# RD Sharma Class 9 Solutions Chapter 17 Constructions Ex 17.1

In this chapter, we provide RD Sharma Solutions for Chapter 17 Constructions Ex 17.1 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Chapter 17 Constructions Ex 17.1 Maths pdf, free RD Sharma Solutions for Chapter 17 Constructions Ex 17.1 Maths book pdf download. Now you will get step by step solution to each question.

### RD Sharma Solutions for Class 9 Chapter 17 Constructions Ex 17.1 Download PDF

Question 1.
Find the area of a triangle whose sides are respectively 150 cm, 120 cm and 200 cm.
Solution:
Sides of triangle are 120 cm, 150 cm, 200 cm

Question 2.
Find the area of a triangle whose sides are 9 cm, 12 cm and 15 cm.
Solution:
Sides of a triangle are 9 cpi, 12 cm, 15 cm

Question 3.
Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Solution:
Perimeter of a triangle = 42 cm
Two sides are 18 cm and 10 cm
Third side = 42 – (18 + 10)
= 42 – 28 = 14 cm

Question 4.
In a ∆ABC, AB = 15 cm, BC = 13 cm and AC = 14 cm. Find the area of ∆ABC and hence its altitude on AC.
Solution:
Sides of triangle ABC are AB = 15 cm, BC = 13 cm, AC = 14 cm

Question 5.
The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the triangle. [NCERT]
Solution:
Perimeter of a triangle = 540 m
Ratio in sides = 25 : 17 : 12
Sum of ratios = 25 + 17 + 12 = 54

Question 6.
The perimeter of a triangle is 300 m. If its sides are in the ratio 3:5:7. Find the area of the triangle. [NCERT]
Solution:
Perimeter of a triangle = 300 m
Ratio in the sides = 3 : 5 : 7
∴ Sum of ratios = 3 + 5 + 7= 15

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:
Perimeter of a triangular field = 240 dm
Two sides are 78 dm and 50 dm
∴ Third side = 240 – (78 + 50)
= 240 – 128 = 112 dm

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:
Sides of a triangle are 35 cm, 54 cm, 61 cm

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:
Ratio in the sides of a triangle = 3:4:5

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:
Perimeter of an isosceles triangle = 42 cm
Base = 32 of its one of equal sides
Let each equal side = x, then 3
Base = 32 x

Question 11.
Find the area of the shaded region in figure.

Solution:
In ∆ABC, AC = 52 cm, BC = 48 cm
and in right ∆ADC, ∠D = 90°
AD = 12 cm, BD = 16 cm
∴ AB²=AD² + BD² (Pythagoras Theorem)
(12)² + (16)² = 144 + 256 = 400 = (20)²
∴ AB = 20 cm

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