Question

In triangle ABC , DE parallel to AB ,DC= 2BD
Area( Triangle CDE)= 20 sqcm then find Area( quadrilateral ABDE)​

Answer 1

Answer:

$\bf ar ( Trap\ ABDE ) =25 \ cm^2$

Step-by-step explanation:

In the ΔABC and ΔCDE

$\ DE \parallel \ AB$, Thus

∠CDE =∠CBA ( Corresponding angles)

∠CED =∠CAB  ( Corresponding angles)

Hence ΔABC $\sim$ ΔCDE

Thus by definition of area of similar triangles

$\frac {ar( \triangle ABC ) }{ar( \triangle CDE)} =( \frac{BC}{CD})^2\\\\ar( \triangle ABC = ar( \triangle CDE) \times ( \frac{2x+x}{2x})^2\\\\=20 \times (\frac32)^2\\\\=\frac{20\times 9}{4}\\\\\bf{ar( \triangle ABC=5\times 9=45 \ cm^2------(1)}$

$Thus\\\\ar ( Trap\ ABDE ) = ar ( \triangle ABC)-ar ( \triangle CDE)\\\\=45-20\\\\\bf ar ( Trap\ ABDE ) =25 \ cm^2$

Answer 2

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