Question

Find the altitude of a rhombus whose diagnols are 24 cm and 36 cm​

Answer 1

Answer:

20 cm

Step-by-step explanation:

For the figure, refer to the attachment.

Let the diagonal AC = 24 cm

and BD = 36 cm.

We know that diagonals of a rhombus bisect each other at right angles.

Thus, AO = CO and BO = DO

So, AC = AO + CO = AO + AO = 2AO

or, AO = 24/2 = 12 cm

Hence, AO = CO = 12 cm

Similarly,

BO = DO = 18 cm

Also,

∠AOB = ∠AOD = ∠BOC = ∠COD = 90°

Now,

In right angled triangle AOB,

By Pythagoras theorem,

AB² = AO² + BO²

⇒ AB = √(12² + 18²)

= √(144 + 324)

= √(468)

= 21.6 cm approx

Since all sides of a rhombus are equal,

AB = BC = CD = AD = 21.6 cm

We know that area of rhombus = 1/2 * diagonal1 * diagonal2

= 1/2 * 24 * 36

= 24 * 18

= 432 cm²

But,

area of rhombus is also equal to Base * alt

= AB * DX

= 21.6 * DX

Therefore,

432 = 21.6 * DX

⇒ DX = 432 ÷ 21.6 = 20 cm

Hence, the length of altitude of rhombus is 20 cm.