RS Aggarwal Solutions for Class 7 Maths Chapter 17 – Constructions are provided here. To cover the entire syllabus in Maths, the RS Aggarwal is an essential material as it offers a wide range of questions that test the students understanding of concepts. Our expert personnel have solved the problems step by step with neat explanations. Students who aspire to score good marks in Maths, then refer RS Aggarwal Class 7 Solutions.

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### Also, access RS Aggarwal Solutions for Class 7 Chapter 17 Exercises

## Exercise 17A Page: 204

**1. Draw a line AB and take a point P outside it. Draw a line CD parallel to AB and passing through the point P.**

**Solution:-**

Steps for construction,

1. Draw a line AB.

2. Take any point Q on AB and a point P outside AB and join PQ.

3. With Q as center and any radius draw an arc to cut AB at E and PQ at F.

4. With P as center and same radius draw an arc to cut QP at G.

5. With as center and radius equal to EF, draw an arc to cut the previous arc at H.

6. Join PH and produce it on both sides to get the required line CD parallel to AB.

**2. Draw a line AB and draw another line CD parallel to AB at a distance of 3.5 cm from it.**

**Solution:-**

Steps for construction,

1. Draw a line AB.

2. Take any two points P and Q on AB.

3. Construct âˆ BPE = 90^{o} and âˆ BQF = 90^{o}.

4. With P as center and radius equal to 3.5 cm, cut PE at R.

5. With Q as center and radius equal to 3.5 cm, cut QF at S.

6. Join CD and produce it on either side to get the required line CD parallel to AB and at a distance of 3 cm from it.

## Exercise 17B Page: 207

**1. Construct a Î”ABC in which BC = 3.6 cm, AB = 5 cm and AC = 5.4 cm. Draw the perpendicular bisector of the side BC.**

**Solution:-**

Steps of construction:

1. Draw a line segment AC = 5.4 cm.

2. With A as a center and radius 5 cm, draw an arc.

3. With C as a center and radius 3.6 cm, draw another arc, cutting the previous arc at B.

4. Join AB and CB.

Then, Î”ABC is the required triangle.

5. With B as center and radius measuring more than half of BC, draw arcs on both sides of BC.

6. With C as center and the same radius as before, draw arcs on both sides of BC, cutting the previous arcs at P and Q, as shown. Join PQ.

Then, PQ is the required perpendicular bisector of BC, meeting BC at M.

**2. Construct a Î”PQR in which QR = 6 cm, PQ = 4.4 cm and PR = 5.3 cm. Draw the bisector of âˆ P.**

**Solution:-**

Steps of construction:

1. Draw a line segment QR = 6 cm.

2. With Q as a center and radius 4.4 cm, draw an arc.

3. With R as a center and radius 5.3 cm, draw another arc, cutting the previous arc at P.

4. Join PQ and PR.

Then, Î”PQR is the required triangle.

5. With B as center and any radius cutting PQ and PR at S and T, respectively

6. With S as center and radius measuring more than half of ST, draw an arc.

7. With T as center and the same radius, draw another arc cutting the previous arc at x.

8. Join P and X

Then, PX is the required bisector of âˆ P.

**3. Construct an equilateral triangle each of whose sides measures 6.2 cm. Measures each one of its angles.**

**Solution:-**

Steps of construction:

1. Draw a line segment AC = 6.2 cm.

2. With A as a center and radius 6.2 cm, draw an arc.

3. With C as a center and radius 6.2 cm, draw another arc, cutting the previous arc at B.

4. Join AB and CB.

Then, Î”ABC is the required equilateral triangle.

5. When we will measure all the angles of triangle by protractor, then all angles are equal to 60^{o}.

**4. Construct a Î”ABC in which AB = AC = 4.8cm and BC = 5.3cm. Measure âˆ B and âˆ C. Draw AD âŠ¥ BC.**

**Solution:-**

Steps of construction:

1. Draw a line segment BC = 5.3 cm.

2. With B as a center and radius 4.8 cm, draw an arc.

3. With C as a center and radius 4.8 cm, draw another arc, cutting the previous arc at A.

4. Join AB and AC.

Then, Î”ABC is the required triangle.

5. When we will measure âˆ B and âˆ C the angles of triangle by protractor, then the measure of angles are 56^{o} and 56^{o} respectively.

