Table of Contents
Chapter 17 – Heron’s Formula Exercise Ex. 17.1
Question 1
Find the area of the triangle whose sides are respectively 150 cm, 120 cm and 200 cm.Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5The perimeter of triangular field is 540 m and its sides are in the ratio 25:17:12. Find the area of the triangle.Solution 5The sides of the triangular field are in the ratio 25:17:12.
Let the sides of triangle be 25x, 17x, and 12x.
Perimeter of this triangle = 540 m
25x + 17x + 12x = 540 m
54x = 540 m
x = 10 m
Sides of triangle will be 250 m, 170 m, and 120 m. Semi-perimeter (s) = By Heron’s formula:
So, area of the triangle is 9000 m2.Question 6
The perimeter of right triangle is 300m. If its sides are in the ratio 3 : 5 : 7. Find the area of the triangle.Solution 6
Question 7
The perimeter of a triangular field is 240 dm. If two of its sides are 78 dm and 50 dm, find the length of the perpendicular on the side of length 50 dm from the opposite vertex.Solution 7
Question 8
A triangle has sides 35 cm, 54 cm and 61 cm long. Find its area. Also, find the smallest of its altitudes.Solution 8
Question 9
The lengths of the sides of a triangle are in the ratio 3 : 4 : 5 and its perimeter is 144 cm. Find the area of the triangle and the height corresponding to the longest side.Solution 9
Question 10
The perimeter of an isosceles triangle is 42 cm and its base is (3/2) times each of the equal sides. Find the length of each side of the triangle, area of the triangle and the height of the triangle.Solution 10
Question 11
Find the area of the shaded region in fig.12.12
Solution 11
Chapter 17 – Heron’s Formula Exercise Ex. 17.2
Question 1Find 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 1
For ABC
AC2 = AB2 + BC2
(5)2 = (3)2 + (4)2
So, ABC is a right angle triangle, right angled at point B.
Area of ABCFor ADC
Perimeter = 2s = AC + CD + DA = (5 + 4 + 5) cm = 14 cm
s = 7 cm
By Heron’s formula
Area of triangle
Area of ABCD = Area of ABC + Area of ACD
= (6 + 9.166) cm2 = 15.166 cm2 = 15.2 cm2 (approximately)
Question 2
Solution 2
Question 3
Solution 3
Question 4A park, in the shape of a quadrilateral ABCD, has = 90o, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?Solution 4Let us join BD.
In BCD applying Pythagoras theorem
BD2 = BC2 + CD2
= (12)2 + (5)2
= 144 + 25
BD2 = 169
BD = 13 m
Area of BCD
For ABD
By Heron’s formula Area of triangle
Area of park = Area of ABD + Area of BCD
= 35.496 + 30 m2 = 65.496 m2 = 65. 5 m2 (approximately)
Question 5
Find the area of a rhombus whose perimeter is 80 m and one of whose diagonal is 24 m.Solution 5
Question 6
A rhombus sheet, whose perimeter is 32 m and whose one diagonal is 10 m long, is painted on both sides at the rate of Rs 5 per m2. Find the cost of painting.Solution 6
Question 7
Solution 7
Question 8
Find the area of a quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm and diagonal BD = 20 cm.Solution 8
Question 09
The adjacent sides of a parallelogram ABCD measure 34 cm and 20 cm, and the diagonal AC measures 42 cm. Find the area of the parallelogram.Solution 09
Question 10
Find the area of the blades of the magnetic compass shown in fig.
(Take √11 = 3.32)
Solution 10
Question 11
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 13 cm, 14 cm and 15 cm and the parallelogram stands on the base 14 cm, find the height of the parallelogram.Solution 11
Question 12
Solution 12
Question 13
Solution 13
Question 14
A hand fan is made by stitching 10 equal size triangular strips of two different types of paper as shown in fig., The dimensions of equal strips are 25 cm, 25 cm and 14 cm. Find the area of each type of paper needed to make the hand fan.
Solution 14
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