Mix Examples - Heron’s Formula Questions in English

(C) The sides of the triangle are $a = 290 \, m$,$b = 290 \, m$,and $c = 400 \, m$.
First,calculate the semi-perimeter $(s)$:
$s = \frac{a + b + c}{2} = \frac{290 + 290 + 400}{2} = \frac{980}{2} = 490 \, m$.
Using Heron's formula,Area = $\sqrt{s(s-a)(s-b)(s-c)}$.
Area = $\sqrt{490(490-290)(490-290)(490-400)}$.
Area = $\sqrt{490 \times 200 \times 200 \times 90}$.
Area = $\sqrt{490 \times 90 \times 200^2}$.
Area = $200 \times \sqrt{44100}$.
Area = $200 \times 210 = 42000 \, m^2$.

(D) Let the sides of the triangle be $a = 13\, cm$,$b = 14\, cm$,and $c = 15\, cm$.
First,calculate the semi-perimeter $s$:
$s = \frac{a + b + c}{2} = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21\, cm$.
Using Heron's formula,the area of the triangle is given by $\sqrt{s(s - a)(s - b)(s - c)}$.
Area $= \sqrt{21(21 - 13)(21 - 14)(21 - 15)}$
Area $= \sqrt{21 \times 8 \times 7 \times 6}$
Area $= \sqrt{(3 \times 7) \times (2^3) \times 7 \times (2 \times 3)}$
Area $= \sqrt{3^2 \times 7^2 \times 2^4}$
Area $= 3 \times 7 \times 2^2 = 21 \times 4 = 84\, cm^2$.

(A) For an equilateral triangle with side length $a = 120 \, m$,the semi-perimeter $s$ is calculated as:
$s = \frac{a + a + a}{2} = \frac{3a}{2} = \frac{3 \times 120}{2} = 180 \, m$.
Using Heron's formula,Area $= \sqrt{s(s-a)(s-b)(s-c)}$.
Since $a = b = c = 120 \, m$,we have:
Area $= \sqrt{180(180-120)(180-120)(180-120)}$
Area $= \sqrt{180 \times 60 \times 60 \times 60}$
Area $= \sqrt{180 \times 60^3} = \sqrt{3 \times 60 \times 60^3} = \sqrt{3 \times 60^4}$
Area $= 60^2 \sqrt{3} = 3600 \sqrt{3} \, m^2$.

(B) Let the sides of the triangle be $a = 33 \, cm$,$b = 34 \, cm$,and $c = 65 \, cm$.
First,calculate the semi-perimeter $s$:
$s = \frac{a + b + c}{2} = \frac{33 + 34 + 65}{2} = \frac{132}{2} = 66 \, cm$.
Using Heron's formula,the area of the triangle is $\sqrt{s(s - a)(s - b)(s - c)}$.
Area $= \sqrt{66(66 - 33)(66 - 34)(66 - 65)}$
Area $= \sqrt{66 \times 33 \times 32 \times 1}$
Area $= \sqrt{(2 \times 33) \times 33 \times (16 \times 2) \times 1}$
Area $= \sqrt{33^2 \times 2^2 \times 4^2} = 33 \times 2 \times 4 = 264 \, cm^2$.

(C) Let the sides of the triangle be $a = 510\,m$,$b = 520\,m$,and $c$.
The perimeter $P = a + b + c = 1560\,m$.
$510 + 520 + c = 1560 \implies 1030 + c = 1560 \implies c = 530\,m$.
The semi-perimeter $s = P/2 = 1560/2 = 780\,m$.
Using Heron's Formula,Area $= \sqrt{s(s-a)(s-b)(s-c)}$.
Area $= \sqrt{780(780-510)(780-520)(780-530)} = \sqrt{780 \times 270 \times 260 \times 250}$.
Area $= \sqrt{(78 \times 10) \times (27 \times 10) \times (26 \times 10) \times (25 \times 10)} = \sqrt{78 \times 27 \times 26 \times 25 \times 10^4}$.
Area $= \sqrt{(3 \times 26) \times (27) \times (26) \times (25) \times 10^4} = \sqrt{26^2 \times 81 \times 25 \times 10^4} = 26 \times 9 \times 5 \times 100 = 1,17,000\,m^2$.
Cost $= \text{Area} \times \text{Rate} = 1,17,000 \times 21 = \text{Rs. } 24,57,000$.

(D) regular hexagon consists of $6$ equilateral triangles with side length $a = 12\,cm$.
Perimeter of a regular hexagon $= 6 \times a = 6 \times 12\,cm = 72\,cm$.
Area of one equilateral triangle $= \frac{\sqrt{3}}{4} \times a^2 = \frac{\sqrt{3}}{4} \times 12^2 = \frac{\sqrt{3}}{4} \times 144 = 36\sqrt{3}\,cm^2$.
Total area of the hexagon $= 6 \times 36\sqrt{3}\,cm^2 = 216\sqrt{3}\,cm^2$.
Thus,the perimeter is $72\,cm$ and the area is $216\sqrt{3}\,cm^2$.

