Jharkhand Board JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 Textbook Exercise Questions and Answers.

## JAC Board Class 10 Maths Solutions Chapter 12 Areas Related to Circles Exercise 12.3

Unless stated otherwise, use π = \(\frac{22}{7}\)

Question 1.

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

Solution:

Given, PQ = 24 cm, PR = 7 cm.

We know any triangle drawn from diameter RQ to the circle is 90°.

Here, ∠RPQ = 90°

In right ΔRPQ, RQ^{2} = PR^{2} + PQ^{2} (By Pythagoras theorem)

RQ^{2} = 7^{2} + 24^{2}

RQ^{2} = 49 + 576

RQ^{2} = 625

RQ = 25 cm

∴ Area of ΔRPQ = \(\frac{1}{2}\) × RP × PQ

= \(\frac{1}{2}\) × 7 × 24 = 84 cm^{2}

∴ Area of semi-circle = \(\frac{\pi r^2}{2}=\frac{22}{7 \times 2}\left(\frac{25}{2}\right)^2\) (∵ r = \(\frac{\mathrm{PQ}}{2}=\frac{25}{2}\) cm)

= \(\frac{11 \times 625}{28}=\frac{6875}{28} \mathrm{~cm}^2\)

∴ Area of the shaded region = Area of the semi-circle – Area of right ΔRPQ

= \(\frac{6875}{28}-84\)

= \(\frac{6875-2352}{28}=\frac{4523}{28}\) cm^{2}.

Question 2.

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

Solution:

R – radius of the bigger circle, r – radius of the smaller circle.

Area of the shaded portion = Area of sector OAC – Area of sector OBD

Question 3.

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

Solution:

Area of the shaded portion = Area of the square – 2 × Area of one semicircle

= (14 × 14) – 2 × \(\frac{\pi r^2}{2}\) [∵ r = 7]

= 14 × 14 – 2 × \(\frac{22}{7} \times \frac{7 \times 7}{2}\)

= 196 – 154 = 42 cm^{2}.

Question 4.

Find the area of the shaded region in the figure, where a circular are 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 portion = Area of the circle of radius 6 cm + Area of equilateral ΔABC – Area of the sector

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

Solution:

Area of the shaded portion = Area of the square – Area of the 4 quadrants – Area of the circle

= (4 × 4) – 4 × area of one quadrant – area of the circle

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 the figure. Find the area of the design.

Solution:

Area of the design (shaded region) = Area of the circle – Area of ΔABC

Area of equilateral triangle ABC = \(\frac{\sqrt{3} a^2}{4}\)

In ΔABC, AL ⊥ BC.

Alternative Method:

Area of shaded region

= 3[Area of segments]

= 3[Area of sector – Area of ΔOBC]

= 3[\(\pi r^2 \frac{\theta}{360^{\circ}}-\frac{1}{2}\) × BC × OL]

Question 7.

In the 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 the square – Area of the 4 quadrants

= Area of the square – 4 × area of one quadrant

= (14)^{2} – 4 × \(\frac{1}{4}\)πr^{2}

= (14)^{2} – 4 × \(\frac{1}{4} \times \frac{22}{7}\) × 7 × 7

= 196 – 154

= 42 cm^{2}.

Question 8.

The figure 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 inner track

ii) Area of the track:

Area of the track = l × b + l × b + 2\(\left[\frac{\pi \mathrm{R}^2}{2}-\frac{\pi \mathrm{r}^2}{2}\right]\) r = 30 mts, R = (30 + 10) = 40 mts.

= 106 × 10 + 106 × 10 + 2 × \(\frac{\pi}{2}\)(R^{2} – r^{2})

= 1060 + 1060 + \(\frac{22}{7}\)[40^{2} – 30^{2}]

= 2120 + \(\frac{22}{7}\)[1600 – 900]

= 2120 + \(\frac{22}{7}\)[700]

= 2120 + 2200

= 4320 m^{2}.

Question 9.

