Mix Examples - Triangles Questions in English

(N/A) Given: $AB + BC = 23$ $(i)$,$BC + AC = 32$ $(ii)$,and $AB + AC = 25$ $(iii)$.
Adding $(i)$,$(ii)$,and $(iii)$,we get: $2(AB + BC + AC) = 23 + 32 + 25 = 80$.
Therefore,$AB + BC + AC = 40$ $(iv)$.
Subtracting $(ii)$ from $(iv)$: $AB = 40 - 32 = 8$.
Subtracting $(iii)$ from $(iv)$: $BC = 40 - 25 = 15$.
Subtracting $(i)$ from $(iv)$: $AC = 40 - 23 = 17$.
Now,check the sides: $AB^2 + BC^2 = 8^2 + 15^2 = 64 + 225 = 289$.
Also,$AC^2 = 17^2 = 289$.
Since $AB^2 + BC^2 = AC^2$,by the converse of the Pythagoras theorem,$\Delta ABC$ is a right-angled triangle with $\angle B = 90^\circ$.

(A) In $\Delta ABC$,$\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$.
By the geometric mean theorem (or properties of similar triangles $\Delta AMB \sim \Delta BMC$),we have $BM^2 = AM \cdot MC$.
Given $AM = 12$ and $BM = 12$,we substitute these values into the equation:
$12^2 = 12 \cdot MC$
$144 = 12 \cdot MC$
$MC = \frac{144}{12} = 12$.
Since $AC = AM + MC$,we have:
$AC = 12 + 12 = 24$.

(A) 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.
According to the geometric mean theorem,the altitude $\overline{YM}$ to the hypotenuse $\overline{XZ}$ satisfies the relation: $YM^2 = XM \cdot ZM$.
Given $XM = \sqrt{12}$ and $ZM = \sqrt{3}$.
Substituting these values into the formula:
$YM^2 = \sqrt{12} \cdot \sqrt{3}$
$YM^2 = \sqrt{12 \cdot 3}$
$YM^2 = \sqrt{36}$
$YM^2 = 6$
Therefore,$YM = \sqrt{6}$.

(B) Let $PD = x$. Since $PR = 34$,then $DR = 34 - x$.
In $\Delta PQR$,$\overline{QD}$ is the altitude to the hypotenuse $PR$. By the Geometric Mean Theorem (or properties of similar triangles),$QD^2 = PD \cdot DR$.
Substituting the given values: $15^2 = x(34 - x)$.
$225 = 34x - x^2$.
$x^2 - 34x + 225 = 0$.
Solving the quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{34 \pm \sqrt{34^2 - 4(1)(225)}}{2} = \frac{34 \pm \sqrt{1156 - 900}}{2} = \frac{34 \pm \sqrt{256}}{2} = \frac{34 \pm 16}{2}$.
Two possible values for $x$ are $x = \frac{50}{2} = 25$ or $x = \frac{18}{2} = 9$.
In $\Delta PQD$,by the Pythagorean theorem,$PQ^2 = QD^2 + PD^2$.
If $PD = 9$,then $PQ^2 = 15^2 + 9^2 = 225 + 81 = 306$. $PQ = \sqrt{306} = \sqrt{9 \cdot 34} = 3 \sqrt{34}$.
If $PD = 25$,then $PQ^2 = 15^2 + 25^2 = 225 + 625 = 850$. $PQ = \sqrt{850} = \sqrt{25 \cdot 34} = 5 \sqrt{34}$.
Comparing with the given options,$5 \sqrt{34}$ is the correct answer.

(C) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$,$\overline{BM}$ is the altitude to the hypotenuse $AC$.
According to the geometric mean theorem for right triangles,the altitude to the hypotenuse divides the triangle into two triangles similar to the original triangle and to each other.
Specifically,$\Delta BMC \sim \Delta AMB$.
From the similarity,we have the property: $BM^2 = AM \cdot CM$.
Given $BM = \sqrt{30}$ and $CM = 3$,we substitute these values into the equation:
$(\sqrt{30})^2 = AM \cdot 3$
$30 = AM \cdot 3$
$AM = 10$.
The length of the hypotenuse $AC$ is the sum of segments $AM$ and $CM$:
$AC = AM + CM = 10 + 3 = 13$.
Therefore,the correct answer is $13$.

