Herons Formula

Class 9 NCERT Maths

NCERT

1   A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is $180 \ cm,$ what will be the area of the signal board?

Solution :

Side of traffic signal board =$ a$$\\$ Perimeter of traffic signal board = $3 * a$$\\$ $2s=3a \Rightarrow s=\dfrac{3}{2}a$$\\$ By Heron’s formula, Area of triangle =$\\$ $\sqrt{s(s-a)(s-b)(s-c)}$$\\$ Area of given triangle=$\\$ $\sqrt{\dfrac{3}{2}a(\dfrac{3}{2}a-a)(\dfrac{3}{2}a-a)(\dfrac{3}{2}a-a)}$$\\$ $=\sqrt{\dfrac{3}{2}a(\dfrac{a}{2})(\dfrac{a}{2})(\dfrac{a}{2})}$$\\$ $=\dfrac{\sqrt{3}}{2}a^2....(1)$$\\$ Perimeter of traffic signal board $= 180 cm$$\\$ Side of traffic signal board $(a) =(\dfrac{180}{3}) cm =60 cm$$\\$ Using Equation (1), area of traffic signal board $=\dfrac{\sqrt{3}}{2}(60 cm)^2$$\\$ $=(\dfrac{3600}{4}\sqrt{3}) cm^2$$\\$ $=900\sqrt{3}cm^2$

2   The triangular side walls of a flyover have been used for advertisements. The sides of the walls are $122 m, 22 m,$ and $120 m$ (see the given figure). The advertisements yield an earning of Rs. $5000 $ per $m^2$ per year. A company hired one of its walls for $3$ months. How much rent did it pay?

Solution :

The sides of the triangle (i.e., a, b, c) are of $122 m, 22 m,$ and $120 m$ respectively.$\\$ Perimeter of triangle = $(122 + 22 + 120) m$$\\$ $2s = 264 m$$\\$ $s = 132 m$$\\$ By Heron’s formula,$\\$ Area of triangle =$\sqrt{s(s-a)(s-b)(s-c)}$$\\$ Area of given triangle =$[\sqrt{132(132-122)(132-22)(132-120)}]m^2$$\\$ $=[\sqrt{132(10)(110)(12)}]$$\\$ Rent of $1 m^2$ area per year = Rs. $5000$$\\$ Rent of $1 m^2$ area per month = Rs.$(\dfrac{5000}{15}*3*1320)$$\\$ $Rs.(5000*330)=Rs 1650000$$\\$ Therefore, the company had to pay Rs. $1650000.$

3   The floor of a rectangular hall has a perimeter $250 m$. If the cost of panting the four walls at the rate of Rs. $10 $ per $m^2$ is Rs.$15000,$ find the height of the hall. [Hint: Area of the four walls = Lateral surface area.]

Solution :

Let length, breadth, and height of the rectangular hall be $l$ m, $b$ m, and $h$ m respectively.$\\$ Area of four walls = $2lh + 2bh = 2(l + b) h$$\\$ Perimeter of the floor of hall =$ 2(l + b) = 250 m$$\\$ $\therefore $ Area of four walls = $2(l + b) h = 250h m^2$$\\$ Cost of painting per $m^2$ area = Rs. $10$$\\$ Cost of painting $250h m^2$ area = Rs.$ (250h × 10) = Rs. 2500h$$\\$ However, it is given that the cost of paining the walls is Rs. $15000.$$\\$ $\therefore 15000 = 2500h$$\\$ $h = 6$$\\$ Therefore, the height of the hall is $6 m.$

4   Find the area of a triangle two sides of which are $18 cm$ and $10 cm$ and the perimeter is $42 cm.$

Solution :

Let the third side of the triangle be $x.$$\\$ Perimeter of the given triangle =$ 42 cm$$\\$ $18 cm + 10 cm + x = 42$$\\$ $x = 14 cm$$\\$ $s=\dfrac{\text{Perimeter}}{2}=\dfrac{42 cm}{2}\\ =21 cm$$\\$ By Heron’s formula, Area of triangle =$\sqrt{s(s-a)(s-b)(s-c)}$$\\$ Area of given triangle = $\\$ $[\sqrt{21(21-18)(21-10)(21-14)}]cm^2$$\\$ $[\sqrt{21(3)(11)(7)}]$$\\$ $=21\sqrt{11}cm^2$$\\$

