Chapter 2 - Pythagoras Theorem- Text Book Solution. Practice Set 2.1 | Q 4 | Page 39
Practice Set 2.1 | Q 4 | Page 39
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\[\sqrt{3}\]
\[ \Rightarrow 6 = \frac{1}{2} \times \text{RP}\]
\[ \Rightarrow 6 \times 2 = \text{RP}\]
\[ \Rightarrow \text{RP} = 12 . . . \left( 1 \right)\]
\[\text{PS} = \frac{\sqrt{3}}{2} \times \text{RP}\]
\[ = \frac{\sqrt{3}}{2} \times 12\]
\[ = 6\sqrt{3} . . . \left( 2 \right)\]
Hence, RP = 12 and PS = 6\[\sqrt{3}\]
Explanation:-
Given, ∆PSR with ∠S = 90<sup>∘</sup>, ∠P = 30<sup>∘</sup>, and ∴ ∠R = 60<sup>∘</sup>.
By 30<sup>∘ </sup>- 60<sup>∘</sup> – 90<sup>∘</sup> theorem, RS = 1/2 x RP.
Therefore, RS = 1/2 x RP 6 = 1/2 x RP 6 x 2 = RP RP = 12 . . . (1)
Also, PS = sqrt(3)/2 x RP = sqrt(3)/2 x 12 = 6sqrt(3) . . . (2)
Hence, RP = 12 and PS = 6√3.
Chapter 2 – Pythagoras Theorem- Text Book Solution.
Practice Set 2.1 | Q 4 | Page 39