(D) In $\Delta ABC$,$\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$.
According to the geometric mean theorem (or properties of similar triangles),in a right-angled triangle,the altitude to the hypotenuse creates two triangles similar to the original triangle and to each other.
Specifically,$\Delta ABM \sim \Delta ACB$.
From the property of the altitude in a right triangle: $AB^2 = AM \cdot AC$.
Given $AB = 2\sqrt{10}$,we have $AB^2 = (2\sqrt{10})^2 = 4 \cdot 10 = 40$.
Given $AM = 5$,we substitute these values into the equation: $40 = 5 \cdot AC$.
Solving for $AC$: $AC = 40 / 5 = 8$.
Since $AC = AM + CM$,we have $8 = 5 + CM$.
Therefore,$CM = 8 - 5 = 3$.

(A) In a right-angled triangle $\Delta ABC$ with altitude $\overline{BM}$ to the hypotenuse $\overline{AC}$,the geometric mean theorem states that $BM^2 = AM \cdot CM$.
Given $AM = x + 7$,$BM = x + 2$,and $CM = x$.
Substituting these values into the formula: $(x + 2)^2 = (x + 7)(x)$.
Expanding both sides: $x^2 + 4x + 4 = x^2 + 7x$.
Subtracting $x^2$ from both sides: $4x + 4 = 7x$.
Rearranging the terms: $4 = 7x - 4x$,which simplifies to $4 = 3x$.
Therefore,$x = \frac{4}{3}$.

(A) In a right-angled triangle,the altitude to the hypotenuse divides the triangle into two triangles similar to the original triangle and to each other.
By the geometric mean theorem,$BM^2 = AM \cdot CM$.
Substituting the given values: $BM^2 = (2x^2)(8x^2) = 16x^4$.
Taking the square root: $BM = 4x^2$.
Using the Pythagorean theorem in $\Delta ABM$: $AB^2 = AM^2 + BM^2 = (2x^2)^2 + (4x^2)^2 = 4x^4 + 16x^4 = 20x^4$.
Thus,$AB = \sqrt{20x^4} = 2\sqrt{5}x^2$.
Using the Pythagorean theorem in $\Delta BCM$: $BC^2 = CM^2 + BM^2 = (8x^2)^2 + (4x^2)^2 = 64x^4 + 16x^4 = 80x^4$.
Thus,$BC = \sqrt{80x^4} = 4\sqrt{5}x^2$.

(C) In a right-angled triangle $\Delta PQR$ where $\angle Q = 90^{\circ}$, $\overline{QM}$ is the altitude to the hypotenuse $PR$.
By the geometric mean theorem for right triangles, the altitude to the hypotenuse divides the triangle into two triangles similar to the original triangle and to each other.
Specifically, $\Delta QMR \sim \Delta PMQ$.
From the similarity $\Delta QMR \sim \Delta PMQ$, we have the ratio of corresponding sides: $\frac{QM}{RM} = \frac{PM}{QM}$.
Substituting the given values: $\frac{14}{7} = \frac{PM}{14}$.
$PM = \frac{14 \times 14}{7} = 2 \times 14 = 28$.
Now, in the right-angled triangle $\Delta PMQ$, by the Pythagorean theorem: $PQ^2 = PM^2 + QM^2$.
$PQ^2 = 28^2 + 14^2$.
$PQ^2 = 784 + 196 = 980$.
$PQ = \sqrt{980} = \sqrt{196 \times 5} = 14 \sqrt{5}$.

(A) In a right-angled triangle $\Delta ABC$ with altitude $\overline{BM}$ to the hypotenuse $\overline{AC}$,we have the geometric mean property: $BM^2 = AM \cdot CM$.
Given $BM = 6$,we have $AM \cdot CM = 6^2 = 36$.
We also know $AM + CM = AC = 13$.
Let $AM = x$ and $CM = y$. Then $x + y = 13$ and $xy = 36$.
These are the roots of the quadratic equation $t^2 - 13t + 36 = 0$.
Factoring the equation: $(t - 9)(t - 4) = 0$,so $t = 9$ or $t = 4$.
Since $AM < CM$,we have $AM = 4$ and $CM = 9$.
In $\Delta ABM$,by the Pythagorean theorem: $AB^2 = AM^2 + BM^2$.
$AB^2 = 4^2 + 6^2 = 16 + 36 = 52$.
$AB = \sqrt{52} = \sqrt{4 \cdot 13} = 2 \sqrt{13}$.

