Question 1

A triangle has a base of 15 cm. If its area is 60 cm², what is the perpendicular height from the opposite vertex to the base?

Accepted Answer

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

Question 2

An equilateral triangle has a side length of 10 cm. What is its area? (Use √3 ≈ 1.73)

Accepted Answer

Step 1: Use the formula for the area of an equilateral triangle: Area = (√3/4) × side². Step 2: Substitute side = 10 cm: Area = (√3/4) × 10² = (√3/4) × 100 = 25√3. Step 3: Calculate: 25 × 1.73 = 43.25 cm². Therefore, the area is 43.25 cm².

Question 3

A right-angled triangle has legs of 6 cm and 8 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 the values: c² = 6² + 8² = 36 + 64 = 100. Step 3: Take the square root: c = √100 = 10 cm. Therefore, the hypotenuse is 10 cm.

Question 4

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 5

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 × (15-5) × (15-12) × (15-13)] = √[15 × 10 × 3 × 2] = √900 = 30 cm². Therefore, the area is 30 cm².

Question 6

The area of a triangle is 24 cm² and its base is 8 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 solve for height: height = (2 × Area) / base. Step 3: Substitute the values: height = (2 × 24) / 8 = 48 / 8 = 6 cm. Therefore, the height is 6 cm.

Question 7

A triangle has sides 5 cm, 5 cm, and 6 cm. Find the length of the altitude drawn to the side of length 6 cm.

Accepted Answer

Step 1: Use Heron's formula. Semi-perimeter s = (5 + 5 + 6)/2 = 8 cm. Step 2: Area = √[8(8-5)(8-5)(8-6)] = √[8 × 3 × 3 × 2] = √144 = 12 cm². Step 3: Using Area = (1/2) × base × height: 12 = (1/2) × 6 × h, so h = 4 cm. Therefore, the altitude is 4 cm.

Question 8

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

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 9

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 10

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, find 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 = 36, so DC = 12 cm.

Question 11

In triangle PQR, the medians from P and Q intersect at the centroid G. If the median from P has length 18 cm, what is the distance PG (in cm)?

Accepted Answer

Step 1: Recall that the centroid divides each median in the ratio 2:1 from the vertex. Step 2: If the median from P is 18 cm, then PG = (2/3) × 18 = 12 cm.

Question 12

A triangle has vertices at A(0, 0), B(12, 0), and C(6, 8). A line parallel to AB passes through the centroid of the triangle and intersects AC at point P and BC at point Q. What is the length of PQ (in cm)?

Accepted Answer

Step 1: Find the centroid G = ((0+12+6)/3, (0+0+8)/3) = (6, 8/3). Step 2: Since PQ is parallel to AB (which lies on the x-axis), PQ is a horizontal line at y = 8/3. Step 3: Find where y = 8/3 intersects AC. Line AC goes from (0,0) to (6,8), so its equation is y = (4/3)x. Setting y = 8/3: 8/3 = (4/3)x → x = 2. So P = (2, 8/3). Step 4: Find where y = 8/3 intersects BC. Line BC goes from (12,0) to (6,8), so its equation is y = -4/3(x-12) = -4x/3 + 16. Setting y = 8/3: 8/3 = -4x/3 + 16 → 8/3 + 4x/3 = 16 → 4x/3 = 40/3 → x = 10. So Q = (10, 8/3). Step 5: PQ = |10 - 2| = 8 cm.

Question 13

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

Accepted Answer

Step 1: Let the sides be 3k, 4k, and 5k. Since 3² + 4² = 9 + 16 = 25 = 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 is 5k = 5 × 4 = 20 cm.

SSC CPO 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