# Heron's Formula : Exercise 12.2 (Mathematics NCERT Class 9th)

Q.1Â Â Â Â Â  A park, in the shape of a quadrilateral ABCD, has $\angle C =$ 90Âº, AB = Â 9 m , BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
Sol.

Area of $\Delta \,BCD = {1 \over 2} \times BC \times CD$
$= \left( {{1 \over 2} \times 12 \times 5} \right){m^2}$
$= 30\,{m^2}$

Using Pythagoras theorem, we have
$B{D^2} = B{C^2} + C{D^2}$ $\Rightarrow$ $B{D^2} = {12^2} + {5^2}$
$\Rightarrow$ $B{D^2} = 144 + 25$ $\Rightarrow$ $B{D^2} = 169$
$\Rightarrow$ $BD = \sqrt {169} = 13\,m$
For :$\Delta \,ABD$:
Let a = 13 m, b = 8 m and c = 9 m
Now, $s = {1 \over 2}\left( {a + b + c} \right) = {1 \over 2}\left( {13 + 8 + 9} \right)m$
$= {1 \over 2} \times 30m = 15m$
$s - a = \left( {15 - 13} \right)m = 2m$
$s - b = \left( {15 - 8} \right)m = 7m$
and $s - c = \left( {15 - 9} \right)m = 6\,m$
Therefore $Area\,of\,\Delta \,ABD = \sqrt {15 \times 2 \times 7 \times 6} \,{m^2}$
$= \sqrt {3 \times 5 \times 2 \times 7 \times 2 \times 3} \,{m^2}$
$= \sqrt {2 \times 2 \times 3 \times 3 \times 5 \times 7} \,{m^2}$
$= 2 \times 3\sqrt {35} \,{m^2}$

$= 6 \times 5.9\,{m^2}\left( {approx} \right)$
$= 35.4\,{m^2}\left( {approx} \right)$
Therefore Required area = $\Delta ABD + \Delta BCD$
$= 35.4\,{m^2} + 30\,{m^2} = 65.4\,{m^2}$

Q.2Â Â Â Â Â  Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
Sol.

Since $A{C^2} = A{B^2} + B{C^2}$
$\left( {as\,{5^2} = {3^2} + {4^2}\,i.e.,\,25 = 9 + 16} \right)$
Therefore $\angle ABC =$ 90Âº

Area of rt $\angle d\,\Delta ABC = {1 \over 2} \times AB \times BC$
$= \left( {{1 \over 2} \times 3 \times 4} \right)c{m^2} = 6\,c{m^2}$
For $\Delta \,ACD$ :
Let a = 5 cm, b = 4cm and c = 5 cm. Then ,
$s = {1 \over 2}\left( {a + b + c} \right) = {1 \over 2}\left( {5 + 4 + 5} \right)cm$
$= {1 \over 2} \times 14\,cm = 7\,cm$
Now, $s - a = \left( {7 - 5} \right)\,cm = 2\,cm$
$s - b = \left( {7 - 4} \right)\,cm = 3\,cm$
and $s - c = \left( {7 - 5} \right)\,cm = 2\,cm$
Area of $\Delta \,ACD = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}$
$= \sqrt {7 \times 2 \times 3 \times 2} \,c{m^2}$
$= 2\sqrt {21} \,c{m^2}$

$= 2 \times 4.6\,c{m^2}\,\left( {approx.} \right)$
$= 9.2\,c{m^2}\,\left( {approx.} \right)$
$Area\,of\,\Delta ABC + Area\,of\,\Delta \,ACD$
$= \left( {6 + 9.2} \right)c{m^2} = 15.2\,c{m^2}\,\,approx$

Q.3Â Â Â Â Â  Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used.

Sol. Â  Â  Â Â Area of region I:

Â  Â  Â  Â  Â  Â  Â  Region I is Enclosed by a triangle of sides a = 5 cm, b = 5 cm and c = 1 cm

Let s be the perimeter of the triangle. Then,
Now, $s = {1 \over 2}\left( {a + b + c} \right)$
$= \frac{1}{2}\left( {5 + 5 + 1} \right)m$
$\Rightarrow s = \frac{{11}}{2}cm$

$\ Area of region I = \sqrt {s(s - a)(s - b)(s - c)} c{m^2}$

$\Rightarrow AreaofregionI = \sqrt {\frac{{11}}{2} \times (\frac{{11}}{2} - 5) \times (\frac{{11}}{2} - 5) \times (\frac{{11}}{2} - 1)} c{m^2}$

$\Rightarrow AreaofregionI = \sqrt {\frac{{11}}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{9}{2}} c{m^2} = \frac{{3\sqrt {11} }}{4}c{m^2}$

$\Rightarrow AreaofregionI = \frac{3}{4} \times 3.32 = 2.49c{m^2}$

Area of region II:

Region II is a rectangle of length 6.5 cm and breadth 1 cm.

