Question

In APQR and AXYZ, it is given that APQR-AXYZ, ZY+ZZ=90° and
XY : XZ = 3:4. Then find the ratio of sides in APQR.​

Answer 1

Answer:

$\boxed{\bf \: PQ : QR : PR \: = \: 3 : 5 : 4 \: }$

Step-by-step explanation:

Given that, In triangle XYZ

$\sf \: \angle \: Y + \angle \: Z = {90}^{ \circ}$

We know, Sum of all interior angles of a triangle is 180°

So,

$\sf \: \angle \: X + \angle \: Y + \angle \: Z = {180}^{ \circ}$

$\sf \: \angle \: X + {90}^{ \circ} = {180}^{ \circ}$

$\sf \: \angle \: X = {180}^{ \circ} - {90}^{ \circ}$

$\implies\sf \: \angle \: X = {90}^{ \circ}$

Further given that,

$\sf \: XY : XZ = 3 : 4$

$\implies\sf \: \dfrac{XY}{XZ} = \dfrac{3}{4}$

Also, given that

$\sf \: \triangle \: PQR \: \sim \: \triangle \: XYZ$

By corresponding parts of similar triangles, we have

$\implies\sf \: \angle \: P = \angle \: X = {90}^{ \circ}$

and

$\sf \: \dfrac{PQ}{PR} = \dfrac{XY}{XZ}$

$\implies\sf \: \dfrac{PQ}{PR} = \dfrac{3}{4}$

Let assume that

$\sf \: PQ = 3k$

$\sf \: PR = 4k$

Now, In right-angle triangle PQR

By Using Pythagoras Theorem, we have

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

$\sf \: {QR}^{2} = {(3k)}^{2} + {(4k)}^{2}$

$\sf \: {QR}^{2} = {9k}^{2} + {16k}^{2}$

$\sf \: {QR}^{2} = {25k}^{2}$

$\sf \: {QR}^{2} = {(5k)}^{2}$

$\implies\sf \: QR = 5k$

Now, Consider

$\sf \: PQ : QR : PR \:$

$\sf \: = \: 3k : 5k : 4k \:$

$\sf \: = \: 3 : 5 : 4 \:$

Hence,

$\implies\sf \: \boxed{\bf \: PQ : QR : PR \: = \: 3 : 5 : 4 \: }$