(A) In a right-angled triangle $\Delta XYZ$ where $\angle Y = 90^{\circ}$,$\overline{YM}$ is the altitude to the hypotenuse $\overline{XZ}$.
According to the geometric mean theorem for right triangles,the altitude to the hypotenuse divides the triangle into two triangles similar to the original triangle and to each other.
Specifically,$YM^2 = XM \cdot MZ$.
Given $YM = 12$ and $XM = 8$,we substitute these values into the equation:
$12^2 = 8 \cdot MZ$
$144 = 8 \cdot MZ$
$MZ = 144 / 8 = 18$.
The length of the hypotenuse $XZ$ is the sum of the segments $XM$ and $MZ$:
$XZ = XM + MZ = 8 + 18 = 26$.
Therefore,the correct option is $A$.

(A) In a right-angled triangle,the altitude to the hypotenuse creates two triangles similar to the original triangle and to each other.
By the geometric mean theorem for the altitude: $BM^2 = AM \cdot CM$.
Substituting the given values: $BM^2 = 4 \cdot 5 = 20$.
Therefore,$BM = \sqrt{20} = 2\sqrt{5}$.
In $\Delta ABM$,by the Pythagorean theorem: $AB^2 = AM^2 + BM^2$.
$AB^2 = 4^2 + (2\sqrt{5})^2 = 16 + 20 = 36$.
Therefore,$AB = \sqrt{36} = 6$.

(C) In a right-angled triangle,the altitude drawn from the right-angle vertex 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,the altitude $\overline{BM}$ to the hypotenuse $\overline{AC}$ satisfies the relation: $BM^2 = AM \times CM$.
Given $AM = 12$ and $CM = 3$.
Substituting the values: $BM^2 = 12 \times 3 = 36$.
Taking the square root of both sides: $BM = \sqrt{36} = 6$.

(A) In a right-angled triangle,the altitude to the hypotenuse creates two triangles similar to the original triangle and to each other. By the geometric mean theorem,$BM^2 = AM \cdot CM$.
Given $AM = 4$ and $CM = 12$,$BM^2 = 4 \cdot 12 = 48$,so $BM = \sqrt{48} = 4\sqrt{3}$.
In $\Delta ABM$,by the Pythagorean theorem,$AB^2 = AM^2 + BM^2 = 4^2 + (4\sqrt{3})^2 = 16 + 48 = 64$. Thus,$AB = \sqrt{64} = 8$.
In $\Delta CBM$,by the Pythagorean theorem,$BC^2 = CM^2 + BM^2 = 12^2 + (4\sqrt{3})^2 = 144 + 48 = 192$. Thus,$BC = \sqrt{192} = 8\sqrt{3}$.

(A) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$,the altitude property states that $BM^2 = AM \cdot CM$.
Given $AM = 9$ and $CM = 16$,we have $BM^2 = 9 \cdot 16 = 144$,so $BM = 12$.
Now,we find the lengths of the sides $AB$ and $BC$ using the Pythagorean theorem in $\Delta AMB$ and $\Delta BMC$:
$AB = \sqrt{AM^2 + BM^2} = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15$.
$BC = \sqrt{CM^2 + BM^2} = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20$.
The hypotenuse $AC = AM + CM = 9 + 16 = 25$.
The perimeter of $\Delta ABC = AB + BC + AC = 15 + 20 + 25 = 60$.

(B) In a right-angled triangle $\Delta ABC$ with altitude $\overline{AM}$ to the hypotenuse $\overline{BC}$,we have the geometric mean theorem: $AM^2 = BM \cdot CM$.
Given $BM = 6$ and $CM = 2$,$AM^2 = 6 \cdot 2 = 12$,so $AM = \sqrt{12} = 2\sqrt{3}$.
Now,calculate the sides of $\Delta ABC$:
$BC = BM + CM = 6 + 2 = 8$.
Using the Pythagorean theorem in $\Delta ABM$: $AB^2 = AM^2 + BM^2 = 12 + 36 = 48$,so $AB = \sqrt{48} = 4\sqrt{3}$.
Using the Pythagorean theorem in $\Delta ACM$: $AC^2 = AM^2 + CM^2 = 12 + 4 = 16$,so $AC = \sqrt{16} = 4$.
Perimeter of $\Delta ABC = AB + AC + BC = 4\sqrt{3} + 4 + 8 = 12 + 4\sqrt{3}$.

