Question 1

An equilateral triangle has a side length of 10 cm. What is the length of its altitude (height)?

Accepted Answer

Step 1: For an equilateral triangle with side 'a', the altitude formula is: h = (√3/2) × a. Step 2: Substitute a = 10: h = (√3/2) × 10 = 5√3 cm. Step 3: Numerically, 5√3 ≈ 5 × 1.732 = 8.66 cm, but the exact answer is 5√3 cm. Therefore, the altitude is 5√3 cm.

Question 2

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: Substitute values: Area = (1/2) × 16 × 12 = (1/2) × 192 = 96 cm². Therefore, the area is 96 cm².

Question 3

The area of a triangle is 60 cm² and its base is 15 cm. What is the height of the triangle?

Accepted Answer

Step 1: Use the triangle area formula: Area = (1/2) × base × height. Step 2: Rearrange to find height: height = (2 × Area) / base. Step 3: Substitute values: height = (2 × 60) / 15 = 120 / 15 = 8 cm. Therefore, the height is 8 cm.

Question 4

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 5

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 6

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 7

The base of a triangle is 14 cm and its height is 10 cm. What is the area of the triangle?

Accepted Answer

Step 1: Recall the formula for area of a triangle = (1/2) × base × height. Step 2: Substitute the values: Area = (1/2) × 14 × 10 = (1/2) × 140 = 70 cm². Therefore, the answer is 70 cm².

Question 8

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 9

A right-angled triangle has legs of length 9 cm and 12 cm. What is the length of its hypotenuse?

Accepted Answer

Step 1: Use the Pythagorean theorem: c² = a² + b², where a and b are the legs and c is the hypotenuse. Step 2: Substitute values: c² = 9² + 12² = 81 + 144 = 225. Step 3: Take the square root: c = √225 = 15 cm. Therefore, the hypotenuse is 15 cm.

Question 10

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 11

In triangle ABC, the median from A to side BC has length 10 cm. If the median from B to side AC has length 12 cm, and these two medians intersect at the centroid G, what is the distance AG?

Accepted Answer

Step 1: The centroid divides each median in the ratio 2:1 from the vertex. Step 2: For the median from A with length 10 cm, the distance from A to the centroid G is (2/3) × 10 = 20/3 cm. Step 3: Therefore, AG = 20/3 ≈ 6.67 cm. However, checking the options, the exact value is 20/3 cm or approximately 6.67 cm. The closest standard answer is 20/3 cm.

Question 12

The sides of a triangle are 13 cm, 14 cm, and 15 cm. What is the area of the triangle (in cm²)?

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 × (21−13) × (21−14) × (21−15)] = √[21 × 8 × 7 × 6] = √[7056] = 84 cm². Therefore, the answer is 84 cm².

Question 13

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 14

A triangle has vertices at A(0, 0), B(8, 0), and C(4, 6). What is the area of the triangle?

Accepted Answer

Step 1: Use the coordinate formula for area: 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 15

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?

Accepted Answer

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

Question 16

Triangle ABC has an area of 60 cm². Point D is on side AB such that AD:DB = 3:2. Point E is on side AC such that AE:EC = 3:2. What is the area of triangle ADE?

Accepted Answer

Step 1: Since D divides AB in ratio 3:2, AD/AB = 3/5. Since E divides AC in ratio 3:2, AE/AC = 3/5. Step 2: Triangles ADE and ABC share angle A. The area ratio is (AD/AB) × (AE/AC) = (3/5) × (3/5) = 9/25. Step 3: Area of ADE = (9/25) × 60 = 21.6 cm². Therefore, the area is 21.6 cm².

Question 17

In triangle ABC, the altitudes from vertices A, B, and C are 12 cm, 15 cm, and 20 cm respectively. If the area of the triangle is 60 cm², find the length of the side opposite to vertex A (i.e., side BC).

Accepted Answer

Step 1: Recall that Area = (1/2) × base × height. If h_a is the altitude from vertex A to side BC, then Area = (1/2) × BC × h_a. Step 2: Given Area = 60 cm² and h_a = 12 cm, we have: 60 = (1/2) × BC × 12. Step 3: Solving for BC: 60 = 6 × BC, so BC = 10 cm. Step 4: Verification using other altitudes: If h_b = 15 cm (altitude to side AC), then 60 = (1/2) × AC × 15, giving AC = 8 cm. If h_c = 20 cm (altitude to side AB), then 60 = (1/2) × AB × 20, giving AB = 6 cm. Step 5: All altitudes are consistent with area = 60 cm², confirming BC = 10 cm.

