Share
Notifications

View all notifications

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

Login
Create free account


      Forgot password?

Question

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\[\sqrt{3}\]
  Is there an error in this question or solution?

APPEARS IN

 Balbharati Solution for Balbharati Class 10 Mathematics 2 Geometry (2018 to Current)
Chapter 2: Pythagoras Theorem
Practice Set 2.1 | Q: 4 | Page no. 39
Solution In the Given Figure. Find Rp and Ps Using the Information Given in ∆Psr. Concept: 30 - 60 - 90 and 45 - 45 - 90 Theorem.
S
View in app×