Question

A Right Angle Triangle is Given as Shown in Picture, with Side AB = 6 m and, BC = 8 m.
A Triangle PQR is drawn on the Sides of Triangle ABC. Dimensions are given in Picture.

Find the Area of ∆PQR.​

Answer 1

AnswEr :

⋆ Reference of Image is in the Attachment :

⋆ Let the ∠A be α° and, ∠C be β°

• In ∆ABC, By Heron's Formula :

↠ AC² = AB² + BC²

↠ AC² = ( 6 )² + ( 8 )²

↠ AC² = 36 + 64

↠ AC² = 100

↠ AC = 10 m

⋆ Now : AP = AC - PC = (10 - 3) m = 7 m

We will find (sin α) and, (sin β) Because Area of Triangle is ( Half of Product of Adjacent Sides and sin θ°, Angle Between them )

$\mapsto\tt{ \sin( \alpha) = \dfrac{BC}{AC} = \cancel\dfrac{8}{10} = \dfrac{4}{5}}$

$\mapsto\tt{ \sin( \beta) = \dfrac{BC}{AC} = \cancel\dfrac{6}{10} = \dfrac{3}{5}}$

$\rule{300}{1}$

If we need to find the Area of ∆ PQR, we can find it by Subtracting the Area of rest of Three Triangles formed i.e. ( ∆ AQP, ∆ BQR and, ∆ CRP ) from the whole Triangle i.e. ( ∆ ABC )

$\rule{300}{2}$

• Let's Head to the Question Now :

$\implies\sf Area{\tiny\triangle ABC} = Area{\tiny\triangle AQP} + Area{\tiny\triangle BQR} + Area{\tiny\triangle CRP}+ Area{\tiny\triangle PQR}$

$\displaystyle \implies \tt \frac{1}{2} \times AB \times BC = \bigg( \frac{1}{2} \times AQ \times QP \times \sin( \alpha)\bigg) + \\ \tt\bigg( \frac{1}{2} \times BQ \times BR\bigg) + \bigg( \frac{1}{2} \times RC \times PC \times \sin( \beta )\bigg) + \sf{Area{\tiny\triangle PQR}}$

$\displaystyle \implies \tt \frac{1}{\cancel2} \times \cancel6 \times 8 = \bigg( \frac{1}{\cancel2} \times \cancel4 \times 7 \times \frac{4}{5} \bigg) + \\ \tt\bigg( \frac{1}{\cancel2} \times \cancel2 \times 4\bigg) + \bigg( \frac{1}{\cancel2} \times \cancel4 \times 3 \times \frac{3}{5} \bigg)+\sf{ Area{\tiny\triangle PQR}}$

$\displaystyle \implies \tt 3 \times 8 = \bigg( 2\times 7 \times \frac{4}{5} \bigg) + 4+ \bigg( 2 \times 3 \times \frac{3}{5} \bigg)+\sf{Area{\tiny\triangle PQR}}$

$\displaystyle \implies \tt 24 - 4 = \bigg(\frac{56}{5}+ \frac{18}{5}\bigg)+\sf{Area{\tiny\triangle PQR}}$

$\displaystyle \implies \tt 20 - \frac{74}{5} = \sf{Area{\tiny\triangle PQR}}$

$\displaystyle \implies \sf{Area{\tiny\triangle PQR}} = \tt\frac{100 - 74}{5}$

$\displaystyle \implies \sf {Area{\tiny\triangle PQR}} = \tt\frac{26}{5}$

$\implies\large\boxed{\sf Area{\tiny\triangle PQR}= 5.2 \: {cm}^{2}}$

⠀

∴ Hence, Area of Triangle PQR is 5.2 cm²

Answer 2

Answer:

Refer to the Attachment For Answer.

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