(A) Let the sides of the triangle be $a = 51 \, cm$,$b = 52 \, cm$,and $c = 101 \, cm$.
First,calculate the semi-perimeter $s$:
$s = \frac{a + b + c}{2} = \frac{51 + 52 + 101}{2} = \frac{204}{2} = 102 \, cm$.
Now,apply Heron's formula for the area of a triangle:
$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$
$\text{Area} = \sqrt{102(102-51)(102-52)(102-101)}$
$\text{Area} = \sqrt{102 \times 51 \times 50 \times 1}$
$\text{Area} = \sqrt{(51 \times 2) \times 51 \times (25 \times 2)}$
$\text{Area} = \sqrt{51^2 \times 2^2 \times 5^2}$
$\text{Area} = 51 \times 2 \times 5 = 510 \, cm^2$.

(B) Let the sides of the triangle be $a = 13 \, cm$,$b = 13 \, cm$,and $c = 10 \, cm$.
First,calculate the semi-perimeter $s = (a + b + c) / 2 = (13 + 13 + 10) / 2 = 36 / 2 = 18 \, cm$.
Using Heron's formula,Area $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{18(18-13)(18-13)(18-10)} = \sqrt{18 \times 5 \times 5 \times 8} = \sqrt{3600} = 60 \, cm^2$.
Now,to find the altitude $h$ corresponding to the base $c = 10 \, cm$,use the formula Area $= (1/2) \times \text{base} \times \text{height}$.
$60 = (1/2) \times 10 \times h \implies 60 = 5h \implies h = 12 \, cm$.

(C) Let the equal sides be $a = 30 \, cm$ and $b = 30 \, cm$.
Given the perimeter $P = 84 \, cm$,the third side $c$ is calculated as $c = P - (a + b) = 84 - (30 + 30) = 84 - 60 = 24 \, cm$.
The semi-perimeter $s = P / 2 = 84 / 2 = 42 \, cm$.
Using Heron's formula,the area $A = \sqrt{s(s-a)(s-b)(s-c)}$.
$A = \sqrt{42(42-30)(42-30)(42-24)}$.
$A = \sqrt{42 \times 12 \times 12 \times 18}$.
$A = 12 \sqrt{42 \times 18} = 12 \sqrt{756}$.
Since $756 = 36 \times 21$,we have $A = 12 \times 6 \sqrt{21} = 72 \sqrt{21} \, cm^2$.

(D) The diagonal $BD$ of quadrilateral $ABCD$ divides it into two non-overlapping triangles,$\Delta ABD$ and $\Delta BCD$.
For $\Delta ABD$,the sides are $a = 40 \, cm$,$b = 29 \, cm$,and $c = 29 \, cm$.
The semi-perimeter $s_1 = \frac{40 + 29 + 29}{2} = \frac{98}{2} = 49 \, cm$.
Area of $\Delta ABD = \sqrt{s_1(s_1 - a)(s_1 - b)(s_1 - c)} = \sqrt{49(49 - 40)(49 - 29)(49 - 29)} = \sqrt{49 \times 9 \times 20 \times 20} = 7 \times 3 \times 20 = 420 \, cm^2$.
For $\Delta BCD$,the sides are $a = 29 \, cm$,$b = 48 \, cm$,and $c = 35 \, cm$.
The semi-perimeter $s_2 = \frac{29 + 48 + 35}{2} = \frac{112}{2} = 56 \, cm$.
Area of $\Delta BCD = \sqrt{s_2(s_2 - a)(s_2 - b)(s_2 - c)} = \sqrt{56(56 - 29)(56 - 48)(56 - 35)} = \sqrt{56 \times 27 \times 8 \times 21} = \sqrt{(8 \times 7) \times (9 \times 3) \times 8 \times (3 \times 7)} = \sqrt{8^2 \times 7^2 \times 3^2 \times 9} = 8 \times 7 \times 3 \times 3 = 504 \, cm^2$.
Area of quadrilateral $ABCD = \text{Area of } \Delta ABD + \text{Area of } \Delta BCD = 420 + 504 = 924 \, cm^2$.

(A) In $\triangle ABD$, $\angle A = 90^{\circ}$, $AB = 15 \text{ cm}$, and $AD = 8 \text{ cm}$.
Using the Pythagorean theorem, $BD = \sqrt{AB^2 + AD^2} = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \text{ cm}$.
Area of right-angled $\triangle ABD = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 15 \times 8 = 60 \text{ cm}^2$.
In $\triangle BCD$, the sides are $a = 17 \text{ cm}$, $b = 25 \text{ cm}$, and $c = 12 \text{ cm}$.
Semi-perimeter $s = \frac{17 + 25 + 12}{2} = \frac{54}{2} = 27 \text{ cm}$.
Using Heron's formula, Area of $\triangle BCD = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{27(27-17)(27-25)(27-12)} = \sqrt{27 \times 10 \times 2 \times 15} = \sqrt{8100} = 90 \text{ cm}^2$.
Area of quadrilateral $ABCD = \text{Area}(\triangle ABD) + \text{Area}(\triangle BCD) = 60 + 90 = 150 \text{ cm}^2$.

