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]
[ = 6sqrt{3} . . . left( 2 right)]
Hence, RP = 12 and PS = 6[sqrt{3}]

Explanation:- 

Given, ∆PSR with ∠S = 90^∘, ∠P = 30^∘, and ∴ ∠R = 60^∘.

By 30^∘- 60^∘ – 90^∘ 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

Q 3


Q 4


Q 5