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In figure, DE||BC and the ratio of the a...

In figure, DE||BC and the ratio of the areas of `triangleADE`and trapezium BDEC is 4:5 . Find the ratio of DE:BC . If BD = 2cm. Then find AD.

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In `triangleADE and triangleABC`
`angle1=angle2` ( corresponding angles as DE||BC)
`angleA= angleA` (common)
`triangleADE~triangleABC` (AA corollary)
`(ar(triangleADE))/(ar(triangleABC))=(AD^(2))/(AB^(2))=(DE^(2))/(BC^(2))=(AE^(2))/(AC^(2)) ...(1)`
ratio of areas of two similar triangles is equal to the ratio squares of their corresponding sides)
Now, `(ar("trapezium"BDEC))/(ar(triangleADE))=5/4`
` Rightarrow " "(ar(triangleABC)-ar(triangleADE))/(ar(triangleADE))=5/4`
`(ar(triangleABC))/(ar(triangleADE))-1=5/4 Rightarrow (ar(triangleABC))/(ar(triangleADE))=9/4`
` Rightarrow" " (ar(triangleADE))/(ar(triangleABC))=4/9` ....(2)
`(DE^(2))/(BC^(2))= 4/9` [ from (1) and (2)]
`(DE)/(BC) = 2/3 `
DE:BC= 2:3
Now from (1) and (2) , we get
`(ar(triangleADE))/(ar(triangleABC))= (AD^(2))/(AB^(2))= 4/9`
` (AD)/(AB)= 2/3 Rightarrow (AD)/(AD+BD)= 2/3`
`(AD + BD)/(AD)= 3/2 Rightarrow 1+(BD)/(AD)= 3/2`
`(BD)/(AD)= 3/2-1= 1/2`
`2/(AD)= 1/2 Rightarrow AD = 4cm `
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