(B) Let $ABCD$ be the plot which is divided by the diagonal $BD = 72 \, m$ into two equal triangles.
Since $ABCD$ is a rhombus,all its sides are equal.
Perimeter $= 4 \times \text{side} = 340 \, m$.
Therefore,side $= \frac{340}{4} = 85 \, m$.
In $\Delta ABD$,the sides are $a = 85 \, m$,$b = 85 \, m$,and $c = 72 \, m$.
Semi-perimeter $(s) = \frac{85 + 85 + 72}{2} = \frac{242}{2} = 121 \, m$.
Using Heron's formula,Area of $\Delta ABD = \sqrt{s(s-a)(s-b)(s-c)}$
$= \sqrt{121(121-85)(121-85)(121-72)}$
$= \sqrt{121 \times 36 \times 36 \times 49}$
$= 11 \times 36 \times 7 = 2772 \, m^2$.
Since the diagonal divides the rhombus into two equal triangles,the total area of the rhombus $ABCD = 2 \times \text{Area of } \Delta ABD$
$= 2 \times 2772 = 5544 \, m^2$.

(C) The quadrilateral $ABCD$ is divided into two triangles: $\triangle ABD$ and $\triangle BCD$.
For $\triangle ABD$,the sides are $a = 9 \, cm$,$b = 12 \, cm$,and $c = 15 \, cm$.
The semi-perimeter $s_1 = (9 + 12 + 15) / 2 = 36 / 2 = 18 \, cm$.
Area of $\triangle ABD = \sqrt{s_1(s_1-a)(s_1-b)(s_1-c)} = \sqrt{18(18-9)(18-12)(18-15)} = \sqrt{18 \times 9 \times 6 \times 3} = \sqrt{2916} = 54 \, cm^2$.
For $\triangle BCD$,the sides are $a = 13 \, cm$,$b = 14 \, cm$,and $c = 15 \, cm$.
The semi-perimeter $s_2 = (13 + 14 + 15) / 2 = 42 / 2 = 21 \, cm$.
Area of $\triangle BCD = \sqrt{s_2(s_2-a)(s_2-b)(s_2-c)} = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84 \, cm^2$.
Total area of quadrilateral $ABCD = 54 + 84 = 138 \, cm^2$.

(D) In parallelogram $ABCD$,the diagonal $AC$ divides the parallelogram into two congruent triangles,$\triangle ABC$ and $\triangle ADC$.
We calculate the area of $\triangle ABC$ using Heron's formula.
The sides of $\triangle ABC$ are $a = 9 \, cm$,$b = 10 \, cm$,and $c = 17 \, cm$.
The semi-perimeter $s = (a + b + c) / 2 = (9 + 10 + 17) / 2 = 36 / 2 = 18 \, cm$.
Area of $\triangle ABC = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{18(18-9)(18-10)(18-17)} = \sqrt{18 \times 9 \times 8 \times 1} = \sqrt{1296} = 36 \, cm^2$.
Since the area of the parallelogram is twice the area of $\triangle ABC$,the total area $= 2 \times 36 = 72 \, cm^2$.

(A) $1$. Divide the quadrilateral $ABCD$ into two triangles by drawing diagonal $AC$.
$2$. In $\triangle ABC$,since $\angle B = 90^{\circ}$,it is a right-angled triangle.
$3$. Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \times 15 = 150 \, cm^2$.
$4$. Using Pythagoras theorem in $\triangle ABC$,$AC^2 = AB^2 + BC^2 = 20^2 + 15^2 = 400 + 225 = 625$. Thus,$AC = 25 \, cm$.
$5$. Now,in $\triangle ADC$,the sides are $a = 25 \, cm$,$b = 12 \, cm$,$c = 17 \, cm$. The semi-perimeter $s = \frac{25 + 12 + 17}{2} = \frac{54}{2} = 27 \, cm$.
$6$. Area of $\triangle ADC = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{27(27-25)(27-12)(27-17)} = \sqrt{27 \times 2 \times 15 \times 10} = \sqrt{8100} = 90 \, cm^2$.
$7$. Total area of quadrilateral $ABCD = \text{Area}(\triangle ABC) + \text{Area}(\triangle ADC) = 150 + 90 = 240 \, cm^2$.

