In this chapter, we provide RS Aggarwal Solutions for Class 7 Chapter 17 Constructions Ex 17C for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 7 Chapter 17 Constructions Ex 17C Maths pdf, free RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17C Maths book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 7 |

Subject | Maths |

Chapter | Chapter 17 |

Chapter Name | Constructions |

Exercise | 17C |

**RS Aggarwal Solutions for Class 7 Chapter 17 Constructions Ex 17C Download PDF**

**Question 1.****Solution:**

(c) Supplement of 45° is 135°

135°+ 45° = 180°

**Question 2.****Solution:**

(b) Complement of 80° is 10°

10° + 80° = 90°

**Question 3.****Solution:**

(b) The angle is its own complement.

The measure of the angles will be 45° (45° + 45° = 90°)

**Question 4.****Solution:**

(a) The angle is one-fifth of its supplement

Let angle be x, then

x + 5x = 180°

⇒ 6x = 180°

⇒ x = 30°

Angle is 30°

**Question 5.****Solution:**

(b) Let angle is x

Then its complement angle=x-24° But x + x- 24° = 90°

⇒ 2x = 90° + 24° = 114°

⇒ x = 57°

The required angle is 57°

**Question 6.****Solution:**

(b) Let required angle = x

Then its supplement angle = x + 32

But x + x + 32° = 180°

⇒ 2x = 180° – 32 = 148°

⇒ x = 74°

Required angle = 74°

**Question 7.****Solution:**

(c) Two supplementary angle are in the ratio = 3 : 2

Let first angle = 3x

Second angle = 2x

But 3x + 2x = 180°

⇒ 5x = 180°

⇒ x = 36°

Smaller angle = 2x = 2 x 36° = 72°

**Question 8.****Solution:**

(b) In the figure ∠BOC = 132°

But ∠AOC + ∠BOC =180° (Linear pair)

⇒ ∠AOC + 132° = 180°

⇒ ∠AOC = 180° – 132° = 48°

**Question 9.****Solution:**

(c) In the figure, ∠AOC = 68°

But ∠AOC + ∠BOC = 180° (Linear pair)

⇒ 68° + x = 180°

⇒ x = 180° – 68° = 112°

**Question 10.****Solution:**

(b) In the figure,

AOB is a straight line

∠AOC + ∠BOC = 180° (Linear pair)

⇒ 2x – 10° + 3x + 15° = 180°

⇒ 5x = 180° + 10° – 15° = 175°

⇒ x = 35°

x = 35

**Question 11.****Solution:**

(d) In the figure,

AOB is a straight line

∠AOC + ∠COD + ∠DOB = 180°

⇒ 55° + x + 45° = 180°

⇒ x + 100° = 180°

⇒ x = 180° – 100° = 80°

**Question 12.****Solution:**

(a) AOB is a straight line

x + y = 180°

But 4x = 5y

**Question 13.****Solution:**

(b) AB and CD intersect each other at O and ∠AOC = 50°

∠BOD = ∠AOC = 50° (Vertically opposite angles)

**Question 14.****Solution:**

(a) AOB is a straight line

∠AOC + ∠COD + ∠DOB = 180°

⇒ 3x – 8° + 50° + x + 10° = 180°

⇒ 4x = 180° + 8° – 50° – 10°

⇒ 4x = 128°

⇒ x = 32°

**Question 15.****Solution:**

(b) In ∆ABC, side BC is produced to D

∠ACD = 132° and ∠A = 54°

Ext. ∠ACD = ∠A + ∠B

⇒ 132° = 54° + ∠B

⇒ ∠B = 132° – 54° = 78°

**Question 16.****Solution:**

(c) In ∆ABC,

Side BC is produced to D

∠A = 45°, ∠B = 55°

Ext. ∠ACD = ∠A + ∠B = 45° + 55° = 100°

**Question 17.****Solution:**

(b) In ∆ABC, side BC is produced to D

∠ABC = 70° and ∠ACD = 120°

Ext. ∠ACD = ∠BAC + ∠ABC

⇒ 120° = ∠BAC + 70°

⇒ ∠BAC = 120° – 70° = 50°

**Question 18.****Solution:**

(c) In the figure,

∠AOB = 50°, ∠BOC = 90°

∠COD = 70°, ∠AOD = x.

But ∠AOB + ∠BOC + ∠COD + ∠DOA = 360° (Angles at a point)

⇒ 50° + 90° + 70° + x = 360°

⇒ 210 + x = 360°

⇒ x = 360° – 210°

⇒ x = 150°

**Question 19.****Solution:**

(c) In the figure,

Side BC of ∆ABC is produced to D

CE || BA is drawn

∠A = 50° and ∠ECD = 60°

AB || CE

∠ABC = ∠ECD (corresponding angle) = 60°

But in ∆ABC,

∠A + ∠B + ∠ACB = 180° (Angles of a triangles)

⇒ 50° + 60° + ∠ACB = 180°

⇒ ∠ACB = 180° – 50° – 60° = 70°

**Question 20.****Solution:**

(b) In ∆ABC,

∠A = 65°, ∠C = 85°

But ∠A + ∠B + ∠C = 180° (Angles of a triangle)

