# In the Given Figure. Find Rp and Ps Using the Information Given in ∆Psr. - Geometry

Sum

In the given figure. Find RP and PS using the information given in ∆PSR.

#### Solution

In ∆PSR,
∠S = 90, ∠P = 30, ∴ ∠R = 60
By 30∘ − 60 − 90 theorem,

$\text{RS} = \frac{1}{2} \times \text{RP}$
$\Rightarrow 6 = \frac{1}{2} \times \text{RP}$
$\Rightarrow 6 \times 2 = \text{RP}$
$\Rightarrow \text{RP} = 12 . . . \left( 1 \right)$
$\text{PS} = \frac{\sqrt{3}}{2} \times \text{RP}$
$= \frac{\sqrt{3}}{2} \times 12$
$= 6\sqrt{3} . . . \left( 2 \right)$
Hence, RP = 12 and PS = 6$\sqrt{3}$
Concept: Property of 30°- 60°- 90° Triangle Theorem
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#### APPEARS IN

Balbharati Mathematics 2 Geometry 10th Standard SSC Maharashtra State Board
Chapter 2 Pythagoras Theorem
Practice Set 2.1 | Q 4 | Page 39