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

##### Solution :-

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

$$\dfrac{AD}{AB}$$ = $$\dfrac{AE}{AC}$$$$\dfrac{1}{4}$$

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

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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