8. The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
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Diagonals intersect at 90° angles
Taking Rhombus ABCD. Diagonals intersect at O.
All 4 triangles formed thus, are congruent.
In triangle AOB, AO2+BO2=AB2
Since diagonals bisect each other,
AO=1/2 of AC, =16/2=8.
BO=1/2 of BD, =30/2=15.
8*8 + 15*15 = AB2
64+225=AB2
289=AB2
AB=√289=17
All 4 sides are equal.
Therefore perimeter is 4*side which is 17*4=68.
Answer is 68 cm.
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