# NCERT Exemplar Problems Class 9 Maths – Heron’s Formula

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## NCERT Exemplar Problems Class 9 Maths – Heron’s Formula

Question 1:
An isosceles right triangle has area 8 cm2. The length of its hypotenuse is
(a) √32 cm (b) √16 cm (c) √48 cm (d) √24
Solution:
(a) Given, area of an isosceles right triangle = 8 cm2
Area of an isosceles triangle = ½ (Base x Height)
=> 8 = ½ (Base x Base)
[∴ base = height, as triangle is an isosceles triangle]
=> (Base)2 =16 => Base= 4 cm In ΔABC, using Pythagoras theore
AC2 = AB2 + BC2 = 42 + 42 = 16 + 16
=> AC2 = 32 => AC = √32 cm
[taking positive square root because length is always positive]
Hence, the length of its hypotenuse is √32 cm.

Question 2:
The perimeter of an equilateral triangle is 60 m. The area is
(a)10√3 m2 (b) 15√3 m2 (c)20√3 m2 (d) 100√3 m2
Thinking Process
(i) First, determine the side of an equilateral by usingformula, perimeter=3x.
(ii) Further, substitute the value of x in the formula, area of an equilateral triangle = √3/4 (a)2 and simplify it.
Solution:
(d) Let each side of an equilateral be x.
Then, perimeter of an equilateral triangle = 60 m
x + x + x = 60 => 3x = 60 => x = 60/3 = 20 m
Area of an equilateral triangle = √3/4 (Side)2 = (√3/4) x 20 x 20 = 100 √3 m2
Thus, the area of triangle is 100√3 m2.

Question 3:
The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then, the area of the triangle is
(a) 1322 cm2 (b) 1311 cm2 (c) 1344 cm2 (d) 1392 cm2
Thinking Process
(i) First, determine the semi-perimeter of a triangle by using the formula, s = (a + b + c)/2
(ii) Further, determine the area of triangle by using the formula, area of triangle (Fieron’s formula) = Solution: Question 4:
The area of an equilateral triangle with side 2√3 cm is
(a) 5.196 cm2 (b) 0.866 cm2 (c) 3.496 cm2 (d) 1.732 cm2
Solution:
(a) Given, side of an equilateral triangle is 2√3 cm.
Area of an equilateral triangle = √3/4 (Side)2
= √3/4 (2√3)2 = (√3/4) x 4 x 3
= 3√3 = 3 x 1.732 = 5.196 cm2
Hence, the area of an equilateral triangle is 5.196 cm2.

Question 5:
The length of each side of an equilateral triangle having an area of 9√3 cm2 is
(a) 8 cm (b) 36 cm (c) 4 cm (d) 6 cm
Solution:
(d) Given, area of an equilateral triangle = 9√3 cm2
∴ Area of an equilateral triangle = √3/4(Side)2
=> √3/4 (Side)2 = 9√3
=> (Side)2 = 36
∴ Side = 6 cm [taking positive square root because side is always positive]
Hence, the length of an equilateral triangle is 6 cm.

Question 6:
If the area of an equilateral triangle is 16√3 cm2, then the perimeter of the triangle is
(a) 48 cm (b) 24 cm (c) 12 cm (d) 36 cm
Solution:
(b) Given, area of an equilateral triangle = 16√3 cm2
Area of an equilateral triangle = √3/4 (side)2
√3/4(Side)2 = 16√3 => (Side)2 = 64
=> Side =8 cm
[taking positive square root because side is always positive]
Perimeter of an equilateral triangle = 3 x Side= 3 x 8 = 24 cm
Hence, the perimeter of an equilateral triangle is 24 cm.