(B) The quadrilateral $PQRS$ is divided into two triangles: $\triangle PQR$ and $\triangle PSR$ by the diagonal $PR = 51 \, cm$.
For $\triangle PQR$,the sides are $a = 27 \, cm$,$b = 30 \, cm$,$c = 51 \, cm$.
The semi-perimeter $s_1 = (27 + 30 + 51) / 2 = 108 / 2 = 54 \, cm$.
Area of $\triangle PQR = \sqrt{s_1(s_1-a)(s_1-b)(s_1-c)} = \sqrt{54(54-27)(54-30)(54-51)} = \sqrt{54 \times 27 \times 24 \times 3} = \sqrt{104976} = 324 \, cm^2$.
For $\triangle PSR$,the sides are $a = 52 \, cm$,$b = 53 \, cm$,$c = 51 \, cm$.
The semi-perimeter $s_2 = (52 + 53 + 51) / 2 = 156 / 2 = 78 \, cm$.
Area of $\triangle PSR = \sqrt{s_2(s_2-a)(s_2-b)(s_2-c)} = \sqrt{78(78-52)(78-53)(78-51)} = \sqrt{78 \times 26 \times 25 \times 27} = \sqrt{1368900} = 1170 \, cm^2$.
Total area of quadrilateral $PQRS = 324 + 1170 = 1494 \, cm^2$.

(C) The quadrilateral $XYZW$ is divided into two triangles by the diagonal $XZ$: $\triangle XYZ$ and $\triangle XZW$.
For $\triangle XYZ$,the sides are $a = 25 \, m$,$b = 60 \, m$,and $c = 65 \, m$.
Semi-perimeter $s_1 = (25 + 60 + 65) / 2 = 150 / 2 = 75 \, m$.
Area of $\triangle XYZ = \sqrt{s_1(s_1-a)(s_1-b)(s_1-c)} = \sqrt{75(75-25)(75-60)(75-65)} = \sqrt{75 \times 50 \times 15 \times 10} = \sqrt{562500} = 750 \, m^2$.
For $\triangle XZW$,the sides are $a = 65 \, m$,$b = 33 \, m$,and $c = 34 \, m$.
Semi-perimeter $s_2 = (65 + 33 + 34) / 2 = 132 / 2 = 66 \, m$.
Area of $\triangle XZW = \sqrt{s_2(s_2-a)(s_2-b)(s_2-c)} = \sqrt{66(66-65)(66-33)(66-34)} = \sqrt{66 \times 1 \times 33 \times 32} = \sqrt{69696} = 264 \, m^2$.
Total area of the quadrilateral $XYZW = \text{Area}(\triangle XYZ) + \text{Area}(\triangle XZW) = 750 + 264 = 1014 \, m^2$.

(D) The quadrilateral $ABCD$ is divided into two triangles,$\triangle ABC$ and $\triangle ADC$,by the diagonal $AC = 48 \, cm$.
For $\triangle ABC$,the sides are $a = 29 \, cm$,$b = 35 \, cm$,and $c = 48 \, cm$.
The semi-perimeter $s_1 = (29 + 35 + 48) / 2 = 112 / 2 = 56 \, cm$.
Area of $\triangle ABC = \sqrt{s_1(s_1-a)(s_1-b)(s_1-c)} = \sqrt{56(56-29)(56-35)(56-48)} = \sqrt{56 \times 27 \times 21 \times 8} = \sqrt{254016} = 504 \, cm^2$.
For $\triangle ADC$,the sides are $a = 50 \, cm$,$b = 14 \, cm$,and $c = 48 \, cm$.
The semi-perimeter $s_2 = (50 + 14 + 48) / 2 = 112 / 2 = 56 \, cm$.
Area of $\triangle ADC = \sqrt{s_2(s_2-a)(s_2-b)(s_2-c)} = \sqrt{56(56-50)(56-14)(56-48)} = \sqrt{56 \times 6 \times 42 \times 8} = \sqrt{112896} = 336 \, cm^2$.
Total area of quadrilateral $ABCD = \text{Area}(\triangle ABC) + \text{Area}(\triangle ADC) = 504 + 336 = 840 \, cm^2$.

(A) $1$. Join $AC$ to divide the quadrilateral $ABCD$ into two triangles: $\triangle ABC$ and $\triangle ADC$.
$2$. In right-angled $\triangle ABC$,by Pythagoras theorem: $AC^2 = AB^2 + BC^2 = 9^2 + 40^2 = 81 + 1600 = 1681$. Thus,$AC = \sqrt{1681} = 41 \, cm$.
$3$. Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 40 \times 9 = 180 \, cm^2$.
$4$. For $\triangle ADC$,the sides are $a = 41 \, cm$,$b = 28 \, cm$,and $c = 15 \, cm$. Semi-perimeter $s = \frac{41 + 28 + 15}{2} = \frac{84}{2} = 42 \, cm$.
$5$. Area of $\triangle ADC = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{42(42-41)(42-28)(42-15)} = \sqrt{42 \times 1 \times 14 \times 27} = \sqrt{(14 \times 3) \times 14 \times (9 \times 3)} = \sqrt{14^2 \times 3^2 \times 3^2} = 14 \times 3 \times 3 = 126 \, cm^2$.
$6$. Total area of quadrilateral $ABCD = \text{Area}(\triangle ABC) + \text{Area}(\triangle ADC) = 180 + 126 = 306 \, cm^2$.