6. With A as the center and any radius, draw an arc cutting BC at M and N.

7. With M as the center and the radius more than half of MN, draw an arc.

8. With N as the center and the same radius, draw another arc cutting the previously drawn.

9. Join AP, cutting BC at D.

Then, AD âŠ¥ BC

**5. Construct a Î”ABC in which AB = 3.8 cm, âˆ A = 60 ^{o} and AC = 5 cm.**

**Solution:-**

Steps of construction:

1. Draw a line segment AB = 3.8 cm.

2. Construct âˆ BAZ = 60^{o}.

3. Along AZ, set off AC = 5cm.

4. Join BC.

Then, Î”ABC is the required triangle.

**6. Construct a Î”ABC in which BC = 4.3 cm, âˆ C = 45 ^{o} and AC = 6 cm.**

**Solution:-**

Steps of construction:

1. Draw a line segment AC = 6 cm.

2. Construct âˆ ACZ = 45^{o}.

3. Along CZ, set off BC = 4.3cm.

4. Join BC.

Then, Î”ABC is the required triangle.

## Exercise 17C Page: 208

**Mark against the correct answer in each of the following:**

**1. The supplement of 45 ^{o} is**

**(a) 45 ^{o} (b) 75^{o} (c) 135^{o} (d) 155^{o}**

**Solution:-**

(c) 135^{o}

Because,

Two angles are said to be supplementary if the sum of their measures is 180^{o}.

The given angle is 45^{o}

Let the measure of its supplement be x^{o}.

Then,

= x + 45 = 180

= x = 180 â€“ 45

= x = 135^{o}

Hence, the supplement of the given angle measures 135^{o}.

**2. The complement of 80 ^{o} is**

**(a) 100 ^{o} (b) 10^{o} (c) 20^{o} (d) 280^{o}**

**Solution:-**

(b) 10^{o}

Because,

Two angles are said to be complementary if the sum of their measures is 90^{o}.

The given angle is 80^{o}

Let the measure of its complement be x^{o}.

Then,

= x + 80^{o} = 90^{o}

= x = 90 â€“ 80

= x = 10^{o}

Hence, the complement of the given angle measures 10^{o}.

**3. An angle is its own complement. The measure of the angle is **

**(a) 30 ^{o} (b) 45^{o} (c) 90^{o} (d) 60^{o}**

**Solution:-**

(b) 45^{o}

Because,

Let the measure of the required angle be x^{o}. Then,

= x + x = 90^{o}

= 2x = 90

= x = 90/2

= x = 45^{o}

Hence, the required angle measures 45^{o}.

**4. An angle is one-fifth of its supplement. The measure of the angle is**

**(a) 30 ^{o} (b) 15^{o} (c) 75^{o} (d) 150^{o}**

**Solution:-**

(a) 30^{o}

Because,

Let the measure of its supplement be x^{o} + (x/5)^{o}.

Then,

= x^{o} + (x/5)^{o} = 180^{o}

= x = (180 â€“ x) / 5

= 5x = 180 â€“ x

= 5x + x = 180

= 6x = 180

= x = 180/6

= x = 30^{o}

Hence, the supplement of the given angle measures 30^{o}.

**5. An angle is 24 ^{o} more than its complement. The measure of the angle is**

**(a) 47 ^{o} (b) 57^{o} (c) 53^{o} (d) 66^{o}**

**Solution:-**

(b) 57^{o}

Because,

Let the measure of its complement be x^{o} + (x+24)^{o}.

Then,

= x^{o} + (x+24)^{o} = 90^{o}

= x = (90 â€“(x + 24))

= x = 90 â€“ x + 24

= x + x = 114

= 2x = 114

= x = 114/2

= x = 57

**6. An angle is 32 ^{o} less than its supplement. The measure of the angle is**

**(a) 37 ^{o} (b) 74^{o} (c) 148^{o} (d) none of these**

**Solution:-**

(b) 74^{o}

Because,

Let the measure of its supplement be x^{o} – (x – 32)^{o}.

Then,

= x^{o} – (x -32)^{o} = 180^{o}

= x = 180 â€“ (x â€“ 32)

= x = 180 â€“ x – 32

= x + x = 148

= 2x = 148

= x = 148/2

= x = 74^{o}

Hence, the supplement of the given angle measures 74^{o}.