Question 18

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 19

A triangle has vertices at A(0, 0), B(8, 0), and C(3, 6). A line parallel to AB passes through the centroid of triangle ABC and intersects AC at point P and BC at point Q. Find the length of PQ (in units).

Accepted Answer

Step 1: The centroid G = ((0+8+3)/3, (0+0+6)/3) = (11/3, 2). Step 2: Since AB lies on the x-axis (y = 0), a line parallel to AB is horizontal: y = 2. Step 3: Line AC has slope 6/3 = 2, equation: y = 2x. Intersection with y = 2: 2 = 2x → x = 1, so P = (1, 2). Step 4: Line BC connects B(8, 0) and C(3, 6). Slope = (6-0)/(3-8) = -6/5. Equation: y - 0 = (-6/5)(x - 8) → y = (-6/5)x + 48/5. Intersection with y = 2: 2 = (-6/5)x + 48/5 → 10 = -6x + 48 → 6x = 38 → x = 19/3, so Q = (19/3, 2). Step 5: PQ = |19/3 - 1| = |19/3 - 3/3| = 16/3 units.

Question 20

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. Find 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 21

In triangle ABC, AB = 10 cm, AC = 15 cm, and angle BAC = 60°. The area of triangle ABC is k√3 cm². Find k.

Accepted Answer

Step 1: Use the formula Area = (1/2) × AB × AC × sin(∠BAC). Step 2: Area = (1/2) × 10 × 15 × sin(60°) = (1/2) × 10 × 15 × (√3/2) = (150√3)/4 = (75√3)/2 cm². Step 3: Comparing with k√3, we have k√3 = (75√3)/2, so k = 75/2 = 37.5.

Question 22

In triangle PQR, the sides are PQ = 7 cm, QR = 8 cm, and PR = 9 cm. The altitude from Q to side PR has length h. Find h² (in cm²).

Accepted Answer

Step 1: Use Heron's formula. s = (7 + 8 + 9)/2 = 12 cm. Step 2: Area = √[12(12-7)(12-8)(12-9)] = √[12 × 5 × 4 × 3] = √720 = 12√5 cm². Step 3: Area = (1/2) × base × height = (1/2) × 9 × h, so 12√5 = 4.5h, giving h = (12√5)/4.5 = (8√5)/3 cm. Step 4: h² = (64 × 5)/9 = 320/9 ≈ 35.56 cm². [Exact: 320/9]

Question 23

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

Accepted Answer

Step 1: The hypotenuse = √(9² + 12²) = √(81 + 144) = √225 = 15 cm. Step 2: For a right-angled triangle, the inradius r = (a + b - c)/2, where a and b are legs and c is the hypotenuse. Step 3: r = (9 + 12 - 15)/2 = 6/2 = 3 cm.

Question 24

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, which states that the angle bisector divides the opposite side in the ratio of the adjacent sides. Step 2: BD/DC = AB/AC, so 9/DC = 18/24. Step 3: 9/DC = 3/4, therefore DC = 9 × 4/3 = 12 cm.

Question 25

The area of a triangle is 60 cm² and its base is 15 cm. What is the height of the triangle?

Accepted Answer

Step 1: Use the formula for the area of a triangle: Area = (1/2) × base × height. Step 2: Rearrange to find height: height = (2 × Area) ÷ base. Step 3: Substitute the values: height = (2 × 60) ÷ 15 = 120 ÷ 15 = 8 cm. Therefore, the height is 8 cm.

Question 26

An isosceles triangle has two equal sides of 13 cm each and a base of 10 cm. What is its perimeter?

Accepted Answer

Step 1: In an isosceles triangle, two sides are equal. Step 2: Perimeter = side₁ + side₂ + base = 13 + 13 + 10 = 36 cm. Therefore, the perimeter is 36 cm.

Question 27

A triangle has sides of length 5 cm, 12 cm, and 13 cm. What is its semi-perimeter?

