the diagonals of a rhombus measure16cm and 30cm. find its perimeter.
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Answered by
1
perimeter is 68cm
we know that diagonals of a rhombus bisect each other at 90 degrees
so acc. to Pythagoras theorm
AC^2 = AB^2 + BC^2
= 64 + 225
= 289
AC = 17cm
So perimeter = 17*4
= 68cm
we know that diagonals of a rhombus bisect each other at 90 degrees
so acc. to Pythagoras theorm
AC^2 = AB^2 + BC^2
= 64 + 225
= 289
AC = 17cm
So perimeter = 17*4
= 68cm
Answered by
0
The diagnols of a rhombus bisect each other and are perpendicular to each other therefore side of rhombus=root of (15^2+8^2)=root of (289)= 17 therefore perimeter of square=68
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