Measuring Space: Perimeter and Area | Exercise 6.2

Question 5

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.

Solution

We use the formula for the area of a rhombus.

Step 1 — Set up the equation

Let's call the shorter diagonal x cm. The longer diagonal will be 2x cm. The area of a rhombus is half the product of its diagonals. We are given the area as 128 cm².

$A = \frac{1}{2} \times d_1 \times d_2$

$128 = \frac{1}{2} \times x \times 2x$

$128 = x^2$

$\boxed{x^2 = 128}$

Diagram 1

Step 2 — Find the shorter diagonal

Now, we find the value of x. We take the square root of 128.

$x = \sqrt{128}$

$x = \sqrt{64 \times 2}$

$x = 8\sqrt{2}$

$\boxed{\text{Shorter diagonal} = 8\sqrt{2} \text{ cm}}$

Answer

(i) The length of the shorter diagonal is $8\sqrt{2} \text{ cm}$ .

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

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