10th Grade > Mathematics > Area
AREAS RELATED TO CIRCLES MCQs
:
Area of a square = side2
Side of the square = 50 cm
Area of square Handkerchief = 502 = 2500 cm2
Area of circles = π×r2
Radius of the circle = 7 cm
Total area of circles = 7 x 227 x 72 = 1078cm2
Remaining area = 2500 - 1078 = 1422 cm2
:
D
Area of a sector with angle θ is given by =θ360° × π × r2 ( where θ is the angle made by the sector )
Required area = Area of sector OABO – Area of sector OCDO
The angle formed by the sector OABO and sector OCDO = 90°
=90360×227×(21)2–90360×227×(14)2
=14×227×{(21)2–(14)2}
=14×227×{(35)(7)} [a2−b2=(a+b)(a−b)]
=192.50 m2
:
B
The circumference i.e , perimeter of a sector of angle P∘ of a circle with radius R is given by
P∘360 ×2πr +2R
=P∘360 × 2π R+2R
=60∘360 × 2π (20)+2(20)
= 20.93 + 40
= 60.93 cm
:
Given that radius r = 6 cm
θ = 70∘
Area of sector of an angle θ with radius r
=θ360 × π × (r)2
= 70360 × 227 × 62
= 22 cm2
:
Here we are asked to find the area of the shaded area. The area of the shaded area should be the difference between the area of the square and the area of the circle.
Here we are given the side of the square, x = 14 cm
The area of the square is = side2 = x2 = 142
= 196cm2
Therefore the radius of the circle will be, r =d2
r = 142 = 7 cm
Area of the circle,
A=π×r2 (use π = 22/7)
= (227)×72 = 154 cm2
Area of the shaded portion = Area of the square - Area of the circle
= 196 - 154 =42 cm2
Therefore the area of the shaded portion will be 42 cm2
:
B
A = πr2
Given r = 7√π cm
A=π×7√π×7√π
=49 cm2
:
B
Area of a circle is directly proportional to the square of the radius of circle.
:
D
Area of a sector of angle p=p360×πR2=p360×22×πR2=p720×2πR2
:
Radius = 42 cm
8 ribs implies angle subtend between consecutive ribs = 3608 = 45∘ .
Area between consecutive ribs = θ360×π×(radius)2
45360 × 227 × 422 = 693 cm2
:
For the quadrant θ = 90°
Area of the quadrant = θ360°×π(r)2
Area of quadrant OACB = 90360 × 227 (7)2 = 772
Area of shaded region = Area of quadrant OACB - area of ΔOAD
= 772 - 12 × 7 × 4
= 24.5 cm2