NCERT Solutions for Class 9 Maths Chapter 12 Herons Formula

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Written by Team Trustudies
Updated at 2021-02-16


NCERT solutions for class 9 Maths Chapter 12 Herons Formula Exercise 12.1

Q1 ) 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?



NCERT Solutions for Class 9 Maths Chapter 12 Herons Formula


Answer :

We know that an equilateral triangle has equal sides.

So, all sides are equal to a.

If Perimeter of triangle = 180 cm (Given)
? a + a + a = 180
? 3a = 180
? a = 60cm
? s=a+a+a2
(Since, 2s = a + b + c)

? s=3a2=1802
? s = 90cm

We also, know that, Area of an equilateral triangle
= s(s?a)(s?a)(s?a)
(Since, Heron's formula = s(s?a)(s?b)(s?c))

? Area
= 90(90?60)(90?60)(90?60)
= 90×30×30×30
= 30×303
?Area of an equilateral triangle = 9003cm2)

Q2 ) The triangular side walls of a flyover have been used for advertisements. The sides of the wails are 122 m, 22 m and 120 m (see figure). The advertisements yield earnings of Rs. 5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?
image



NCERT Solutions for Class 9 Maths Chapter 12 Herons Formula


Answer :

Let a = 122 m, b = 22 m, c = 120 m
Also, we have,

b2+c2=(22)2+(120)2=484+14400=14884=(122)2=a2

Thus, we observe that the side walls are in right triangular shape.

Thus, the area of the triangular side walls
= 12×a×c
?= 12×22×120
? = 11×120=1320m2

We know that, yearly rent = Rs.5000 per m2

Therefore, yearly rent
= Rs.5000 × 12 per m2

Now, the company has hired one of its walls for 3 months.

Thus, rent paid by the company for 3 months
= 1320 × 500012 × 3
= 110 × 5000 × 3
= Rs. 1650000

Therefore, rent paid by the company for 3 months = Rs. 1650000.

Q3 ) There is a slide in a park. One of its side walls has been painted in same colour with a message “KEEP THE PARK GREEN AND CLEAN” (see figure). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.
image



NCERT Solutions for Class 9 Maths Chapter 12 Herons Formula


Answer :

The given figure formed a triangle whose sides are:
a = 15m,b = 11m, c = 6m

s=15+11+62
(Since, 2s = a + b + c)

? s=322
? s = 16m

Therefore, area painted in colour
= 16(16?15)(16?11)(16?6)
(Since, Heron's formula [area = s(s?a)(s?b)(s?c)])

? Area
= 16×1×5×10
= 2×2×2×2×5×5×2
= 400×2m2
? the area painted in colour is 202m2)

Q4 ) Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.



NCERT Solutions for Class 9 Maths Chapter 12 Herons Formula


Answer :

Let the sides of a triangle a = 18 cm, b = 10 cm and c

We have, perimeter = 42 cm

So, a + b + c = 42

By substituting the values,

? 18 + 10 + c = 42
? 28 + c = 42cm
? c = 42 - 28 cm
? c = 14cm

Now, we know that,
s=a+b+c2

? s=18+10+142=422=21cm

Now, Area of triangle
= 21(21?18)(21?10)(21?14)
(Since, Heron's formula [area = s(s?a)(s?b)(s?c)])

= 21×3×11×7
= 7×3×3×11×7
= 2111cm2

Q5 ) Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.



NCERT Solutions for Class 9 Maths Chapter 12 Herons Formula


Answer :

Suppose that the sides in cm, are 12x, 17x and 25x.

Then, we know that

12x + 17x + 25x = 540
(Given, Perimeter of the triangle)

54x = 540
? x = 10

So, the sides of the triangle are 12 × 10cm, 17 × 10cm, 25 × 10cm ? 120cm, 170cm and 250cm.