(C) In $\Delta ABC$,the sum of angles is $m \angle A + m \angle B + m \angle C = 180^{\circ}$.
Given $m \angle A = m \angle B + m \angle C$,substituting this into the sum gives $m \angle A + m \angle A = 180^{\circ}$,so $2m \angle A = 180^{\circ}$,which means $m \angle A = 90^{\circ}$.
Since $\overline{AM}$ is an altitude to $\overline{BC}$,$\Delta ABM$ and $\Delta ACM$ are right-angled triangles at $M$.
Let $BM = x$. Then $MC = 8 - x$.
In $\Delta ABM$,$AB^2 = AM^2 + BM^2 = 12 + x^2$.
In $\Delta ACM$,$AC^2 = AM^2 + MC^2 = 12 + (8 - x)^2$.
In $\Delta ABC$,by the Pythagorean theorem,$AB^2 + AC^2 = BC^2$,so $(12 + x^2) + (12 + (8 - x)^2) = 8^2$.
$24 + x^2 + 64 - 16x + x^2 = 64$.
$2x^2 - 16x + 24 = 0$.
$x^2 - 8x + 12 = 0$.
$(x - 6)(x - 2) = 0$.
Thus,$x = 6$ or $x = 2$.

(D) In a right-angled triangle $\Delta PQR$ with $\angle Q = 90^{\circ}$,$\overline{QM}$ is the altitude to the hypotenuse $\overline{PR}$.
By the geometric mean theorem (or properties of similar triangles),we have $QM^2 = PM \cdot MR$.
Let $PM = x$. Since $PR = 10$,then $MR = 10 - x$.
Given $QM = 4$,we substitute these values into the equation:
$4^2 = x(10 - x)$
$16 = 10x - x^2$
$x^2 - 10x + 16 = 0$
Factoring the quadratic equation:
$(x - 2)(x - 8) = 0$
Thus,$x = 2$ or $x = 8$.
Therefore,$PM$ can be $2$ or $8$.

(A) Let $AM = x$. Since $AC = 25$,then $MC = 25 - x$.
In $\Delta ABC$,$\overline{BM}$ is the altitude to the hypotenuse $\overline{AC}$. By the Geometric Mean Theorem,$BM^2 = AM \cdot MC$.
Substituting the values: $12^2 = x(25 - x) \implies 144 = 25x - x^2$.
Rearranging gives the quadratic equation: $x^2 - 25x + 144 = 0$.
Factoring the equation: $(x - 16)(x - 9) = 0$,so $x = 16$ or $x = 9$.
Given $AM > CM$,we choose $AM = 16$ and $MC = 9$.
Using the Pythagorean theorem in $\Delta ABM$: $AB^2 = AM^2 + BM^2 = 16^2 + 12^2 = 256 + 144 = 400 \implies AB = 20$.
Using the Pythagorean theorem in $\Delta CBM$: $BC^2 = MC^2 + BM^2 = 9^2 + 12^2 = 81 + 144 = 225 \implies BC = 15$.
Thus,$AB = 20$ and $BC = 15$.

(B) In a right-angled triangle,the altitude to the hypotenuse creates two triangles that are similar to the original triangle and to each other.
By the Geometric Mean Theorem (or properties of similar triangles),we have the relationship: $YZ^2 = ZM \cdot XZ$.
First,calculate the total length of the hypotenuse $XZ$:
$XZ = XM + ZM = 5 + 4 = 9$.
Now,substitute the values into the formula:
$YZ^2 = 4 \cdot 9 = 36$.
Taking the square root of both sides:
$YZ = \sqrt{36} = 6$.
Thus,the length of $YZ$ is $6$.

(C) In a right-angled triangle,if an altitude is drawn from the right-angle vertex to the hypotenuse,the triangle on either side of the altitude is similar to the original triangle and to each other.
Specifically,$\Delta RMQ \sim \Delta RQP$.
From the property of similar triangles,the ratio of corresponding sides is equal:
$\frac{RM}{RQ} = \frac{RQ}{RP}$.
Given $QR = 9$ and $PR = 13.5$.
Substituting the values into the equation:
$\frac{RM}{9} = \frac{9}{13.5}$.
$RM = \frac{9 \times 9}{13.5}$.
$RM = \frac{81}{13.5}$.
$RM = 6$.

