Textbook - Heron’s Formula Questions in English

(A) Given,the perimeter of the triangle is $32 \, cm$. Let the sides be $a = 8 \, cm$ and $b = 11 \, cm$.
The third side $c$ is calculated as:
$c = 32 \, cm - (8 + 11) \, cm = 32 \, cm - 19 \, cm = 13 \, cm$.
The semi-perimeter $s$ is:
$s = \frac{\text{Perimeter}}{2} = \frac{32}{2} \, cm = 16 \, cm$.
Now,calculate $(s - a)$,$(s - b)$,and $(s - c)$:
$s - a = 16 - 8 = 8 \, cm$
$s - b = 16 - 11 = 5 \, cm$
$s - c = 16 - 13 = 3 \, cm$
Using Heron's formula,the area of the triangle is:
$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$
$\text{Area} = \sqrt{16 \times 8 \times 5 \times 3} \, cm^2$
$\text{Area} = \sqrt{16 \times 120} \, cm^2$
$\text{Area} = 4 \times \sqrt{4 \times 30} \, cm^2$
$\text{Area} = 4 \times 2 \sqrt{30} \, cm^2 = 8 \sqrt{30} \, cm^2$.

(B) To find the area of the park,we use Heron's formula.
First,calculate the semi-perimeter $s$:
$2s = 50\,m + 80\,m + 120\,m = 250\,m$
$s = 125\,m$
Now,calculate the differences:
$s - a = 125\,m - 120\,m = 5\,m$
$s - b = 125\,m - 80\,m = 45\,m$
$s - c = 125\,m - 50\,m = 75\,m$
Area of the park $= \sqrt{s(s-a)(s-b)(s-c)}$
$= \sqrt{125 \times 5 \times 45 \times 75}\,m^2$
$= \sqrt{2109375}\,m^2 = 375\sqrt{15}\,m^2$
Next,calculate the cost of fencing:
Perimeter of the park $= 250\,m$
Length of wire needed $= 250\,m - 3\,m = 247\,m$
Cost of fencing $= 247\,m \times Rs. 20/m = Rs. 4940$

(C) Let the sides of the triangle be $3x, 5x$ and $7x$ meters.
The perimeter of the triangle is given as $300\, m$.
Therefore,$3x + 5x + 7x = 300$.
$15x = 300$,which gives $x = 20$.
Thus,the sides of the triangle are $3 \times 20 = 60\, m$,$5 \times 20 = 100\, m$,and $7 \times 20 = 140\, m$.
Using Heron's formula,the semi-perimeter $s$ is:
$s = \frac{60 + 100 + 140}{2} = \frac{300}{2} = 150\, m$.
The area of the triangle is $\sqrt{s(s-a)(s-b)(s-c)}$.
Area $= \sqrt{150(150 - 60)(150 - 100)(150 - 140)}\, m^2$.
Area $= \sqrt{150 \times 90 \times 50 \times 10}\, m^2$.
Area $= \sqrt{6750000}\, m^2$.
Area $= 1500\sqrt{3}\, m^2$.

(D) For an equilateral triangle with side $a$,the area is given by $\frac{\sqrt{3}}{4} a^2$.
Given that the perimeter of the equilateral triangle is $180 \, cm$.
Since all three sides are equal,$3a = 180 \, cm$,which gives $a = 60 \, cm$.
Using Heron's formula,the semi-perimeter $s = \frac{\text{Perimeter}}{2} = \frac{180}{2} = 90 \, cm$.
The area of the triangle is $\sqrt{s(s-a)(s-b)(s-c)}$.
Substituting the values: $\text{Area} = \sqrt{90(90-60)(90-60)(90-60)} \, cm^2$.
$\text{Area} = \sqrt{90 \times 30 \times 30 \times 30} \, cm^2$.
$\text{Area} = \sqrt{3 \times 30 \times 30 \times 30 \times 30} \, cm^2$.
$\text{Area} = \sqrt{3 \times (30)^2 \times (30)^2} \, cm^2$.
$\text{Area} = 30 \times 30 \times \sqrt{3} \, cm^2 = 900 \sqrt{3} \, cm^2$.

