RS Aggarwal Class 7 Solutions Chapter 17 Constructions CCE Test Paper

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.

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