Students can Download Class 10 Maths Chapter 5 Areas Related to Circles Ex 5.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 10 Maths Chapter 5 Areas Related to Circles Ex 5.3

{Unless stated otherwise, use }

Question 1.

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

Answer:

PQ = 24 cm, PR = 7 cm,

‘O’ is the centre of circle.

Angle in semicircle,

∠RPQ = 90°

In ⊥∆RPQ, ∠P = 90°

RQ = RP^{2} + PQ^{2}

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

= 49 + 576

RQ^{2} = 625

∴ RQ = 25 cm.

Diameter, RQ = 25 cm.

∴ Radius, OR = OQ = cm.

Area of shaded part :

= Area of semicircle – Area of ∆RPQ

Question 2.

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

Answer:

Area of shaded region

= Are of sector AOC – area of sector OBD

= 51.33cm^{2}

Question 3.

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

Answer:

Radius of semicircle 7 cm & side of square = 14cm.

Area of shaded region = Area of square ABCD – Areas of semicircles (APD + BPC)

= 196 – \(\frac{22}{7}\) × (7)^{2}

= 196 – \(\frac{22}{7}\) × 7 × 7

= 196 – 154

= 42 cm^{2}

Question 4.

Find the area of the shaded region in the given 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.

Answer:

Radius of circle 6 cm & side of equilateral ∆^{le} = 12cm

∴ Area of the circular portion = Area of circle – Area of the sector

= \(\frac{22}{7}\) ×(6)^{2} × \(\frac{5}{6}\)

= \(\frac{22 \times 30}{7}\) = \(\frac{660}{7}\) 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 the given figure. Find the area of the remaining portion of the square.

Answer:

i) Each side of square ABCD = 4 cm.

∴ Area of Square:

= (Side)^{2}

= (4)^{2}

= 16 sq.cm.

(ii) Area of circle in the corner A is 1 cm.

Radius, r = 1 cm.

θ = 90°

∴ Area of quadrant of circle:

= 0.78 sq.cm.

iii) Radius of circle having diameter 2 cm, r = 1 cm.

∴ Area of circle:

= 3.14 sq. cm.

∴ Area of remaining part of Square :

= Area of Square – Area of 4 quadrants – Area of Circle.

= 16 – 4 × 0.78 – 3.14

= 16 – 3.12 – 3.14

= 16 – 6.26

= 9.74 sq.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 the given figure. Find the area of the design.

Answer:

In ∆ABC is an equilateral triangle. Let O be the circumcentre of circum-circle, radius = r = 32cm.

Area of circle = πr^{2}

= 3 × \(\frac{1}{2}\) × 32 × 32 × sin 120°

= 3 × 16 × 32 × \(\frac{\sqrt{3}}{2}\)

= 3 × 16 × 16 × \(\sqrt{3}\)

= 768 \(\sqrt{3}\) cm^{2
}∴ Area of the design = Area of the circle – area of ∆ ABC.

= 3218.28 – 1330.176

= 1888.7cm^{2}

Question 7.

In the Figure given below, 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.

Answer:

i) Each side of square ABCD = 14 cm.

∴ Area of square ABCD

= (Side)^{2}

= (14)^{2}

= 196 sq.cm.

ii) Measure of radii of 4 circles, r = \(\frac{14}{2}=\) = 7 cm.

∵ Distance between tangents which touches circles externally,

d = R + r = 7 + 7 = 14 cm.

Area of the segment with a centre, A =?

r = 7 cm, θ = 90°

∴ Total area of 4 sectors

= 4 × 38.5 = 154 sq.cm.

iii) The Area of track :

= (Area of shaded portion – Area of ABCD)

= 196 – 154

= 42 sq.cm.

Question 8.

The 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.

Answer:

OE = O’G = 3 cm

AE = CG = 10 m

OA = O’C(30 + 10) m = 40 m

AC = EG = FH = BD = 106 m.

(i) Distance around the track along its inner edge = EG + FH + 2 x (circumference of semi-crcle of radius OE)

ii) Area of track = Area of the shaded region = Area of rectangle AEGC + Area of rectangle BFHD.+ 2 (Area of the semi circle of radius 40m – Area of the semicircle with radius 30m)

Question 9.

In the figure given below, 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.