(A) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$,$\overline{BM}$ is the altitude to the hypotenuse $\overline{AC}$.
By the geometric mean theorem for right triangles,we have the relation $BM^2 = AM \cdot CM$.
Given $BM = 12$ and $CM = 18$,we substitute these values into the equation:
$12^2 = AM \cdot 18$
$144 = AM \cdot 18$
$AM = \frac{144}{18} = 8$.
Now,consider $\Delta ABM$,which is a right-angled triangle at $M$ $(\angle AMB = 90^{\circ})$.
Using the Pythagorean theorem: $AB^2 = AM^2 + BM^2$.
$AB^2 = 8^2 + 12^2$
$AB^2 = 64 + 144$
$AB^2 = 208$
$AB = \sqrt{208} = \sqrt{16 \cdot 13} = 4 \sqrt{13}$.
Thus,the correct option is $A$.

(A) Let the length of the ladder be the hypotenuse $c = 6 \,m$.
Let the height reached on the wall be one leg $a = 3.6 \,m$.
Let the distance from the base of the wall to the lower end of the ladder be $b$.
According to the Pythagorean theorem for a right-angled triangle formed by the wall,the ground,and the ladder:
$a^2 + b^2 = c^2$
$(3.6)^2 + b^2 = 6^2$
$12.96 + b^2 = 36$
$b^2 = 36 - 12.96$
$b^2 = 23.04$
$b = \sqrt{23.04} = 4.8 \,m$.
Thus,the distance is $4.8 \,m$.

(B) Given that in $\Delta PQR$,$m \angle P = m \angle Q + m \angle R$.
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 $m \angle Q + m \angle R = m \angle P$,we get $m \angle P + m \angle P = 180^\circ$,which implies $2(m \angle P) = 180^\circ$,so $m \angle P = 90^\circ$.
Thus,$\Delta PQR$ is a right-angled triangle with the hypotenuse $QR = 25$ and one side $PQ = 7$.
Using the Pythagorean theorem,$PQ^2 + PR^2 = QR^2$.
$7^2 + PR^2 = 25^2$.
$49 + PR^2 = 625$.
$PR^2 = 625 - 49 = 576$.
$PR = \sqrt{576} = 24$.
The perimeter of $\Delta PQR = PQ + QR + PR = 7 + 25 + 24 = 56$.

(C) Let the side length 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 $d = 8$,we have $8 = a \sqrt{2}$.
To find the side length $a$,we divide both sides by $\sqrt{2}$:
$a = \frac{8}{\sqrt{2}}$.
Rationalizing the denominator:
$a = \frac{8}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{8 \sqrt{2}}{2} = 4 \sqrt{2}$.
Thus,the length of the side of the square is $4 \sqrt{2}$.

(D) $1$. The perimeter of a rhombus is given by $4 \times \text{side}$. Let the side length be $s$. So, $4s = 116$, which gives $s = 29$.
$2$. In a rhombus, the diagonals bisect each other at right angles $(90^{\circ})$.
$3$. Let the diagonals $AC$ and $BD$ intersect at point $O$. Then $AO = AC / 2 = 42 / 2 = 21$.
$4$. In the right-angled triangle $\triangle AOB$, by the Pythagorean theorem: $AO^2 + BO^2 = AB^2$.
$5$. Substituting the values: $21^2 + BO^2 = 29^2$.
$6$. $441 + BO^2 = 841$.
$7$. $BO^2 = 841 - 441 = 400$.
$8$. $BO = \sqrt{400} = 20$.
$9$. Since the diagonals bisect each other, $BD = 2 \times BO = 2 \times 20 = 40$.

