Mix Examples - Triangles Questions in English

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

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

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

(B) In $\Delta ABC$,we are given $m \angle A = 90^{\circ}$ and $m \angle B = 30^{\circ}$.
Since the sum of angles in a triangle is $180^{\circ}$,$m \angle C = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$.
In a right-angled triangle,the side opposite to the $30^{\circ}$ angle is half the length of the hypotenuse.
The side opposite to $\angle B$ $(30^{\circ})$ is $AC$,and the hypotenuse is $BC$.
Therefore,$AC = \frac{1}{2} BC$.

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

(C) In $\Delta ABC$ and $\Delta ADB$:
$1$. $m \angle ABC = m \angle ADB = 90^{\circ}$ (Given).
$2$. $m \angle BAC = m \angle DAB$ (Common angle).
By the $AA$ (Angle-Angle) similarity criterion,the triangles are similar.
Therefore,the correspondence $ABC \leftrightarrow ADB$ is a similarity.

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

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

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

(D) In $\Delta ABC$,$m \angle A = 90^{\circ}$ and $\overline{AD}$ is an altitude to the hypotenuse $\overline{BC}$.
According to the geometric mean theorem for right-angled triangles,$AB^{2} = BD \cdot BC$.
Substituting the given values: $(\sqrt{5})^{2} = 2 \cdot BC$.
$5 = 2 \cdot BC$.
$BC = \frac{5}{2} = 2.5$.
Therefore,the length of the hypotenuse $\overline{BC}$ is $2.5$.

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

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

(A) In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $AC$.
According to the geometric mean theorem for right-angled triangles,$AB^2 = AM \cdot AC$.
Given $AB = \sqrt{10}$ and $AM = 2.5$.
Substituting the values: $(\sqrt{10})^2 = 2.5 \cdot AC$.
$10 = 2.5 \cdot AC$.
$AC = \frac{10}{2.5} = 4$.
Since $AC = AM + MC$,we have $4 = 2.5 + MC$.
$MC = 4 - 2.5 = 1.5$.

(C) In $\Delta PQR$,since $m\angle Q = 90^{\circ}$,the triangle is a right-angled triangle.
By the Pythagoras theorem,$PR^{2} = PQ^{2} + QR^{2}$.
Therefore,$PQ^{2} = PR^{2} - QR^{2}$.
Substituting the given values:
$PQ^{2} = (a^{2} + b^{2})^{2} - (a^{2} - b^{2})^{2}$.
Using the algebraic identity $(x + y)^{2} - (x - y)^{2} = 4xy$,where $x = a^{2}$ and $y = b^{2}$:
$PQ^{2} = 4(a^{2})(b^{2}) = 4a^{2}b^{2}$.
Taking the square root on both sides:
$PQ = \sqrt{4a^{2}b^{2}} = 2ab$.

(B) Let $AB = AC = x$.
In $\Delta ABC$,$m \angle A = 90^\circ$.
By the Pythagorean theorem,$AB^2 + AC^2 = BC^2$.
Substituting the given values: $x^2 + x^2 = (\sqrt{2} a)^2$.
$2x^2 = 2a^2$.
$x^2 = a^2$,so $x = a$ (since $a \in R^+$).
Thus,$AB = AC = a$.
The area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times AC$.
Area $= \frac{1}{2} \times a \times a = \frac{1}{2} a^2$.

(C) In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$.
By the properties of right-angled triangles with an altitude to the hypotenuse,we have:
$AB^{2} = AM \cdot AC$ and $BC^{2} = MC \cdot AC$
Subtracting these two equations:
$AB^{2} - BC^{2} = AM \cdot AC - MC \cdot AC$
$AB^{2} - BC^{2} = AC(AM - MC)$
Given that $AB^{2} - BC^{2} = 175$ and $AM - MC = 7$,we substitute these values:
$175 = AC(7)$
$AC = \frac{175}{7}$
$AC = 25$

