Math, asked by urvashi6892, 1 year ago

find the area of the rhombus one side of measures 20 cm and one of whose diagonals is 24cm​

Answers

Answered by abhi569
0

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.

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