Question 1

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, where c is the hypotenuse and a, b are the legs. Step 2: Substitute: c² = 6² + 8² = 36 + 64 = 100. Step 3: Take the square root: c = √100 = 10 cm. Therefore, the hypotenuse is 10 cm.

Question 2

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 length = 3 × 10 = 30 cm. Therefore, the perimeter is 30 cm.

Question 3

A triangle has a base of 12 cm and a height of 8 cm. What is its area?

Accepted Answer

Step 1: Use the formula for the area of a triangle: Area = (1/2) × base × height. Step 2: Substitute the values: Area = (1/2) × 12 × 8 = (1/2) × 96 = 48 cm². Therefore, the area is 48 cm².

Question 4

The area of a triangle is 24 cm² and its base is 6 cm. What is the height of the triangle?

Accepted Answer

Step 1: Use the area formula: Area = (1/2) × base × height. Step 2: Rearrange to find height: height = (2 × Area) / base. Step 3: Substitute: height = (2 × 24) / 6 = 48 / 6 = 8 cm. Therefore, the height is 8 cm.

Question 5

A triangle has sides of length 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 6

In a triangle, the sum of any two sides must be greater than the third side. If two sides of a triangle are 7 cm and 10 cm, which of the following can be the length of the third side?

Accepted Answer

Step 1: Apply the triangle inequality theorem. For sides a = 7, b = 10, and c (unknown): (1) 7 + 10 > c → c < 17, (2) 7 + c > 10 → c > 3, (3) 10 + c > 7 (always true for positive c). Step 2: Combine conditions: 3 < c < 17. Step 3: Check the options: 5 cm satisfies 3 < 5 < 17. Therefore, the third side can be 5 cm.

Question 7

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: Equilateral triangle perimeter = 3 × 8 = 24 cm. Area of triangle = (√3/4) × 8² = (√3/4) × 64 = 16√3 cm². Step 2: Square with perimeter 24 cm has side = 24/4 = 6 cm. Area of square = 6² = 36 cm². Step 3: Ratio = 16√3 : 36 = 4√3 : 9. Numerically, 16√3 ≈ 27.71 and 36, so ratio ≈ 27.71/36 ≈ 0.77 or 4√3/9. Therefore, the ratio is 4√3 : 9.

Question 8

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

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. Therefore, the area is 84 cm².

Question 9

In triangle ABC, the altitude from vertex A to side BC is 12 cm. If the area of triangle ABC is 96 cm², what is the length of side BC?

Accepted Answer

Step 1: Use the area formula for a triangle: Area = (1/2) × base × height. Step 2: Substitute known values: 96 = (1/2) × BC × 12. Step 3: Simplify: 96 = 6 × BC. Step 4: Solve for BC: BC = 96/6 = 16 cm. Therefore, the length of side BC is 16 cm.

Question 10

A triangle has sides of length 13 cm, 14 cm, and 15 cm. Using Heron's formula, find its area.

Accepted Answer

Step 1: Calculate semi-perimeter s = (13 + 14 + 15)/2 = 42/2 = 21 cm. Step 2: Apply Heron's formula: Area = √[s(s-a)(s-b)(s-c)] = √[21 × 8 × 7 × 6] = √7056 = 84 cm². Therefore, the area is 84 cm².

Question 11

An equilateral triangle has a side length of 8 cm. Find the ratio of its area to its perimeter (in numerical form, without units).

Accepted Answer

Step 1: Area of equilateral triangle = (√3/4) × side² = (√3/4) × 64 = 16√3 cm². Step 2: Perimeter = 3 × 8 = 24 cm. Step 3: Ratio = (16√3)/24 = (2√3)/3. Step 4: Numerically, (2 × 1.732)/3 ≈ 3.464/3 ≈ 1.155. Rounding to two decimal places: 1.15 (or exact form: 2√3/3).

Question 12

In triangle ABC, angle B = 90°, AB = 5 cm, and BC = 12 cm. A line parallel to BC intersects AB at point D such that AD = 2 cm. Find the length of DE, where E is on AC and DE is parallel to BC.

Accepted Answer

Step 1: Since DE is parallel to BC, triangle ADE is similar to triangle ABC. Step 2: The ratio of similarity = AD/AB = 2/5. Step 3: By similar triangles, DE/BC = AD/AB, so DE/12 = 2/5 → DE = (2 × 12)/5 = 24/5 = 4.8 cm. Therefore, DE = 4.8 cm.

Question 13

A triangle with vertices at coordinates A(0, 0), B(8, 0), and C(4, 6) is given. Find its area using the coordinate formula.

Accepted Answer

Step 1: Use the coordinate area formula: Area = (1/2)|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|. Step 2: Substitute A(0,0), B(8,0), C(4,6): Area = (1/2)|0(0 - 6) + 8(6 - 0) + 4(0 - 0)| = (1/2)|0 + 48 + 0| = (1/2) × 48 = 24 cm². Therefore, the area is 24 cm².

Question 14

A triangle has an area of 60 cm² and a base of 15 cm. If the base is increased by 20% and the height is decreased by 20%, what is the new area?

Accepted Answer

Step 1: Original area = 60 cm², base = 15 cm. Using Area = (1/2) × base × height: 60 = (1/2) × 15 × h, so h = 8 cm. Step 2: New base = 15 × 1.2 = 18 cm. New height = 8 × 0.8 = 6.4 cm. Step 3: New area = (1/2) × 18 × 6.4 = (1/2) × 115.2 = 57.6 cm². Therefore, the new area is 57.6 cm².

