**Maths Class 10
Exercise 12.3**

**1. Find the area of the shaded region in
figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.**

**Solution:**∠RPQ = 90° [Angle in semicircle is 90°]

In right-angled ∆PQR, we have

RQ^{2} = PR^{2} + PQ^{2 }[By
Pythagoras theorem]

= (7)^{2} + (24)^{2}

= 49 + 576

= 625

RQ
= 25 cm

Diameter
of the circle = 25 cm

Radius
of the circle = 25/2 cm

Area
of the semicircle = ½ × Ï€r^{2}

=
6875/28 cm^{2}

Area
of right-angled ∆PQR = ½ × PQ × PR

=
½ × 24 × 7

=
84 cm^{2}

Area
of the shaded region = Area of semicircle – Area of right-angled ∆PQR

^{2}

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

**Solution:**Area of the shaded region = Area of sector OAC – Area of sector OBD

=
154/3 cm^{2} ^{ }

**3. Find the area of the shaded region in
figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.**

**Solution:**Area of the shaded region = Area of square ABCD – (Area of semicircle APD + Area of semicircle BPC)

= 196
– 22/7 × 7 × 7

=
196 – 154 = 42 cm^{2}

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

**Solution:**Area of the shaded region = Area of circle + Area of equilateral triangle OAB – Area of common part to the circle and the triangle (i.e., a sector)

=
36Ï€ + 36√3 – 6Ï€

=
30Ï€ + 36√3

=
30 × 22/7 + 36√3

**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 figure. Find the area of the remaining portion of the square.**

**Solution:**Area of the remaining portion of the square = Area of square – (4 × Area of a quadrant + Area of a circle)

=
16 – 2 × 22/7

=
68/7 cm^{2}

**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 figure. Find the area of the design.**

**Solution:**Area of the design = Area of circular table cover – Area of the equilateral triangle ABC

Since G is the centroid of the equilateral
triangle.

Therefore, radius of the circumscribed
circle = 2h/3 cm

According
to the question, 2h/3 = 32

h = 48 cm

Now,
a^{2} = h^{2} + (a/2)^{2}

a^{2}
= h^{2} + a^{2}/4

a^{2}
– a^{2}/4 = h^{2}

3a^{2}/4
= h^{2}

a^{2}
= 4h^{2}/3

a^{2}
= 4(48)^{2}/3

a^{2}
= 3072

Hence,
the required area =

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

**Solution:**Area of the shaded region = Area of square – 4 × Area of a sector

=
196 – 154

= 42
cm^{2}

**8. Figure given below depicts a
racing track whose left and right ends are semicircular.**

**The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find:****(i) the distance around the track along its
inner edge.**

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

**Solution: (i) **Distance around the track along its
inner edge

=
106 + 106 + 2 × ½ × 2Ï€r

=
212 + 2 × 22/7 × 30

=
212 + 1320/7

=
2804/7 m

=
2120 + 22/7 × 700

= 4320
m^{2}

**9. In figure, 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.**

**Solution:**Area of the shaded region = Area of the smaller circle + Area of semicircle ACB – Area of ∆ACB

=
77/2 + 187 – 49

=
133/2

=
66.5 cm^{2}

**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 the length of the side of the triangle (see
figure). Find the area of the shaded region. (Use **

**Ï€**

**= 3.14 and √3 = 1.73205)**

**Solution:**Area of equilateral triangle = √3a

^{2}/4 = 17320.5

*a* = 200 cm

Area of the shaded region = Area of ∆ABC –

= 17320.5 – 15700

= 1620.5 cm^{2}

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

**Solution:**Area of the remaining portion of the handkerchief = Area of square ABCD – Area of 9 circular designs

=
1764 – 1386

= 378
cm^{2}

**12. In figure, 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: (i) **Area of the quadrant OACB =

^{2}

**(ii)** Area of the shaded region = Area
of the quadrant OACB – Area of ∆OBD

=
77/8 – ½ × OB × OD

=
77/8 – ½ × 3.5 × 2

= 77/8 – 35/10

= 49/8 cm^{2}

**13. In figure, a square OABC is inscribed in a
quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use
****Ï€**** = 3.14) **

= √2 OA = √2 × 20 = 20√2 cm

Area
of the shaded region = Area of the quadrant OPBQ – Area of square OABC

= 1/4
× 3.14 × 800 – 400

= 200
× 3.14 – 400

= 228
cm^{2}

**14. AB and CD are respectively arcs of two
concentric circles of radii 21 cm and 7 cm and centre O (see figure). If **∠**AOB = ****30°,**** find the area of the
shaded region.**

=
231/2 – 77/6

=
(692 – 77)/6

=
616/6

=
308/3 cm^{2}

**15. In figure, ABC is a quadrant of a circle
of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of
the shaded region.**

^{2}= AB

^{2}+ AC

^{2}[By Pythagoras theorem]

BC^{2}
= (14)^{2} + (14)^{2} = 2(14)^{2}

BC
= 14√2 cm

Radius
of the semicircle = 14√2/2 = 7√2 cm

Area
of the shaded region = Area of the semicircle BCQB – Area of segment BCPB

= Area of the semicircle BCQB – (Area of the
quadrant BACPB – Area of ∆ABC)

=
154 – (154 – 98)

=
98 cm^{2}

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

**Solution:**In right-angled triangle ADC, AC

^{2}= AD

^{2}+ CD

^{2}[By Pythagoras theorem]

AC^{2}
= (8)^{2} + (8)^{2} = 2(8)^{2}

AC
= 8√2 cm

Draw
BM perpendicular to AC.

Then
AM = MC = ½ AC = ½ × 8√2 = 4√2 cm

In
right-angled triangle AMB,

AB^{2}
= AM^{2} + BM^{2} [By Pythagoras theorem]

(8)^{2
}= (4√2)^{2} + BM^{2}

BM^{2 }= 64 – 32 = 32

BM
= 4√2 cm

Area
of ∆ABC = ½ × AC × BM

=
32 cm^{2}

=
16 × 22/7 – 32

=
128/7 cm^{2}

Area
of the designed region = 2 × 128/7 = 256/7 cm^{2}

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