find the area of the rhombus one side of measures 20 cm and one of whose diagonals is 24cm
Answers
Given,
One side of rhombus measures 20 cm and one diagonal is of 24 cm.
We know that the diagonals of rhombus bisects each other at 90°. And therefore by pythagoras theorem : -
= > sum of squares of half of both the diagonals of rhombus = square of one side of rhombus
In the question length of one diagonal is 24 cm, so half of this diagonal should be 24 cm / 2 i.e. 12 cm.
Thus,
= > ( half of 1st diagonal )^2 + ( half of 2nd diagonal )^2 = side^2
Let the length of other diagonal be 2a, and half of this diagonal should be 2a / 2 i.e. a,
Hence,
= > ( 12 cm )^2 + ( a )^2 = ( 20 cm )^2
= > 144 cm^2 + a^2 = 400 cm^2
= > a^2 = 400 cm^2 - 144 cm^2
= > a^2 = 256 cm^2
= > a^2 = ( 16 cm )^2
= > a = 16 cm
Hence the required length of other diagonal should be 2a = 2 x 16 cm = 32 cm.
From the properties of rhombus,
- Area of rhombus = 1 / 2 x product of diagonals
Therefore,
= > Area of this rhombus should be =
1 / 2 x 24 cm x 32 cm
= > Area of this rhombus should be = 12 cm x 32 cm
= > Area of this rhombus should be = 384 cm^2
Answer : The required area of this rhombus is 384 cm^2.