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

(A) Given that $\square PQRS$ is a rectangle.
Area of rectangle $PQRS = PQ \cdot QR = 12$.
Sum of adjacent sides $PQ + QR = 7$.
In $\triangle PQR$,$\angle Q = 90^\circ$.
By Pythagoras theorem,$PR^2 = PQ^2 + QR^2$.
We know that $(PQ + QR)^2 = PQ^2 + QR^2 + 2(PQ \cdot QR)$.
Therefore,$PQ^2 + QR^2 = (PQ + QR)^2 - 2(PQ \cdot QR)$.
Substituting the given values: $PR^2 = (7)^2 - 2(12)$.
$PR^2 = 49 - 24 = 25$.
Taking the square root on both sides,$PR = 5$.

(A) Let $AB = x$ and $BC = y$.
Given that $x + y = 23$ and $xy = 120$.
We can form a quadratic equation $t^2 - (x+y)t + xy = 0$,which becomes $t^2 - 23t + 120 = 0$.
Factoring the quadratic equation: $t^2 - 15t - 8t + 120 = 0$.
$t(t - 15) - 8(t - 15) = 0$.
$(t - 15)(t - 8) = 0$.
So,the roots are $t = 15$ and $t = 8$.
Since $AB > BC$,we assign $AB = 15$ and $BC = 8$.
Therefore,$AB = 15$.

(B) Given that in $\Delta ABC$,$m \angle A = m \angle B + m \angle C$.
We know that the sum of angles in a triangle is $180^{\circ}$,so $m \angle A + m \angle B + m \angle C = 180^{\circ}$.
Substituting $m \angle B + m \angle C$ with $m \angle A$,we get $m \angle A + m \angle A = 180^{\circ}$,which implies $2 m \angle A = 180^{\circ}$,so $m \angle A = 90^{\circ}$.
Since $\Delta ABC$ is a right-angled triangle with the right angle at $A$,$BC$ is the hypotenuse.
By the Pythagorean theorem,$BC^2 = AB^2 + AC^2$.
Given $AB = 7$ and $BC = 25$,we have $25^2 = 7^2 + AC^2$.
$625 = 49 + AC^2$,so $AC^2 = 625 - 49 = 576$.
Thus,$AC = \sqrt{576} = 24$.
The perimeter of $\Delta ABC = AB + BC + AC = 7 + 25 + 24 = 56$.

(D) Let $\overline{AC}$ represent the ladder and $\overline{AB}$ represent the height of the wall.
In $\Delta ABC$,$\angle B = 90^{\circ}$.
By the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.
Given $AC = 6.5 \, m$ and $AB = 6 \, m$.
$6.5^2 = 6^2 + BC^2$
$42.25 = 36 + BC^2$
$BC^2 = 42.25 - 36 = 6.25$
$BC = \sqrt{6.25} = 2.5 \, m$.
Thus,the lower end of the ladder is $2.5 \, m$ away from the wall.

(C) In $\Delta ABC$,$D$ is the midpoint of $\overline{BC}$. Therefore,$\overline{AD}$ is a median.
By Apollonius' theorem:
$AB^2 + AC^2 = 2(AD^2 + BD^2)$
$7^2 + 5^2 = 2(5^2 + BD^2)$
$49 + 25 = 2(25 + BD^2)$
$74 = 2(25 + BD^2)$
$37 = 25 + BD^2$
$BD^2 = 12$
$BD = \sqrt{12} = 2\sqrt{3}$
Since $D$ is the midpoint of $\overline{BC}$,$BC = 2 \times BD = 2 \times 2\sqrt{3} = 4\sqrt{3}$.

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

(C) In $\Delta ABC$,$\overline{AD}$ is a median to side $\overline{BC}$.
According to Apollonius's Theorem,$AB^2 + AC^2 = 2(AD^2 + BD^2)$.
Given $AB^2 + AC^2 = 148$ and $AD = 7$.
Substituting the values: $148 = 2(7^2 + BD^2)$.
$148 = 2(49 + BD^2)$.
$74 = 49 + BD^2$.
$BD^2 = 74 - 49 = 25$.
$BD = \sqrt{25} = 5$.
Since $\overline{AD}$ is a median,$D$ is the midpoint of $\overline{BC}$,so $BC = 2 \times BD$.
$BC = 2 \times 5 = 10$.