Now, we know that,
s=a+b+c2
? s=120+170+2502=5402=270cm

Now, Area of triangle
= 270(270?120)(270?170)(270?250)
(Since, Heron's formula [area = s(s?a)(s?b)(s?c)])

= 270×150×100×20
= 27×10×3×15×10×100×20
= 10027×15×10×2cm2
= 1009×3×3×5×10×2cm2
= 1009×3×3×10×10cm2
= 100 × 10 × 9

Therefore, area of the given triangle is 9000cm2

Q6 ) An isosceles triangle has perimeter 30 cm arid each of the equal sides is 12 cm. Find the area of the triangle.



NCERT Solutions for Class 9 Maths Chapter 12 Herons Formula


Answer :

Let ?ABC be an isosceles triangle, in which,

We have,
AB = AC = 12cm ...(Given)

image

Now, AB + AC + BC = 30
(Given, perimeter = 30cm)

? 12 + 12 + BC = 30
? BC = 30 - 24
? BC = 6cm

Now, we know that,
s=a+b+c2
? s=302=15cm

Now, Area of triangle
= 15(15?12)(15?6)(15?12)
(Since, Heron's formula [area = s(s?a)(s?b)(s?c)])

= 15×3×9×3
= 3×5×9×9
= 915cm2

NCERT solutions for class 9 Maths Chapter 12 Herons Formula Exercise 12.2

Q1 ) A park, in the shape of a quadrilateral ABCD, has ?C = 90? , AB = 9m, BC = 12m, CD = 5m and AD = 8m. How much area does it occupy?



NCERT Solutions for Class 9 Maths Chapter 12 Herons Formula


Answer :

In the quadrilateral ABCD,
we have, right angled triangle ?BCD,

image

We have,
BD2=BC2+CD2
(By Pythagoras theorem)

= 122+52=144+25=169
? BD2=132
? BD = 13m

Also, In ?ABD, we have,

AB = 9m, BD = 13m, DA = 8m

Now, we know that,
s=a+b+c2
? s=9+13+82=302=15m

Now, Area of triangle
= 15(15?9)(15?13)(15?8)
(Since, Heron's formula [area = s(s?a)(s?b)(s?c)])

= 15×6×2×7
= 3×5×3×2×2×7
= 3×25×7m2
= 635m2 = 6 × 5.9m2 = 35.4m2(approx) ...(i)

Since, ?BCD is an right angled triangle,

Area of ?BCD = 12 × BC × CD
(Since, Area of triangle = 12 × Base × Height)

= 12×12×5=30m2 ...(ii)

Area of quadrilateral ABCD
= Area of ?ABD + Area of ?BCD

Hence from (i) and (ii),

Area of quadrilateral ABCD = 35.4m2+30m2=65.4m2

Q2 ) Find the area of a quadrilateral ABCD in which AB = 3cm, BC = 4cm, CD = 4cm, DA = 5cm and AC = 5cm.



NCERT Solutions for Class 9 Maths Chapter 12 Herons Formula


Answer :

In the quadrilateral ABCD,
we have, ?ABC,

We have, AB = 3cm, BC = 4cm, CA = 5cm
image

We have, AC2=AB2+BC2
(By Pythagoras theorem)

= 32+42=9+16=25
? AC2=52
? AC = 5cm

Hence, ?ABC is a right triangle ...(i)

Now in ?ABD, we have,
AC = 5cm, CD = 4cm, DA = 5cm
We know that,
s=a+b+c2
? s=5+4+52=142=7cm

Now, Area of triangle
= 7(7?5)(7?4)(7?5)
(Since, Heron's formula [area = s(s?a)(s?b)(s?c)])

= 7×2×3×2
= 27×3cm2
= 221cm2 = 2 × 4.6cm2 = 9.2cm2(approx) ...(ii)

Since, ?ABC is an right angled triangle, ...(from(i))

Area of ?ABC = 12 × AB × BC
(Since, Area of triangle = 12 × Base × Height)

= 12×3×4=6cm2 ...(iii)

Area of quadrilateral ABCD = Area of ?ABC + Area of ?ACD

Hence from (ii) and (iii),

Area of quadrilateral ABCD = 9.2cm2+6m2=15.2cm2

Q3 ) Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used.
image



NCERT Solutions for Class 9 Maths Chapter 12 Herons Formula


Answer :

For part I :
It is a triangle with sides 5 cm, 5 cm and 1 cm.