(A) In a rectangle,the diagonals are equal in length,so $XZ = YW$. Given $XZ + YW = 26$,we have $2XZ = 26$,which implies $XZ = 13$.
In the right-angled triangle $\triangle XYZ$,by the Pythagorean theorem,$XY^2 + YZ^2 = XZ^2 = 13^2 = 169$.
We are given $XY + YZ = 17$. Squaring both sides,we get $(XY + YZ)^2 = 17^2 = 289$.
Expanding this,$XY^2 + YZ^2 + 2(XY \cdot YZ) = 289$.
Substituting $XY^2 + YZ^2 = 169$,we get $169 + 2(XY \cdot YZ) = 289$,so $2(XY \cdot YZ) = 120$,which means $XY \cdot YZ = 60$.
We need two numbers that add to $17$ and multiply to $60$. These are $12$ and $5$.
Since $XY > YZ$,we have $XY = 12$ and $YZ = 5$.

(B) In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.
Here,$\overline{BE}$ is the median to the hypotenuse $\overline{AC}$ because $\angle B = 90^{\circ}$.
Therefore,$BE = \frac{1}{2} AC$.
Given $BE = 8.5$,we have $8.5 = \frac{1}{2} AC$,which implies $AC = 17$.
In $\Delta ABC$,by the Pythagorean theorem,$AB^2 + BC^2 = AC^2$.
Substituting the known values: $15^2 + BC^2 = 17^2$.
$225 + BC^2 = 289$.
$BC^2 = 289 - 225 = 64$.
$BC = \sqrt{64} = 8$.

(C) Let the two equal sides of the isosceles right-angled triangle be $x$.
According to the Pythagorean theorem, $x^2 + x^2 = (\text{hypotenuse})^2$.
Given the hypotenuse is $24$, we have $2x^2 = 24^2$.
$2x^2 = 576$.
$x^2 = 288$.
The area of a right-angled triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
Here, base $= x$ and height $= x$, so Area $= \frac{1}{2} \times x \times x = \frac{1}{2}x^2$.
Substituting the value of $x^2$, we get Area $= \frac{1}{2} \times 288 = 144$.

(N/A) Let the two equal sides of the isosceles right-angled triangle be $x$. According to the Pythagorean theorem, $x^2 + x^2 = (20)^2$.
$2x^2 = 400$, which implies $x^2 = 200$.
Therefore, $x = \sqrt{200} = 10\sqrt{2}$.
The perimeter of the triangle is the sum of all sides: $P = x + x + 20 = 2x + 20 = 2(10\sqrt{2}) + 20 = 20\sqrt{2} + 20$.
The area of the triangle is given by $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times x \times x = \frac{1}{2} \times x^2$.
Substituting $x^2 = 200$, we get Area $= \frac{1}{2} \times 200 = 100$.

(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 $8.5$.

(A) Given that $m \angle A + m \angle B = m \angle C$.
Since the sum of angles in a triangle is $180^{\circ}$,we have $m \angle A + m \angle B + m \angle C = 180^{\circ}$.
Substituting $m \angle A + m \angle B = m \angle C$,we get $m \angle C + m \angle C = 180^{\circ}$,which implies $2m \angle C = 180^{\circ}$,so $m \angle C = 90^{\circ}$.
Thus,$\Delta ABC$ is a right-angled triangle with the hypotenuse $AB = 17.5$.
Given $AC : BC = 3 : 4$,let $AC = 3x$ and $BC = 4x$.
By the Pythagorean theorem,$AC^2 + BC^2 = AB^2$.
$(3x)^2 + (4x)^2 = (17.5)^2$.
$9x^2 + 16x^2 = 306.25$.
$25x^2 = 306.25$.
$x^2 = 12.25$,so $x = 3.5$.
Thus,$AC = 3(3.5) = 10.5$ and $BC = 4(3.5) = 14$.
The perimeter of $\Delta ABC = AC + BC + AB = 10.5 + 14 + 17.5 = 42$.

(B) Since $\Delta ABC$ is a right-angled triangle with $m \angle A = 90^{\circ}$,by the Pythagorean theorem,we have:
$AB^2 + AC^2 = BC^2$
Substituting the given expressions:
$(3x - 2)^2 + (5x + 4)^2 = (6x + 2)^2$
Expanding the squares:
$(9x^2 - 12x + 4) + (25x^2 + 40x + 16) = (36x^2 + 24x + 4)$
Combining like terms on the left side:
$34x^2 + 28x + 20 = 36x^2 + 24x + 4$
Rearranging the terms to one side:
$36x^2 - 34x^2 + 24x - 28x + 4 - 20 = 0$
$2x^2 - 4x - 16 = 0$
Dividing the entire equation by $2$:
$x^2 - 2x - 8 = 0$
Factoring the quadratic equation:
$(x - 4)(x + 2) = 0$
This gives $x = 4$ or $x = -2$.
Since lengths must be positive,$3x - 2 > 0$ implies $x > 2/3$. Thus,$x = 4$ is the only valid solution.