(B) In $\Delta ADC$,$\angle D = 90^{\circ}$.
By Pythagoras theorem,$AC^{2} = AD^{2} + CD^{2}$.
$25^{2} = AD^{2} + 7^{2}$.
$625 = AD^{2} + 49$.
$AD^{2} = 625 - 49 = 576$.
$AD = \sqrt{576} = 24$.
The perimeter of rectangle $ABCD = 2(AD + CD) = 2(24 + 7) = 2(31) = 62$.

(B) In rhombus $XYZW$,let the diagonals $XZ$ and $YW$ intersect at point $O$.
The diagonals of a rhombus bisect each other at right angles.
Therefore,$OX = \frac{1}{2} XZ = \frac{1}{2} \times 14 = 7$ and $OY = \frac{1}{2} YW = \frac{1}{2} \times 48 = 24$.
In the right-angled triangle $\Delta XOY$,by the Pythagorean theorem:
$XY^2 = OX^2 + OY^2$
$XY^2 = 7^2 + 24^2$
$XY^2 = 49 + 576 = 625$
$XY = \sqrt{625} = 25$.

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

(C) Given that in $\Delta PQR$,$m \angle Q : m \angle R : m \angle P = 1 : 2 : 1$.
Let the angles be $x, 2x, x$.
Since the sum of angles in a triangle is $180^{\circ}$,we have $x + 2x + x = 180^{\circ}$,which gives $4x = 180^{\circ}$,so $x = 45^{\circ}$.
Thus,$m \angle Q = 45^{\circ}$,$m \angle R = 90^{\circ}$,and $m \angle P = 45^{\circ}$.
Since $m \angle P = m \angle Q$,the sides opposite to these angles are equal,so $QR = PR$.
In the right-angled $\Delta PQR$ (where $m \angle R = 90^{\circ}$),by the Pythagorean theorem:
$PQ^2 = PR^2 + QR^2$
Since $QR = PR$,we have $PQ^2 = PR^2 + PR^2 = 2PR^2$.
Given $PQ = 2 \sqrt{6}$,so $PQ^2 = (2 \sqrt{6})^2 = 4 \times 6 = 24$.
Therefore,$24 = 2PR^2$,which implies $PR^2 = 12$.
Taking the square root,$PR = \sqrt{12} = 2 \sqrt{3}$.

(D) In $\Delta ABC$,$\overline{AB} \cong \overline{AC}$,which means $\Delta ABC$ is an isosceles triangle.
In an isosceles triangle,the median to the base is also the altitude to the base.
Therefore,$\overline{AD} \perp \overline{BC}$,so $\overline{AD}$ is the height of the triangle with respect to base $\overline{BC}$.
Area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height}$
Area $= \frac{1}{2} \times BC \times AD$
Area $= \frac{1}{2} \times 18 \times 12$
Area $= 9 \times 12 = 108$.

(A) In $\Delta ABC$,the sum of angles is $m \angle A + m \angle B + m \angle C = 180^\circ$.
Given $m \angle A + m \angle C = m \angle B$,substituting this into the sum gives $m \angle B + m \angle B = 180^\circ$,so $2 m \angle B = 180^\circ$,which means $m \angle B = 90^\circ$.
Since $\Delta ABC$ is a right-angled triangle at $B$,by the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.
Let $AC = 17x$ and $AB = 15x$. Given $BC = 8$,we have $(17x)^2 = (15x)^2 + 8^2$.
$289x^2 = 225x^2 + 64$.
$64x^2 = 64$,so $x^2 = 1$,which gives $x = 1$.
Thus,$AB = 15(1) = 15$ and $AC = 17(1) = 17$.
The area of the right-angled triangle is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BC \times AB$.
Area $= \frac{1}{2} \times 8 \times 15 = 4 \times 15 = 60$.

