If diagonal of rhombus are 16 cm and 30 cm find perimeter
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Answered by
5
Solution :
Let d1 and d2 are two diagonals of a
Rhombus .
d1 = 16 cm , d2 = 30 cm
side of a Rhombus = √( d2/2 )² + ( d1/2 )²
= √( 30/2 )² + ( 16/2 )²
= √ 15² + 8²
=√225 + 64
= √289
= √17²
= 17 cm
Therefore ,
Perimeter of a Rhombus = 4 × side
= 4 × 17
= 68 cm
••••
Let d1 and d2 are two diagonals of a
Rhombus .
d1 = 16 cm , d2 = 30 cm
side of a Rhombus = √( d2/2 )² + ( d1/2 )²
= √( 30/2 )² + ( 16/2 )²
= √ 15² + 8²
=√225 + 64
= √289
= √17²
= 17 cm
Therefore ,
Perimeter of a Rhombus = 4 × side
= 4 × 17
= 68 cm
••••
Answered by
0
Answer:
Step-by-step explanation:
let a rhombus ABCD be there
the diagonal are perpendicular bisectors of each other...
let the diagonal intersected at point O in the rhombus..
D=16 cm
d=30 cm
in triangle AOB
(D/2)^2 + (d/2)^2= (AB)^2
8^2+ 15^2=(AB)^2
AB= root 64+225
AB= root 289
AB= 17 cm
perimeter of the rhombus= 4*AB = 4*17 = 68 cm
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