(B) Step $1$: Calculate the semi-perimeter $(s)$ of the triangle with sides $a = 10 \, cm$,$b = 17 \, cm$,and $c = 21 \, cm$.
$s = (a + b + c) / 2 = (10 + 17 + 21) / 2 = 48 / 2 = 24 \, cm$.
Step $2$: Calculate the area of the triangle using Heron's formula: $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$.
$\text{Area} = \sqrt{24(24-10)(24-17)(24-21)} = \sqrt{24 \times 14 \times 7 \times 3} = \sqrt{7056} = 84 \, cm^2$.
Step $3$: The area of the parallelogram is equal to the area of the triangle,so $\text{Area} = 84 \, cm^2$.
The formula for the area of a parallelogram is $\text{base} \times \text{height}$.
Given $\text{base} = 12 \, cm$,we have $12 \times \text{height} = 84$.
$\text{height} = 84 / 12 = 7 \, cm$.

(C) parallelogram is divided into two congruent triangles by its diagonal. Let the sides of the triangle be $a = 12 \, cm$,$b = 17 \, cm$,and $c = 25 \, cm$.
The semi-perimeter $s$ is calculated as $s = \frac{a + b + c}{2} = \frac{12 + 17 + 25}{2} = \frac{54}{2} = 27 \, cm$.
Using Heron's formula,the area of one triangle is $\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{27(27-12)(27-17)(27-25)} = \sqrt{27 \times 15 \times 10 \times 2} = \sqrt{8100} = 90 \, cm^2$.
The total area of the parallelogram is $2 \times 90 = 180 \, cm^2$.
The cost of painting at the rate of Rs. $3$ per $cm^2$ is $180 \times 3 = \text{Rs. } 540$.

(D) $1$. The perimeter of the rhombus is $680 \, m$. Since all sides of a rhombus are equal,the side length $a = 680 / 4 = 170 \, m$.
$2$. Let the diagonals be $d_1 = 300 \, m$ and $d_2$. In a rhombus,diagonals bisect each other at right angles.
$3$. Considering a right-angled triangle formed by the sides of the rhombus and half of the diagonals,we have $(d_1/2)^2 + (d_2/2)^2 = a^2$.
$4$. Substituting the values: $(300/2)^2 + (d_2/2)^2 = 170^2$,which gives $150^2 + (d_2/2)^2 = 28900$.
$5$. $22500 + (d_2/2)^2 = 28900$,so $(d_2/2)^2 = 6400$,which means $d_2/2 = 80 \, m$.
$6$. Thus,$d_2 = 160 \, m$.
$7$. The area of the rhombus is $(1/2) \times d_1 \times d_2 = (1/2) \times 300 \times 160 = 24000 \, m^2$.

(A) kite $ABCD$ is composed of two triangles,$\triangle ABC$ and $\triangle ADC$,joined at the diagonal $AC$.
Since $AB = AD = 12 \, cm$ and $CB = CD = 17 \, cm$,$\triangle ABC \cong \triangle ADC$ by $SSS$ congruence.
For $\triangle ABC$,sides are $a = 12 \, cm, b = 17 \, cm, c = 25 \, cm$.
Semi-perimeter $s = (12 + 17 + 25) / 2 = 54 / 2 = 27 \, cm$.
Area of $\triangle ABC = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{27(27-12)(27-17)(27-25)} = \sqrt{27 \times 15 \times 10 \times 2} = \sqrt{8100} = 90 \, cm^2$.
Total area of kite $ABCD = 2 \times 90 = 180 \, cm^2$.
Let $BD$ intersect $AC$ at $O$. $AC \perp BD$. Area of $\triangle ABC = (1/2) \times AC \times BO = 90$.
$(1/2) \times 25 \times BO = 90 \implies BO = 180 / 25 = 7.2 \, cm$.
Since $BD = 2 \times BO$,$BD = 2 \times 7.2 = 14.4 \, cm$.

(B) $1$. Draw a line $CE$ parallel to $AD$ such that $E$ lies on $AB$. Since $AD \parallel CE$ and $AE \parallel CD$, $AECD$ is a parallelogram.
$2$. Thus, $AE = CD = 20 \, cm$ and $CE = AD = 9 \, cm$.
$3$. The length $EB = AB - AE = 30 - 20 = 10 \, cm$.
$4$. In $\triangle CEB$, the sides are $17 \, cm$, $9 \, cm$, and $10 \, cm$. The semi-perimeter $s = (17 + 9 + 10) / 2 = 18 \, cm$.
$5$. Using Heron's formula, Area of $\triangle CEB = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{18(18-17)(18-9)(18-10)} = \sqrt{18 \times 1 \times 9 \times 8} = \sqrt{1296} = 36 \, cm^2$.
$6$. The area of $\triangle CEB$ is also given by $(1/2) \times \text{base} \times \text{height} = (1/2) \times 10 \times h = 36$, so $h = 7.2 \, cm$.
$7$. The area of trapezium $ABCD = (1/2) \times (AB + CD) \times h = (1/2) \times (30 + 20) \times 7.2 = 25 \times 7.2 = 180 \, cm^2$.

