## RS Aggarwal Class 7 Solutions Chapter 17 Constructions CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 17 Constructions CCE Test Paper.

**Other Exercises**

- RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17A
- RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17B
- RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17C
- RS Aggarwal Solutions Class 7 Chapter 17 Constructions CCE Test Paper

**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.

Hope given RS Aggarwal Solutions Class 7 Chapter 17 Constructions CCE Test Paper are helpful to complete your math homework.

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