**RD Sharma Solutions Class 9 Maths Chapter 17 Constructions:** Each question is solved using figures for easy as well as better understanding. RD Sharma Solutions Class 9 Maths Chapter 17 builds a good foundation for students. They can secure excellent marks in the exams for answering more difficult questions accurately by practicing these solutions. All answers to the questions of this chapter constructions are accurate & reliable.

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RD Sharma Class 9 Solutions Chapter 17

**Exercise-wise RD Sharma Solutions Class 9 Maths Chapter 17 Constructions**

RD Sharma class 9 chapter 17 exercise 17a |

RD Sharma class 9 chapter 17 exercise 17b |

RD Sharma class 9 chapter 17 exercise 17c |

**Access answers of R****D Sharma Solutions Class 9 Maths Chapter 17 Constructions**

### RD Sharma Solutions Class 9 Chapter 17 Constructions Ex 17.1

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

### RD Sharma Class 9 Solution Chapter 17 Constructions Ex 17.2

Question 1.

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

Solution:

In the quadrilateral, AC is the diagonal which divides the figure into two triangles

Now in ∆ABC, AB = 3 cm, BC = 4cm, AC = 5cm

Question 2.

The sides of a quadrangular field taken in order are 26 m, 27 m, 7 m and 24 m respectively. The angle contained by the last two sides is a right angle. Find its area.

Solution:

In quad. ABCD, AB = 26 m, BC = 27 m CD = 7m, DA = 24 m, ∠CDA = 90°

Join AC,

Question 3.

The sides of a quadrilateral taken in order are 5, 12, 14 and 15 metres respectively, and the angle contained by the first two sides is a right angle. Find its area.

Solution:

In quad. ABCD,

AB = 5m, BC = 12 m, CD = 14m,

DA = 15 m and ∠ABC = 90°

Join AC,

Now in right ∆ABC,

AC² = AB² + BC² = (5)² + (12)²

= 25 + 144 = 169 = (13)²

∴ AC = 13 m

Now area of right ∆ABC

Question 4.

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

Solution:

In quadrilateral ABCD,

AB = 9m, BC = 12m, CD = 5m and

DA = 8m, ∠C = 90°

Join BD,

Now in right ∆BCD,

BD² = BC²+ CD² = (12)² + (5)²

= 144 + 25 = 169 = (13)²

∴ BD = 13m

Question 5.

Find the area of a rhombus whose perimeter is 80m and one of whose diagonal is 24m.

Solution:

Perimeter of rhombus ABCD = 80 m

Question 6.

A rhombus sheet whose perimeter = 32 m and whose one diagonal is 10 m long, is painted on both sides at the rate of ₹5 per m². Find the cost of painting.

Solution:

Perimeter of the rhombus shaped sheet = 32 m

∴ Length of each side = 324 = 8m

and length of one diagonal AC = 10 m

In ∆ABC, sides are 8m, 8m, 10m

Question 7.

Find the area of a quadrilateral ABCD in which AD = 24 cm, ∠BAD = 90° and BCD forms an equilateral triangle whose each side is equal to 26 cm. (Take 3–√ = 1.73 )

Solution:

In quadrilateral ABCD, AD = 24cm, ∠BAD = 90°

BCD is an equilateral triangle with side 26cm

In right ∆ABD,

BD² = AB²+ AD²

(26)² = AB² + (24)²

⇒ 676 = AB² + 576

AB² = 676 – 576 = 100 = (10)²

∴ AB = 10cm

Now area of right ∆ABD,

Question 8.

Find the area of a quadrilateral ABCD in which AB = 42cm, BC = 21cm, CD = 29 cm, DA = 34 cm and diagonal BD = 20 cm.

Solution:

In quadrilateral ABCD,

AB = 42 cm, BC = 21 cm, CD = 29cm DA = 34 cm, BD = 20 cm

Question 9.

The adjacent sides of a parallelogram ABCD measures 34 cm and 20 cm, and the diagonal AC measures 42 cm. Find the area of the parallelogram.

Solution:

In ||gm ABCD,

AB = 34cm, BC = 20 cm

and AC = 42 cm

∵ The diagonal of a parallelogram divides into two triangles equal in area,

Now area of ∆ABC,

Question 10.

Find the area of the blades of the magnetic compass shown in figure. (Take 11−−√ = 3.32).

Solution:

ABCD is a rhombus with each side 5cm and one diagonal 1cm

Diagonal BD divides into two equal triangles Now area of ∆ABD,

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:

Area of a triangle with same base and area of a 11gm with equal sides of triangle are 13, 14, 15 cm

Question 12.