Question 7:
The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude
(a) 1675 cm (b) 1o75 cm (c) 2475 cm (d) 28 cm
Thinking Process
(i) First, determine the semi-perimeter, s and then determine the area of triangle by using Heron’s formula.
(ii) For the longest altitude, take base as the smallest side. Apply the formula,
Area = ½ x Base x Altitude
(iii) Equate the area obtained using the two formula’s and obtain the required height.
Solution:  Question 8:
The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is Solution: Question 9:
The edges of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting it at the rate of 9 paise per cm2 is
(a)Rs. 2.00 (b) Rs. 2.16 (c) Rs. 2.48 (d) Rs. 3.00
Solution: Exercise 12.2: Very Short Answer Type Questions

Question 1:
The area of a triangle with base 4 cm and height 6 cm is 24 cm2.
Solution:
False
We know that, area of a triangle=½ (Base x Height)
Here Base = 4 cm and Height = 6 cm
∴ Area of a triangle =½ x 4 x 6= 12 cm2

Question 2:
The area of a ΔABC is 8 cm2 in which AB = AC = 4 cm and ∠A = 90°.
Solution:
True
We have, ABC is a right angled triangle at A, where sides are given AB = AC = 4 cm
∴ Area of a triangle ABC = ½ (Base x Height) = ½ x AC x AB=½ x 4 x 4=8 cm2 Question 3:
The area of the isosceles triangle is (5/4) √11 cm2 if the perimeter is 11 cm and the base is 5 cm.
Solution: Question 4:
The area of the equilateral triangle is 20 √3 cm2 whose each side is 8 cm.
Solution:
False
Given, side of an equilateral triangle be 8 cm.
Area of the equilateral triangle = √3/4 (Side)2
= (√3/4) x (8)2 = (64/4) √3 [∴ side = 8 cm]
= 16 √3 cm2

Question 5:
If the side of a rhombus is 10 cm and one diagonal is 16 cm, then area of the rhombus is 96 cm2.
Solution:
True
Given, side of a rhombus PQRS is 10 cm and one of the diagonal is 16 cm.
i.e., PQ = QR = RS =SP = 10 cm and PR = 16 cm In ΔPOQ, PQ2 = OP2 + OQ2 [by Pythagoras theorem]
[ since, the diagonal of rhombus bisects each other at 90°] => OQ2 = PQ2 – OP2 = (10)2 – (8)2
=> OQ2 =100 – 64 = 36
=> OQ = 6 cm
[taking positive square root because length is always positive] SQ =2 x OP = 2 x 6= 12 cm
Area of the rhombus = ½ (Product of diagonals)
= ½ (OS x PR) =½ x 12 x 16 = 96 cm2.

Question 6:
The base and the corresponding altitude of a parallelogram are 10 cm and 3.5 cm, respectively. The area of the parallelogram is 30 cm2.
Solution:
False
Given, parallelogram in which base = 10 cm and altitude = 3.5 cm
Area of a parallelogram = Base x Altitude = 10 x 3.5 = 35 cm2.

Question 7:
The area of regular hexagon of side a is the sum of the areas of the five equilateral triangles with side a.
Solution:
False
We know that regular hexagon is divided into six equilateral triangles. Area of regular hexagon of side a = Sum of area of the six equilateral triangles.

Question 8:
The cost of levelling the ground in the form of a triangle having the sides 51 m, 37 m and 20 m at the rate of Rs. 3 per m2 is Rs. 918.
Solution: Question 9:
In a triangle, the sides are given as 11 cm, 12 cm and 13 cm. The length of the altitude is 10.25 cm corresponding to the side having 12 cm.
Solution: Exercise 12.3: Short Answer Type Questions

Question 1:
Find the cost of laying grass in a triangular field of sides 50 m, 65 m and 65 m at the rate of Rs. 7 per m2.
Thinking Process Solution:  Question 2:
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 13 m, 14 m and 15 m. The advertisements yield an earning of Rs. 2000 per m2 a year. A company hired one of its walls for 6 months. How much rent did it pay?
Solution: Question 3:
From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The lengths of the perpendiculars are 14 cm, 10 cm and 6 cm. Find the area of the triangle.
Thinking Process
(i) From a interior point a triangle, three triangle will be formed Determine area of each triangle by using the formula =½ (Base x Height).
(ii) Determine the area of an equilateral triangle by using the Heron’s formula i.e., (iii) Further, equate the area of triangle by Heron’s formula and the sum of all three triangle. Simplify it to get the side of D. Hence, find area of the triangle.
Solution:  Question 4:
The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. Find the area of the triangle.
Solution: Question 5:
Find the area of a parallelogram given in the figure. Also, find the length of the altitude from vertex A on the side DC. Thinking process
(i) Determine the area of ABCD by using Heron’s formula.
(ii) Using relation, area of parallelogram ABCD =2 (Area of ΔBCD)
(iii) Also, determine the area of parallelogram by using the formula Base x Altitude.
(iv) Further, equating the area of parallelogram in (ii) and (iii). Obtain the required length of the altitude.
Solution: =3 x 3 x 5 x 2 cm2
Area of parallelogram ABCD = 2 x 90
= 180 cm2 …(ii)
Let altitude of a parallelogram be h.
Also, area of parallelogram ABCD =Base x Altitude
=> 180 = DC x h [from Eq. (ii)]
=> 180 = 12 x h
∴ h = 180/12= 15 cm
Hence, the area of parallelogram is 180 cm2 and the length of altitude is 15 cm.