Thus, We know that,

s=a+b+c2
? s=5+5+12=112cm

Now, Area of part I triangle
= 112(112?5)(112?5)(112?1)
(Since, Heron's formula [area = s(s?a)(s?b)(s?c)])

= 112×12×12×92
= 3411cm2
= 34×3.31cm2 = 3 × 0.829cm2 = 2.487cm2(approx)

For part II :

It is a rectangle with sides 6.5 cm and 1 cm

? Area of part II = 6.5 × 1
(Since, Area of rectangle = Lenght × Breadth)

= 6.5cm2

For part III :

It is a trapezium ABCD.

image

?EBC is an equilateral with side 1 cm.
? Area of ?EBC = 12 × EB × CF = 34×12
(Since, Area of triangle = frac12 × Base × Height and Area of equilateral triangle = frac34×(side)2)

? 12 × 1 × CF = 34
? CF = 32cm

Now, Area of trapezium
= 12 × Sum of parallel sides × Height
= 12 × (AB + CD) × CF
= 12 × (2 + 1) × 32
= 34 × 3
= 3 × 0.433
= 1.299 cm2

Therefore, Area of part III is 1.299 cm2

For part IV & V :

Both the parts IV and V are the same.

? it is a right triangle with sides 6 cm and 1.5 cm.

Area of part IV & V = 12 × 1.5 × 6 = 92 = 4.5 cm2

So, we get,

Total area of paper used = Area of (I + II + III + IV + V)
= (2.487 + 6.5 + 1.299 + 4.5 + 4.5)cm2
= 19.286cm2
= 19.3cm2(approx.)

Q4 ) 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?



NCERT Solutions for Class 9 Maths Chapter 12 Herons Formula


Answer :

Let ABC be a triangle with sides AB = 26 cm, BC = 28 cm, CA = 30 cm

image

Now, we know that,

s=AB+BC+CA2
? s=26+28+302=842=42cm

Now, Area of triangle
= 42(42?26)(42?28)(42?30)
(Since, Heron's formula [area = s(s?a)(s?b)(s?c)])

= 42×16×14×12
= 7×6×4×4×2×7×2×6
= 7×6×4×2cm2
= 336cm2 ...(i)

We know that,

Area of parallelogram = Base × Height ...(ii)

We also have,
Area of parallelogram = Area of ?ABC ...(Given)

Thus, from (i) and (ii),

? Base × Height = 336
? 28 × Height = 336 ...(Given)
? Height = 12cm

Therefore, the height of the parallelogram is 12cm.

Q5 ) 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?



NCERT Solutions for Class 9 Maths Chapter 12 Herons Formula


Answer :

image

Let ABCD be a rhombus.

Area of the rhombus ABCD
= 2 × area of ?ABD ... (i)

As we know that, in a rhombus, diagonals divides it in two equal parts.

In ?ABD, we have,
AB = 30 m, BD = 48 m, DA=30 m

Now, we know that,

s=AB+BD+AD2
? s=30+48+302=1082=54m

Now, Area of triangle
= 54(54?30)(54?48)(54?30)
(Since, Heron's formula [area = s(s?a)(s?b)(s?c)])

= 54×24×6×24
= 9×6×4×6×6×4×6
= 3×6×6×4m2
= 432m2

Thus, from Equation (i),

Area of rhombus
= 2 × 432m
= 864 m

Also, Number of cows = 18

So, therefore, Area of grass field per cow
= 86418 = 48m2.

Q6 ) 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?
image



NCERT Solutions for Class 9 Maths Chapter 12 Herons Formula


Answer :

Each triangular piece in an umbrella is an isosceles triangle with sides 50 cm, 50 cm, 20 cm.