(A) The sides of the triangular wall are $a = 122\, m, b = 120\, m, c = 22\, m$.
The semi-perimeter $s$ is given by:
$s = \frac{a + b + c}{2} = \frac{122 + 120 + 22}{2}\, m = \frac{264}{2}\, m = 132\, m$.
The area of the triangle using Heron's formula is $\sqrt{s(s-a)(s-b)(s-c)}$.
Area $= \sqrt{132(132 - 122)(132 - 120)(132 - 22)}\, m^2$
$= \sqrt{132 \times 10 \times 12 \times 110}\, m^2$
$= \sqrt{12 \times 11 \times 10 \times 12 \times 11 \times 10}\, m^2$
$= \sqrt{12^2 \times 11^2 \times 10^2}\, m^2 = 12 \times 11 \times 10\, m^2 = 1320\, m^2$.
Rent for $1$ year ($12$ months) per $m^2 = Rs. 5000$.
Rent for $3$ months per $m^2 = Rs. 5000 \times \frac{3}{12} = Rs. 1250$.
Total rent for $1320\, m^2$ for $3$ months $= 1320 \times 1250 = Rs. 16,50,000$.

(B) The sides of the triangular wall are $a = 15 \, m$,$b = 11 \, m$,and $c = 6 \, m$.
First,we calculate the semi-perimeter $(s)$:
$s = \frac{a + b + c}{2} = \frac{15 + 11 + 6}{2} \, m = \frac{32}{2} \, m = 16 \, m$.
Now,using Heron's formula,the area of the triangular surface is:
Area $= \sqrt{s(s - a)(s - b)(s - c)}$
$= \sqrt{16(16 - 15)(16 - 11)(16 - 6)} \, m^2$
$= \sqrt{16 \times 1 \times 5 \times 10} \, m^2$
$= \sqrt{16 \times 50} \, m^2$
$= \sqrt{800} \, m^2$
$= \sqrt{400 \times 2} \, m^2 = 20 \sqrt{2} \, m^2$.
Thus,the required area painted in colour is $20 \sqrt{2} \, m^2$.

(C) Let the sides of the triangle be $a = 18 \, cm$,$b = 10 \, cm$,and $c$ be the third side.
Given that the perimeter $(2s) = 42 \, cm$.
Therefore,the semi-perimeter $s = \frac{42}{2} = 21 \, cm$.
The third side $c = 42 - (18 + 10) = 42 - 28 = 14 \, cm$.
Using Heron's formula,the area of a triangle $= \sqrt{s(s-a)(s-b)(s-c)}$.
Substituting the values: Area $= \sqrt{21(21-18)(21-10)(21-14)} \, cm^2$.
Area $= \sqrt{21 \times 3 \times 11 \times 7} \, cm^2$.
Area $= \sqrt{(3 \times 7) \times 3 \times 11 \times 7} \, cm^2 = \sqrt{3^2 \times 7^2 \times 11} \, cm^2$.
Area $= 3 \times 7 \sqrt{11} \, cm^2 = 21 \sqrt{11} \, cm^2$.

(D) Perimeter of the triangle $= 540 \, cm$.
Semi-perimeter $s = \frac{540}{2} = 270 \, cm$.
Let the sides be $12x, 17x,$ and $25x$.
$12x + 17x + 25x = 540 \Rightarrow 54x = 540 \Rightarrow x = 10$.
Thus,the sides are $a = 120 \, cm, b = 170 \, cm, c = 250 \, cm$.
Using Heron's formula: Area $= \sqrt{s(s-a)(s-b)(s-c)}$.
$s-a = 270 - 120 = 150 \, cm$.
$s-b = 270 - 170 = 100 \, cm$.
$s-c = 270 - 250 = 20 \, cm$.
Area $= \sqrt{270 \times 150 \times 100 \times 20} = \sqrt{81,000,000} = 9,000 \, cm^2$.

