Question

in a right-angled triangle PQR, ∠R=90⁰, if PQ = 30cm, PR = 18cm, find QR​

Answer 1

(i) Given:

PQ = 8 cm

QR = 6 cm

PR = ?

∠PQR = 90°

According to Pythagoras Theorem,

(PR)2 = (PQ)2 + (QR)2

PR2 = 82 + 62

PR2 = 64 + 36

PR2 = 100

∴ PR = √100 = 10 cm

(ii) Given :

PR = 34 cm

QR = 30 cm

PQ = ?

∠PQR = 90°

According to Pythagoras Theorem,

(PR)2 = (PQ)2 + (QR)2

(34)2 = PQ2 + (30)2

1156 = PQ2 + 900

1156 - 900 = PQ2

256 = PQ2

∴ PQ = 16 cm

Answer 2

$\sf \huge{Given - }$

$\sf\angle \: R = 90 \degree , \: PQ = 30cm \: and \: PR = 18cm$

$\huge \sf \: find \: -$

$\sf \: QR = ?$

$\sf \huge{Solution - }$

$\sf \angle \: PRQ = 90 \degree$

$\sf \: by \: pythagoras \: theorem..$

$\sf \: {PQ}^{2} = {PR}^{2} + {QR}^{2}$

$\rightarrow\sf{30}^{2} = {18}^{2} + {QR}^{2}$

$\sf \rightarrow \: 900 = 324 + {QR}^{2}$

$\sf \rightarrow\: 900 - 324 = {QR}^{2}$

$\sf \rightarrow \: 576 = {QR}^{2}$

$\sf \rightarrow \: \sqrt{576} = {QR}$

$\sf \rightarrow \: 24 = QR$

$\sf \therefore \: QR \: is \: \bold \green{24 \: cm.}$