## CBSE Class 10 Maths Chapter-12 Areas Related to Circles – Free PDF Download

Free PDF download of Important Questions with Answers for CBSE Class 10 Maths Chapter 12 – Areas Related to Circles prepared by expert Maths teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.

CBSE Class 10 Maths Chapter-12 Areas Related to Circles Important Questions

**CBSE Class 10 Maths Important Questions Chapter 12 – Areas Related to Circles**

**4 Mark Questions**

**1. The area of an equilateral is 1732.5 cm ^{2} with each centre of the triangle as vertex of circle is drawn with radius equal to half the length of the side of triangle, find the area of the shaded region. **

**Ans.**Area of shaded region = area of (Area of sector BPR)

Let ‘a’ be the side of the equilateral

Required area =

**2. The given figure depicts a racing track whose left and right ends are semi-circular. The difference between the two inner parallel line segments is 60m and they are each 106m long. If the track is 10m wide, find:**

**(i) The distance around the track along its inner edge,**

**(ii) The area of the track.**

**Ans. (i)** The distance around the track along the inner edge

**(ii) **The area of the track

**3. Three horses are tethered with 7m long ropes at the three corners of a triangular field having sides 20m, 34m and 42 m. Find the area of the plot which can be grazed by the horses. Also, find the area of the plot which remains ungrazed.**

**Ans.** Let

Area which can be grazed by three horses = Area of sector with central angle and radius 7 cm + Area of sector with central angle and radius 7 cm + Area of sector with central angle and radius 7 cm

(Sum of three angles of a )

Sides of plot ABC are = 20m, b = 34m and c = 42m

Semi perimeter,

Area of triangular plot

= Area of

Area grazed by the horses = 77m^{2}

Ungrazed area = (336 – 77)

= 259 m^{2}

**4. In given figure, two circular flower beds have been shown on two sides of a square lawn ABCD of side 56m. If the corner of each circular flower bed is the point of enter section O of the diagonals of the square lawn. Find the sum of the areas of the lawn and the flower beds.**

**Ans.** Area of the square lawn ABCD =

Let OA = OB =meter

By Pythagoras theorem,

Again, area of sector OAB =

Also, area of

[Here,Since square ABCD is divided into 4 right triangles]

Area of flowerbed

Similarly, area of the other flowerbed = 448m^{2}

Total area of the lawn and the flowerbeds

=

**5. An elastic belt is placed round the rein of a pulley of radius 5cm. One point on the belt is pulled directly away from the centre O of the pulley until it is at P, 10cm from O. Find the length of the best that is in contact**

**with the rim of the pulley. Also find the shaded area.**

**Ans.**

Arc

Length of the belt that is in contact with the rim of the pulley

= Circumference of the rim – length of arc AB

Now, area of sector OAQB =

Area of quadrilateral OAPB

Hence, shaded area

**6. A round table cover has six equal designs as shown in figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Ts. 0.35 per cm ^{2}. **

**Ans.**= 28 cm and

Area of minor sector =

=

= = 410.67 cm

^{2}

For, Area of AOB,

Draw OMAB.

In right triangles OMA and OMB,

OA = OB [Radii of same circle]

OM = OM [Common]

OMAOMB [RHS congruency]

AM = BM [By CPCT]

AM = BM = AB and AOM

= BOM = AOB =

In right angled triangle OMA,

OM = cm

Also,

AM = 14 cm

2 AM = 2 x 14 = 28 cm

AB = 28 cm

Area of AOB = x AB x OM

= =

= = 333.2 cm

^{2}

Area of minor segment = Area of minor sector – Area of AOB

= 410.67 – 333.2 = 77.47 cm

^{2}

Area of one design = 77.47 cm

^{2}

Area of six designs = 77.47 x 6 = 464.82 cm

^{2}

Cost of making designs = 464.82 x 0.35 = Rs. 162.68