 392$.
$x^2 = 196$.
$x = 14$.
Thus,the base $BC = 14$ and the height $AB = 14$.
The area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 14 \times 14 = 98$ square units.

(B) In a right-angled triangle,if an altitude is drawn from the right-angle vertex to the hypotenuse,the altitude is the geometric mean of the segments of the hypotenuse.
According to the geometric mean theorem: $BM^2 = AM \cdot CM$.
Given $AM = 2x$,$BM = 3x + 5$,and $CM = 8x$.
Substituting these values into the equation: $(3x + 5)^2 = (2x)(8x)$.
Expanding the left side: $9x^2 + 30x + 25 = 16x^2$.
Rearranging the terms to form a quadratic equation: $7x^2 - 30x - 25 = 0$.
Factoring the quadratic equation: $7x^2 - 35x + 5x - 25 = 0$.
$7x(x - 5) + 5(x - 5) = 0$.
$(7x + 5)(x - 5) = 0$.
This gives $x = 5$ or $x = -5/7$.
Since $x$ represents a length,it must be positive,so $x = 5$.

(C) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$,if $\overline{BM}$ is the altitude to the hypotenuse $\overline{AC}$,then by the geometric mean theorem (or property of similar triangles $\Delta AMB \sim \Delta ABC$),we have the relation: $AB^2 = AM \cdot AC$.
Given values are $AB = x - 2$,$AM = x - 6$,and $AC = 2x - 4$.
Substituting these into the equation: $(x - 2)^2 = (x - 6)(2x - 4)$.
Expanding both sides: $x^2 - 4x + 4 = 2x^2 - 4x - 12x + 24$.
Simplifying: $x^2 - 4x + 4 = 2x^2 - 16x + 24$.
Rearranging the terms to form a quadratic equation: $x^2 - 12x + 20 = 0$.
Factoring the quadratic equation: $(x - 10)(x - 2) = 0$.
This gives $x = 10$ or $x = 2$.
If $x = 2$,then $AB = 2 - 2 = 0$,which is impossible for a triangle side length. Thus,$x = 10$ is the only valid solution.

(D) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$,if $\overline{BM}$ is the altitude to the hypotenuse $\overline{AC}$,then by the geometric mean theorem,$BM^2 = AM \cdot CM$.
Substituting the given values: $(x + 1)^2 = (x - 1)(x + 4)$.
Expanding both sides: $x^2 + 2x + 1 = x^2 + 4x - x - 4$.
Simplifying the equation: $x^2 + 2x + 1 = x^2 + 3x - 4$.
Subtracting $x^2$ from both sides: $2x + 1 = 3x - 4$.
Rearranging the terms: $3x - 2x = 1 + 4$.
Thus,$x = 5$.

(B) In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.
Let the diagonals $AC$ and $BD$ intersect at point $O$.
Since $AC = 24$,the length of $AO = AC / 2 = 24 / 2 = 12$.
In the right-angled triangle $\triangle AOB$,by the Pythagorean theorem,$AB^2 = AO^2 + BO^2$.
Given $AB = 13$ and $AO = 12$,we have $13^2 = 12^2 + BO^2$.
$169 = 144 + BO^2$.
$BO^2 = 169 - 144 = 25$.
$BO = \sqrt{25} = 5$.
Since the diagonals bisect each other,$BD = 2 \times BO = 2 \times 5 = 10$.

(C) In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.
Let the diagonals $PR$ and $QS$ intersect at point $O$.
Since the diagonals bisect each other,$PO = OR = \frac{PR}{2} = \frac{9}{2} = 4.5$ and $QO = OS = \frac{QS}{2} = \frac{12}{2} = 6$.
In the right-angled triangle $\triangle POQ$,by the Pythagorean theorem:
$PQ^2 = PO^2 + QO^2$
$PQ^2 = (4.5)^2 + (6)^2$
$PQ^2 = 20.25 + 36 = 56.25$
$PQ = \sqrt{56.25} = 7.5$.
Since all sides of a rhombus are equal,the perimeter is $4 \times PQ$.
Perimeter $= 4 \times 7.5 = 30$.

(B) The perimeter of an equilateral triangle is given by $P = 3a$, where $a$ is the side length.
Given $P = 30$, we have $3a = 30$, which implies $a = 10$.
The area of an equilateral triangle is given by the formula $Area = \frac{\sqrt{3}}{4} a^2$.
Substituting $a = 10$ into the formula:
$Area = \frac{\sqrt{3}}{4} (10)^2$
$Area = \frac{\sqrt{3}}{4} \times 100$
$Area = 25 \sqrt{3}$.

(A) Let $AB = x$.
Given $BC - AB = 4$,so $BC = x + 4$.
Given $AC - BC = 4$,so $AC = BC + 4 = (x + 4) + 4 = x + 8$.
Since $\Delta ABC$ is a right-angled triangle with $\angle B = 90^{\circ}$,by the Pythagorean theorem: $AB^2 + BC^2 = AC^2$.
Substituting the values: $x^2 + (x + 4)^2 = (x + 8)^2$.
$x^2 + x^2 + 8x + 16 = x^2 + 16x + 64$.
$x^2 - 8x - 48 = 0$.
$(x - 12)(x + 4) = 0$.
Since $x$ must be positive,$x = 12$.
Thus,$AB = 12$,$BC = 12 + 4 = 16$,and $AC = 16 + 4 = 20$.