5   Sides of a triangle are in the ratio of $12:17:25 $ and its perimeter is $540 cm$. Find its area.

Solution :

Let the common ratio between the sides of the given triangle be $x$.$\\$ Therefore, the side of the triangle will be$ 12x, 17x,$ and $25x.$$\\$ Perimeter of this triangle$ = 540 cm$$\\$ $12x + 17x + 25x = 540 cm$$\\$ $54x = 540 cm$$\\$ $x = 10 cm$$\\$ Sides of the triangle will be $120 cm, 170 cm,$ and $250 cm.$$\\$ $s=\dfrac{\text{Perimeter of triangle}}{2}\\ \dfrac{540 cm}{2}\\ =270 cm$$\\$ By Heron’s formula,$\\$ Area of triangle =$\\$ $\sqrt{s ( s - a )( s - b )( s - c )}$$\\$ Area of given triangle = $\sqrt{270 ( 270 -120 )(270 - 170 )( 270 - 250 )} cm ^2$$\\$ $\sqrt{ 270 ( 150 )(100 )( 20 )} cm^ 2$$\\$ $9000cm^2$$\\$ Therefore, the area of this triangle is $9000 cm ^2$ .

6   An isosceles triangle has perimeter $30 cm$ and each of the equal sides is $12 cm.$ Find the area of the triangle.$\\$ $\sqrt{s ( s - a )( s - b )( s -c )}$

Solution :

Let the third side of this triangle be $x.$ $\\$ Perimeter of triangle =$ 30 cm$ $\\$ $12 cm + 12 cm + x = 30 cm $ $\\$ $x = 6 cm $ $\\$ $s=\dfrac{\text{Perimeter of triangle }}{2}=\dfrac{30 cm}{2}=15 cm $ $\\$ By Heron’s formula, Area of triangle =$\sqrt{s ( s - a )( s - b )( s -c )}$ $\\$ Area of given triangle =[$\sqrt{15(15-12)(15-12)(15-6)}]cm^2$ $\\$ $=[\sqrt{15(3)(3)(9)}]cm^2 $ $\\$ $=9\sqrt{15}cm^2$

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

Solution :

Let us join $BD.$ $\\$ In $\Delta BCD,$ applying Pythagoras theorem,$\\$ $BD ^2 = BC ^2 + CD ^2$ $\\$ $= (12) ^2 + (5)^ 2$$\\$ $= 144 + 25$ $\\$ $BD ^2 = 169$ $\\$ $BD = 13 m$ $\\$

Area of $\Delta BCD = \dfrac{1}{2}* BC*CD=(\dfrac{1}{2}*12*5)m^2=30m^2$ $\\$ For $\Delta ABD,$ $\\$ $s=\dfrac{\text{Perimeter}}{2}=\dfrac{(9+8+13)m}{2}=15 m $ $\\$ By Heron’s formula, Area of triangle =$\sqrt{s(s-a)(s-b)(s-c)}$ $\\$ Area of $\Delta ABD =\sqrt{15(15 -9)(15 - 8)(15 - 13) m ^2}$ $\\$ $=\sqrt{15(6)(7)(2) }m^ 2$ $\\$ $= 6\sqrt{ 35} m ^2$ $\\$ $= (6 * 5.916) m^ 2$ $\\$ $=35.496 m ^2$ $\\$ Area of the park = Area of $\Delta ABD$ + Area of $\Delta BCD$ $\\$ $= 35.496 + 30 m^ 2$ $\\$ $= 65.496 m ^2 = 65.5 m ^2$ (approximately).

8   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.$

Solution :

For $\Delta ABC,$ $\\$ $AC^ 2 = AB ^2 + BC^ 2$ $\\$ $(5)^ 2 = (3)^ 2 + (4)^ 2$ $\\$ Therefore, $\Delta ABC$ is a right-angled triangle, right-angled at point $B.$$\\$ Area of $\Delta ABC = \dfrac{1}{2}*AB * BC=(\dfrac{1}{2}*3*4)=6 cm^2$ $\\$ For $\Delta ADC,$ $\\$ Perimeter=$ 2s=AC+CD+DA=(5+4+5)cm=14 cm$ $\\$ $ s=7 cm $ $\\$ By Heron’s formula, Area of triangle =$\sqrt{s ( s - a )( s - b )( s - c )}$ $\\$ Area of $\Delta ADC = \sqrt{7(7 - 5)(7 - 5)(7 - 4) }cm ^2$ $\\$ $=\sqrt{7(2)(2)(3)} cm ^2$ $\\$ $= 2\sqrt{ 21 }cm ^2$ $\\$ $=(2 * 4.583) cm ^2$ $\\$ $= 9.166 cm^2$ $\\$ Area of $ABCD = $Area of $\Delta ABC +$ Area of $\Delta ACD$ $\\$ $= (6 + 9.166) cm ^2$ $\\$ $= 15.166 cm^ 2 = 15.2 cm ^2$ (approximately).