In the 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:

Given,

OA = 7 cm

∴ OD = 7 cm

Now, area of smaller circle whose diameter (OD) = 7 cm is

Now,

Area of ΔABC = \(\frac{1}{2}\) × AB × OC

= \(\frac{1}{2}\) × 2 × OA × OC

= \(\frac{1}{2}\) × 14 × 7 (∵ OA = OC)

= 49 cm^{2}.

Area of semi-circle ABCA = \(\frac{\pi r^2}{2}=\frac{22}{7 \times 2}(7)^2\)

= 77 cm^{2}

∴ Area of segments BC and AC = Area of semi-circle – Area of AABC

= 77 – 49 = 28 cm^{2}

∴ Area of total shaded region = Area of small circle + Area of segments BC and AC

= \(\frac{77}{2}\) + 28

= 38.5 + 28

= 66.5 cm^{2}.

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 the length of the side of the triangle. Find the area of the shaded region. (Use π = 3.14 and \(\sqrt{3}\) = 1.73205).

Solution:

Area of the shaded portion = Area of equilateral ΔABC – Area of sector Axy – Area of sector Bxz – Area of sector Cyz.

π = 3.14, θ = 60°, r = ?, r = \(\frac{a}{2}\)

Area of the shaded portion = Area of the equilateral Δ – 3 × Area of one sector

Area of the shaded region = Area of ΔABC – \(\frac{3 \times \pi r^2 \theta}{360}\)

= 17320.5 – \(\frac{3 \times 3.14 \times 100 \times 100 \times 60}{360}\)

= 17320.5 – 1.57 × 10000

= 17320.5 – 15700.0

= 1620.5 sq.cm.

Question 11.

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

Solution:

Number of circular designs = 9

Radius of the circular design = 7 cm

There are three circles in one side of the square handkerchief.

∴ Side of the square = 3 × diameter of circle = 3 × 14 = 42 cm

Area of the square = 42 × 42 cm^{2} = 1764 cm^{2}

Area of the circle = πr^{2} = \(\frac{22}{7}\) × 7 × 7 = 154 cm^{2}

Total area of the design = 9 × 154 = 1386 cm^{2}

Area of the remaining portion of the handkerchief = Area of the square – Total area of the design

= 1764 – 1386

= 378 cm^{2}.

Question 12.

In the 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:

Question 13.

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

Solution:

Radius of the quadrant = Diagonal of the square (OB)

OB^{2} = OA^{2} + AB^{2}

= 20^{2} + 20^{2}

= 400 + 400

= 800

OB2 = 400 × 2

OB = \(\sqrt{400 \times 2}\) = 20\(\sqrt{2}\)

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

= \(\frac{1}{4}\) πr^{2} – (OA)^{2}

= \(\frac{1}{4}\) × 3.14 × 20\(\sqrt{2}\) × 20\(\sqrt{2}\) – (20)^{2}

= \(\frac{1}{4}\) × 3.14 × 400 × \(\sqrt{4}\) – 400

= \(\frac{1}{4}\) × 3.14 × 400 × 2 – 400

= 100 × 2 × 3.14 – 400

= 100 × 2(3.14 – 2)

= 200 × 1.14

= 228 sq.cm.

Question 14.

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

Solution:

R = 21, r = 7, θ = 30°, π = \(\frac{22}{7}\)

Area of the shaded region = Area of sector OAB – Area of sector OCD

Question 15.

In the 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.

Solution:

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

BC^{2} = 196 + 196 = 392

BC = \(\sqrt{392}\) = \(\sqrt{196 \times 2}\) = 14\(\sqrt{2}\) = d

r = \(\frac{14 \sqrt{2}}{2}=7 \sqrt{2}\)

Radius of the sector = 14.

Area of the shaded region Area of the semicircle BEC – Area of the segment BDCB

Question 16.

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

Solution:

Area of the design = Area of sector DXB + Area of ΔDCB

Area of the segment DXB = Area of the sector DXBC – Area of ΔDCB