**7. Two supplementary angles are in the ratio 3: 2. The smaller angle measures**

**(a) 108 ^{o} (b) 81^{o} (c) 72^{o} (d) none of these**

**Solution:-**

(c) 72^{o}

Because,

Let the measure of its supplement be 3x and 2x

Then,

= 3x + 2x = 180

= 5x = 180

= x = 180/5

= x = 36^{o}

Hence, the smaller angle measures 2x = (2 Ã— 36) = 72^{o}

**8. In the given figure, AOB is a straight line and the ray OC stands on it. If âˆ BOC = 132 ^{o}, then âˆ AOC =?**

**(a) 68 ^{o} (b) 48^{o} (c) 42^{o} (d) none of these**

**Solution:-**

(b) 48^{o}

Because,

âˆ AOC =?

= âˆ AOC + âˆ BOC = 180^{o} â€¦ [âˆµ Linear pair]

= âˆ AOC + 132^{O} = 180^{o}

= âˆ AOC = 180 â€“ 132

= âˆ AOC = 48^{o}

**9. In the given figure, AOB is a straight line, âˆ AOC = 68 ^{o} and âˆ BOC =x^{O}.**

**(a) 32 (b) 22 (c) 112 (d) 132**

**Solution:-**

(c) 112

Because,

= âˆ BOC + âˆ AOC = 180^{o} â€¦ [âˆµ Linear pair]

= x^{O} + 68^{O} = 180^{o}

= x^{O} = 180 â€“ 68

= x^{O} = 112

**10. In the adjoining figure, what value of x will make AOB a straight line?**

**(a) x = 30 (b) x = 35 (c) x = 25 (d) x = 40**

**Solution:-**

(b) x = 35

Because,

It is given that the angles of the side be (2x â€“ 10)^{o} and (3x + 15)^{o}.

= (2x â€“ 10) + (3x + 15) = 180 â€¦ [âˆµ Linear pair]

= 2x â€“ 10 + 3x + 15 = 180

= 5x + 5 = 180

= 5x = 180 – 5

= 5x = 175

= x = 175/5

= x = 35

**11. In the given figure, what value of x will make AOB a straight line?**

**(a) x = 50 (b) x = 100 (c) x = 60 (d) x = 80**

**Solution:-**

(d) x = 80

Because,

It is given that the angles of the side be 55^{o}, 45^{o} and (x)^{o}.

= x + 55^{o} + 45^{o} = 180^{o }â€¦ [âˆµ Linear pair]

= x = 180 â€“ 100

= x = 80

**12. In the given figure, it is given that AOB is a straight line and 4x = 5y. What is the value of x?**

**(a) 100 (b) 105 (c) 110 (d) 115**

**Solution:-**

(a) 100

Because,

Let the measure of its supplement be 4x and 5y.

Then,

= x + (4/5)x = 180^{o}

= (5x + 4x)/5 = 180

= 9x = 180 Ã— 5

= 9x = 900

= x = 900/9

= x = 100

**13. In the given figure, two straight lines AB and CD intersect at a point O and âˆ AOC = 50 ^{o}. Then âˆ BOD =?**

**(a) 40 ^{o} (b) 50^{o} (c) 130^{o} (d) 60^{o}**

**Solution:-**

(c) 130^{o}

Because,

= âˆ AOC + âˆ BOD = 180^{o} â€¦ [âˆµ Linear pair]

= 50^{o} + âˆ BOD = 180^{o}

= âˆ BOD = 180 â€“ 50

= âˆ BOD = 130^{o}

**14. In the given figure, AOB is a straight line, âˆ AOC = (3x â€“ 8) ^{o}, âˆ AOC = 50^{o} and âˆ BOD = (x + 10)^{o}. The value of x is**

**(a) 32 (b) 42 (c) 36 (d) 52**

**Solution:-**

(a) 32

Because,

It is given that the angles of the side be (3x â€“ 8)^{o}, 50^{o} and (x + 10)^{o}.

