Question

P is a point inside the ∆ABC. Three straight lines parallel to AB, BC and AC are drawn which divides the triangle into six regions. If the area of three triangles formed in these regions are 9,16 and 25 square units, then the area (in sq, units of ∆ABC is​
(Spams will be reported)

Answer 1

$\large\underline{\sf{Solution-}}$

P is a point inside the ∆ ABC, such that three straight lines DE, HK and GD are drawn parallel to AB, BC and AC which divides the triangle into six regions.

Now,

• Area of ∆ PGH = 9 square units

• Area of ∆ PDK = 16 square units

• Area of ∆ PEF = 25 square units.

Let assume that

• BE = a units, EF = b units and FC = c units.

As, HK || BC ⇛ HP || BE ( part of parallel lines )

Also, DE || AB ⇛ PE || HB ( part of parallel lines )

⇛ BEPH is a parallelogram

⇛ BE = HP = a units.

Similarly, PK = FC = b units

Now, In ∆PEF and ∆ABC

$\rm \: \angle 2 = \angle 3 \: \: \{corresponding\:angles \}$

$\rm \: \angle 4 = \angle 5 \: \: \{corresponding\:angles \}$

⇛ ∆ PEF ~ ∆ ABC [ AA Similarity Rule ]

So, By area ratio theorem, we have

$\rm \: \dfrac{ar(\triangle PEF)}{ar(\triangle ABC)} = \dfrac{ {EF}^{2} }{ {BC}^{2} }$

$\rm \: \dfrac{25}{ar(\triangle ABC)} = \dfrac{ {b}^{2} }{ {BC}^{2} }$

$\rm \: \dfrac{5}{ \sqrt{ ar(\triangle ABC)}} = \dfrac{ {b}}{ {BC}} - - - (1)$

Now, In ∆ GHP and ∆ ABC

$\rm \: \angle 6 = \angle 7 \: \: \{corresponding\:angles \}$

$\rm \: \angle 1 = \angle 2 \: \: \{corresponding\:angles \}$

⇛ ∆ GHP ~ ∆ ABC [ AA Similarity Rule ]

So, using Area Ratio Theorem, we have

$\rm \: \dfrac{ar(\triangle GHP)}{ar(\triangle ABC)} = \dfrac{ {HP}^{2} }{ {BC}^{2} }$

$\rm \: \dfrac{9}{ar(\triangle ABC)} = \dfrac{ {a}^{2} }{ {BC}^{2} }$

$\rm \: \dfrac{3}{ \sqrt{ ar(\triangle ABC)}} = \dfrac{ {a}}{ {BC}} - - - (2)$

Similarly,

$\rm \: \dfrac{4}{ \sqrt{ ar(\triangle ABC)}} = \dfrac{ {c}}{ {BC}} - - - (3)$

Now, on adding equation (1), (2) and (3), we get

$\rm \: \dfrac{4}{ \sqrt{ ar(\triangle ABC)}} + \dfrac{3}{ \sqrt{ ar(\triangle ABC)}} + \dfrac{5}{ \sqrt{ ar(\triangle ABC)}} = \dfrac{ {c}}{ {BC}} + \dfrac{ {b}}{ {BC}} + \dfrac{ {a}}{ {BC}}$

$\rm \: \dfrac{4 + 3 + 5}{ \sqrt{ ar(\triangle ABC)}} = \dfrac{ {c + a + b}}{ {BC}}$

$\rm \: \dfrac{12}{ \sqrt{ ar(\triangle ABC)}} = \dfrac{ {BC}}{ {BC}}$

$\rm \: \dfrac{12}{ \sqrt{ ar(\triangle ABC)}} = 1$

$\rm \: \sqrt{ ar(\triangle ABC)} = 12$

On squaring both sides, we get

$\rm \: ar(\triangle ABC) = {12}^{2}$

$\rm \: ar(\triangle ABC) = 144$

Hence,

$\color{green}\rm\implies \:\boxed{ \rm{ \:\rm \: ar(\triangle ABC) = 144 \: sq. \: units \: \: }}$

$\rule{190pt}{2pt}$

Short Cut Trick :-

P is a point inside the ∆ABC. Three straight lines parallel to AB, BC and AC are drawn which divides the triangle into six regions. If the area of three triangles formed in these regions are a, b and c square units, then the area (in sq. units) of ∆ABC is

$\rm \: ar(\triangle ABC) = {\bigg( \sqrt{a} + \sqrt{b} + \sqrt{c} \bigg)}^{2} \: sq. \: units$