Question

Find the area of the shaded region in Fig. 12.19, if
PQ = 24 cm. PR = 7 cm and O is the centre of the
circle.​

Answer 1

Answer:

Step-by-step explanation:

Given :  

PQ= 24 cm ,PR = 7 cm

We know that any angle made by the diameter QR in the semicircle is 90°.

∠RPQ = 90°

In right angled ∆RPQ

RQ² = PQ² + PR²

[By pythagoras theorem]

RQ² = 24² + 7²

RQ² = 576 + 49  

RQ² = 625

RQ = √625cm

RQ= 25 cm

radius of the circle (OQ)= 25 / 2 cm

Area of right ∆ RPQ= ½ × Base × height

Area of right ∆ RPQ= ½ × RP × PQ

Area of right ∆ RPQ = ½ × 7 × 24 = 7 × 12 = 84 cm²

Area of right ∆ RPQ = 84 cm²

Area of semicircle= πr²/2

= (22/7) × (25/2)² / 2

= (22 × 25 × 25)/ (7× 2 × 2 × 2)

= 11 × 625 /28 = 6875/28 cm²

Area of semicircle = 6875/28 cm²

Area of the shaded region = Area of semicircle - Area of right ∆ RPQ

=  (6875/28  - 84 )cm²

= (6875 - 2532)/ 28

Area of the shaded region = 4523 / 28= 161.54 cm²

Hence, the area of the shaded region = 161.54 cm²

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Answer 2

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Questioη:-

• Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle..?

$\sf\star \: \underbrace\red{Solution} \: \star$

Given:-

• PQ = 24 cm, PR = 7 cm.
• Point O is the centre of the circle.

$\text{\large\underline{\pink{To find:-}}}$

• The area of the shaded region.?

$\text{\large\underline{\red{Explaination:-}}}$

Point P is in the semi-circle,

• $\sf { \implies}$P = 90°

Here, Hypotenuse of the circle =

Here, Hypotenuse of the circle =Diameter of the circle.

• $\sf { \implies}$QR = Diameter

In AQPR,

• Angle P is 90°

According to the Pythagoras theorem,

• $\sf{\implies}$QR² = PR² +PQ²
• $\sf{\implies}$QR² = 7² + 24²
• $\sf{\implies}$QR² = 625
• $\sf{\implies}$QR = 25 cm

$\text{\large\underline{\orange{We know that }}}$

Radius = Diameter/2

• $\sf{\implies}$25/2
• $\sf{\implies}$ 12.5 cm

As we know that:-

Area of the semicircle = πR²/2

• $\sf{\implies}$ 3.14 (12.5)²/2
• $\sf{\implies}$ 3.14 x 156.25/2
• $\sf{\implies}$490.625/2
• $\sf{\implies}$245.3125 cm²

Now,

Area of APQR = ½ Base Height

• $\sf{\implies}$½ 7 * 24
• $\sf{\implies}$7 x 12
• $\sf{\implies}$84 cm²

so,

Area of shaded region = Area of the

Area of shaded region = Area of thesemicircle - Area of APQR

• $\sf{ \implies}$245.3125 - 84
• $\sf{\implies}$ 161.3125 cm²

Therefore,

The area of shaded region is 161.3125 cm²

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