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NCERT Solutions for Class 9 Maths chapter-12 Herons Formula
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NCERT Solutions for Class 9 Maths chapter-12 Herons Formula
NCERT solutions for Class 9 Maths
chapter 12 Herons Formula is prepared by academic team of Physics Wallah. We have prepared
NCERT Solutions
for all exercise of chapter 12. Given below is step by step solutions of all questions given in NCERT textbook for chapter 12. Read chapter 12 theory make sure you have gone through the theory part of chapter 12 from NCERT textbook and you have learned the formula of the given chapter. Physics Wallah prepared a detail notes and additional questions for class 9 maths with short notes of all maths formula of class 9 maths. Do read these contents before moving to solve the exercise of NCERT chapter 12.
NCERT Solutions for Class 9 Maths Exercise 12.1
Question 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:
Let each side of the equilateral triangle be a.
Semi-perimeter of the triangle,
Question 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 figure). The advertisements yield an earning of ₹5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?
Solution:
Let the sides of the triangular will be
a = 122m, b = 12cm, c = 22m
Semi-perimeter, s =
(
)m =
m = 132m
The area of the triangular side wall
Rent for 1 year (i.e. 12 months) per m2 = Rs. 5000
∴ Rent for 3 months per m2 = Rs. 5000 x
= Rent for 3 months for 1320 m2
= Rs. 5000 x
x 1320 = Rs. 16,50,000.
Question 3. There is a slide in a park. One of its side Company hired one of its walls for 3 months.walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN” (see figure). If the sides of the wall are 15 m, 11 m and 6m, find the area painted in colour.
Solution:
Let the sides of the wall be
a = 15m, b = 11m, c = 6m
Semi-perimeter,
Thus, the required area painted in colour
= 20√2 m2
Question 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 sides of the triangle be a
=18 cm, b = 10 cm and c = x cm
Since, perimeter of the triangle
= 42 cm
∴ 18cm + 10 cm + xcm = 42
x = [42 – (18 + 10)cm = 14cm
Now, semi-permimeter, s =
cm = 21 cm
Thus, the required area of the triangle
= 21
cm2
Question 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 sides of the triangle be
a = 12x cm, b = 17x cm, c = 25x cm
Perimeter of the triangle = 540 cm
Now, 12x + 17x + 25x = 540
⇒ 54x = 54 ⇒ x = 10
∴ a = (12 x10)cm = 120cm,
b = (17 x 10) cm = 170 cm
and c = (25 x 10)cm = 250 cm
Now, semi-perimeter, s =
cm = 270 cm
Question 6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Solution:
Let the sides of an isosceles triangle be
a = 12cm, b = 12cm,c = x cm
Since, perimeter of the triangle = 30 cm
∴ 12cm + 12cm + x cm = 30 cm
⇒ x = (30 – 24) = 6
Now, semi-perimeter, s =
cm =15 cm
Thus, the required area of the triangle
= 9√15 cm2
NCERT Solutions for Class 9 Maths Exercise 12.2
Question 1. A park, in the shape of a quadrilateral ABCD has
C =
AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
Solution:
Since BD divides quadrilateral ABCD in two triangles:
(i) Right triangle BCD and (ii)
ABD.
In right triangle BCD, right angled at C,
therefore, Base = CD = 5 m and Altitude = BC = 12 m
Area of
BCD =
=
In
ABD, AB = 9 m, AD = 8 m
And BD =
[Using Pythagoras theorem]
BD =
=
=
= 13 m
Now, Semi=perimeter of
ABD =
= 15 m
Using Heron’s formula,
Area of
ABD =
=
=
=
=
(approx.)
Area of quadrilateral ABCD = Area of
BCD + Area of
ABD
= 30 + 35.4
Question 2. 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:
In quadrilateral ABCE, diagonal AC divides it in two triangles,
ABC and
ADC.
In
ABC, Semi-perimeter of
ABC =
= 6 cm
Using Heron’s formula,
Area of
ABC =
=
=
Again, In
ADC, Semi-perimeter of
ADC =
= 7 cm
Using Heron’s formula, Area of
ABC =
=
=
= 2
(approx.)
Now area of quadrilateral ABCD = Area of
ABC + Area of
ADC
= 6 + 9.2
Question 3. Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used.
