RS Aggarwal Solutions for Class 7 Maths Exercise 17C Chapter 17 Constructions in simple PDF are available here. This exercise is of completely objective type questions. It includes all topics present in the RS Aggarwal Solutions for Class 7 Maths Chapter 17 Constructions. After solving these objective type questions, students can absorb more knowledge from the chapter and consequently, score high-grades in Maths.

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**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}