Accepted Answer

Step 1: Calculate the perimeter: P = 5 + 12 + 13 = 30 cm. Step 2: Semi-perimeter (s) = P / 2 = 30 / 2 = 15 cm. Therefore, the semi-perimeter is 15 cm.

Question 28

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

Question 29

The area of a triangle is 54 cm² and its base is 12 cm. What is its height?

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 × 54) / 12 = 108 / 12 = 9 cm. Therefore, the height is 9 cm.

Question 30

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 answer is 30 cm.

Question 31

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 32

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 33

In triangle ABC, the altitude from vertex A to side BC is 12 cm. If the area of triangle ABC is 90 cm², and a line parallel to BC intersects AB at point P and AC at point Q such that AP:PB = 2:1, then the area of triangle APQ is:

Accepted Answer

Step 1: Find BC using area formula. Area = (1/2) × base × height, so 90 = (1/2) × BC × 12, giving BC = 15 cm. Step 2: Since PQ is parallel to BC and AP:PB = 2:1, we have AP:AB = 2:3. Step 3: When a line parallel to the base divides two sides in ratio m:n, the triangle formed has area in ratio m²:(m+n)². Here, triangle APQ has area = (2/3)² × 90 = (4/9) × 90 = 40 cm².

Question 34

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 35

The area of an equilateral triangle is 36√3 cm². Find the length of its side.

Accepted Answer

Step 1: The formula for area of equilateral triangle with side a is: Area = (√3/4) × a². Step 2: Set up equation: (√3/4) × a² = 36√3. Step 3: Divide both sides by √3: a²/4 = 36. Step 4: Multiply by 4: a² = 144. Step 5: Take square root: a = 12 cm. Therefore, the side length is 12 cm.

Question 36

A triangle has vertices at A(0, 0), B(8, 0), and C(4, 6). Find the area of the triangle using the coordinate formula.

Accepted Answer

Question 37

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

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. Therefore, the altitude is 4√3 cm.

Question 38

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 39

In triangle ABC, the sides are in the ratio 3:4:5. If the area of the triangle is 96 cm², find the length of the longest side (in cm).

Accepted Answer

Step 1: Let the sides be 3k, 4k, and 5k. Since 3² + 4² = 5², this is a right triangle with legs 3k and 4k. Step 2: Area = (1/2) × 3k × 4k = 6k² = 96. Step 3: k² = 16, so k = 4. Step 4: The longest side = 5k = 5 × 4 = 20 cm.

Question 40

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, meeting it at point P. Find 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. Area = √[21(21-13)(21-14)(21-15)] = √[21 × 8 × 7 × 6] = √7056 = 84 cm². Step 2: Using Area = (1/2) × base × height, where base = 14 cm. 84 = (1/2) × 14 × h. Therefore h = 12 cm.

Question 41

A triangle has vertices at A(0, 0), B(8, 0), and C(4, 6). A line parallel to AB passes through the centroid of the triangle and intersects sides AC and BC at points P and Q respectively. Find the length of PQ (in cm).

Accepted Answer

Step 1: The centroid G = ((0+8+4)/3, (0+0+6)/3) = (4, 2). Step 2: AB lies on the x-axis (y = 0). A line parallel to AB through G has equation y = 2. Step 3: Line AC has equation y = (6/4)x = 1.5x. Setting y = 2: 2 = 1.5x → x = 4/3. So P = (4/3, 2). Step 4: Line BC passes through B(8, 0) and C(4, 6). Slope = (6-0)/(4-8) = -1.5. Equation: y - 0 = -1.5(x - 8) → y = -1.5x + 12. Setting y = 2: 2 = -1.5x + 12 → x = 20/3. So Q = (20/3, 2). Step 5: PQ = |20/3 - 4/3| = 16/3 ≈ 5.33 cm.

Question 42

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. Find 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 43

In triangle ABC, the medians from vertices A and B intersect at the centroid G. If the median from A has length 18 cm and the median from B has length 24 cm, and these medians are perpendicular to each other at G, find the area of triangle ABC (in cm²).

Accepted Answer

Step 1: When two medians of a triangle are perpendicular, use the formula: Area of triangle = (2/3) × m₁ × m₂, where m₁ and m₂ are the lengths of the two perpendicular medians. Step 2: Area = (2/3) × 18 × 24 = (2/3) × 432 = 288 cm².

RBI Grade B Triangles — Area & Properties

RBI Grade B 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