Measuring Space: Perimeter and Area | Exercise 6.2

Question 3

Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.

Solution

We will use Heron's formula to find the area of the triangle.

Step 1 — Find the third side

Let the two given sides be $a = \textbf{8 cm}$ and $b = \textbf{11 cm}$ . The perimeter of the triangle is $\textbf{32 cm}$ . Let the third side be $c$ . The perimeter is the sum of all three sides.

$c = \text{Perimeter} - (a + b)$

$c = 32 - (8 + 11)$

$c = 32 - 19$

$\boxed{c = 13 \text{ cm}}$

Diagram 1

Step 2 — Calculate semi-perimeter

The sides of the triangle are $\textbf{8 cm}$ , $\textbf{11 cm}$ , and $\textbf{13 cm}$ . The semi-perimeter is half of the perimeter. Let's denote the semi-perimeter as $s$ .

$s = \frac{\text{Perimeter}}{2}$

$s = \frac{32}{2}$

$\boxed{s = 16 \text{ cm}}$

Step 3 — Apply Heron's formula

Heron's formula helps us find the area of a triangle. The formula is $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$ . Let's substitute the values we found.

$\text{Area} = \sqrt{16(16-8)(16-11)(16-13)}$

$\text{Area} = \sqrt{16 \times 8 \times 5 \times 3}$

$\text{Area} = \sqrt{1920}$

$\text{Area} = \sqrt{64 \times 30}$

$\text{Area} = 8\sqrt{30}$

$\boxed{\text{Area} = 8\sqrt{30} \text{ cm}^2}$

Answer

(i) The area of the triangle is $8\sqrt{30} \text{ cm}^2$ .

More questions in Exercise 6.2

Find the area of triangle ADE in Fig. 6.31.

The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are both equal, each being 26 cm, find the area of the trapezium.

Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.

The sides of a triangular plot are in the ratio 3: 5: 7; its perimeter is 300 m. Find its area.

One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area 128 cm², find the length of the shorter diagonal.

ABCD is a parallelogram. P and Q are any two points on side AB. What can you say about the ratio area (ΔPCD): area (ΔQCD)?

O is any point on the diagonal PR of a parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal.

If the mid-points of the sides of a 4-gon (also known as a quadrilateral, but we prefer to call it a ‘4-gon’) are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon. (You may wonder whether the 4-gon thus formed is always a parallelogram, and if so, why? These questions will be tackled and answered in the chapter on quadrilaterals.)

In $\Delta$ ABC, the midpoint of BC is D (Fig. 6.32). Median AD is drawn. P is any point on AD. Show that area ( $\Delta$ ABP) = area ( $\Delta$ ACP).

Q10

Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD (Fig. 6.33). What is the ratio of the areas of the red region ( $\Delta$ PAB and $\Delta$ PCD) and the green region ( $\Delta$ PBC and $\Delta$ PDA)?

Q11

In $\Delta$ ABC, D is the midpoint of AB. P is any point on BC, and Q is a point on AB such that CQ || PD. PQ is joined (Fig. 6.34). Prove that

$\text{Area }(\Delta\text{BPQ}) = \frac{1}{2} \text{Area }(\Delta\text{ABC})$