$\ Area of region II = 6.5 \times 1 = 6.5 c{m^2}$

Area of region III:

Region III is an isosceles trapezium.

InÂ Î”ABE, We have

Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  AB2Â  = AE2 + BE2

â‡’ Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  1 Â  Â = Â  0.25 +Â  BE2

â‡’ Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  BE Â =Â $\sqrt {0.75} = \sqrt {\frac{3}{4}}$

$Area of region III = \frac{1}{2}(AD + BC) \times BE = \frac{1}{2}(2 + 1) \times \sqrt {\frac{3}{4}} c{m^2}$

$\Rightarrow Area of region III = \frac{{3\sqrt 3 }}{4}c{m^2} = 1.3c{m^2}$

Area of region IV:

Region IV forms a right triangle whose two sides are of lengths 6 cm and 1.5 cm.

$Area of region IV = \frac{1}{2} \times 6 \times 1.5c{m^2} = 4.5c{m^2}$

Area of region V:

Region IV and V are congruent.

Area of region V = 4.5 cm2

Hence, total area of the paper used = (2.49 + 6.5 + 1.3 + 4.5 + 4.5) cm2

Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  = 19.29 cm2

Q.4Â Â Â  A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Sol.

For the triangle :
Let its sides be a, b and c such that
a = 26 cm, b = 28 cm and c = 30. Then,
$s = {1 \over 2}\left( {26 + 28 + 30} \right)cm$
$= {1 \over 2} \times 84cm = 42\,cm$
Now, $s - a = \left( {42 - 26} \right)cm = 16\,cm$
$s - b = \left( {42 - 28} \right)cm = 14\,cm$
and $s - c = \left( {42 - 30} \right)cm = 12\,cm$
Area of the triangle
$= \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}$
$= \sqrt {42 \times 16 \times 14 \times 12} \,c{m^2}$
$= \sqrt {2 \times 3 \times 7 \times 4 \times 4 \times 2 \times 7 \times 2 \times 2 \times 3} \,c{m^2}$
$= \sqrt {2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 4 \times 4 \times 7 \times 7} \,c{m^2}$
$= \left( {2 \times 2 \times 3 \times 4 \times 7} \right)\,c{m^2} = 336\,c{m^2}$
For the parallelogram :
Area = base Ã— height
Therefore $Height = {{Area} \over {Base}}$
[Since Area of parallelogram = Area of $\Delta$ (given)
Therefore Area of parallelogram = 336 cm2 and its base = 28 cm]

$= \left( {{{336} \over {28}}} \right)\,cm$
$= 12\,cm$

Q.5Â Â Â Â  A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?
Sol.

We know that the diagonals of the rhombus bisect each other at right angles. Using Pythagoras theorem, we have $OD = \sqrt {A{D^2} - A{O^2}} = \sqrt {{{30}^2} - {{24}^2}} \,m$
$= \sqrt {\left( {30 + 24} \right)\left( {30 - 24} \right)} \,m$

$= \sqrt {54 \times 6} \,m$
$= \sqrt {9 \times 6 \times 6} \,m$
$= \left( {3 \times 6} \right)\,m = 18\,m$
Area of one $\Delta \,AOD = \left( {{1 \over 2} \times 24 \times 18} \right){m^2} = 216\,{m^2}$
Therefore Area of rhombus $= 4 \times \Delta \,AOD$
$= \left( {4 \times 216} \right){m^2} = 864\,{m^2}$
Grass area for 18 cows $= 864\,{m^2}$
Grass area for 1 cow $= \left( {{{864} \over {18}}} \right)\,{m^2} = 48\,{m^2}$

Q.6Â Â Â Â Â  An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

Sol.

In one triangular piece, let a = 20 cm, b = 50 cm and c = 50 cm
Now, $s = {1 \over 2}\left( {a + b + c} \right) = {1 \over 2}\left( {20 + 50 + 50} \right)cm$
$= {1 \over 2} \times 120\,cm = 60\,cm$
$s - a = \left( {60 - 20} \right)cm = 40\,cm$
$s - b = \left( {60 - 50} \right)cm = 10\,cm$
and $s - c = \left( {60 - 50} \right)cm = 10\,cm$
Therefore Area of one triangular piece
$= \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}$
$= \sqrt {60 \times 40 \times 10 \times 10} \,c{m^2} = 200\sqrt 6 \,c{m^2}$
Therefore Area of 5 red triangles $= 1000\sqrt 6 \,c{m^2}$
and , area of 5 green triangles $= 1000\sqrt 6 \,c{m^2}$

Q.7Â Â Â Â  A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. How much paper of each shade has been used in it ?
Sol.

