# PSEB Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3

PSEB Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3

## PSEB 10th Class Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3

Question 1.

Find the area of the shaded region in Fig., If PQ = 24 cm, PR =7 cm and O is the centre of the circle.

Question 2.

Find the area of the shaded region in Fig., if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Question 3.

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

Question 4.

Find the area of the shaded region in fig., where a circular arc of radius 6 cm has been drawn iith vertex O of an equilateral triangle OAB of side 12 cm as centre.

Shaded Area = Area of equilateral triangle OAB + Area of major sector of circle

= 62.28 + 94.28 = 62.28 cm^{2}

Shaded Area = Area of equilateral triangle OAR + Area of major sectoç of circle

= 62.28 + 94.28 = 156.56

Shaded Area = 156.56 cm^{2}.

Question 5.

From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in fig. Find the area of the remaining portion of the square.

Solution:

Side of square = 4 cm

Radius of each semi circle cut out (r) = 1 cm

Question 6.

In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as

shown in fig. Find the area of the design (shaded region).

Question 7.

In fig., ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

Question 8.

Fig. depicts a racing track whose left and right ends are semicircular. The distahce between the two inner parallel line segments is 60 m and they are each 106 m long. 1f the track is 10 m wide, find

(i) the distance around the track along its inner edge.

(ii) the area of the track.

(i) Here AB = DC = 106 m

AF = BE = CG = HD = 10m

Diameter of inner semicircle (APD and BRC) =60 m

∴ Radius of inner semicircle (APD (r) = 30 m

Radius of outer semicircle (R) = r + 10 = 30 + 10 = 40 m

Distance around the track along inner edge = AB + circumference of semi circle BRC + CD + circumference of semi circle DPA

= 2 AB + 2 [circumference of semi circle BRC]

= 2 (106) + 2((2πr/2))

= 212 + 2πr

= 212 + 2 × 22/7 × 30

= 212 + 60×22/7

= 212 + 188.57 = 400.57 m.

∴ Distance around the track along its inner edge = 400.57 m

(ii) Area of track = Area of rectangle ABEF + Area of region BEMGCRB + Area of rectangle CGHD + area of region.

= 2 Area of rectangle ABCD + 2 Area of region (II)

= 2 (AB × AF) + 2

[Area of semi circle with Radius 60 cm – Area of semi circle with radius 30 cm]

= 2 [106 × 10] + 2 [latex]\frac{\pi \mathrm{R}^{2}}{2}-\frac{\pi r^{2}}{2}[/latex]

= 2 × 1060 + 2π/2 [R^{2} – r^{2}]

= 2120 + 22/7 (40^{2} – 30^{2})

= 2120 + 22/7 [1600 – 900]

= 2120 + 22/7 [700]

= 2120 + 2200 = 4320 m^{2}

Area of track = 4320 m^{2}

Question 9.

In Fig., AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Question 10.

The area of an equilateral triangle ABC is 17320.5 cm^{2}. With each vertex of the triangle as centre, a circle is drawn with radius equal to half of the length of the side of the triangle (see Fig.). Find the area of the shaded region.

(Use n = 3.14 and ,√3 = 1.73205)

Solution:

Area of equilateral triangle ABC = 17320.5 cm^{2}

Area of three sector = 3 × 15700/3 cm^{2}

∴ Required shaded Area = Area of triangle – Area of three sectors

= 17320.5 – 15700 = 1620.5 cm^{2}

∴ Hence, Required shaded Area = 1620.5 cm^{2}

Question 11.

On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig). Find the area of the remaining portion of the handkerchief.

Question 12.

In Fig., OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the

(i) quadrant OACB,

(ii) shaded region.

Solution:

∴ Shaded Area = Area of quadrant OACB – Area of ∆ODB

= 9.625 – 3.5 = 6.125 cm^{2}

∴ Hence, Shaded Area = 6.125 cm^{2}.

Question 13.

In fig., a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region.

Solution:

Side of square ABCO = 20 cm

∠AOC = 90°

AB = OA

Question 14.

AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O. If ZAOB = 30°, find the area of the shaded region.

Area of smaller sector (ODC) = 12.83 cm^{2}

Now, Shaded Area = Area of bigger sector OAB – Area of smaller sector OCD

= 115.5 – 12.83 = 102.66

Hence, Shaded Area = 102.66 cm^{2}.

Question 15.

In fig., ABC is a quadrant of a circle of radius 14 cm and a semi circle is drawn with BC as diameter. Find the area of the shaded region.

Solution:

Radius of quadrant ACPB (r) = 14 cm

Sector angle (θ) = 90°

AB = AC = 14 cm

Required Area = Area of semi circle – [Area of sector – Area of ∆BAC]

= 154 – [154 – 98]

= (154 – 56) cm^{2} = 98 cm^{2}

Hence, Shaded Area = 98 cm^{2}.

Question 16.

Calculate the area of the designed region in fig. common between the two quadrants of circles of radius 8 cm each.

Area of sector = 50.28 cm^{2}

Area of ∆ABD = 1/2 × AB × AD

= 1/2 × 8 × 8

= 32 cm^{2}

∴ Area of segment DMBPD = Area of sector ∆BPD – Area of ∆ABD

= 50.28 – 32 = 18.28 cm^{2}

Hence, Shaded area = 2 area of segment DMBPD = 2 (18.28) = 36.56 cm^{2}

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