Question 15

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: 9/DC = 18/24. Step 3: Simplify: 9/DC = 3/4. Step 4: Cross-multiply: 3 × DC = 9 × 4 = 36, so DC = 12 cm.

Question 16

In triangle ABC, the incircle touches side BC at point P, side AC at point Q, and side AB at point R. If AB = 13 cm, BC = 14 cm, and CA = 15 cm, what is the length of AR (in cm)?

Accepted Answer

Step 1: Calculate the semi-perimeter: s = (13 + 14 + 15)/2 = 21 cm. Step 2: By the property of tangent segments from an external point, AR = AQ = s - a, where a is the side opposite to vertex A (i.e., BC). Step 3: AR = s - BC = 21 - 14 = 7 cm.

Question 17

A triangle has sides of length 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 18

In triangle ABC, the median from vertex A to side BC has length 13 cm. If AB = 12 cm and AC = 14 cm, what is the length of BC (in cm)?

Accepted Answer

Step 1: Use the median length formula: m_a² = (2b² + 2c² - a²)/4, where m_a is the median from A, and a, b, c are the sides opposite to vertices A, B, C respectively. Step 2: Here, m_a = 13, b = AC = 14, c = AB = 12, and a = BC (unknown). Step 3: 13² = (2 × 14² + 2 × 12² - BC²)/4. Step 4: 169 = (2 × 196 + 2 × 144 - BC²)/4 = (392 + 288 - BC²)/4 = (680 - BC²)/4. Step 5: 676 = 680 - BC², so BC² = 4, thus BC = 2 cm.

Question 19

A triangle has a base of 12 cm and a height of 8 cm. What is its area?

Accepted Answer

Step 1: Use the formula for the area of a triangle: Area = (1/2) × base × height. Step 2: Substitute the values: Area = (1/2) × 12 × 8 = (1/2) × 96 = 48 cm². Therefore, the area is 48 cm².

Question 20

A triangle has sides of length 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 21

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 length = 3 × 10 = 30 cm. Therefore, the perimeter is 30 cm.

Question 22

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, where c is the hypotenuse and a, b are the legs. Step 2: Substitute: c² = 6² + 8² = 36 + 64 = 100. Step 3: c = √100 = 10 cm. Therefore, the hypotenuse is 10 cm.

Question 23

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

Question 24

A triangle has an area of 24 cm² and a base of 8 cm. What is its height?

Accepted Answer

Step 1: Use the area formula rearranged for height: Area = (1/2) × base × height, so height = (2 × Area) ÷ base. Step 2: Substitute the values: height = (2 × 24) ÷ 8 = 48 ÷ 8 = 6 cm. Therefore, the height is 6 cm.

Question 25

An equilateral triangle has a side length of 8 cm. What is the ratio of its area to its perimeter (in cm)?

Accepted Answer

Step 1: Perimeter of equilateral triangle = 3 × 8 = 24 cm. Step 2: Area of equilateral triangle = (√3/4) × side² = (√3/4) × 64 = 16√3 cm². Step 3: Ratio = Area/Perimeter = 16√3/24 = (2√3)/3 cm. Numerically, this equals approximately 1.155 cm, but the exact answer is (2√3)/3. Therefore, the ratio is (2√3)/3 or approximately 1.15 cm.

Question 26

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 27

An equilateral triangle has a side length of 8 cm. What is the length of its altitude in cm?

Accepted Answer

Step 1: For an equilateral triangle with side a, the altitude h = (√3/2) × a. Step 2: h = (√3/2) × 8 = 4√3 cm. Step 3: 4√3 ≈ 4 × 1.732 = 6.928 ≈ 6.93 cm.

Question 28

A triangle has sides of length 13 cm, 14 cm, and 15 cm. Find its area using Heron's formula.

Accepted Answer

Step 1: Calculate semi-perimeter s = (13 + 14 + 15)/2 = 42/2 = 21 cm. Step 2: Apply Heron's formula: Area = √[s(s-a)(s-b)(s-c)] = √[21 × 8 × 7 × 6] = √7056 = 84 cm². Therefore, the area is 84 cm².

Question 29

A right-angled triangle has legs of 9 cm and 12 cm. A circle is inscribed in this triangle. What is the radius of the inscribed circle in cm?

Accepted Answer

Step 1: For a right triangle with legs a = 9 and b = 12, hypotenuse c = √(81 + 144) = √225 = 15 cm. Step 2: Inradius formula for right triangle: r = (a + b - c)/2 = (9 + 12 - 15)/2 = 6/2 = 3 cm.

Question 30

A right-angled triangle has legs of 20 cm and 21 cm. A circle is inscribed in this triangle. What is the radius of the inscribed circle (in cm)?

Accepted Answer

Step 1: For a right-angled triangle with legs a = 20 and b = 21, the hypotenuse c = √(20² + 21²) = √(400 + 441) = √841 = 29 cm. Step 2: The inradius of a right triangle is given by r = (a + b - c)/2. Step 3: r = (20 + 21 - 29)/2 = 12/2 = 6 cm.

Question 31

A triangle has sides of length 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 32

A triangle has sides of length 13 cm, 14 cm, and 15 cm. If 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.

RRB NTPC Triangles — Area & Properties

RRB NTPC Triangles — Area & Properties — Past Exam Questions

Triangles — Area & Properties— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Triangles — Area & Properties — Practice Questions

60-Second Revision — Triangles — Area & Properties