(B) The perimeter of an equilateral triangle is given by the formula: $P = 3 \times \text{side}$.
Given $P = 12$,we have $12 = 3 \times \text{side}$.
Therefore,the side length is $\text{side} = \frac{12}{3} = 4$.
The area of an equilateral triangle is given by the formula: $\text{Area} = \frac{\sqrt{3}}{4} \times \text{side}^2$.
Substituting the side length: $\text{Area} = \frac{\sqrt{3}}{4} \times 4^2 = \frac{\sqrt{3}}{4} \times 16 = 4\sqrt{3}$.

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

(A) In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $AC$.
According to the geometric mean theorem for right-angled triangles,the altitude to the hypotenuse divides the triangle into two triangles similar to the original and to each other.
Thus,$BM^2 = AM \cdot CM$.
Given $AC = 13$ and $CM = 9$,we have $AM = AC - CM = 13 - 9 = 4$.
Substituting these values into the formula: $BM^2 = 4 \cdot 9 = 36$.
Therefore,$BM = \sqrt{36} = 6$.

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

(B) Let the side of the square be $a$.
We know that 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 perimeter of a square is given by the formula $P = 4 \times a$.
Substituting the value of $a$,we get $P = 4 \times 8 = 32$.

(D) The area of a square can be calculated using the length of its diagonal $d$ with the formula: $\text{Area} = \frac{d^2}{2}$.
Given the length of the diagonal $d = 6$.
Substituting the value into the formula:
$\text{Area} = \frac{6^2}{2} = \frac{36}{2} = 18$.

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

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

(A) In rectangle $ABCD$,we know that opposite sides are equal,so $CD = AB$ and $DA = BC$.
Also,$\angle B = 90^{\circ}$.
Given the equation: $AB^{2} + BC^{2} + CD^{2} + DA^{2} = 338$.
Substituting the equal sides: $AB^{2} + BC^{2} + AB^{2} + BC^{2} = 338$.
$2(AB^{2} + BC^{2}) = 338$.
$AB^{2} + BC^{2} = 169$.
In $\triangle ABC$,by Pythagoras theorem: $AC^{2} = AB^{2} + BC^{2}$.
Therefore,$AC^{2} = 169$.
$AC = \sqrt{169} = 13$.

(B) In $\Delta ABC$,$m \angle B = 90^{\circ}$.
Using the Pythagorean theorem,$AC^2 = AB^2 + BC^2 = 8^2 + 6^2 = 64 + 36 = 100$.
Therefore,$AC = \sqrt{100} = 10$.
The area of $\Delta ABC$ can be calculated in two ways: $\text{Area} = \frac{1}{2} \times AB \times BC$ or $\text{Area} = \frac{1}{2} \times AC \times BM$.
Equating the two expressions: $\frac{1}{2} \times AB \times BC = \frac{1}{2} \times AC \times BM$.
$AB \times BC = AC \times BM$.
$8 \times 6 = 10 \times BM$.
$48 = 10 \times BM$.
$BM = \frac{48}{10} = 4.8$.

(C) In $\Delta AMB$,$m \angle M = 90^{\circ}$ and $m \angle B = 45^{\circ}$.
Since the sum of angles in a triangle is $180^{\circ}$,$m \angle MAB = 180^{\circ} - 90^{\circ} - 45^{\circ} = 45^{\circ}$.
Since $m \angle MAB = m \angle B = 45^{\circ}$,$\Delta AMB$ is an isosceles right triangle.
Therefore,$AM = BM = 4$.
Given $M \in \overline{BC}$,we have $BC = BM + MC$.
$7 = 4 + MC$,which implies $MC = 7 - 4 = 3$.
In right-angled $\Delta AMC$,by the Pythagorean theorem:
$AC^{2} = AM^{2} + MC^{2}$
$AC^{2} = 4^{2} + 3^{2} = 16 + 9 = 25$
$AC = \sqrt{25} = 5$.

