Question

IN TRIANGLE QSR M ANGLE Q=45 DEGREE S=90 SR= 4 THEN FIND QS

Answer 1

Given, ∆PQR in which ∠Q = 90°, QS ⊥ PR and PQ = 6cm. PS = 4cm In ∆SQP and ∆SRQ, ∠PSQ = ∠RSQ [each equal to 90°] ∠SPQ = ∠SQR [each equal to 90°-∠R] ∴ ∆SQP ∼ ∆SRQ Then, = SQ/PS = SR/SQ ⇒ SQ2 = PSXSR …..(i) In right angled ∆PSQ, PQ2 = PS2 + QS2 [by Pythagoras theorem] = (6)2 = (4)2 + QS2 = 36 = 16 + QS2 = QS2 = 36-16 = 20 ∴ QS = √20 = 2√5 cm On putting the value of QS in Eq. (i), we get, In right angled ∆QSR, QR2 = QS2 + SR2 = QR2 = (2√5)2 + (5)2 = QR2 = 20 + 25 ∴ QR = √45 = 3√5cm Hence, QS = 2√5 cm, RS = 5 cm and QR = 3√5 cm.Read more on Sarthaks.com - https://www.sarthaks.com/883287/in-given-figure-pqr-right-triangle-right-angled-and-qs-pr-if-pq-cm-and-ps-cm-then-find-qs-rs-and

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Answer 2

Given:

IN TRIANGLE QSR M ANGLE Q=45 DEGREE S=90 SR= 4.

To Find:

Angle QS.

Solution:

To find the angle QS we will follow the following step:

As we know,

The Sum of all the angles of a triangle is equal to 180°.

In triangle QSR.

Angle Q + Angle S+ Angle R = 180°.

By ASA(angle side angle) rule:

$\frac{angle \: Q}{angle \: S } = \frac{QR}{QS}$

Now,

$\frac{45}{90} = \frac{QS}{4}$

$QS = \frac{45}{90} \times 4$

$QS = 2$

Henceforth, QS = 2.