(A) The equal sides of the triangle are $12 \, cm$ each.
Let the third side be $x \, cm$.
Since the perimeter is $30 \, cm$,we have:
$12 \, cm + 12 \, cm + x \, cm = 30 \, cm$
$24 \, cm + x = 30 \, cm$
$x = 30 - 24 = 6 \, cm$.
Now,the semi-perimeter $s = \frac{\text{Perimeter}}{2} = \frac{30}{2} = 15 \, cm$.
Using Heron's formula,the area of the triangle is:
$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$
$\text{Area} = \sqrt{15(15-12)(15-12)(15-6)} \, cm^2$
$\text{Area} = \sqrt{15 \times 3 \times 3 \times 9} \, cm^2$
$\text{Area} = \sqrt{15 \times 9 \times 9} \, cm^2 = 9 \sqrt{15} \, cm^2$.
Thus,the area of the triangle is $9 \sqrt{15} \, cm^2$.

(A) For the triangular field with sides $a = 200 \, m, b = 240 \, m, c = 360 \, m$:
Semi-perimeter $s = \frac{200 + 240 + 360}{2} = 400 \, m$.
Area of the wheat field = $\sqrt{400(400-200)(400-240)(400-360)} = \sqrt{400 \times 200 \times 160 \times 40} = \sqrt{512000000} = 16000\sqrt{2} \, m^2 \approx 22627.4 \, m^2 \approx 2.26 \, \text{hectares}$.
For the second triangular field with sides $240 \, m, 320 \, m, 400 \, m$:
Semi-perimeter $s = \frac{240 + 320 + 400}{2} = 480 \, m$.
Area of the second field = $\sqrt{480(480-240)(480-320)(480-400)} = \sqrt{480 \times 240 \times 160 \times 80} = \sqrt{1474560000} = 38400 \, m^2 = 3.84 \, \text{hectares}$.
The median divides the triangle into two triangles of equal area.
Area for potatoes = Area for onions = $\frac{3.84}{2} = 1.92 \, \text{hectares}$ each.

(C) Since $AB = 9 \, m$ and $BC = 40 \, m$ with $\angle B = 90^{\circ}$,we use the Pythagorean theorem to find $AC$:
$AC = \sqrt{AB^2 + BC^2} = \sqrt{9^2 + 40^2} = \sqrt{81 + 1600} = \sqrt{1681} = 41 \, m$.
The first group cleans the area of the right-angled triangle $ABC$:
Area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 40 \times 9 = 180 \, m^2$.
The second group cleans the area of triangle $ACD$ with sides $41 \, m, 15 \, m$ and $28 \, m$. Using Heron's formula,the semi-perimeter $s = \frac{41 + 15 + 28}{2} = \frac{84}{2} = 42 \, m$.
Area of $\Delta ACD = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{42(42 - 41)(42 - 15)(42 - 28)} = \sqrt{42 \times 1 \times 27 \times 14} = \sqrt{15876} = 126 \, m^2$.
The first group cleaned $180 \, m^2$ and the second group cleaned $126 \, m^2$. The first group cleaned more area by $(180 - 126) = 54 \, m^2$.
Total area cleaned by the students $= 180 + 126 = 306 \, m^2$.

(D) Join $B$ and $D$ to divide the quadrilateral $ABCD$ into two triangles: $\Delta BCD$ and $\Delta ABD$.
$1$. In the right-angled $\Delta BCD$ (since $\angle C = 90^{\circ}$):
Area of $\Delta BCD = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times 5 = 30 \, m^{2}$.
$2$. Calculate the length of the diagonal $BD$ using the Pythagoras theorem:
$BD^{2} = BC^{2} + CD^{2} = 12^{2} + 5^{2} = 144 + 25 = 169$.
$BD = \sqrt{169} = 13 \, m$.
$3$. For $\Delta ABD$,the sides are $a = 9 \, m$,$b = 8 \, m$,and $c = 13 \, m$.
Semi-perimeter $s = \frac{9 + 8 + 13}{2} = \frac{30}{2} = 15 \, m$.
Using Heron's formula,Area of $\Delta ABD = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{15(15-9)(15-8)(15-13)} = \sqrt{15 \times 6 \times 7 \times 2} = \sqrt{1260} = 6\sqrt{35} \, m^{2}$.
Using $\sqrt{35} \approx 5.916$,Area $\approx 6 \times 5.916 = 35.496 \approx 35.5 \, m^{2}$.
$4$. Total area of quadrilateral $ABCD = \text{Area}(\Delta BCD) + \text{Area}(\Delta ABD) = 30 + 35.5 = 65.5 \, m^{2}$.

