Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

Class 10 Mathematics by VP Classes

Class 10 : Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

The document Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.
All you need of Class 10 at this link: Class 10

VERY SHORT ANSWER TYPEQUESTIONS

Q1. PQRS is a diameter of a circle of radius 6 cm. The equal lengths PQ, QR and RS are drawn on PQ and QS as diameters, as shown in figure. Find the perimeter of the shaded region.

Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

Sol. Diameter PS = 12 cm      [∵ Radius OS = 6 cm]

Since PQ, QR and RS are three equal parts of diameter,

Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

Now, the total required perimeter

Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

Q2. A sheet of paper is in the form of a rectangle ABCD in which AB = 40 cm, and BC = 28 cm. A semi-circlular portion with BC as diameter is cut off. Find the area of the remaining paper.

Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

Sol. Length of the paper = 40 cm, width of the paper = 28 cm.
∴ Area of the rectangle =length × breadth = 40 × 28 cm2 = 1120 cm2
Again, diameter of semi-circle = 28 cm.
⇒ Radius of the semi-circle = 28/2 = 14 cm.

∴ Area of the semi-circle 

Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

∴ Area of the remaining paper = 1120 cm2 − 308 cm2 = 812 cm2.

Q3. Find the area of the shaded region in the figure, if ABCD is a square of side 14 cm and APD and BPC are semi-circles.

Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

Sol. Side of the square = 14 cm
∴ Area of the square = 14 × 14 cm2 = 196 cm2

Also, diameter of each semi-circle = side of the square = 14 cm.
⇒ Radius =  14/2 = = 7 cm.
Area of 1 semi-circle = 1/2 πr2

Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

⇒ Area of both semi-circles = 2 × 77 cm2
= 154 cm2.
Now, the area of the shaded region = [Area of the square ABCD] − [Area of both the semi-circles]
= (196 − 154) cm2 = 42 cm2.

Q4. Find the perimeter of the shaded region, if ABCD is a square of side 21 cm and APB and CPD are semi-circles. (use π = 22/7)

Sol. Here diameter = 21 cm ⇒  r = 21/2

Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

Thus the required perimeter of shaded region is 108 cm.

Q5. A park is in the form of a rectangle 120 m long and 100 m wide. At the centre, there is a circular lawn of radius  Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev Find the area of the park excluding the lawn.

[Take π = 22/7]

Sol. Length of the park l = 120 m
Breadth of the park b = 100 m

Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev
Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

∴ Area of the park excluding the central park

= 12000 − 3300 m2
= 8700 m2.

Q6. In the given figure, OAPB is a sector of a circle of radius 3.5 cm with the centre at O and ∠AOB = 120°. Find the length of OAPBO.

Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

Sol. Here, the major sector angle is given by θ = 360° − 120° = 240°

Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

∴ Circumference of the sector APB
Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev
∴ Perimeter of OAPBO = [Circumference of sector AOB] + OA + OB]

Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev
Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

Q7. Find the area of the shaded region of the following figure, if the diameter of the circle with centre O is 28 cm and AQ = 1/4 AB.

Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

Sol. We have AB = 28 cm

∴ Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev
⇒ BQ = 28 − 7 = 21 cm
∴ Area of the semi-circle having diameter as 21 cm
Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

∴ Also area of the semi-circle having diameter as 7 cm

Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

Thus the area of the shaded region

Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

= 192.5 cm2.

Q8. PQRS is a square land of side 28 m. Two semi-circular grass covered postions are to be made on two of its opposite sides as shown in the figure. How much area will be left uncovered? [Take π = 22/7]

Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

Sol. Side of the square = 28 m
∴ Area of the square PQRS = 28 × 28 m2
Diameter of a semi-circle = 28 m

⇒ Radius of a semi-circle = 28/2 = 14m

∴ Area of 1 semi-circle = 1/2πr2  =  Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

⇒ Area of both the semi-circles = 2 × 308 m2 = 616 m2

∴ Area of the square left uncovered = (28 × 28) − 616 m2 = 784 − 616 m2 = 168 m2.

Q9. Find the area of a square inscribed in a circle of radius 10 cm.

Sol. Let ABCD be the square such that

Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

AB = BC = 10 cm
∴ AC2 = AB2 + BC2
AB2 + BC2 = (10 × 2)
⇒ x2 + x2 = (20)2
[Let AB = BC = x]
⇒ 2x2 = 400

⇒  Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

∴ Area of the square = 200 cm2.

Q10. In the given figure, O is the centre of a circular arc and AOB is a straight line. Find the perimeter of the shaded region.

Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

Sol. O is the centre of the circle.
∴ AB is its diameter.
In right Δ ABC,
AC2 + BC2 = AB2
⇒ 122 + 162 = AB2
⇒ 144 + 256 = AB2 ⇒ AB2 = 400

⇒ Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev
∴ Circumference of semi-circle ACB
Very Short Answer Type Questions- Areas Related to Circles Class 10 Notes | EduRev

∴ Perimeter of the shaded region = 22/7 cm + 12 cm + 16 cm
= 31.43 cm + 12cm + 16cm
= 59.43 cm

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