9   Radha made a picture of an aeroplane with coloured papers as shown in the given figure. Find the total area of the paper used.

Solution :

$\text{For triangle I}$$\\$ This triangle is an isosceles triangle$\\$ Perimeter = $2s = (5 + 5 + 1) cm = 11 cm $ $\\$ $s =\dfrac{11 cm}{2}=5.5 cm $ $\\$ Area of the triangle = $ \sqrt{s ( s - a )( s - b )( s - c )}$ $\\$ $= \sqrt{5.5(5.5 -5)(5.5 - 5)(5.5 - 1)} cm ^2$ $\\$ $= \sqrt{5.5(0.5)(0.5)(4.5)} cm ^2$ $\\$ $=0.75 \sqrt{11} cm ^2$ $\\$ $= (0.75 * 3.317) cm ^2$ $\\$ $=2.488 cm ^2$ (approximately)$\\$ $\text{For quadrilateral II}$ $\\$ This quadrilateral is a rectangle.$\\$ Area $= l × b = (6.5 × 1) cm ^2 = 6.5 cm^ 2$ $\\$ $ \text{For quadrilateral III}$$\\$ This quadrilateral is a trapezium.$\\$ Perpendicular height of parallelogram =$\sqrt{ 1^ 2 - (0.5)^ 2} cm$ $\\$ $=\sqrt{0.75 }cm = 0.866 cm$ $\\$ Area = Area of parallelogram + Area of equilateral triangle $\\$ $=(0.866)1+\dfrac{\sqrt{3}}{4}(1)^2$ $\\$ $= 0.866 + 0.433 = 1.299 cm ^2$$\\$ Area of triangle (IV) = Area of triangle in (V)$\\$ $=(\dfrac{1}{2}*1.5*6)cm^2=4.5 cm^2$ $\\$ Total area of the paper used =$ 2.488 + 6.5 + 1.299 + 4.5 × 2$ $\\$ $= 19.287 cm ^2$ $\\$

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

Solution :

Perimeter of triangle =$ (26 + 28 + 30) cm = 84 cm$ $\\$ $2s = 84 cm$ $\\$ $s = 42 cm $ $\\$ By Heron’s formula, Area of triangle =$\sqrt{s ( s - a )( s - b )( s - c )}$ $\\$ Area of given triangle $= \sqrt{42(42 -26)(42 - 28)(42 - 30)} cm ^2$ $\\$ $=\sqrt{42(16)(14)(12)} cm ^2 = 336 cm ^2$ $\\$ Let the height of the parallelogram be $h.$ $\\$ Area of parallelogram = Area of triangle$\\$ $h × 28 cm = 336 cm^ 2 $ $\\$ $h = 12 cm$ $\\$ Therefore, the height of the parallelogram is $12 cm.$

11   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?

Solution :

Let $ABCD$ be a rhombus-shaped field.$\\$ For $\Delta BCD,$ $\\$ Semi-perimeter, $s=\dfrac{(48+30+30)m}{2}=54 m $ $\\$ By Heron’s formula, Area of triangle = $\sqrt{s ( s - a )( s - b )( s - c )}$ $\\$ Therefore, area of $\Delta BCD = 54(54 - 48)(54 - 30)(54 - 30) m^ 2$ $\\$ $=\sqrt{54(6)(24)(24)}=3*6*24=432 m^2$ $\\$ Area of field $= 2 ×$ Area of $\Delta BCD$ $\\$ $= (2 × 432) m^ 2 = 864 m^ 2$ $\\$ Area for grazing for $1$ cow = $\dfrac{864}{18}=48 m^2$ $\\$ Each cow will get $48 m^ 2$ area of grass field.

