Question

if AB = 2, BC = 6, AE = 6, BF = 8, CE = 7, and CF = 7, compute the ratio of the area of quadrilateral ABDE to the area of ΔCDF.

Answer 1

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Answer 2

Answer: 2: 1.

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Given:

AB = 2, BC = 6, AE = 6, BF = 8, CE = 7, and CF = 7

Solution:

Concept:

Δ BCD ~ ΔACE ~ Δ BCF

Let us take Δ BCD,

Using Linear Scale factor (LSF),

          $\frac{8}{6} = \frac{4}{3}  \\\rightarrow \frac {8 (BF)} { BD} = \frac{4}{3}$

24 = 4 BD

BD =$\frac{ 24}{4} = 6$

$\frac{CE}{CD} =\frac{ 8}{6} \\\rightarrow \frac{7}{(CD)} = \frac{8}{6}$

42 = 8 CD

CD $= \frac{42}{8} = 5.25$

Now, to find the area of △ACE, using Heron’s relation:

A =$\sqrt { S(S-a)(S-b)(S-c) }  whereas, S = \frac{(a + b + c)}{ 2} = 10.5$

A = $\sqrt {10.5(10.5 - 6) (10.5 - 7) (10.5 - 8)} = \sqrt {413.4375} = 20.33$

Area of Δ ACE = Area of Δ BCF = 20.33.

Area of Δ BCD,

S$=\frac{ (6 + 6 + 5.25)} {2} = 8.625$

Thus, the area is:

A = $\sqrt {8.625(8.625 - 6) (8.625 - 6) (8.625 - 5.25)} = \sqrt{200.58} = 14.16$

Area of CDF:

S = $\frac{(5.25 + 7 + 2)} { 2} = 7.125$

A = $\sqrt{ 7.125(7.125 - 5.25) (7.125 - 7) (7.125 - 2)} = \sqrt{8.558} = 2.925$

Area of ABDE. 20.33 - 14.16 = 6.17

Now the ratio:

       $\frac{ABDE}{CDF} = \frac{6.17}{2.925 }$

Which is approximately equal to 6: 3 or 2: 1.