1.

ᴘʟᴇᴀsᴇ sᴏʟᴠᴇ ɴᴏ. 13 ✌︎​

Answer»

ᴇɴ-:

In ∆ABC,D and E are the midpoints of AB and AC respectively.

Therefore,DE || BC (By converse of mid point theoram)

Also,DE = \sf\frac{1}{2BC}

In ∆ADE and ∆ABC

∠ADE = ∠B (corresponding angles)

∠DAE = ∠BAC (COMMON)

∆ADE = ∆ABC (By AA SIMILARITY)

We know that the ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides.

☞\sf\frac{area(∆ADE)}{area(∆ABC} = \sf\frac{AD^2}{AB^2}

[AB = 2AB as D is the mid point]

☞\sf\frac{area(∆ADE)}{area(∆ABC} = \sf\frac{1^2}{2^2}

☞\sf\frac{area(∆ADE)}{area(∆ABC} = \sf\frac{1}{4}

Hence,the ratio of the areas ∆ADE and ∆ABC is area(∆ADE) : area(∆ABC) = 1:4


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