Olympiad Test: Triangle - 1 - Free MCQ with solutions Class 7 Mathematics

Let the measure of each unknown angle be x°. We know that the sum of the angles of a triangle is 180°. 
∴ x + x + 70 = 180° 
⇒ 2x = (180° – 70°)
⇒ 2x = 110°
⇒ x = 55°
Hence each unknown angle is 55°.

We know that the sum of the angles of a triangle is 180°.

Let the third angle be x°.
We know that the sum of the angles of a triangle is 180°. 
∴ 2x + 2x + x = 180° 
⇒ 5x = 180° 
⇒ x = 36°
So, the angles are (2 × 36)°, (2 × 36)° and (x)°
Hence, the angles of the triangles are 72°, 72°, 36°. 

Let the measure of the given angles of the triangle be (4x)°, (3x)° and (2x)° respectively.
4x + 3x + 2x = 180°
⇒ 9x = 180°

So, the angles measure (4×20)°, (3 × 20)° and (2 × 20)° 
Hence, the angles of the triangle are 80°, 60°, 40°.

Let O be the initial position of the man. Let he cover OA = 24m due east and then AB = 10m due north.

Finally, he reaches the point B. Join OB.
OB² = (OA² + OB²) = {(24)² + (10)²} m²
= (576 + 100) m² = 676 m²

Hence, the man is at a distance at 26m from his initial position. 

(i)
Here a  = 8 cm, b = 5cm  and c = 10c,
The largest side is c = 10cm.
a² + b² = {(8)² + (5)²} cm²
= m (64 + 25) cm² = 89 cm² ≠ (10)² cm²
a² + b² ≠ c²
∴ Given triangle is not right angled.
(ii) Here a = 7 cm, b = 24 cm and c = 25 cm
The largest side is c = 25 cm
a² + b² = {(7)² + (24)²} cm² = (49 + 576) cm²
= 625 cm² = (25 cm)² = c²
⇒ a² + b² = c²
Given triangle is right angled. 

Let AB and CD be the given poles such that
AB = 9 m, CD = 14 m and AC = 12 m. Join BD.
From B, draw BL ⊥ CD.
DL = (CD – CL) = (CD – AB)
= (14 – 9) m = 5 m

BL = AC = 12 m
Now, in right ∆BLD, by Pythagoras theorem.
We have
BD² = BL² + DL²  = {(12)² + (5)²}m²
= (144 + 25) m² = 169 m²

We know that the sum of the angles of  a triangle is 180°.

From 2∠A = 3∠B, ∠A = (3/2)·∠B. Since ∠B = 6·∠C, we have ∠C = ∠B/6. The angle‐sum in ΔABC gives:
(3/2)·∠B + ∠B + (∠B/6) = (3/2 + 1 + 1/6)·∠B = (8/3)·∠B = 180°
Hence ∠B = 180° × (3/8) = 67.5°.