Question

The vertices of ∆ABC are A(4,6),B(1,5) and C(7,2).A line segment DE is drawn to intersect the sides AB and AC at D and E respectively such that AD/AB=AE/AC=1/3.Calculate the area of ∆ADE and compare it with area of ∆ABC.

Answer 1

Answer:

Given:

The vertices of ∆ABC are A(4,6),B(1,5) and C(7,2).A line segment DE is drawn to intersect the sides AB and AC at D and E respectively such that AD/AB=AE/AC=1/3.

To prove :

Compare area of ∆ADE and ∆ABC .

$Let \: A(4,6) = (x_{1},y_{1}),\:B(1,5) = (x_{2},y_{2},\:\\C(7,2) = (x_{3},y_{3})$

$Area \: of \triangle \: ABC\\ =\frac{1}{2} | x_{1}(y_{2} - y_{3})+x_{2}(y_{3} - y_{1})+x_{3}(y_{1} - y_{2})|</p><p>$

$= \frac{1}{2}| 4(5-2)+1(2-6)+7(6-5)|$

$= \frac{1}{2} | 4\times 3 - 1\times 4+7\times 1|$

$=\frac{1}{2} | 12 - 4+7 |$

$= \frac{1}{2} | 15|$

$area \: of \: ABC = \frac{15}{2} \: ---(1)$

$\frac{Area\: \triangle ADE}{Area \:\triangle ABC} =\frac{AD^{2}}{AB^{2}}$

$\implies \frac{Area\: \triangle ADE}{\frac{15}{2}} =\left(\frac{AD}{AB}\right)^{2}$

$= \left(\frac{1}{3}\right)^{2}\\= \frac{1}{9}$

$\implies Area\: \triangle ADE = \frac{1}{9}\times \frac{15}{2}\\= \frac{5}{6}\: ---(2)$

$\frac{Area\: \triangle ADE}{Area \:\triangle ABC} = \frac{\frac{3}{2}}{\frac{15}{2}}\\= \frac{1}{5}$

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Answer 2

Answer:

Step-by-step explanation: