Question

If PQ = 7 cm and QR = 3RS = 6 cm, find the perimeter of the cyclic quadrilateral PQRS.​

Answer 1

Step-by-step explanation:

24 cm

In △PQR,

∠PQR=90

∘

(Angle in a semi circle)

PR

2

=PQ

2

+QR

2

(Pythagoras theorem)

PR

2

=7

2

+6

2

PR=

85

Now, In △PRS

∠PSR=90

∘

(Angle is a semi circle)

PR

2

=PS

2

+SR

2

85=PS

2

+4

PS=

81

=9

Thus, perimeter of the quadrilateral = PS+SR+QR+PQ = 9+2+6+7=24 cm

Answer 2

Step-by-step explanation:

☘$Question:-$

The following figure shows a circle with PR as its diameter. If PQ = 7 cm and QR = 3RS = 6 cm, find the perimeter of the cyclic quadrilateral PQRS.

☘$Answer:-$

Given:-

• PQ = 7 cm and
• QR = 3RS = 6 cm

To find:-

• the perimeter of the cyclic quadrilateral PQRS.

Solution:-

$In \: \: the \: \: figure,$

$PQRS \: \: is \: \: a \: \: cyclic \: \: quadrilateral \: \\ in \: \: which \: \: PR \: \: is \: \: a \: \: diameter$

$PQ \: = \: 7 cm$

$QR \: = \: 3 RS \: = \: 6 cm$

$3 RS \: = \: 6 cm \\ RS \: = \: 2 cm$

$Now \: \: in \: \: ΔPQR,$

$∠Q = 90° \: \: (Angles \: in \: a \: semi \: circle)$

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

$(Pythagoras \: \: theorem )$

$= 7 {}^{2} \: + \: 6 {}^{2}$

$= 49 + 36$

$= 85$

$Again \: \: in \: \: right \: \: Δ PSQ, \: PR {}^{2} = \: PS {}^{2} \: + RS {}^{2}$

$= > \: 85 \: = \: PS {}^{2} \: + \: 2 {}^{2}$

$= > \: PS { }^{2} \: = 85 - 4 \\ = 81 \\ = (9) {}^{2}$

$∴ \: PS \: = 9cm$

$Now, \: \: perimeter \: \: of \: \: quad \: \: PQRS \:$

$= PQ \: + \: QR \: + \: RS \: + \: SP$

$= \: (7 \: + \: 9 \: + \: 2 \: + \: 6) \: cm$

= $24cm$

✿ Hence, the perimeter of the cyclic quadrilateral PQRS is 24 cm.