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$(Numbers indicate the lengths of the sides of the triangles) If the area of $\bigtriangleup$ PQR is k2sq. units what is the area of the shades portion?$

(Numbers indicate the lengths of the sides of the triangles) If the area of $\bigtriangleup$ PQR is k2sq. units what is the area of the shades portion?

$\frac{5}{9}$ k² sq. units

$\frac{1}{3}$ k² sq. units

$\frac{8}{9}$ k² sq. units

$\frac{7}{9}$ k² sq. units

$\frac{2}{3}$ k² sq. u

View answer & explanation

Correct answer is A

Area of shaded portion = Area of triangle PQR - Area of inner triangle

Area of triangle given 3 sides a, b, c = $\sqrt{s(s - a)(s - b)(s - c)}$

where $s = \frac{a + b + c}{2}$

Area of PQR :

$s = \frac{3 + 5 + 6}{2} = \frac{14}{2} = 7$

Area = $\sqrt{7(7 - 3)(7 - 5)(7 - 6)}$

= $\sqrt{7(4)(2)(1)} = \sqrt{56}$

$\implies K^{2} = \sqrt{56}$

Area of inner triangle :

$s = \frac{2 + 4 + \frac{10}{3}}{2} = \frac{14}{3}$

Area = $\sqrt{\frac{14}{3} (\frac{14}{3} - 2)(\frac{14}{3} - 4)(\frac{14}{3} - \frac{10}{3})}$

= $\sqrt{\frac{14}{3} (\frac{8}{3})(\frac{2}{3})(\frac{4}{3})}$

= $\sqrt{\frac{896}{81}}$

= $\sqrt{\frac{16}{81}} \times \sqrt{56}$

= $\frac{4}{9} K^{2}$

$\therefore \text{The area of the shaded portion} = K^{2} - \frac{4}{9}K^{2} = \frac{5}{9}K^{2}$

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(Numbers indicate the lengths of the sides of the triangles) If the area of △\bigtriangleup PQR is k2sq. units what is the area of the shades portion?

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(Numbers indicate the lengths of the sides of the triangles) If the area of $\bigtriangleup$ PQR is k2sq. units what is the area of the shades portion?