Area of two similar triangles are9sqcm and 16sqcm. Then the ratio of their corresponding sides are ———————-
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let the 2 triangles be ∆ABC and ∆PQR
A(∆ABC) /A(∆PQR) =AB^2/PQ^2
Thus, 9/16=AB^2/PQ^2
3/4=AB/PQ...(Taking square root)
thus the ratio of the corresponding sides are
3/4
(B.T.W. the sum is solved using the theorem on areas of similar triangles)
A(∆ABC) /A(∆PQR) =AB^2/PQ^2
Thus, 9/16=AB^2/PQ^2
3/4=AB/PQ...(Taking square root)
thus the ratio of the corresponding sides are
3/4
(B.T.W. the sum is solved using the theorem on areas of similar triangles)
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