(C) Let the equal sides be $3x$ and the base be $2x$.
The perimeter of the triangle is the sum of all sides: $3x + 3x + 2x = 32 \, cm$.
$8x = 32 \implies x = 4$.
Thus,the sides are $a = 12 \, cm$,$b = 12 \, cm$,and $c = 8 \, cm$.
The semi-perimeter $s = \frac{32}{2} = 16 \, cm$.
Using Heron's formula,Area $= \sqrt{s(s-a)(s-b)(s-c)}$.
Area $= \sqrt{16(16-12)(16-12)(16-8)} = \sqrt{16 \times 4 \times 4 \times 8}$.
Area $= \sqrt{16 \times 16 \times 8} = 16 \times 2\sqrt{2} = 32\sqrt{2} \, cm^2$.

(D) $1$. The perimeter of the rhombus is $40 \, cm$. Since all sides of a rhombus are equal,the side length $a = 40 / 4 = 10 \, cm$.
$2$. Let the diagonals be $d_1 = 12 \, cm$ and $d_2$. In a rhombus,diagonals bisect each other at right angles. Using the Pythagorean theorem on one of the four right-angled triangles formed by the diagonals,we have $(d_2 / 2)^2 + (d_1 / 2)^2 = a^2$.
$3$. $(d_2 / 2)^2 + (12 / 2)^2 = 10^2 \implies (d_2 / 2)^2 + 36 = 100 \implies (d_2 / 2)^2 = 64 \implies d_2 / 2 = 8 \implies d_2 = 16 \, cm$.
$4$. The area of the rhombus is $(d_1 \times d_2) / 2 = (12 \times 16) / 2 = 96 \, cm^2$.
$5$. Since the sheet is printed on both sides,the total area to be printed is $96 \times 2 = 192 \, cm^2$.
$6$. The cost of printing is $192 \times 5 = Rs. \, 960$.

(A) To find the area of the trapezium $PQRS$,draw a line $ST$ perpendicular to $PQ$ at $T$.
Since $PQ \parallel SR$ and $QR \perp PQ$,$QR \perp SR$ as well. Thus,$TQRS$ is a rectangle.
Given $PQ = 12 \, m$ and $SR = 7 \, m$,we have $TQ = SR = 7 \, m$.
Therefore,$PT = PQ - TQ = 12 \, m - 7 \, m = 5 \, m$.
In the right-angled triangle $\triangle PTS$,by Pythagoras theorem:
$TS^2 = PS^2 - PT^2 = 13^2 - 5^2 = 169 - 25 = 144$.
So,$TS = \sqrt{144} = 12 \, m$.
Since $TQRS$ is a rectangle,$QR = TS = 12 \, m$.
The area of the trapezium is given by the formula:
$\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$
$\text{Area} = \frac{1}{2} \times (PQ + SR) \times QR$
$\text{Area} = \frac{1}{2} \times (12 + 7) \times 12 = 19 \times 6 = 114 \, m^2$.

(A) $(1)$ False. The perimeter of an equilateral triangle with side $a = 10 \, cm$ is $3a = 30 \, cm.$ The semi-perimeter $s$ is $\frac{30}{2} = 15 \, cm,$ not $30 \, cm.$
$(2)$ False. An isosceles triangle has two equal sides of $20 \, cm.$ Without the third side,the semi-perimeter cannot be determined as $20 \, cm.$ Even if the third side were $0$ (which is impossible),the semi-perimeter would be $\frac{20+20+0}{2} = 20 \, cm.$ Since the third side must be greater than $0,$ the semi-perimeter must be greater than $20 \, cm.$
$(3)$ False. In a right-angled triangle with sides $AB = 6 \, cm$ and $BC = 8 \, cm,$ the hypotenuse $AC = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, cm.$ The perimeter is $6 + 8 + 10 = 24 \, cm.$ The semi-perimeter $s = \frac{24}{2} = 12 \, cm.$ Wait,the statement is True.

(A) $(1)$ True: The diagonal $d$ of a square with side $a$ is given by $d = a \sqrt{2}$. For $a = 20 \, cm$,$d = 20 \sqrt{2} \, cm$. Thus,the statement is true.
$(2)$ False: The area of an equilateral triangle with side $a$ is $A = \frac{\sqrt{3}}{4} a^2$. If the side is doubled $(a' = 2a)$,the new area $A'$ becomes $A' = \frac{\sqrt{3}}{4} (2a)^2 = \frac{\sqrt{3}}{4} (4a^2) = 4 \times A$. The area becomes $4$ times the original area,not twice.

(D) The sides of the triangle are given as $a = 15 \text{ cm}$,$b = 17 \text{ cm}$,and $c = 20 \text{ cm}$.
The perimeter of a triangle is the sum of its sides: $P = a + b + c = 15 + 17 + 20 = 52 \text{ cm}$.
The semi-perimeter $(s)$ is defined as half of the perimeter: $s = \frac{P}{2} = \frac{52}{2} = 26 \text{ cm}$.
Therefore,the semi-perimeter of the triangle is $26 \text{ cm}$.

