The vertices of a ∆ 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 \( \dfrac{ AD}{AB}= \dfrac{AE}{AC }= \dfrac{1}{4}\)

Calculate the area of the ∆ ADE and compare it with area of ∆ ABC. 

Asked by Pragya Singh | 1 year ago |  201

1 Answer

Solution :-

Given: The vertices of a ∆ ABC are A (4, 6), B (1, 5) and C (7, 2)

NCERT Solutions for Class 10 Chapter 7- 31

\( \dfrac{AD}{AB}\) = \( \dfrac{AE}{AC}\)\( \dfrac{1}{4}\)

\( \dfrac{AD}{(AD + BD)}\) = \( \dfrac{ AE}{(AE + EC)}\)\( \dfrac{1}{4}\)

Point D and Point E divide AB and AC respectively in ratio 1 : 3.

Coordinates of D can be calculated as follows:

x = (m1x2 + m2x1)/(m1 + m2) and y = (m1y2 + m2y1)/(m1 + m2)

Here m1 = 1 and m2 = 3

Consider line segment AB which is divided by the point D at the ratio 1:3.

x = [3(4) + \( \dfrac{1(1)}{4}\)=\( \dfrac{13}{4}\)

y = [3(6) + \( \dfrac{1(5)}{4}\)\( \dfrac{23}{4}\)

Similarly, Coordinates of E can be calculated as follows:

x = [1(7) + \( \dfrac{3(6)}{4}\)\( \dfrac{19}{4}\)

y = [1(2) + \( \dfrac{3(6)}{4}\)\( \dfrac{20}{4}\) = 5

Find Area of triangle:

Using formula: Area of a triangle = 

\( \dfrac{1}{2}\) × [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

Area of triangle ∆ ABC can be calculated as follows:

\( \dfrac{1}{2}\) [4(5 – 2) + 1( 2 – 6) + 7( 6 – 5)]

\( \dfrac{1}{2}\) (12 – 4 + 7) =\( \dfrac{15}{2}\) sq unit

Area of ∆ ADE can be calculated as follows:

=\( \dfrac{1}{2}\) [4(\( \dfrac{23}{4}\) – 5) + \( \dfrac{13}{4}\) (5 – 6) + \( \dfrac{19}{4}\) (6 – \( \dfrac{23}{4}\))]

\( \dfrac{1}{2}\) (3 –\( \dfrac{13}{4}\) + \( \dfrac{19}{16}\))

=\( \dfrac{1}{2}\) (\( \dfrac{15}{16}\) ) = \( \dfrac{15}{32}\)sq unit

Hence, ratio of area of triangle ADE to area of triangle ABC = 1 : 16.

Answered by Abhisek | 1 year ago

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