⇒ 65° +∠B+ 85° = 180°

⇒ 150° + ∠B = 180°

⇒ ∠B = 180° – 150° = 30°

**Question 21.****Solution:**

(d) Sum of angles of a triangle = 180°

**Question 22.****Solution:**

(c) Sum of angles of a quadrilateral = 360°

**Question 23.****Solution:**

(b) In the figure, AB || CD

∠OAB = 150°, ∠OCD = 120°

From O, draw OE || AB or CD

AB || DE

∠OAB + ∠AOE =180°

⇒ 150° + ∠AOE = 180°

⇒ ∠AOE = 180° – 150° = 30°

Similarly DE || CD

∠EOC + ∠OCD = 180°

⇒ ∠EOC + 120° = 180°

⇒ ∠EOC = 180° – 120° = 60°

Now ∠AOC = ∠AOE + ∠EOC = 30° + 60° = 90°

**Question 24.**

**Solution:**

(a) In the given figure,

PQ || RS,

∠PAB = 60° and ∠ACS = 100°

PQ || RS

∠ABC = ∠PAB (alternate angles) = 60°

But Ext. ∠ACS = ∠BAC + ∠ABC

⇒ 100° = ∠BAC + 60°

⇒ ∠BAC = 100° – 60° = 40°

**Question 25.****Solution:**

(c) In the figure, AB || CD || EF

∠ABG =110° and ∠GCD = 100°

∠BGC = x°

AB || EF

∠ABG + ∠BGE = 180°

⇒ 110° + ∠BGE = 180°

⇒ ∠BGE = 180° – 110° = 70°

Similarly CD || EF

∠GCD + ∠CGF = 180°

⇒ 100° + ∠CGF = 180°

⇒ ∠CGF = 180° – 100° = 80°

But ∠BGE + ∠BGC + ∠CGF = 180°

⇒ 70° + x + 80° = 180°

⇒ 150° + x = 180°

⇒ x = 180° – 150° = 30°

**Question 26.****Solution:**

(d) Sum of any two sides of a triangle is always greater than the third side

**Question 27.****Solution:**

(d) The diagonals of a rhombus always bisect each other at right angles.

**Question 28.****Solution:**

(c) In ∆ABC, ∠B = 90°

AB = 5 cm and AC = 13 cm

But AC² = AB² + BC² (By Pythagoras Theorem)

⇒ (13)² = (5)² + BC²

⇒ 169 = 25 + BC2

⇒ BC² = 169 – 25 = 144 = (12)²

BC = 12 cm

**Question 29.****Solution:**

(c) In ∆ABC, ∠B = 37°, ∠G = 29°

But ∠A + ∠B + ∠C = 180° (angles of a triangle)

⇒ ∠A + 37° + 29° = 180°

⇒ ∠A + 66° = 180°

⇒ ∠A = 180° – 66° = 114°

**Question 30.****Solution:**

(c) The ratio of angles of a triangle is 2 : 3 : 7

But sum of angles of a triangle = 180°

**Question 31.****Solution:**

In ∆ABC,

Let 2∠A = 3∠B = 6∠C = x

**Question 32.****Solution:**

(a) In ∆ABC,

∠A + ∠B = 65°, ∠B + ∠C = 140°

∠A = 65°

∠C = 140° – ∠B

But ∠A + ∠B + ∠C = 180° (Angles of a triangle)

⇒ 65° – ∠B + 140° – ∠B + ∠B = 180°

⇒ 205° – ∠B = 180°

⇒ ∠B = 205° – 180° = 25°

**Question 33.****Solution:**

(b) In ∆ABC, ∠A – ∠B = 33°

and ∠B – ∠C = 18°

∠A = 33° + ∠B and ∠C = ∠B – 18°

But ∠A + ∠B + ∠C = 180°

⇒ 33° + ∠B + ∠B + ∠B – 18° = 180°

⇒ 3∠B = 180° – 33° + 18° = 165°

⇒ ∠B = 55°

**Question 34.****Solution:**

(c) In ∆ABC

∠A + ∠B + ∠C= 180° (Sum of angles of a triangle)

But angles are (3x)°, (2x – 7)° and (4x – 11)°

3x + (2x – 7) + (4x – 11)° = 180°

⇒ 3x + 2x – 7 + 4x – 11° = 180°

⇒ 9x – 18° = 180°

⇒ 9x = 180° + 18° = 198°

⇒ x = 22°

**Question 35.****Solution:**

(c) ∆ABC is a right angled, ∠A = 90°

AB = 24 cm, AC = 7 cm

but BC² = AB² + AC²

⇒ BC² = (24)² + (7)² = 576 + 49 = 625 = (25)²

BC = 25 cm

**Question 36.****Solution:**

(b) Let AB is a ladder and A is the window

BC = 15 m, AC = 20 m

Now in right ∆ABC

AB² = BC² + AC² = (15)² + (20)² = 225 + 400 = 625 = (25)²

AB = 25 m

Length of ladder = 25 m

**Question 37.****Solution:**

(a) Let AB and CD are two poles such that

AB = 6 m, CD = 11 m

and distance between two poles BD = 12m

From A, draw AE || BD

AE = BD = 12m

CE = CD – ED = 11 – 6 = 5 m

Now in right ∆AEC

AC² = AE² + CE² = (12)² + (5)² = 144 + 25 = 169 = (13)²

AC = 13 m

Distance between tops of poles = 13 m

**Question 38.****Solution:**

(d) ∆ABC is an isosceles triangle

∠C = 90°,

AC = 5 cm

BC = AC = 5 cm

In right ∆ABC

AB² = AC² + BC² = (5)² + (5)² = 25 + 25 = 50 = 2 x 25

AB = √(2 x 25) = 5√2 cm

**All Chapter RS Aggarwal Solutions For Class 7 Maths**

—————————————————————————–**All Subject NCERT Exemplar Problems Solutions For Class** **7**

**All Subject NCERT Solutions For Class 7**

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rsaggarwalsolutions.in to your friends.