In this chapter, we provide RS Aggarwal Solutions for Class 7 Chapter 17 Constructions CCE Test Paper 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 CCE Test Paper Maths pdf, free RS Aggarwal Solutions Class 7 Chapter 17 Constructions CCE Test Paper 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 | CCE Test Paper |

**RS Aggarwal Solutions for Class 7 Chapter 17 Constructions CCE Test Paper Download PDF**

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

Given, ∠ABO = 60°

∠CDO = 40°

⇒ ∠ABO = ∠BOC = 60° [alternate angles]

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

Here, AB || EC

∠BAC = ∠ACE = 70° (alternate angles)

⇒ ∠BCA = 180° – ∠BAC

⇒ ∠BCA = 180°- 120°

⇒ ∠BCA = 60°

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

(i) ∠AOC = ∠BOD = 50° [vertically opposite angles]

(ii) ∠BOC = 180° – 50° (linear pair)

= 130°

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

Here, 3x + 20 + 2x – 10 = 180

⇒ 5x + 10 = 180

⇒ 5x = 170

⇒ x = 34

∠AOC = (3 x 34 + 20)° = (102 + 20)° = 122°

∠BOC = (2 x 34 – 10)° = (68 – 10)° = 58°

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

In ∆ABC, ∠A + ∠B + ∠C = 180°

⇒ 65° + 45° + ∠C= 180°

⇒ ∠C = 180° – 110° = 90°

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

Let x = 2k and y = 3k

2k + 3k = 120° [Exterior angle property]

⇒ 5k = 120°

⇒ k = 24°

x = 2 x 24° = 48° and y = 3 x 24° = 72°

In ∆ABC :

∠A + ∠B + ∠C = 180°

⇒ 48° + 72° + ∠C = 180°

⇒ ∠C = 180°- 120°

⇒ ∠C = 60°

z = 60°

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

Since it is a right triangle, by using the Pythagoras theorem:

Length of the hypotenuse = √(8² + 15²) = √(64 + 225) = √289 = ± 17 cm

The length of the side can not be negative.

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

Given:

∠BAD = ∠DAC …..(i)

To show that ∆ABC is isosceles, we should show that ∠B = ∠C

AD ⊥ BC, ∠ADB = ∠ADC = 90°

∠ADC = ∠ADB

∠BAD + ∠ABD = ∠DAC + ∠ACD (exterior angle property)

∠DAC + ∠ABD = ∠DAC + ∠ ACD [from equation (i)]

∠ABD = ∠ACD

This is because opposite angles of a triangle ∆ABC are equal.

Hence, ∆ABC is an isosceles triangle.

**Mark (✓) against the correct answer in each of the following :****Question 9.****Solution:**

(c) 145°

The supplement of 35° = 180° – 35° = 145°

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

(d) 124

x° + 56° = 180° (linear pair)

⇒ x = 180° – 56°

⇒ x = 124

x = 124°

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

(c) 65°

∠ACD = 125°

∠ACD = ∠CAB + ∠ABC (the exterior angles are equal to the sum of its interior opposite angles)

∠ABC = 125° – 60° = 65°

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

(c) 105°

∠A + ∠B + ∠C = 180°

⇒ ∠A = 180° – (40° + 35°)

⇒ ∠A = 105°

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

(c) 60°

Given:

2∠A = 3∠B

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

(b) 55°

In ∆ABC :

A + B + C = 180° …(i)

Given, A – B = 33°

A = 33° + B …(ii)

B – C = 18°

C = B + 18° …(iii)

Putting the values of A and B in equation (i):

⇒ B + 33° + B + B – 18° = 180°

⇒ 3B = 180°

⇒ B = 55°

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

(b) 3√2 cm

Here, AB = AC

In right angled isosceles triangle:

BC² = AB² + AC²

⇒ BC² = AB² + AB²

⇒ BC² = 2AB²

⇒ 36 = 2AB²

⇒ AB² = 18

⇒ AB = √18

⇒ AB = 3√2

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

(i) The sum of the angles of a triangle is 180°.

(ii) The sum of any two sides of a triangle is always greater than the third side.

(iii) In ∆ABC, if ∠A = 90°, then BC² = (AB²) + (BC²)

(iv) In ∆ABC :

AB = AC

AD ⊥ BC

Then, BD = DC

This is because in an isosceles triangle, the perpendicular dropped from the vertex joining the equal sides, bisects the base.

(v) In the given figure, side BC of ∆ABC has produced to D and CE || BA.

If ∠ABC = 50°, then ∠ACE = 50°

AB || CE

∠BAC = ∠ACE = 50° (alternate angles)

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

(i) True

(ii) True

(iii) False. Each acute angle of an isosceles right triangle measures 45°.

(iv) True.

**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.