(A) The quadrilateral $ABCD$ is divided into two triangles,$\Delta ABC$ and $\Delta ACD$,by the diagonal $AC$.
For $\Delta ABC$:
The sides are $a = 3 \, cm, b = 4 \, cm, c = 5 \, cm$.
Semi-perimeter $s_1 = \frac{3 + 4 + 5}{2} = 6 \, cm$.
Area of $\Delta ABC = \sqrt{s_1(s_1 - a)(s_1 - b)(s_1 - c)} = \sqrt{6(6 - 3)(6 - 4)(6 - 5)} = \sqrt{6 \times 3 \times 2 \times 1} = \sqrt{36} = 6 \, cm^2$.
For $\Delta ACD$:
The sides are $a = 5 \, cm, b = 4 \, cm, c = 5 \, cm$.
Semi-perimeter $s_2 = \frac{5 + 4 + 5}{2} = 7 \, cm$.
Area of $\Delta ACD = \sqrt{s_2(s_2 - a)(s_2 - b)(s_2 - c)} = \sqrt{7(7 - 5)(7 - 4)(7 - 5)} = \sqrt{7 \times 2 \times 3 \times 2} = \sqrt{84} = 2\sqrt{21} \, cm^2$.
Using $\sqrt{21} \approx 4.58$,Area $\approx 2 \times 4.58 = 9.16 \, cm^2$.
Total area of quadrilateral $ABCD = \text{Area}(\Delta ABC) + \text{Area}(\Delta ACD) = 6 + 9.16 = 15.16 \, cm^2 \approx 15.2 \, cm^2$.

(B) Area of surface $I$:
It is an isosceles triangle with sides $a = 5\, cm, b = 5\, cm, c = 1\, cm$.
Semi-perimeter $s = \frac{5+5+1}{2} = 5.5\, cm$.
Area $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{5.5(5.5-5)(5.5-5)(5.5-1)} = \sqrt{5.5 \times 0.5 \times 0.5 \times 4.5} = \sqrt{6.1875} \approx 2.48\, cm^{2}$.
Area of surface $II$:
It is a rectangle with length $6.5\, cm$ and breadth $1\, cm$.
Area $= 6.5 \times 1 = 6.5\, cm^{2}$.
Area of surface $III$:
It is a trapezium with parallel sides $1\, cm$ and $2\, cm$. The height $h$ is calculated using Pythagoras theorem: $h = \sqrt{1^{2} - (0.5)^{2}} = \sqrt{0.75} \approx 0.866\, cm$.
Area $= \frac{1}{2} \times (1 + 2) \times 0.866 = 1.5 \times 0.866 \approx 1.3\, cm^{2}$.
Area of surface $IV$ and $V$:
These are two congruent right-angled triangles with base $6\, cm$ and height $1.5\, cm$.
Area of $IV = \frac{1}{2} \times 6 \times 1.5 = 4.5\, cm^{2}$.
Area of $V = 4.5\, cm^{2}$.
Total area $= 2.48 + 6.5 + 1.3 + 4.5 + 4.5 = 19.28\, cm^{2} \approx 19.3\, cm^{2}$.