= (3x â€“ 8) + 50^{o} + (x + 10) = 180 â€¦ [âˆµ Linear pair]

= 3x â€“ 8 + 50 + x + 10 = 180

= 4x + 52 = 180

= 4x = 180 – 52

= 4x = 128

= x = 128/4

= x = 32

**15. In Î”ABC, side BC has produced to D. If âˆ ACD = 132 ^{o} and âˆ A = 54^{o}, then âˆ B =?**

**(a) 48 ^{o} (b) 78^{o} (c) 68^{o} (d) 58^{o}**

**Solution:-**

(b) 78^{o}

Because,

Consider the Î”ABC,

We know that the exterior angle of a triangle is equal to the sum of its interior opposite angles.

âˆ´ âˆ ABC + âˆ BAC = âˆ ACD

= âˆ ABC + 54^{o} = 132^{o}

= âˆ ABC = 132^{o} â€“ 54^{o}

= âˆ ABC = 78^{o}

**16. In Î”ABC, side BC has been produced to D. if âˆ BAC = 45 ^{o} and âˆ ABC = 55^{o}, then âˆ ACD =?**

**(a) 80 ^{o} (b) 90^{o} (c) 100^{o} (d) 110^{o}**

**Solution:-**

(c) 100^{o}

Because,

We know that the exterior angle of a triangle is equal to the sum of its interior opposite angles.

âˆ´ âˆ ABC + âˆ BAC = âˆ ACD

= 55^{o} + 45^{o} = âˆ ACD

= âˆ ACD = 100^{o}

**17. In the given figure, side BC of Î”ABC is produced to D such that âˆ ABC = 70 ^{o} and âˆ ACD = 120^{o}. Then, âˆ BAC =?**

**(a) 60 ^{o} (b) 50^{o} (c) 70^{o} (d) 35^{o}**

**Solution:-**

(b) 50^{o}

Because,

We know that the exterior angle of a triangle is equal to the sum of its interior opposite angles.

âˆ´ âˆ ABC + âˆ BAC = âˆ ACD

= 70^{o} +âˆ BAC = 120^{o}

= âˆ BAC = 120^{o} â€“ 70^{o}

= âˆ BAC = 50^{o}

**18. In the given figure, rays OA, OB, OC and OD are such that âˆ AOB = 50 ^{o}, âˆ BOC = 90^{o}, âˆ COD = 70^{o} and âˆ AOD = x^{o}.**

**(a) 50 ^{o} (b) 70^{o} (c) 150^{o} (d) 90^{o}**

**Solution:-**

(c) 150^{o}

Because,

We know that the complete angle is equal to the sum of all angles is equal to 360^{o}.

âˆ´ âˆ AOB + âˆ BOC + âˆ COD + âˆ AOD = 360^{o}

= 50^{o} + 90^{o} + 70^{o} + x^{o} = 360^{o}

= x^{o} + 210^{o} = 360^{o}

= x^{o} = 360^{o} â€“ 210^{o}

= x^{o} = 150^{o}

**19. In the given figure, âˆ A = 50 ^{o}, CE âˆ¥ BA and âˆ ECD = 60^{o}. Then, âˆ ACB =?**

**(a) 50 ^{o} (b) 60^{o} (c) 70^{o} (d) 80^{o}**

**Solution:-**

(c) 70^{o}

Because,

Here,

= âˆ ACE = âˆ BAC = 50^{o} [Alternate angles]

= âˆ ACB + âˆ ACE + âˆ DCE = 180^{o} [Linear pair]

= âˆ ACB = 180^{o} â€“ (50^{o} + 60^{o})

= âˆ ACB = 180^{0 }â€“ 110^{o}

= âˆ ACB = 70^{o}

## RS Aggarwal Solutions for Class 7 Maths Chapter 17 – Constructions

Chapter 17 – Constructions contains 3 exercises and the RS Aggarwal Solutions available on this page provide solutions to the questions present in the exercises. Now, let us have a look at some of the concepts discussed in this chapter.

- Construction of Parallel Lines
- Construction of Triangles
- SSS Triangle Construction
- SAS Triangle Construction
- ASA Triangle Construction
- RHS Triangle Construction

### Chapter Brief of RS Aggarwal Solutions for Class 7 Maths Chapter 17 – Constructions

RS Aggarwal Solutions for Class 7 Maths Chapter 17 – Constructions refers to a precise way of drawing using two tools, an unmarked straight line and a compass. In this chapter, students are going to learn about the construction of parallel lines and the construction of different types of triangles.