(C) In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.
First,calculate the length of the hypotenuse $PR$ using the Pythagorean theorem: $PR^2 = PQ^2 + QR^2$.
$PR^2 = 20^2 + 21^2 = 400 + 441 = 841$.
$PR = \sqrt{841} = 29$.
Since $\overline{QM}$ is the median to the hypotenuse $PR$,its length is $QM = \frac{1}{2} \times PR$.
$QM = \frac{29}{2} = 14.5$.

(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 $XZ = a^{2} + b^{2}$.

(B) Let the length of the ladder be the hypotenuse $h = 4 \,m$.
Let the height reached on the wall be the perpendicular $p = 3.2 \,m$.
Let the distance between the base of the wall and the lower end of the ladder be the base $b$.
According to the Pythagorean theorem,$h^2 = p^2 + b^2$.
Substituting the values: $4^2 = (3.2)^2 + b^2$.
$16 = 10.24 + b^2$.
$b^2 = 16 - 10.24 = 5.76$.
$b = \sqrt{5.76} = 2.4 \,m$.

(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,$BD = \frac{1}{2} \times AC = \frac{1}{2} \times 13 = 6.5$.

(A) Let $AB = x$ and $BC = y$. Given $x + y = 47$ and $x^2 + y^2 = AC^2 = 37^2 = 1369$.
We know that $(x + y)^2 = x^2 + y^2 + 2xy$.
Substituting the values: $47^2 = 1369 + 2xy$.
$2209 = 1369 + 2xy$,which gives $2xy = 840$,so $xy = 420$.
Now,we have $x + y = 47$ and $xy = 420$. These are the roots of the quadratic equation $t^2 - 47t + 420 = 0$.
Solving the quadratic equation: $t^2 - 35t - 12t + 420 = 0$.
$t(t - 35) - 12(t - 35) = 0$,so $(t - 35)(t - 12) = 0$.
The roots are $t = 35$ and $t = 12$.
Since $AB > BC$,we have $AB = 35$ and $BC = 12$.

(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 $= 4 \times 19.5 = 78$.

(D) In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.
Let the diagonals $AC$ and $BD$ intersect at point $O$.
Given $AB = 25$ and $AC = 14$.
Since diagonals bisect each other,$AO = OC = AC / 2 = 14 / 2 = 7$.
In the right-angled triangle $\triangle AOB$,by the Pythagorean theorem:
$AO^2 + OB^2 = AB^2$
$7^2 + OB^2 = 25^2$
$49 + OB^2 = 625$
$OB^2 = 625 - 49 = 576$
$OB = \sqrt{576} = 24$.
Since $BD = 2 \times OB$,we have $BD = 2 \times 24 = 48$.

(A) rhombus is a quadrilateral where all sides are equal. Let the side length be $s$.
Given the perimeter is $68$,we have $4s = 68$,which implies $s = 17$.
In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.
Let the diagonals $AC$ and $BD$ intersect at point $O$.
Since $AC = 30$,the length $AO = AC / 2 = 15$.
In the right-angled triangle $\triangle AOB$,by the Pythagorean theorem,$AO^2 + BO^2 = AB^2$.
Substituting the values: $15^2 + BO^2 = 17^2$.
$225 + BO^2 = 289$.
$BO^2 = 289 - 225 = 64$.
$BO = \sqrt{64} = 8$.
Since the diagonals bisect each other,$BD = 2 \times BO = 2 \times 8 = 16$.

(B) The perimeter of a square is given by $4 \times \text{side}$.
Given,$4 \times \text{side} = 20$.
Therefore,$\text{side} = 20 / 4 = 5$.
In a square $ABCD$,the diagonal $AC$ forms a right-angled triangle $ABC$ with sides $AB$ and $BC$.
By the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.
Since $AB = BC = 5$,we have $AC^2 = 5^2 + 5^2 = 25 + 25 = 50$.
Thus,$AC = \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$.

(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: $P = AB + BC + AC$.
$P = 11 + 60 + 61 = 132$.