(C) For the given triangle,the sides are $a = 28 \, cm$,$b = 30 \, cm$,and $c = 26 \, cm$.
The semi-perimeter $s$ is calculated as:
$s = \frac{a + b + c}{2} = \frac{28 + 30 + 26}{2} \, cm = \frac{84}{2} = 42 \, cm$.
Using Heron's formula,the area of the triangle is:
$\text{Area} = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{42(42 - 28)(42 - 30)(42 - 26)} \, cm^2$
$= \sqrt{42 \times 14 \times 12 \times 16} \, cm^2$
$= \sqrt{(6 \times 7) \times (2 \times 7) \times (6 \times 2) \times 16} \, cm^2$
$= \sqrt{6^2 \times 7^2 \times 2^2 \times 4^2} \, cm^2 = 6 \times 7 \times 2 \times 4 \, cm^2 = 336 \, cm^2$.
Since the area of the parallelogram is equal to the area of the triangle:
$\text{Area of the parallelogram} = 336 \, cm^2$.
The formula for the area of a parallelogram is $\text{base} \times \text{height}$.
$28 \, cm \times h = 336 \, cm^2$
$h = \frac{336}{28} \, cm = 12 \, cm$.
Thus,the height of the parallelogram is $12 \, cm$.

(D) Here,each side of the rhombus $= 30\, m$.
One of the diagonals $= 48\, m$.
Since a diagonal divides the rhombus into two congruent triangles,the sides of the first triangle are $a = 30\, m, b = 30\, m, c = 48\, m$.
The semi-perimeter $s = \frac{a + b + c}{2} = \frac{30 + 30 + 48}{2}\, m = \frac{108}{2} = 54\, m$.
Area of triangle $I = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{54(54 - 30)(54 - 30)(54 - 48)}\, m^{2}$.
$= \sqrt{54 \times 24 \times 24 \times 6}\, m^{2} = \sqrt{(9 \times 6) \times 24 \times 24 \times 6}\, m^{2} = \sqrt{9 \times 36 \times 24^{2}}\, m^{2} = 3 \times 6 \times 24\, m^{2} = 432\, m^{2}$.
Area of the second triangle is also $432\, m^{2}$.
Total area of the rhombus $= 432\, m^{2} + 432\, m^{2} = 864\, m^{2}$.
Thus,the area of grass for $18$ cows $= 864\, m^{2}$.
Area of grass for $1$ cow $= \frac{864}{18}\, m^{2} = 48\, m^{2}$.

(N/A) The sides of each triangular piece are $a = 20\, cm$,$b = 50\, cm$,and $c = 50\, cm$.
The semi-perimeter $s$ is calculated as:
$s = \frac{a + b + c}{2} = \frac{20 + 50 + 50}{2}\, cm = \frac{120}{2}\, cm = 60\, cm$.
Using Heron's formula,the area of each triangular piece is:
Area $= \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{60(60 - 20)(60 - 50)(60 - 50)}\, cm^2$
$= \sqrt{60 \times 40 \times 10 \times 10}\, cm^2 = \sqrt{240000}\, cm^2 = 200\sqrt{6}\, cm^2$.
Since there are $10$ triangular pieces in total,there are $5$ pieces of each colour.
Area of cloth required for each colour $= 5 \times 200\sqrt{6}\, cm^2 = 1000\sqrt{6}\, cm^2$.

(N/A) Area of triangle $I$:
Diagonal $= 32\, cm$.
Since the diagonals of a square bisect each other at $90^\circ$,the height of triangle $I$ (which is half the diagonal) $= \frac{1}{2} \times 32\, cm = 16\, cm$.
The base of the triangle is the other diagonal of the square,which is also $32\, cm$.
Area of triangle $I = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 32\, cm \times 16\, cm = 256\, cm^2$.
Area of triangle $II$:
Since the diagonal of a square divides it into two congruent triangles,the area of triangle $II$ is equal to the area of triangle $I$.
Area of triangle $II = 256\, cm^2$.
Area of triangle $III$:
The triangle at the base has sides $a = 8\, cm, b = 6\, cm, c = 6\, cm$.
Semi-perimeter $s = \frac{a + b + c}{2} = \frac{8 + 6 + 6}{2} = 10\, cm$.
Using Heron's formula,Area $= \sqrt{s(s-a)(s-b)(s-c)}$.
Area $= \sqrt{10(10-8)(10-6)(10-6)} = \sqrt{10 \times 2 \times 4 \times 4} = \sqrt{320} = 8\sqrt{5}\, cm^2$.
Taking $\sqrt{5} \approx 2.24$,Area $\approx 8 \times 2.24 = 17.92\, cm^2$.
Thus,the area of paper used for each shade is:
Shade $I = 256\, cm^2$,Shade $II = 256\, cm^2$,Shade $III = 17.92\, cm^2$.