ABCD is a square such that AC = BC = 32 cm and AEF is an isosceles triangle in which AE = AF = 6 cm
For the area of shades I and II
Clearly from the figure
Area of shade I = Area of shade II = Area of $\Delta$ CDB
$= {1 \over 2} \times DB \times CM$ [Since ABCD is in square shapes]
$= {1 \over 2} \times 32 \times 16$ sq cm

$= \,256\,sq\,cm$
For the area of shade III (i.e., $\Delta$ AEF)
$EL = LF = {1 \over 2}EF$
$= {1 \over 2} \times 8 = 4\,cm$
and $AE = 6\,cm\,\left( {given} \right)$
Therefore $AL = \sqrt {A{E^2} - E{L^2}}$
$= \sqrt {36 - 16} = \sqrt {20} = 2\sqrt 5 \,cm$
Area of shade III $= {1 \over 2} \times EF \times AL$
$= {1 \over 2} \times 8 \times 2\sqrt 5 \,sq\,cm$
$= 8\sqrt 5 \,sq\,cm$
$= 8 \times 2.24\,sq\,cm\,\left( {approx} \right)$
$= 17.92\,sq\,cm\,\left( {approx} \right)$

Q.8Â Â Â Â  A floral design on floor is made up of 16 tiles which areÂ triangularÂ , the sides of the triangle being 9 cm, 28 cm and 35Â cm (see figure). Find the cost of polishing the tiles at the rate of 50 p per $c{m^2}$.

Sol.

For one triangular tile :
Let a = 9 cm, b = 28 cm and c = 35 cm
Now , $s = {1 \over 2}\left( {a + b + c} \right) = {1 \over 2}\left( {9 + 28 + 35} \right)cm$
$= {1 \over 2} \times 72\,cm = 36\,cm$
$s - a = \left( {36 - 9} \right)cm = 27\,cm$
$s - b = \left( {36 - 28} \right)cm = 8\,cm$
and $s - c = \left( {36 - 35} \right)cm = 1\,cm$
Area of one tile $= \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}$
$= \sqrt {36 \times 27 \times 8 \times 1} \,c{m^2}$
$= \sqrt {9 \times 4 \times 9 \times 3 \times 4 \times 2} \,c{m^2}$
$= \sqrt {2 \times 3 \times 4 \times 4 \times 9 \times 9} \,c{m^2} = 4 \times 9\sqrt 6 \,c{m^2}$
$= 36\sqrt 6 \,c{m^2} = 36 \times 2.45\,c{m^2}\,approx.$
$= 88.2\,c{m^2}\,approx.$
Therefore Area of 16 such tiles $= \left( {16 \times 88.2} \right)c{m^2}$
$= 1411.2\,c{m^2}\,approx.$
Cost of polishing @ 50 P per $c{m^2} = Rs\left( {1411.2 \times {{50} \over {100}}} \right)$
$= Rs\,705.60\,\,approx.$

Q.9Â Â Â Â  A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non- parallel sides are 14 m and 13 m. Find the area of the field.
Sol.Â Â Â Â Â Â Â Â Â  Various length are as marked in the figure.

Clearly DM = CN
$\Rightarrow$ $D{M^2} = C{N^2}$
[Distance between parallel sides are always equal]
$\Rightarrow$ ${\left( {13} \right)^2} - {x^2} = {14^2} - {\left( {15 - x} \right)^2}$

$\Rightarrow$ $169 - {x^2} = 196 - \left( {225 - 30x + {x^2}} \right)$
$\Rightarrow$ $50x = 169 - 196 + 625 = 598$
$\Rightarrow$ $x = {{598} \over {50}} = 11.96$
Now $DM = \sqrt {A{D^2} - A{M^2}} = \sqrt {{{13}^2} - {{\left( {11.96} \right)}^2}}$
$= \sqrt {\left( {13 + 11.96} \right)\left( {13 - 11.96} \right)}$
$= \sqrt {24.96 \times 1.04} = \sqrt {25.9584} = 5.09$
Area of the trapezium $= {1 \over 2} \times \left( {AB + CD} \right) \times DM$
$= {1 \over 2} \times \left( {25 + 10} \right) \times 5.09\,sq.m$
$= {1 \over 2} \times 35 \times 5.09\,\,sq.m$
$= 89.075\,\,sq.m$

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