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.**

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

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

Now, the total required perimeter

**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.**

**Sol.** Length of the paper = 40 cm, width of the paper = 28 cm.

âˆ´ Area of the rectangle =length Ã— breadth = 40 Ã— 28 cm^{2} = 1120 cm^{2}

Again, diameter of semi-circle = 28 cm.

â‡’ Radius of the semi-circle = 28/2 = 14 cm.

âˆ´ Area of the semi-circle

âˆ´ Area of the remaining paper = 1120 cm^{2} âˆ’ 308 cm^{2} = 812 cm^{2}.

**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.**

**Sol.** Side of the square = 14 cm

âˆ´ Area of the square = 14 Ã— 14 cm^{2} = 196 cm^{2}

Also, diameter of each semi-circle = side of the square = 14 cm.

â‡’ Radius = 14/2 = = 7 cm.

Area of 1 semi-circle = 1/2 Ï€r^{2}

â‡’ Area of both semi-circles = 2 Ã— 77 cm2

= 154 cm^{2}.

Now, the area of the shaded region = [Area of the square ABCD] âˆ’ [Area of both the semi-circles]

= (196 âˆ’ 154) cm^{2} = 42 cm^{2}.

**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 Ï€ = 22/7)**

**Sol. **Here diameter = 21 cm â‡’ r = 21/2

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 Find the area of the park excluding the lawn.**

**[Take Ï€ = 22/7]**

**Sol.** Length of the park l = 120 m

Breadth of the park b = 100 m

âˆ´ Area of the park excluding the central park

= 12000 âˆ’ 3300 m^{2}

= 8700 m^{2}.

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

**Sol.** Here, the major sector angle is given by Î¸ = 360Â° âˆ’ 120Â° = 240Â°

âˆ´ Circumference of the sector APB

âˆ´ Perimeter of OAPBO = [Circumference of sector AOB] + OA + OB]

**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.**

**Sol. **We have AB = 28 cm

âˆ´

â‡’ BQ = 28 âˆ’ 7 = 21 cm

âˆ´ Area of the semi-circle having diameter as 21 cm

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

Thus the area of the shaded region

= 192.5 cm^{2}.

**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 Ï€ = 22/7]**

**Sol.** Side of the square = 28 m

âˆ´ Area of the square PQRS = 28 Ã— 28 m^{2}

Diameter of a semi-circle = 28 m

â‡’ Radius of a semi-circle = 28/2 = 14m

âˆ´ Area of 1 semi-circle = 1/2Ï€r^{2} =

â‡’ Area of both the semi-circles = 2 Ã— 308 m^{2} = 616 m^{2}

âˆ´ Area of the square left uncovered = (28 Ã— 28) âˆ’ 616 m^{2} = 784 âˆ’ 616 m^{2} = 168 m^{2}.

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

**Sol.** Let ABCD be the square such that

AB = BC = 10 cm

âˆ´ AC^{2} = AB^{2} + BC^{2}

AB^{2} + BC^{2} = (10 Ã— 2)

â‡’ x^{2} + x^{2} = (20)2

[Let AB = BC = x]

â‡’ 2x^{2} = 400

â‡’

âˆ´ Area of the square = 200 cm^{2}.

**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.**

Sol. O is the centre of the circle.

âˆ´ AB is its diameter.

In right Î” ABC,

AC^{2} + BC^{2} = AB^{2}

â‡’ 12^{2} + 16^{2} = AB^{2}

â‡’ 144 + 256 = AB^{2} â‡’ AB^{2} = 400

â‡’

âˆ´ Circumference of semi-circle ACB

âˆ´ Perimeter of the shaded region = 22/7 cm + 12 cm + 16 cm

= 31.43 cm + 12cm + 16cm

= 59.43 cm

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

132 docs