Now, we know that,

s=a+b+c2
? s=50+50+202=1202=60cm

Now, Area of triangle
= 60(60?50)(60?50)(60?20)
(Since, Heron's formula [area = s(s?a)(s?b)(s?c)])

= 60×10×10×40
= 3×2×10×10×10×10×2×2
= 2×1006cm2
= 2006cm2

Thus, from Equation (i),

Area of each triangular piece
= 2 × 432 cm = 864 cm

Since, there are 10 triangular pieces, out of which 5 & 5 are of different colours.

Hence, total area of cloth of each colour = 5×2006cm2 = 10006cm2

Q7 ) 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?
image



NCERT Solutions for Class 9 Maths Chapter 12 Herons Formula


Answer :

Since, the kite is in the shape of a square.

Each diagonal of square = 32 cm ...(Given)

We know that, the diagonals of a square bisect each other at right angle.

image

For Part I :
Area of Part I
= 12 × Base × Height
= 12 × 32 × 16
= 16 × 16 = 256cm2

For Part II :
Area of Part II
= 12 × Base × Height
= 12 × 32 × 16
= 16 × 16 = 256cm2

For Part III :
It is a triangle with sides 6 cm, 6 cm and 8 cm.
Now, we know that,

s=a+b+c2
? s=6+6+82=202=10cm

Now, Area of triangle
= 10(10?6)(10?6)(10?8)
(Since, Heron's formula [area = s(s?a)(s?b)(s?c)])
= 10×4×4×2

= 5×2×4×4×2
= 2×45cm2
= 85cm2
= 8 × 2.24 cm = 17.92 cm2

Hence, the area of colour used for Paper I = 256 cm2

The area of colour used for Paper II = 256 cm2

The area of colour used for Paper III = 17.92 cm2

Q8 ) 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 figure). Find the cost polishing the tiles at the rate of Rs.50 paise per cm2.
image



NCERT Solutions for Class 9 Maths Chapter 12 Herons Formula


Answer :

Given,
the sides of a triangular tile are 9 cm, 28 cm and 35 cm.

For each triangular tile, we have,
s=a+b+c2
? s=9+28+352=722=36cm

Now, Area of triangle
= 36(36?9)(36?28)(36?35)
(Since, Heron's formula [area = s(s?a)(s?b)(s?c)])

= 36×27×8×1
= 6×6×9×3×4×2×1
= 6×3×26cm2
= 366cm2
? Total area of 16 such titles
= 16 × 366
= 16 × 36 × 2.45 cm2 = 1411.20 cm2

Total cost of ploshing the titles at the rate of Rs.50 paise per cm2
= Rs. 50100×1411.20 = Rs.705.60

Q9 ) 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.



NCERT Solutions for Class 9 Maths Chapter 12 Herons Formula


Answer :

Here, ABCD is a trapezium and AB || DC.

Here, ABCD is a trapezium and CE || DA and CF?AB.

image

Now, AB = 25 m, BC = 14 m, CD = 10 m, DA = 13 m, AE = 10 m and CE = 13m
? EB = 25 - 10 = 15 m

Now, for ?EBC, we have,
s=a+b+c2
? s=15+14+132=422=21m

Now, Area of triangle
= 21(21?15)(21?14)(21?13)
(Since, Heron's formula [area = s(s?a)(s?b)(s?c)])
= 21×6×7×8

= 7×3×3×2×7×4×2
= 7×3×4m2
= 84cm2

Also, Area of triangle ?EBC
= frac12 × Base × Height
= 12 × 15 × CF
? frac12 × 15 × CF = 84
? CF = 84×215=16815=11.2m

Now, area of the trapezium ABCD
= 12 × (Sum of parallel sides) × (Distance between parallel sides)
= 12 × (AB + CD) × CF
= 12 × (25 + 10) × 11.2
= 12 × 35 × 11.2
= 35 × 5.65

Therefore, the area of the field is 196 m2.