Answer:

Area of sector OBC = \(\frac{θ}{360}\) × πr^{2
}= Area of sector AOC – area of sector OBD

= \(\frac{90}{360} \times \frac{22}{7}\) (7)^{2
}= \(\frac{1}{4} \times \frac{22}{7} \times 7 \times 7=\frac{77}{2}\) cm^{2}

and Area of ∆ OBC = \(\frac{1}{2}\) × OC × OB

= \(\frac{1}{2}\) × 7 × 7 = \(\frac{49}{2}\) cm^{2
}∴ Area of the segment BQC = Area cf sector OBC – area of ∆ OBC

= \(\frac{77}{2}-\frac{49}{2}=\frac{28}{2}\) = 14 cm^{2
}Area of the segment APC = 14cm^{2} The area of the circle with OD as

diameter = πr^{2
}= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}=\frac{77}{2}\) cm^{2
}Total 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 the length of the side of the triangle (see the Figure). Find the area of the shaded region.

(Use π = 314 and \(\sqrt{3}\) = 1.73205).

Answer:

i) Area of an equilateral ∆ABC :

= 17320.5 cm^{2}

Radius of each circle = \(\frac{200}{2}\) = 100 cm

Area of shaded region = Area of MBC – 3 × Area of a sector of angle 60° & radius 100cm

= 17320.5 – 3 × \(\frac{θ}{360}\) × πr^{2
}= 17320.5 – 3 × \(\frac{60}{360}\) × 3.14 ×(100)^{2}

= 17320.5 – \(\frac{1}{2}\) × 3.14 × 10000

= 17320 – 3.14 × 5000

= 17320.5 – 15700

= 1620.5 cm^{2}

Question 11.

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

Answer:

i) Total area of 9 circular designs each of radius 7 cm

ii) All the circles touches externally.

∴ Sum of the diameter of 3 circles in first Row =

14 + 14 + 14 = 42 cm.

∴ Length of each side of square ABCD,

a = 42 cm.

∴ Area of square ABCD = a^{2}

= (42)^{2}

= 1764 sq.cm.

Area of remaining part of handkerchief:

= Area of square – Area of 9 circles.

= 1764 – 1386

= 378 sq.cm.

Question 12.

In the given 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.

Answer:

i) Area of Quadrant OACB = \(\frac{1}{4}\) πr^{2
}= \(\frac{1}{4} \times \frac{22}{7} \times(3.5)^{2}=\frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\)

= \(\frac{77}{8}=9 \frac{5}{8}\) cm^{2}

ii) Area of ∆OBD = \(\frac{1}{2}\) × OB × OD

= \(\frac{1}{2}\) × 3.5 × 2 = 3.5 cm^{2}

Area of shaded region – Area of Quadrant OACB – Area of ∆OBD

Question 13.

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

Answer:

Area of shaded region = Area of Quadrant OPBQ – area of square OABC

= \(\frac{1}{4}\) πr^{2} – (side)^{2}

= \(\frac{1}{4}\) π (20√2)2 – (20)^{2}

= \(\frac{1}{4}\) × π × 800 – 400

= 200 × 3.14 – 400 = 628 – 400

= 228 cm^{2}

Question 14.

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

Answer:

Radius = r_{1} = 21cm & r_{2} = 7cm

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

Question 15.

In the given figure (i), 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.

Answer:

Area of shaded region = Area of Quadrant OPBQ – area of square OABC

= \(\frac{1}{4}\) πr^{2} – (side)^{2}

= \(\frac{1}{4}\) π (20√2)^{2} – (20)^{2}

= \(\frac{1}{4}\) × π × 800 – 400

= 200 × 3.14 – 400 = 628 – 400

= 228 cm^{2
}area of segment BPC Area of sector ABPC – Area of ∆ABC

= (154 – 98) cm^{2} = 56cm^{2}

radius of semi – circle BQC

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

Area of semi circle = \(\frac{1}{2}\) πr^{2}

= \(\frac{1}{2} \times \frac{22}{7} \times 7 \sqrt{2} \times 7 \sqrt{2}\)

= 154 cm^{2
}Area of shaded region = Area of the semi circle BQC – Area of the segment BPC

= 154 – 56 = 98cm^{2}.

Question 16.

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

Answer:

radius of each Quadrant = 8cm.

Sum of areas of Quadrants = 2 × \(\frac{1}{4}\) πr^{2}

= \(\frac{1}{2}\) × \(\frac{22}{7}\) × (8)^{2}

= \(\frac{11}{7}\) × 64 = \(\frac{704}{7}\) cm^{2}

area of square = (side)^{2} = (8)^{2}

= 64 cm^{2}

Area of designed region = Area of shaded region = sum of areas of quadrants – Area of the square

= \(\frac{704}{7}-\frac{64}{1}\)

= \(\frac{704-448}{7}=\frac{256}{7}\) cm^{2}

= 36.57 cm^{2}