. The vertices of a triangle ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and


AC at D and E respectively, such that AD/AB = AE/AC = 1/4 Calculate the area of theΔ ADE and compare it with the area of Δ ABC (Recall Theorem 6.2 and Theorem 6.6)

Best Answer

Solution:Given that

VERTICES OF A TRiangle

vertices of traingle

Clearly, the ratio between the areas of ΔADE and ΔABC is 1:16

We know that if a line segment in a triangle divides its two sides in the same ratio, then the line segment is parallel tothe third side of the triangle.

These two triangles so formed (here ΔADE and ΔABC) will be similar to each other.

Hence, the ratio between the areas of these two triangles will be the square of the ratio between the sides of these two triangles

And, ratio between the areas of ΔADE and ΔABC = (1/4)2

= 1/16

 

 

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