Question 1

An equilateral triangle has a side length of 10 cm. What is its perimeter?

Accepted Answer

Step 1: An equilateral triangle has all three sides equal. Step 2: Perimeter = 3 × side = 3 × 10 = 30 cm. Therefore, the answer is 30 cm.

Question 2

A triangle has sides of 5 cm, 12 cm, and 13 cm. What is its area using Heron's formula?

Accepted Answer

Step 1: Calculate the semi-perimeter: s = (5 + 12 + 13) ÷ 2 = 30 ÷ 2 = 15 cm. Step 2: Apply Heron's formula: Area = √[s(s-a)(s-b)(s-c)] = √[15 × 10 × 3 × 2] = √900 = 30 cm². Therefore, the area is 30 cm².

Question 3

A triangle has a base of 20 cm. If its area is 150 cm², what is the height of the triangle?

Accepted Answer

Step 1: Use the area formula: Area = (1/2) × base × height. Step 2: Rearrange for height: height = (2 × Area) ÷ base. Step 3: height = (2 × 150) ÷ 20 = 300 ÷ 20 = 15 cm. Therefore, the height is 15 cm.

Question 4

A triangle has a base of 16 cm and a height of 12 cm. What is its area?

Accepted Answer

Step 1: Use the formula for area of a triangle: Area = (1/2) × base × height. Step 2: Area = (1/2) × 16 × 12 = (1/2) × 192 = 96 cm². Therefore, the answer is 96 cm².

Question 5

A right-angled triangle has legs of length 6 cm and 8 cm. What is the length of its hypotenuse?

Accepted Answer

Step 1: Use the Pythagorean theorem: c² = a² + b². Step 2: c² = 6² + 8² = 36 + 64 = 100. Step 3: c = √100 = 10 cm. Therefore, the hypotenuse is 10 cm.

Question 6

In a triangle, the sum of two angles is 110°. What is the measure of the third angle?

Accepted Answer

Step 1: Recall that the sum of all angles in a triangle is 180°. Step 2: If two angles sum to 110°, then the third angle = 180° - 110° = 70°. Therefore, the third angle is 70°.

Question 7

In triangle ABC, the altitude from vertex A to side BC is 12 cm. If the area of triangle ABC is 90 cm², what is the length of side BC in cm?

Accepted Answer

Step 1: Use the area formula for a triangle: Area = (1/2) × base × height. Step 2: Substitute known values: 90 = (1/2) × BC × 12. Step 3: Simplify: 90 = 6 × BC. Step 4: Solve: BC = 90/6 = 15 cm.

Question 8

A triangle has sides of length 13 cm, 14 cm, and 15 cm. What is its area in cm²?

Accepted Answer

Step 1: Use Heron's formula. Semi-perimeter s = (13 + 14 + 15)/2 = 21 cm. Step 2: Area = √[s(s-a)(s-b)(s-c)] = √[21 × 8 × 7 × 6] = √7056 = 84 cm².

Question 9

An equilateral triangle has a side length of 8 cm. What is the ratio of its area to the area of a square with the same perimeter?

Accepted Answer

Step 1: Triangle perimeter = 3 × 8 = 24 cm. Square perimeter = 24 cm, so square side = 6 cm. Step 2: Triangle area = (√3/4) × 8² = 16√3 cm². Step 3: Square area = 6² = 36 cm². Step 4: Ratio = 16√3 : 36 = 4√3 : 9 ≈ 6.93 : 9 or simplified as 4√3 : 9.

Question 10

Two similar triangles have corresponding sides in the ratio 3:5. If the area of the smaller triangle is 36 cm², what is the area of the larger triangle in cm²?

Accepted Answer

Step 1: For similar triangles, the ratio of areas equals the square of the ratio of corresponding sides. Step 2: Ratio of sides = 3:5, so ratio of areas = 3²:5² = 9:25. Step 3: If smaller area is 36 cm², then 36/A = 9/25 ⟹ A = 36 × 25/9 = 100 cm².

Question 11

In an isosceles triangle, the two equal sides are each 25 cm, and the base is 30 cm. A line parallel to the base cuts the triangle such that the area of the smaller triangle (at the top) is 1/4 of the original triangle's area. At what distance from the apex is this parallel line drawn (in cm)?

Accepted Answer

Step 1: Find the height of the original triangle. Using the formula for isosceles triangle: h = √(25² - 15²) = √(625 - 225) = √400 = 20 cm. Step 2: If the smaller triangle has area 1/4 of the original, then its linear dimensions are √(1/4) = 1/2 of the original. Step 3: The height of the smaller triangle = (1/2) × 20 = 10 cm. Step 4: The parallel line is at distance 10 cm from the apex.

Question 12

A triangle has sides 13 cm, 14 cm, and 15 cm. A perpendicular is drawn from the vertex opposite the 14 cm side to that side. What is the length of this perpendicular (in cm)?

Accepted Answer

Step 1: Use Heron's formula to find the area. Semi-perimeter s = (13 + 14 + 15)/2 = 21 cm. Step 2: Area = √[21(21-13)(21-14)(21-15)] = √[21 × 8 × 7 × 6] = √7056 = 84 cm². Step 3: Using Area = (1/2) × base × height, we have 84 = (1/2) × 14 × h, so h = 12 cm.

Question 13

In triangle ABC, the angle bisector from vertex A meets side BC at point D. If AB = 18 cm, AC = 24 cm, and BD = 9 cm, what is the length of DC (in cm)?

Accepted Answer

Step 1: Apply the Angle Bisector Theorem: BD/DC = AB/AC. Step 2: Substitute values: 9/DC = 18/24. Step 3: Simplify: 9/DC = 3/4. Step 4: Cross-multiply: 3 × DC = 36, so DC = 12 cm.

Question 14

In triangle PQR, the sides are PQ = 26 cm, QR = 28 cm, and RP = 30 cm. The incircle of the triangle touches side QR at point X. What is the length of QX (in cm)?

Accepted Answer

Step 1: Use the property that tangent segments from an external point to a circle are equal. Let the incircle touch PQ at Y, QR at X, and RP at Z. Step 2: Then QY = QX, PY = PZ, and RX = RZ. Step 3: We have QY + PY = PQ = 26, so QX + PZ = 26. Also, PZ + RZ = RP = 30, so PZ + RX = 30. And RX + QX = QR = 28, so RZ + QX = 28. Step 4: From the three equations: QX + PZ = 26, PZ + RX = 30, RX + QX = 28. Adding all three: 2(QX + PZ + RX) = 84, so QX + PZ + RX = 42. Step 5: The semi-perimeter s = (26 + 28 + 30)/2 = 42 cm. Step 6: Using the formula QX = s - PQ = 42 - 26 = 16 cm.

IBPS Clerk Triangles — Area & Properties

Triangles — Area & Properties— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Triangles — Area & Properties — Practice Questions

60-Second Revision — Triangles — Area & Properties