Question 6:
A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. Find the area of the parallelogram. Solution: Question 7:
The perimeter of a triangular field is 420 m and its sides are in the ratio 6:7:8. Find the area of the triangular field.
Solution:
Given, perimeter of a triangular field is 420 m and its sides are in the ratio 6:7:8.
Let sides of a triangular field be a = 6x, b = 7x and c = 8x.
Perimeter of a triangular field, 2s = a + b + c => 420 = 6x + 7x + 8x => 420 = 21x
=> x = 420/21 = 20 m Question 8:
The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. Find its area.
Solution:
Given ABCD is a quadrilateral having sides AB = 6 cm, BC = 8 cm, CD = 12 cm and DA = 14 cm. Now, join AC. Question 9:
A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs. 5 per cm2. Find the cost of painting.
Solution: Question 10:
Find the area of the trapezium PQRS with height PQ given in the figure given below Solution: Exercise 12.4: Long Answer Type Questions

Question 1:
How much paper of each shade is needed to make a kite given in figure, in which ABCD is a square with diagonal 44 cm. Solution: Question 2:
The perimeter of a triangle is 50 cm. One side of a triangle is 4 cm longer than the smaller side and the third side is 6 cm less than twice the smaller side. Find the area of the triangle.
Solution: Question 3:
The area of a trapezium is 475 cm2 and the height is 19 cm. Find the lengths of its two parallel sides, if one side is 4 cm greater than the other.
Solution:  Question 4:
A rectangular plot is given for constructing a house having a measurement of 40 m long and 15 m in the front. According to the laws, a minimum of 3 m, wide space should be left in the front and back each and 2 m wide space on each of other sides. Find the largest area where house can be constructed.
Solution:
Let ABCD is a rectangular plot having a measurement of 40 m long and 15 m front. ∴ Length of inner-rectangle, EF = 40 – 3 – 3 = 34 m and breadth of inner-rectangle, FG =15 – 2 – 2 = 11 m
∴ Another rectangle EFGH will be formed inside the rectangle ABCD
∴ Area of innerrectangle, EFGH = Length x Breadth
= EF x FG = 34 x 11 = 374 m2
[∴ area of a rectangle = length x breadth]
Hence, the largest area where the house can be constructed in 374 m2.

Question 5:
A field is in the shape of a trapezium having parallel sides 90 m and 30 m. These sides meet the third side at right angles. The length of the fourth side is 100 m. If it Rs. 4 costs to plough 1 m2 of the field, find the total cost of ploughing the field.
Solution:
In trapezium ABCD, we draw a perpendicular line CE to the line AB.  Question 6:
In figure, ΔABC has sides AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC a parallelogram DBCE of same area as that of ΔABC is constructed. Find the height OF of the parallelogram. Thinking process
(i) First, determine the area of AABC by using Heron’s formula i.e., (ii) Second, determine the area of parallelogram BCED by using the formula Base x Height.
(iii) Further, equate the area of triangle and area of parallelogram, to get the height of the parallelogram.
Solution:  Question 7:
The dimensions of a rectangle ABCD are 51 cm x 25 cm. A trapezium PQCD with its parallel sides QC and PD in the ratio 9 : 8, is cut off from the rectangle as shown in the figure. If the area of the trapezium PQCD is 5/6 th part of the area of the rectangle. Find the lengths QC and PD. Solution: Question 8:
A design is made on a rectangular tile of dimensions 50 cm x 17 cm as shown in figure. The design shows 8 triangle, each of sides 26 cm, 17 cm and 25 cm. Find the total area of the design and the remaining area of the tiles. Solution: +