Question

in an isoceles triangle PQR QR is 10 cm pq is 6 cm and PR is 6 cm if RS is perpendicular to PQ FIND RS

Answer 1

solution.

We know that Altitude of triangle bisect the opposite side.

Here ,we can apply Pythagoras theorems

$= > {5}^{2} + {h}^{2} = {6}^{2} \\ = > 25 + {h}^{2} = 36 \\ = > {h}^{2} = 9 \\ = > h = \sqrt{9} \\ \\ so \: height \: = 3 cm$
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Answer 2

In ΔPQR where RS = 5 cm

Step-by-step explanation:

Given : ΔPQR is an isosceles triangle where QR = 10 cm, PQ = 6 cm, PR = 6 cm , RS ⊥ PQ

To find : RS

By using Pythagoras theorem,

In ΔPQS , ∠S = 90°

PQ² = PS² + QS²

6² = 4² + QS²

36 = 16 + QS²

QS² =  36 -16 = 20

QS = 2$\sqrt{5}$ cm

In Δ PQS , ∠S = 90°

tan Θ = $\frac{2\sqrt{5} }{4}$

In ΔPQR , ∠Q = 90°

tan θ =$\frac{QR}{6}$

QR/6  = 3$\sqrt{5}$/6

QR = 3$\sqrt{5}$ cm

Using Pythagoras theorem

In Δ RQS , ∠S =90°

RQ² = RS² + QS²

(3√5)² = RS² + (2√5)²

RS² = 45 - 20

RS = √45-20

RS = √25 =  5 cm