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In the given figure angle $Q$ is a right angle, $PS:QS = 3:1, RT:QT = 5:2$ and $PU:UR = 1:1. $ If area of triangle $QTS$ is $20cm^{2},$ then the area of triangle $PQR$ in $cm^{2}$ is ______

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$\frac{RT}{QT}= \frac{5x}{2x} \Rightarrow RQ=7x$

$\frac{PS}{QS}= \frac{3y}{1y} \Rightarrow PQ=4y$

Area of Triangle $QTS =\frac{1}{2} \times QS \times QT =20 cm^2$

$\frac{1}{2} \times 1y \times 2x =20 cm^2$

$\Rightarrow xy =20 cm^2$

Area of Triangle $PQR = \frac{1}{2} \times PQ \times RQ$

$\frac{1}{2} \times 4y \times 7x$

$\quad = 14xy = 14\times 20 cm^2 =280cm^2$
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$$\def\o{\overline}\frac{\o{PS}}{\o{QS}} = \frac 3 1 \implies \frac{\o{PS} + \o{QS}}{\o{QS}} = \frac{3+1}{1} = 4 \implies \o{PQ} = 4\cdot \o{QS}$$

Similarly, $$\frac{\o{RT}}{\o{QT}} = \frac 5 2 \implies \frac{\o{RT} + \o{QT}}{\o{QT}} = \frac{5+2}{2} = \frac 7 2 \implies \o{RQ} = \frac 7 2 \cdot \o{QT}$$

Now, $$\text{Area of } \triangle PQR = \frac 1 2 \cdot \o{PQ} \cdot \o{RQ} = \frac 1 2 \cdot \left ( 4 \cdot \o{QS} \right ) \cdot \left ( \frac 7 2 \cdot {QT} \right ) \\= \frac 1 2 \cdot \left ( \o{QS} \cdot \o{QT} \right ) \cdot 4 \cdot \frac 7 2 = 20 \text{ cm}^2 \cdot 14 = 280 \text{ cm}^2$$

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lets PS/QS =3x/x and QT/TR=2y/5y PU/ZR=Z/Z

given:

area of triangle QST is 20 cm^2 and angle Q is 90 .now we know are of a right angled triangle is 1/2 * height * base

so it can be re written as 1/2 * SQ*QT=20 =>1/2 * x*2y=20 =>xy=20................(i)

required :

area of QPR which is 1/2 *PQ*QR =1/2*(3x+x)*(5y+2y) =>(1/2)*4x*7y => (1/2) * 28 xy

now from eqn 1 we replace xy=20 to get 1/2 * 28* 20 =280 cm^2
Answer:

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