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prove that AB is parallel to the side of the inner triangle
70+110=180
thus Co interior angles are supplementary
also<C = <C ---------(common)
thus the two triangles are similar by
AA similarity criterion
BC/DC = 6/3 = 2:1
Then AB/DE = 2:1--(since sides of similar triangles are in equal ratio)
5:DE = 2:1
2DE = 5
DE = 2.5
70+110=180
thus Co interior angles are supplementary
also<C = <C ---------(common)
thus the two triangles are similar by
AA similarity criterion
BC/DC = 6/3 = 2:1
Then AB/DE = 2:1--(since sides of similar triangles are in equal ratio)
5:DE = 2:1
2DE = 5
DE = 2.5
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