Question

.....
In fig. 12.30 , OABC is quadrant of a circle with centre o and the radius is 3.5cm .If OD = 2cm, find the area of the

i) quadrant OACB
ii) shaded region .

Answer 1

⏩hello sissy⏪

☺here is ur answer☺

:Given → OD = 2cm
→radius(r)=3.5 cm

( ¡ )

:to find → area of Quadrant OACB.


soln →area of quadrant OACB = ¼πr²

=¼ × 22/7 × {3.5}²


=11×35×35
________cm²
14×10×10

=77
___cm²
8

area of quadrant OABC is 77/8cm²

________________

( ¡¡ )

:to find →shaded area ACBD


soln→shaded portion = area of quadrant OACB-area of ∆BDO

=(77/8)-½×b×h

=(77/8)-½×3.5×2cm²

=(77/8)-3.5cm²

=49
___cm²
8

area of shaded portion is (49/8)cm²

!@!@!@!@!@!@!@!@!@!

hope it helps☑___________with reGarDs ================ SnehaG☑

Answer 2

Answer:

9.625 cm², 6.125 cm²

Step-by-step explanation:

Given, radius of Quadrant r = 3.5 cm.

Angle of sector = 90°

(i) Area of Quadrant OACB:

We know that Area of sector = (θ/360°) * πr²

= (90/360) * (22/7) * (3.5)²

= (1/4) * (22/7) * 12.25

= 77/8

= 9.625 cm²

Therefore, Area of Quadrant OACB = 9.625 cm²

(ii) Area of ΔOBD

Given, OB = 3.5 cm and OD = 2 cm.

Area = (1/2) * OB * OD

        = (1/2) * 3.5 * 2

        = 3.5 cm²

(iii) Area of shaded region:

= Area of quadrant OACB - Area of ΔOBD

= 9.625 - 3.5

= 6.125 cm²

Therefore, Area of shaded region = 6.125 cm²

Hope it helps!