Question

Find the altitude of the rhombus if lengths of it's two diagonals are 12 cm and 16 cmrespectively

Answer 1

♧♧HERE IS YOUR ANSWER♧♧

(SEE THE ATTACHMENT***)

Let, ABCD is a rhombus, whose diagonals are AC and BD.

We consider :

AC = 16 cm and BD = 12 cm.

Let, the diagonals AC and BD bisect each other at P.

Then,

AP = PC = 8 cm

and

BP = PD = 6 cm.

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

Then, triangle APB is right-angled.

By Pythagorous theorem, we write :

AP² + BP² = AB²

=> 8² + 6² = AB²

=> AB² = 64 + 36 = 100 = 10²

So, AB = 10 cm, since length is never negative.

Thus, each side of the rhombus is

= AB = BC = CD = DA = 10 cm

Now, area of the rhombus ABCD (using length of the diagonals)

Δ = ½(AC × BD)

=  ½(16 × 12) cm²

= 96 cm

Again, area of the rhombus ABCD (using length of the side and altitude)

G' = AB × DR, where DR is the altitude of the rhombus.

Δ' = 10 × DR cm²

So, Δ = Δ' gives :

96 = DR × 10

=> DR = 96/10

=> DR = 9.6

Therefore, length of the altitude DR is 9.6 cm.

♧♧HOPE THIS HELPS YOU♧♧