Two parallel sides of a trapezium are 60cm and 77 cm and other sides are 25 cm and 26 cm. Find the area of the trapezium.

Solution:

In trapezium ABCD, AB || DC

AB = 77cm, BC = 26 cm, CD 60cm DA = 25 cm

Through, C, draw CE || DA meeting AB at E

∴ AE = CD = 60 cm and EB = 77 – 60 = 17 cm,

CE = DA = 25 cm

Now area of ∆BCE, with sides 17 cm, 26 cm, 25 cm

Question 13.

Find the perimeter and area of the quadrilateral ABCD in which AB = 17 cm, AD = 9cm, CD = 12cm, ∠ACB = 90° and AC = 15cm.

Solution:

In right ΔABC, ∠ACB = 90°

AB² = AC² + BC²

(17)² = (15)²+ BC² = 289 = 225 + BC²

Question 14.

A hand fan is made by stitching 10 equal size triangular strips of two different types of paper as shown figure. The dimensions of equal strips are 25 cm, 25 cm and 14 cm. Find the area of each types of paper needed to make the hand fan.

Solution:

In the figure, a hand fan has 5 isosceles and triangle. With sides 25 cm, 25 cm and 14 cm each.

### Class 9 RD Sharma Solutions Chapter 17 Constructions VSAQS

Question 1.

Find the area of a triangle whose base and altitude are 5 cm and 4 cm respectively.

Solution:

In ∆ABC,

Base BC = 5cm

Altitude AD = 4cm

Question 2.

Find the area of a triangle whose sides are 3 cm, 4 cm and 5 cm respectively.

Solution:

Sides of triangle are 3 cm, 4cm and 5cm

Question 3.

Find the area of an isosceles triangle having the base x cm and one side y cm.

Solution:

In isosceles ∆ABC,

AB = AC = y cm

BC = x cm

Question 4.

Find the area of an equilateral triangle having each side 4 cm.

Solution:

Each side of equilateral triangle (a) = 4cm

Question 5.

Find the area of an equilateral triangle having each side x cm.

Solution:

Question 6.

The perimeter of a triangular field is 144 m and the ratio of the sides is 3 : 4 : 5. Find the area of the field.

Solution:

Perimeter of the field = 144 m

Ratio in the sides = 3:4:5

Sum of ratios = 3 + 4 + 5 = 12

Question 7.

Find the area of an equilateral triangle having altitude h cm.

Solution:

Altitude of an equilateral triangle = h

Let side of equilateral triangle = x

Question 8.

Let ∆ be the area of a triangle. Find the area of a triangle whose each side is twice the side of the given triangle.

Solution:

Let a, b, c be the sides of the original triangle

Hence area of new triangle = 4 x area of original triangle.

Question 9.

If each side of a triangle is doubled, then find percentage increase in its area.

Solution:

Sides of original triangle be a, b, c

Question 10.

If each side of an equilateral triangle is tripled then what is the percentage increase in the area of the triangle?

Solution:

Let the sides of the original triangle be a, b, c and area ∆, then

### RD Sharma Solutions Class 9 Chapter 17 Constructions MCQS

Mark the correct alternative in each of the following:

Question 1.

The sides of a triangle are 16 cm, 30 cm, 34 cm. Its area is

(a) 225 cm²

(b) 2253–√ cm²

(c) 2252–√ cm²

(d) 240 cm²

Solution:

Sides of triangle and 16 cm, 30 cm, 34 cm

Question 2.

The base of an isosceles right triangle is 30 cm. Its area is

(a) 225 cm²

(b) 2253–√ cm²

(c) 2252–√ cm²

(d) 450 cm²

Solution:

Base of isosceles triangle ∆ABC = 30cm

Let each of equal sides = x

Then AB = AC = x

Now in right ∆ABC,

Question 3.

The sides of a triangle are 7cm, 9cm and 14cm. Its area is

(a) 125–√ cm²

(b) 123–√ cm²

(c) 24 5–√ cm²

(d) 63 cm²

Solution:

Question 4.

The sides of a triangular field are 325 m, 300 m and 125 m. Its area is

(a) 18750 m²

(b) 37500 m²

(c) 97500 m²

(d) 48750 m²

Solution:

Sides of a triangular field are 325m, 300m, 125m

Question 5.