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

(C) In parallelogram $ABCD$,let $\overline{AC} \cap \overline{BD} = \{O\}$.
Since the diagonals of a parallelogram bisect each other,$O$ is the midpoint of $\overline{AC}$ and $\overline{BD}$.
Therefore,$BO = \frac{1}{2} BD = \frac{1}{2}(12) = 6$.
In $\Delta ABD$,$\overline{AO}$ is the median to side $\overline{BD}$.
By Apollonius' theorem,$AB^{2} + AD^{2} = 2(AO^{2} + BO^{2})$.
Substituting the given values: $200 = 2(AO^{2} + 6^{2})$.
$100 = AO^{2} + 36$.
$AO^{2} = 100 - 36 = 64$.
$AO = \sqrt{64} = 8$.
Since $O$ is the midpoint of $\overline{AC}$,$AC = 2 AO = 2 \times 8 = 16$.

(B) In a parallelogram $ABCD$,the sum of the squares of the diagonals is equal to the sum of the squares of its four sides.
This is known as the Parallelogram Law: $AC^{2} + BD^{2} = 2(AB^{2} + BC^{2})$.
Given $AB^{2} + BC^{2} = 260$ and $AC = 18$.
Substituting the values into the formula: $18^{2} + BD^{2} = 2(260)$.
$324 + BD^{2} = 520$.
$BD^{2} = 520 - 324$.
$BD^{2} = 196$.
$BD = \sqrt{196} = 14$.

(C) In a square,the length of a diagonal is given by the formula: $\text{Diagonal} = \sqrt{2} \times \text{side}$.
Given the side length $AB = 4 \sqrt{2}$.
Substituting the value of the side into the formula:
$\text{Diagonal} = \sqrt{2} \times (4 \sqrt{2})$
$= 4 \times (\sqrt{2} \times \sqrt{2})$
$= 4 \times 2 = 8$.
Therefore,the length of the diagonal is $8$.

(D) The perimeter of an equilateral triangle is given by the formula $P = 3 \times \text{side}$.
Given $P = 18$,we have $18 = 3 \times \text{side}$,which implies $\text{side} = 6$.
The altitude $(h)$ of an equilateral triangle is given by the formula $h = \frac{\sqrt{3}}{2} \times \text{side}$.
Substituting the value of the side,we get $h = \frac{\sqrt{3}}{2} \times 6 = 3 \sqrt{3}$.

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

(D) triplet $(a, b, c)$ is a Pythagorean triplet if it satisfies the condition $a^2 + b^2 = c^2$,where $c$ is the largest number.
For option $A$: $7^2 + 24^2 = 49 + 576 = 625 = 25^2$. This is a Pythagorean triplet.
For option $B$: $20^2 + 21^2 = 400 + 441 = 841 = 29^2$. This is a Pythagorean triplet.
For option $C$: $11^2 + 60^2 = 121 + 3600 = 3721 = 61^2$. This is a Pythagorean triplet.
For option $D$: $13^2 + 35^2 = 169 + 1225 = 1394$,while $37^2 = 1369$. Since $1394 \neq 1369$,this is not a Pythagorean triplet.

(B) $1.$ In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse. Thus,$BM = \frac{1}{2} AC$ $(1-d)$.
$2.$ In a right-angled triangle $\Delta ABC$ with altitude $\overline{AD}$ to the hypotenuse,by Apollonius theorem or properties of similar triangles,the relation $AC^2 = CD \cdot BC$ is derived from $\Delta ACD \sim \Delta BCA$ $(2-c)$.
$3.$ In a $30^\circ-60^\circ-90^\circ$ triangle,the side opposite to $30^\circ$ is half the hypotenuse. Here,$BC$ is opposite to $30^\circ$,so $BC = \frac{1}{2} AB$ $(3-b)$.
$4.$ Apollonius theorem states that for any triangle $\Delta ABC$ with median $\overline{BD}$,$AB^2 + BC^2 = 2(BD^2 + CD^2)$ $(4-a)$.
Therefore,the correct match is $(1-d), (2-c), (3-b), (4-a)$.