(A) According to Heron's formula,the area of a triangle with sides $a$,$b$,and $c$ is given by the expression $\sqrt{s(s-a)(s-b)(s-c)}$,where $s$ is the semi-perimeter of the triangle.
The semi-perimeter $s$ is calculated as $s = \frac{a+b+c}{2}$.
Therefore,the correct formula is $\sqrt{s(s-a)(s-b)(s-c)}$.

(B) In an equilateral triangle with side length $a$,the altitude divides the base into two equal parts of length $\frac{a}{2}$.
By applying the Pythagorean theorem in one of the resulting right-angled triangles:
$h^{2} + (\frac{a}{2})^{2} = a^{2}$
$h^{2} + \frac{a^{2}}{4} = a^{2}$
$h^{2} = a^{2} - \frac{a^{2}}{4}$
$h^{2} = \frac{3a^{2}}{4}$
Taking the square root of both sides:
$h = \frac{\sqrt{3}}{2} a$
Thus,the length of the altitude is $\frac{\sqrt{3}}{2} a$.

(C) The perimeter of a triangle with sides $a, b,$ and $c$ is the sum of its sides,which is $a+b+c$.
By definition,the semiperimeter $(s)$ is half of the perimeter.
Therefore,$s = \frac{a+b+c}{2}$.

(D) First,observe that the sides $6 \, cm$,$8 \, cm$,and $10 \, cm$ form a right-angled triangle because $6^2 + 8^2 = 36 + 64 = 100 = 10^2$.
Alternatively,using Heron's formula:
Semi-perimeter $s = (6 + 8 + 10) / 2 = 24 / 2 = 12 \, cm$.
Area $= \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{12(12 - 6)(12 - 8)(12 - 10)}$.
Area $= \sqrt{12 \times 6 \times 4 \times 2} = \sqrt{576} = 24 \, cm^{2}$.

(A) For an equilateral triangle with side length $a$,the semi-perimeter $s$ is given by $s = \frac{a+a+a}{2} = \frac{3a}{2}$.
Heron's formula for the area of a triangle is $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$.
Since $a = b = c$,the formula becomes $\text{Area} = \sqrt{s(s-a)(s-a)(s-a)} = \sqrt{s(s-a)^3}$.
This can be simplified as $\text{Area} = \sqrt{s(s-a)^2 \cdot (s-a)} = (s-a) \sqrt{s(s-a)}$.

(B) We are given the semi-perimeter $s = \frac{a + b + c}{2}$.
We have the equations:
$s - a = 10 \implies a = s - 10$
$s - b = 15 \implies b = s - 15$
$s - c = 12 \implies c = s - 12$
Adding these three equations:
$(s - a) + (s - b) + (s - c) = 10 + 15 + 12$
$3s - (a + b + c) = 37$
Since $a + b + c = 2s$,we substitute this into the equation:
$3s - 2s = 37$
$s = 37$
Thus,the semi-perimeter $s$ is $37 \, cm$.

(C) Let the sides of the triangle be $3x$,$4x$,and $6x$.
The perimeter of the triangle is the sum of its sides: $P = 3x + 4x + 6x = 13x$.
The semiperimeter $s$ is given by $s = P / 2 = 13x / 2$.
Given that the semiperimeter $s = 39 \, cm$,we have:
$13x / 2 = 39$
$13x = 78$
$x = 78 / 13 = 6$.
The sides of the triangle are $3(6) = 18 \, cm$,$4(6) = 24 \, cm$,and $6(6) = 36 \, cm$.
The longest side is $36 \, cm$.

(D) Let the sides of the right-angled triangle be $a$,$b$,and $c$,where $c$ is the hypotenuse.
Given: $c = 29\,cm$ and $a = 21\,cm$.
Using the Pythagorean theorem: $a^2 + b^2 = c^2$.
$21^2 + b^2 = 29^2$.
$441 + b^2 = 841$.
$b^2 = 841 - 441 = 400$.
$b = \sqrt{400} = 20\,cm$.
The perimeter of the triangle is $P = a + b + c = 21 + 20 + 29 = 70\,cm$.
The semi-perimeter $s$ is given by $s = \frac{P}{2} = \frac{70}{2} = 35\,cm$.

(A) In an isosceles triangle $\Delta ABC$,let $AB = AC = 13 \, cm$ and $BC = 24 \, cm$.
Draw an altitude $AD$ from vertex $A$ to the base $BC$. Since the triangle is isosceles,the altitude $AD$ bisects the base $BC$.
Therefore,$BD = DC = \frac{BC}{2} = \frac{24}{2} = 12 \, cm$.
In the right-angled triangle $\Delta ABD$,by the Pythagorean theorem:
$AB^2 = AD^2 + BD^2$
$13^2 = AD^2 + 12^2$
$169 = AD^2 + 144$
$AD^2 = 169 - 144 = 25$
$AD = \sqrt{25} = 5 \, cm$.
The area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BC \times AD$.
Area $= \frac{1}{2} \times 24 \times 5 = 12 \times 5 = 60 \, cm^2$.

