Math, asked by rajnigoyal380, 1 year ago

In ABC, D and E are midpoints of side AB and AC respectively.Show that ar (ADE) =1/4 ar ( ABC)

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Answered by Anonymous
6

Step-by-step explanation:

Draw line BE.

Consider triangles ABC and ABE.  Since A, E, C lie in the same line, these triangles have the same perpendicular height from B to the base (AE or AC).  Since the areas are half x base x height, the ratio between their areas is equal to the ratio between their bases.  That is

area(ABC) / area(ABE) = AC / AE = 2, since E is the midpoint.

Similarly, by considering the triangle ABE and ADE which have the same perpendicular height from E, we have

area(ABE) / area(ADE) = AB / AD = 2, since D is the midpoint.

So

area(ABC) / area(ADE)

= ( area(ABC) / area(ABE) ) x ( area(ABE) / area(ADE) )

= 2 x 2 = 4.

That is, area(ABC) is 4 times greater than area(ADE).

Or in other words, area(ADE) = (1/4) area(ABC)

Answered by sautrikc1425
0

Answer:

proved

Step-by-step explanation:

Draw line BE.

Consider triangles ABC and ABE.  Since A, E, C lie in the same line, these triangles have the same perpendicular height from B to the base (AE or AC).  Since the areas are half x base x height, the ratio between their areas is equal to the ratio between their bases.  That is

area(ABC) / area(ABE) = AC / AE = 2, since E is the midpoint.

Similarly, by considering the triangle ABE and ADE which have the same perpendicular height from E, we have

area(ABE) / area(ADE) = AB / AD = 2, since D is the midpoint.

So

area(ABC) / area(ADE)

= ( area(ABC) / area(ABE) ) x ( area(ABE) / area(ADE) )

= 2 x 2 = 4.

That is, area(ABC) is 4 times greater than area(ADE).

Or in other words, area(ADE) = (1/4) area(ABC)

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