The sides of a triangle are 50 cm, 78 cm and 112 cm. The smallest altitude is

(a) 20 cm

(b) 30 cm

(c) 40 cm

(d) 50 cm

Solution:

The sides of a triangle are 50 cm, 78 cm, 112cm

Question 6.

The sides of a triangle are 11m, 60m and 61m. Altitude to the smallest side is

(a) 11m

(b) 66 m

(c) 50 m

(d) 60 m

Solution:

Sides of a triangle are 11m, 60m and 61m

Question 7.

The sides of a triangle are 11 cm, 15 cm and 16 cm. The altitude to the largest side is

Solution:

Sides of a triangle are 11 cm, 15 cm, 16 cm

Question 8.

The base and hypotenuse of a right triangle are respectively 5cm and 13cm long. Its area is

(a) 25 cm²

(b) 28 cm²

(c) 30 cm²

(d) 40 cm²

Solution:

In a right triangle base = 5 cm

base hypotenuse = 13 cm

Question 9.

The length of each side of an equilateral triangle of area 4 3–√ cm², is

Solution:

Area of an equilateral triangle = 43–√ cm²

Let each side be = a

Question 10.

If the area of an isosceles right triangle is 8cm, what is the perimeter of the triangle.

(a) 8 + 2–√ cm²

(b) 8 + 42–√ cm²

(c) 4 + 82–√ cm²

(b) 122–√ cm²

Solution:

Let base = x

ABC an isosceles right triangle, which has 2 sides same

⇒ Height = x

Question 11.

The length of the sides of ∆ABC are consecutive integers. If ∆ABC has the same perimeter as an equilateral triangle with a side of length 9cm, what is the length of the shortest side of ∆ABC?

(a) 4

(b) 6

(c) 8

(d) 10

Solution:

Side of an equilateral triangle = 9 cm

Its perimeter = 3 x 9 = 27 cm

Now perimeter of ∆ABC = 27 cm

and let its sides be x, x + 1, x +2

Question 12.

In the figure, the ratio of AD to DC is 3 to 2. If the area of ∆ABC is 40cm2, what is the area of ∆BDC?

(a) 16 cm²

(b) 24 cm²

(c) 30 cm²

(d) 36 cm²

Solution:

Ratio in AD : DC = 3:2

and area ∆ABC = 40 cm²

Question 13.

If the length of a median of an equilateral triangle is x cm, then its area is

Solution:

∵ The median of an equilateral triangle is the perpendicular to the base also,

∴ Let side of the triangle = a

Question 14.

If every side of a triangle is doubled, then increase in the area of the triangle is

(a) 1002–√ %

(b) 200%

(c) 300%

(d) 400%

Solution:

Let the sides of the original triangle be a, b, c

Question 15.

A square and an equilateral triangle have equal perimeters. If the diagonal of the square is 1272 cm, then area of the triangle is

Solution:

A square and an equilateral triangle have equal perimeter

**Detailed Exercise-wise Explanation with Listing of Important Topics in the Exercise**

**RD Sharma class 9 chapter 17 exercise 17a**: This exercise includes concepts related to construction & also include problems based on it. In RD Sharma class 9 chapter 17 exercise 17a, the students will study how to draw a perpendicular bisector for a given line segment. These study material assist the students to obtain higher marks in the exam. The solutions for all questions are provided as per the CBSE syllabus.**RD Sharma class 9 chapter 17 exercise 17b:**This exercise includes topics on the construction of the bisector of a given angle. Each question is prescribed as per the syllabus and guidelines of CBSE. RD Sharma class 9 chapter 17 exercise 17b enables the students to learn how to construct different angles by using ruler and compass only. The students can definitely acquire better marks in the exams by solving questions from RD Sharma textbook.**RD Sharma class 9 chapter 17 exercise 17c:**This exercise includes word problems on constructions. The students can consider these detailed solutions to understand the in-depth topics of constructions of different kinds of triangles. By practicing the textbook questions with the help of RD Sharma solutions, a student’s problem-solving abilities are also enhanced.

## Important Topics from Class 9 Maths Chapter 17 Constructions

The students will study some important concepts on RD Sharma class 9 solutions chapter 17 constructions which are listed below:

- Introduction of Construction
- Construction of Bisector of a Line Segment
- Construction of the Bisector of a Given Angle
- Constructions of Triangles

RD Sharma CBSE class 9 solutions chapter 17 constructions are explained in detail & in an easy language to foster quick knowledge among students. If you have any doubts regarding the Class 9 Maths exam, ask in the comments.

## RD Sharma Solutions Class 9 Maths Chapter 17 Constructions

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