(B) In $\Delta ABC$,$\angle B = 90^{\circ}$ and $BM \perp AC$.
By the geometric mean theorem in right triangles,$BM^2 = AM \cdot CM$.
Given $BM = 10$ and $CM = 5$,we have $10^2 = AM \cdot 5$,which gives $100 = 5 \cdot AM$,so $AM = 20$.
The hypotenuse $AC = AM + CM = 20 + 5 = 25$.
Using the Pythagorean theorem in $\Delta BMC$,$BC^2 = BM^2 + CM^2 = 10^2 + 5^2 = 100 + 25 = 125$,so $BC = \sqrt{125} = 5\sqrt{5}$.
Using the Pythagorean theorem in $\Delta AMB$,$AB^2 = BM^2 + AM^2 = 10^2 + 20^2 = 100 + 400 = 500$,so $AB = \sqrt{500} = 10\sqrt{5}$.
The perimeter of $\Delta ABC = AB + BC + AC = 10\sqrt{5} + 5\sqrt{5} + 25 = 25 + 15\sqrt{5}$.

(A) Given that $\Delta ABC$ is an isosceles triangle with $AB = AC$. Let $AB = AC = x$ and $BC = y$.
The perimeter is $AB + AC + BC = 50$,so $2x + y = 50$,which implies $y = 50 - 2x$.
Since $\overline{AM}$ is an altitude to the base $\overline{BC}$ in an isosceles triangle,it bisects the base. Thus,$BM = MC = y/2 = (50 - 2x)/2 = 25 - x$.
In right-angled $\Delta ABM$,by the Pythagorean theorem: $AM^2 + BM^2 = AB^2$.
Substituting the values: $15^2 + (25 - x)^2 = x^2$.
$225 + 625 - 50x + x^2 = x^2$.
$850 - 50x = 0$,so $50x = 850$,which gives $x = 17$.
Now,find the base $y$: $y = 50 - 2(17) = 50 - 34 = 16$.
The area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BC \times AM$.
Area $= \frac{1}{2} \times 16 \times 15 = 8 \times 15 = 120$.

(B) In $\Delta ABC$,$AB = AC = 37$,which means $\Delta ABC$ is an isosceles triangle.
Since $\overline{AM}$ is the median to the base $BC$,it is also the altitude to the base $BC$ in an isosceles triangle.
Therefore,$\overline{AM} \perp \overline{BC}$ and $M$ is the midpoint of $BC$.
Since $BC = 70$,$BM = MC = \frac{70}{2} = 35$.
In the right-angled triangle $\Delta ABM$,by the Pythagorean theorem:
$AB^2 = AM^2 + BM^2$
$37^2 = AM^2 + 35^2$
$1369 = AM^2 + 1225$
$AM^2 = 1369 - 1225 = 144$
$AM = \sqrt{144} = 12$.

(A) Let the length of the ladder be $L$. The ladder, the wall, and the ground form a right-angled triangle.
According to the Pythagorean theorem, $L^2 = (\text{height})^2 + (\text{base})^2$.
In the first case, height $= 8 \, m$ and base $= 6 \, m$.
So, $L^2 = 8^2 + 6^2 = 64 + 36 = 100$.
Thus, $L = \sqrt{100} = 10 \, m$.
In the second case, the same ladder of length $L = 10 \, m$ is used, and the new height is $6 \, m$.
Let the new base be $x$.
Using the Pythagorean theorem again: $L^2 = (\text{new height})^2 + x^2$.
$10^2 = 6^2 + x^2$.
$100 = 36 + x^2$.
$x^2 = 100 - 36 = 64$.
$x = \sqrt{64} = 8 \, m$.
Therefore, the distance of the lower end from the base of the wall is $8 \, m$.

(A) rhombus is a quadrilateral where all sides are equal and diagonals bisect each other at right angles $(90^{\circ})$.
Let the diagonals be $d_1 = AC = 24$ and $d_2 = BD = 70$.
The diagonals bisect each other at point $O$. Thus,the segments are $AO = OC = \frac{24}{2} = 12$ and $BO = OD = \frac{70}{2} = 35$.
In the right-angled triangle $\triangle AOB$,by the Pythagorean theorem:
$AB^2 = AO^2 + BO^2$
$AB^2 = 12^2 + 35^2$
$AB^2 = 144 + 1225 = 1369$
$AB = \sqrt{1369} = 37$.
The perimeter of the rhombus is $4 \times \text{side} = 4 \times 37 = 148$.

(N/A) In $\Delta PQR$,$\angle Q = 90^{\circ}$ and $QS \perp PR$.
By the property of geometric mean in a right-angled triangle,$PQ^2 = PS \cdot PR$.
Given $PQ = 6$ and $PS = 4$,we have $6^2 = 4 \cdot PR$,which implies $36 = 4 \cdot PR$,so $PR = 9$.
Since $PR = PS + RS$,we have $9 = 4 + RS$,so $RS = 5$.
Using the altitude theorem,$QS^2 = PS \cdot RS = 4 \cdot 5 = 20$,so $QS = \sqrt{20} = 2\sqrt{5}$.
In $\Delta QSR$,by the Pythagorean theorem,$QR^2 = QS^2 + RS^2 = 20 + 5^2 = 20 + 25 = 45$.
Thus,$QR = \sqrt{45} = 3\sqrt{5}$.