Solution:
Area of triangular part I: Here, Semi-perimeter
= 5.5 cm
Therefore, Area =
=
=
=
Area of triangular part II = Length x Breadth
Area of triangular part III (trapezium):
=
(AB + DC)
=
(1 + 2)
=
=
Area of triangular parts IV & V:
Total area = 2.4825 + 6.2 + 1.299 + 9
Question 4. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 29 cm and 30 cm and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Solution:
Semi-perimeter of triangle
=
= 42 cm
Using Heron’s formula,
Area of triangle =
=
=
According to question, Area of parallelogram = Area of triangle
Base x Corresponding height = 336
= 336
Height = 12 cm
Question 5. 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, grass of how much area of grass field will each cow be getting?
Solution:
Here, AB = BC = CD = DA = 30 m and Diagonal AC = 48 m which divides the rhombus ABCD in two congruent triangle.
Area of
ABC = Area of
ACD
Now, Semi-perimeter of
ABC
=
= 54 m
Now Area of rhombus ABCD = Area of
ABC + Area of
ACD
= 2
Area of
ABC [
Area of
ABC = Area of
ACD]
=
[ Using Heron’s formula]
=
=
=
Field available for 18 cows to graze the grass
Field available for 1 cow to graze the grass =
Question 6. 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?
Solution:
Here, sides of each of 10 triangular pieces of two different colours are 20 cm, 50 cm and 50 cm.
Semi-perimeter of each triangle
=
= 60 cm
Now, Area of each triangle =
=
=
=
According to question, there are 5 pieces of red colour and 5 pieces of green colour.
Cloth required for 5 red pieces =
=
And Cloth required to 5 green pieces =
=
Question 7. A kite is 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 side has been used in it?
Solution:
Let ABCD is a square of side
cm and diagonals AC = BD = 32 cm
In right triangle ABC,
[Using Pythagoras theorem]
= 512
Area of square
[Area of square =
]
Diagonal BD divides the square in two equal triangular parts I and II.
Area of shaded part I = Area of shaded part II
=
Now, semi-perimeter of shaded part III
= 10 cm
Area of shaded part III
=
=
=
=
Question 8. 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 of polishing the tiles at the rate of 50 paise per cm2.
Solution:
Here, Sides of a triangular shaped tile area 9 cm, 28 cm and 35 cm.
Semi-perimeter of tile
=
= 36 cm
Area of triangular shaped tile =
=
=
=
(approx.)
Area of 16 such tiles
(Approx.)
Cost of polishing
of tile = Rs. 0.50
Cost of polishing
of tile
=
= Rs. 705.60 (Approx.)
Question 9. 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:
Let ABCD be a field in the shape of trapezium and parallel side AB = 10 m & CD = 25 m
And Non-parallel sides AD = 13 m and BC = 14 m
Draw BM
DC and BE
AD so that ABED is a parallelogram.
BE = AD = 13 m and DE = AB = 10 m
Now in
BEC, Semi-perimeter
= 21 m
Area of
BEC =
=
=
=
And Area of
BEC =
= 84
= 84
BM =
= 11.2 m
Now area of trapezium ABCD =
=
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)m =
m = 132m
cm = 21 cm
cm2
cm = 270 cm
cm =15 cm
C =
AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
ABD.
Area of
=
[Using Pythagoras theorem]
BD =
=
=
= 13 m
= 15 m
=
=
(approx.)
Area of quadrilateral ABCD = Area of
= 6 cm
=
= 7 cm
=
= 2
(approx.)
= 5.5 cm
=
(AB + DC)
=
=
=
= 42 cm
Base x Corresponding height = 336
= 336
=
= 54 m
Area of
Area of
[ Using Heron’s formula]
=
Field available for 18 cows to graze the grass
= 60 cm
=
=
cm and diagonals AC = BD = 32 cm
[Using Pythagoras theorem]
= 512
[Area of square =
]
= 10 cm
=
=
= 36 cm
=
(approx.)
(Approx.)
Cost of polishing
of tile = Rs. 0.50
of tile
= Rs. 705.60 (Approx.)
DC and BE
AD so that ABED is a parallelogram.
=
= 84
= 84
= 11.2 m
