10th Grade > Mathematics > Area
AREAS RELATED TO CIRCLES MCQs
:
B
The circumference of a circle with radius r is given by
2πr.
Hence,
2πr=528
r=5282π =84 cm
(π =227)
:
Here we need to find the area of the shaded part of the square. For finding the area of the shaded region we can find the area of the bigger square and the smaller square and then subtract the area of the smaller square from area of the larger square.
Area of the larger square =x2
= 82
= 64cm2
Area of the smaller square = b2
= 42
= 16 cm2
The area of the shaded region will be = Area of the bigger square – Area of the smaller square
= 64 – 16
= 48 cm2
Therefore the area of the shaded region will be 48cm2.
:
The area of the sector = 120360×π×122 = 150.8 cm2
Perpendicular from chord is drawn to the centre bisects the chord. The angle subtended by each triangle at the centre is 60∘.
Height of perpendicular = r x cos 60∘ = 12 x 0.5 = 6 cm.
Length of chord = 2 × r × sin 60∘ = 24 × √32= 20.6 cm
The area of triangle = 0.5 × 20.6 × 6 = 61.8 cm2
The area of segment = 150.8 - 61.8 = 89 cm2
:
D
Area of a circle of radius r
=πr2
Area
=π x 7√π x 7√π=49 cm2
:
B
Area of a circle
=πr2
From Figure, the diameter of circle is 14 cm. Two semi-circles make one full circle.
∴ The area of one full circle is
=227×72=154 cm2
The total area of square
=142=196 cm2
The area of shaded portion = [Area of square- Area of full circle]
= 196 - 154 = 42cm2.
Hence, area of shaded region
=42 cm2
:
Area of shaded region = area of ΔABC - 3 (Area of sector BPR)
Let 'a' be the side of the equilateral ΔABC.
Using area of an equilateral triangle = √34a2,
√34a2 = 17320.5
Solving, a2=17320.5×4√3
⟹a2=17320.5×41.73205
⟹a2=17320.5×417320.5×10−4
⟹a=2×102
⟹ a = 200 cm.
Radius of the circles = 12×200 = 100 cm
Now, using area of a sector when the degree measure of the angle at the centre is θ = θ360πr2
∴ Required area =17320.5 - 3[60360×3.14×1002 ]
∴ Required area = 1620.5cm2
:
A
Circumference of the circle
=2πr =528 cm
(where r is the radius of the circle)
⇒r=5282π =84 cm
Now, area of the circle = πr2
= 227 x 842 = 22176 cm2
:
B
Area of a square plate
=side2
Given length of the side of the square plate = 40 cm
Area of square plate
=402
=1600 cm2
Area of a circle
=π r2
There are 336 holes of radius 1 cm each.
Total area of circles
=336× 227 × 12
=1056 cm2
Remaining area = [Area of square plate- Total area of circles]
=1600−1056
=544 cm2
∴ Area of remaining square plate
=544 cm2
:
B
Area of the sector of angle θ = θ360 × πr2
A = 60360 × 3.14 × (62) = 18.85 cm2
:
D
Perimeter of sector = 2r + Length of the arc of sector
From the given data,
Length of arc of sector = 27.2 - 2 × 5.7
Length of arc of sector = 27.2 - 11.4 = 15.8 m.