Maharashtra BoardEnglish MediumSTD 10MathsP-2 Pythagoras Theorem2 Marks
Question
See figure 2.19. Find RP and PSusing the information given in ΔPSR.
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Answer
$\text { In } \triangle P S R, \angle P S R=90^0$
$\text { So } P S^2+S R^2=R P^2$
$\Rightarrow 6^2+\left(R P \cos \left(30^{\circ}\right)\right)^2=R P^2$
$\Rightarrow 6^2+R P^2 \times \frac{3}{4}=R P^2$
$\Rightarrow 6^2=\frac{R P^2}{4}$
$\Rightarrow R P^2=4 \times 36$
Thus RP $=12$.
$P S=R P \cos \left(30^{\circ}\right)$
$\Rightarrow P S=12 \times \frac{\sqrt{3}}{2}$
$P S=6 \sqrt{ } 3$.
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