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

Solution :

For each triangular piece,$\\$ Semi-perimeter, $s =\dfrac{(20 + 50 + 50) cm}{2}= 60 cm$ $\\$ By Heron’s formula,$\\$ Area of triangle = $\sqrt{s ( s - a )( s - b )( s - c )}$ $\\$ Area of each triangular piece $= \sqrt{ 60(60 - 50)(60 - 50)(60 - 20) }cm ^2$ $\\$ $=\sqrt{ 60(10)(10)(40)}= 200\sqrt{ 6} cm ^2$ $\\$ Since there are $5$ triangular pieces made of two different coloured cloths,$\\$ Area of each cloth required $=(5 * 200\sqrt{ 6}) cm^ 2 -1000\sqrt{6} cm ^2$

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

Solution :

We know that$\\$ Area of square $=\dfrac{1}{2}(diagonal) ^2 $ $\\$ Area of the given kite $=\dfrac{1}{2} (32 cm^ 2 ) = 512 cm^ 2$ $\\$ Area of $1^{ st}$ shade = Area of $2^{ nd}$ shade $\\$ $=\dfrac{512 cm ^2}{2}= 256 cm^ 2$ $\\$ Therefore, the area of paper required in each shape is $256 cm ^2 .$ $\\$ For IIIrd triangle$\\$ Semi-perimeter,$ s =\dfrac{(6 + 6 + 8) cm}{2}=10 cm$ $\\$ By Heron’s formula,$\\$ Area of triangle =$\sqrt{s ( s - a )( s - b )( s - c )}$ $\\$ Area of III rd triangle =$\sqrt{10(10 - 6)(10 - 6)(10 - 8) }cm^ 2$ $\\$ $=\sqrt{10(4)(4)(2)} = (4 * 2\sqrt{ 5}) cm ^2$ $\\$ $= 8 \sqrt{5} cm ^2$ $\\$ $= (8 * 2.24) cm ^2$ $\\$ $= 17.92 cm ^2$ $\\$ Area of paper required for III rd shade = $17.92 cm^ 2$

14   A floral design on a floor is made up of $16$ tiles which are triangular, the sides of the triangle being $9 cm, 28 cm $ and $35 cm$ (see the given figure). Find the cost of polishing the tiles at the rate of $50p$ per $cm ^2 .$

Solution :

It can be observed that $\\$ Semi-perimeter of each triangular-shaped tile, $s =\dfrac{ (35 + 28 + 9) cm}{2} = 36 cm$ $\\$ By Heron’s formula,$\\$ Area of triangle =$\sqrt{s ( s - a )( s - b )( s - c )}$ $\\$ Area of each tile $=\sqrt{36(36 - 35)(36 - 28)(36 - 9) }cm ^2$ $\\$ $=\sqrt{ 36(1)(8)(27) }= 36\sqrt{ 6} cm ^2$ $\\$ $= (36 × 2.45) cm ^2$ $\\$ $= 88.2 cm ^2$ $\\$ Area of 16 tiles =$ (16 × 88.2) cm ^2 = 1411.2 cm^ 2$ $\\$ Cost of polishing per $cm^ 2$ area =$ 50 p$ $\\$ Cost of polishing $1411.2 cm ^2$ area = Rs. $(1411.2 × 0.50) $= Rs. $705.60$ $\\$ Therefore, it will cost Rs.$ 705.60 $ while polishing all the tiles.

15   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.

Solution :

Draw a line $BE$ parallel to $AD$ and draw a perpendicular $BF$ on $CD.$ $\\$ It can be observed that $ABED$ is a parallelogram.$\\$ $BE = AD = 13 m$ $\\$ $ED = AB = 10 m$ $\\$ $EC = 25 - ED = 15 m$ $\\$ For $\Delta BEC,$ $\\$ Semi-perimeter, $s =\dfrac{ (13 + 14 + 15) m}{2} = 21 m$ $\\$ By Heron’s formula, Area of triangle =$\sqrt{s ( s - a )( s - b )( s - c )}$ $\\$ Area of $\Delta BEC = \sqrt{21(21 - 13)(21 - 14)(21 - 15)} m^ 2$ $\\$ $=\sqrt{21(8)(7)(6) }m^ 2 = 84 m ^2$ $\\$ Area of $\Delta BEC =\dfrac{1}{2}*CE*BF$ $\\$ $BF=(\dfrac{168}{15})=11.2 m$ $\\$ Area of $ABED = BF × DE = 11.2 × 10 = 112 m ^2$ $\\$ Area of the field =$ 84 + 112 = 196 m ^2$