Triangles & Angle Properties MCQ Test — Class 7 CBSE

Solution:

An acute angle is an angle that measures less than 90°.

An acute-angled triangle is defined as a triangle where all three of its interior angles are acute.

Therefore, in an acute-angled triangle, every angle must be less than 90°.

Solution:

An obtuse angle is an angle greater than 90°.

If a triangle had two obtuse angles, their sum would be greater than 90° + 90° = 180°.

However, the angle sum property of a triangle states that the sum of all three interior angles must be exactly 180°.

Therefore, it is impossible for a triangle to have two obtuse angles, as their sum alone would exceed the total allowed sum for the triangle.

Solution:

The sum of the interior angles in any triangle is 180°. — ∠P + ∠Q + ∠R = 180°

Substitute the given values: 45° + 65° + ∠R = 180°.

Calculate the sum of the known angles: 110° + ∠R = 180°.

Solve for ∠R: ∠R = 180° - 110° = 70°.

Solution:

The exterior angle property states that an exterior angle of a triangle is equal to the sum of its two interior opposite angles.

Let the exterior angle be E = 120° and one interior opposite angle be A = 70°.

Let the other interior opposite angle be B. So, E = A + B. — 120° = 70° + B

Solve for B: B = 120° - 70° = 50°.

Solution:

An isosceles triangle is defined as a triangle with at least two sides of equal length. In most contexts, it means exactly two sides are equal.

If all three sides were equal, it would be an equilateral triangle (which is a special type of isosceles triangle).

If all three angles were equal, it would also be an equilateral triangle.

An isosceles triangle can be acute-angled, right-angled, or obtuse-angled; it does not necessarily have a right angle.

Solution:

First, check the angle sum: 30° + 60° + 90° = 180°, so it is a valid triangle.

A triangle with one angle exactly 90° is classified as a right-angled triangle.

An acute-angled triangle requires all three angles to be less than 90°. Since one angle is 90°, it cannot be acute-angled.

For a triangle to be isosceles, at least two of its angles must be equal. Since 30°, 60°, and 90° are all different, it is not an isosceles triangle (it's a scalene triangle).

Rahul's primary error is classifying it as acute-angled when the presence of a 90° angle makes it a right-angled triangle.

Solution:

According to the exterior angle property, the exterior angle at Z is equal to the sum of the interior opposite angles ∠X and ∠Y.

So, Exterior angle at Z = ∠X + ∠Y.

Substitute the given values: 115° = 70° + ∠Y.

Solve for ∠Y: ∠Y = 115° - 70° = 45°.

Solution:

Let the angles of the triangle be 2x, 3x, and 4x.

The sum of the interior angles of a triangle is 180°. — 2x + 3x + 4x = 180°

Combine like terms: 9x = 180°.

Solve for x: x = 180° / 9 = 20°.

The angles are: 2 × 20° = 40°, 3 × 20° = 60°, and 4 × 20° = 80°. The largest angle is 80°.

Solution:

The exterior angle at vertex C is equal to the sum of the two interior opposite angles, ∠A and ∠B.

Exterior angle at C = ∠A + ∠B.

Substitute the given values: Exterior angle at C = 50° + 60°.

Calculate the sum: Exterior angle at C = 110°.

Solution:

Apply the triangle inequality theorem (a + b > c) to each student's set of stick lengths.

For Anil (3 cm, 4 cm, 8 cm): Check 3 + 4 > 8. This simplifies to 7 > 8, which is False. So, Anil cannot form a triangle.

For Bala (5 cm, 5 cm, 5 cm): Check 5 + 5 > 5. This simplifies to 10 > 5, which is True. Bala can form a triangle (an equilateral one).

For Chitra (6 cm, 7 cm, 10 cm): Check 6 + 7 > 10 (13 > 10, True); 6 + 10 > 7 (16 > 7, True); 7 + 10 > 6 (17 > 6, True). Chitra can form a triangle.

Therefore, only Bala and Chitra can form a triangle with their sticks.