Measuring Space: Perimeter and Area | Exercise 6.2

Question 4

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

Solution

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

Step 1 — Find side lengths

Let's represent the sides. The sides are in the ratio 3:5:7. So, we can write them as $3x$ , $5x$ , and $7x$ . The perimeter is given as 300 m. The sum of the sides equals the perimeter.

$3x + 5x + 7x = 300$

$15x = 300$

$x = \frac{300}{15}$

$\boxed{x = 20}$

Now, we find the actual side lengths. The first side is $3x$ .

$3 \times 20 = 60 \text{ m}$

The second side is $5x$ .

$5 \times 20 = 100 \text{ m}$

The third side is $7x$ .

$7 \times 20 = 140 \text{ m}$

Diagram 1

Step 2 — Calculate semi-perimeter

We need the semi-perimeter for Heron's formula. The semi-perimeter 's' is half of the perimeter.

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

$s = \frac{300}{2}$

$\boxed{s = 150 \text{ m}}$

Step 3 — Apply Heron's formula

Heron's formula gives the area of a triangle. The formula is $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$ . Here, $a = 60 \text{ m}$ , $b = 100 \text{ m}$ , $c = 140 \text{ m}$ . We found $s = 150 \text{ m}$ .

$\text{Area} = \sqrt{150(150-60)(150-100)(150-140)}$

$\text{Area} = \sqrt{150 \times 90 \times 50 \times 10}$

$\text{Area} = \sqrt{6750000}$

$\text{Area} = \sqrt{225 \times 3 \times 10000}$

$\text{Area} = \sqrt{225} \times \sqrt{3} \times \sqrt{10000}$

$\text{Area} = 15 \times \sqrt{3} \times 100$

$\boxed{\text{Area} = 1500\sqrt{3} \text{ m}^2}$

Answer

The area of the triangular plot is $1500\sqrt{3} \text{ m}^2$ .

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

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