(B) Let the side of the square $PQRS$ be $a \, cm$.
In a square,the diagonal $d$ is related to the side $a$ by the formula $d = a\sqrt{2}$.
Given that the diagonal $PR = 10 \, cm$,we have $a\sqrt{2} = 10$.
Solving for $a$,we get $a = \frac{10}{\sqrt{2}} = 5\sqrt{2} \, cm$.
The area of a square is given by the formula $\text{Area} = a^2$.
Substituting the value of $a$,we get $\text{Area} = (5\sqrt{2})^2 = 25 \times 2 = 50 \, cm^2$.
Alternatively,the area of a square can be calculated using the diagonal $d$ as $\text{Area} = \frac{1}{2} \times d^2$.
Substituting $d = 10 \, cm$,we get $\text{Area} = \frac{1}{2} \times (10)^2 = \frac{1}{2} \times 100 = 50 \, cm^2$.

(C) The area of a quadrilateral can be calculated by dividing it into two triangles using a diagonal.
Area of quadrilateral $ABCD = \text{Area}(\triangle ABC) + \text{Area}(\triangle ADC)$.
Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \times 8 = 80 \, cm^2$.
Area of $\triangle ADC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \times 12 = 120 \, cm^2$.
Total Area $= 80 + 120 = 200 \, cm^2$.

(D) Let the side of the rhombus be $a$. Since the perimeter is $100\,cm$,we have $4a = 100\,cm$,which gives $a = 25\,cm$.
Let the diagonals of the rhombus be $d_1 = 48\,cm$ and $d_2$. The diagonals of a rhombus bisect each other at right angles.
This forms a right-angled triangle with the side of the rhombus as the hypotenuse and half of the diagonals as the other two sides.
Using the Pythagorean theorem: $(a)^2 = (d_1/2)^2 + (d_2/2)^2$.
Substituting the values: $(25)^2 = (48/2)^2 + (d_2/2)^2$.
$625 = (24)^2 + (d_2/2)^2$.
$625 = 576 + (d_2/2)^2$.
$(d_2/2)^2 = 625 - 576 = 49$.
$d_2/2 = \sqrt{49} = 7$.
Therefore,$d_2 = 7 \times 2 = 14\,cm$.

(A) The area of a parallelogram is given by the formula: $\text{Area} = \text{base} \times \text{height}$.
Given,base $AB = 16 \, cm$ and the corresponding altitude $DM = 12 \, cm$.
Therefore,$\text{Area} = 16 \, cm \times 12 \, cm = 192 \, cm^{2}$.

(B) First,observe that $AB^2 + BC^2 = 6^2 + 8^2 = 36 + 64 = 100 = 10^2 = AC^2$.
Since the sum of the squares of two sides equals the square of the third side,$\Delta ABC$ is a right-angled triangle with the right angle at $B$.
The area of $\Delta ABC$ can be calculated as $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 6 \times 8 = 24 \, cm^2$.
Alternatively,using $AC$ as the base and $BM$ as the altitude,the area is $\frac{1}{2} \times AC \times BM = \frac{1}{2} \times 10 \times BM = 5 \times BM$.
Equating the two areas: $5 \times BM = 24$.
Therefore,$BM = \frac{24}{5} = 4.8 \, cm$.

(C) The area of a square is calculated by multiplying the length of its side by itself. Therefore, the formula for the area of a square is $\text{Area} = (\text{side})^{2}$.

(D) The area of an equilateral triangle with side length $a$ is given by the formula:
Area $= \frac{1}{2} \times \text{base} \times \text{height}$.
For an equilateral triangle,the base is $a$ and the height $h$ is calculated using the Pythagorean theorem: $h = \sqrt{a^2 - (a/2)^2} = \sqrt{3a^2/4} = \frac{\sqrt{3}}{2}a$.
Substituting these values: Area $= \frac{1}{2} \times a \times \frac{\sqrt{3}}{2}a = \frac{\sqrt{3}}{4}a^2$.
Therefore,the correct option is $D$.

(A) The area of a rectangle is defined as the product of its length and breadth.
Therefore,Area of a rectangle $=$ $\text{length} \times \text{breadth}$.

(A) The area of a rhombus is calculated by the formula: $\text{Area} = \frac{1}{2} \times d_1 \times d_2$,where $d_1$ and $d_2$ are the lengths of the two diagonals of the rhombus.
Therefore,the area of a rhombus is $\frac{1}{2} \times$ product of its diagonals.

(B) The area of a square can be calculated using its side length as $Area = side^2$. However,if the length of the diagonal $(d)$ is given,the area is calculated as $Area = \frac{d^2}{2}$. Therefore,the missing term is the diagonal.

(C) The area of a parallelogram is calculated by the product of its base and the corresponding altitude (height) drawn to that base. Therefore, $\text{Area} = \text{base} \times \text{corresponding altitude}$.