(C) There are $16$ equal triangular tiles.
Sides of the triangle are $a = 9\, cm, b = 28\, cm, c = 35\, cm$.
Semi-perimeter $s = \frac{a + b + c}{2} = \frac{9 + 28 + 35}{2} = \frac{72}{2} = 36\, cm$.
Area of one triangle using Heron's formula $= \sqrt{s(s - a)(s - b)(s - c)}$.
Area $= \sqrt{36(36 - 9)(36 - 28)(36 - 35)} = \sqrt{36 \times 27 \times 8 \times 1} = \sqrt{7776} = 36\sqrt{6}\, cm^2$.
Using $\sqrt{6} \approx 2.449$,Area $\approx 36 \times 2.449 = 88.164\, cm^2$.
Total area of $16$ tiles $= 16 \times 88.164 = 1410.624\, cm^2$.
Cost of polishing at $50p$ (or $Rs. 0.50$) per $cm^2$ $= 1410.624 \times 0.50 = Rs. 705.312$.
Rounding to the nearest value provided in the options,the cost is $Rs. 705.60$.

(D) The given field is in the form of a trapezium $ABCD$ such that parallel sides are $AB = 10\, m$ and $DC = 25\, m$.
Non-parallel sides are $AD = 13\, m$ and $BC = 14\, m$.
Draw $BE \parallel AD$,such that $E$ lies on $DC$. Then $ABED$ is a parallelogram.
$DE = AB = 10\, m$,so $EC = DC - DE = 25\, m - 10\, m = 15\, m$.
In $\Delta BCE$,the sides are $a = 13\, m$,$b = 14\, m$,and $c = 15\, m$.
Semi-perimeter $s = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21\, m$.
Area of $\Delta BCE = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84\, m^{2}$.
Let $h$ be the height of $\Delta BCE$ corresponding to base $EC = 15\, m$.
Area of $\Delta BCE = \frac{1}{2} \times \text{base} \times \text{height} = 84\, m^{2}$.
$\frac{1}{2} \times 15 \times h = 84 \Rightarrow h = \frac{168}{15} = 11.2\, m$.
Area of parallelogram $ABED = \text{base} \times \text{height} = AB \times h = 10 \times 11.2 = 112\, m^{2}$.
Total area of the field = Area of $\Delta BCE +$ Area of parallelogram $ABED = 84 + 112 = 196\, m^{2}$.

(A) Let $ABCD$ be the rhombus-shaped field.
Perimeter of the rhombus $= 400\, m$.
Since all sides of a rhombus are equal,each side $= 400\, m / 4 = 100\, m$.
Thus,$AB = BC = CD = DA = 100\, m$.
Let the diagonal $BD = 160\, m$.
The diagonal divides the rhombus into two congruent triangles,$\triangle ABD$ and $\triangle BCD$.
For $\triangle ABD$,the sides are $a = 100\, m$,$b = 100\, m$,and $c = 160\, m$.
The semi-perimeter $s$ is given by:
$s = (100 + 100 + 160) / 2 = 360 / 2 = 180\, m$.
Using Heron's formula,the area of $\triangle ABD$ is:
Area $= \sqrt{s(s-a)(s-b)(s-c)}$
Area $= \sqrt{180(180-100)(180-100)(180-160)}$
Area $= \sqrt{180 \times 80 \times 80 \times 20}$
Area $= \sqrt{180 \times 20 \times 80^2}$
Area $= \sqrt{3600 \times 6400}$
Area $= 60 \times 80 = 4800\, m^2$.
Since the land is divided into two equal parts,each of them will get an area of $4800\, m^2$.