In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 12 Areas Related to Circles Ex 12.3. These solutions are based on new NCERT Syllabus.

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

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

Question 1.

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

Solution:

Since O is the centre of the circle,

∴ QOR is a diameter.

⇒ ∠RPQ = 90° [Angle in a semicircle]

Now, in right ∆RPQ, RQ^{2} = PQ^{2}+ PR^{2} [Pythagoras theorem]

⇒ RQ^{2} = 24^{2} + 7^{2} = 576 + 49 = 625

⇒ RQ = \(\sqrt{625}\) = 25 cm

∴ Radius of a circle (r) = \(\frac{25}{2}\) cm

∴ Area of ∆RPQ

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

= 12 × 7 cm^{2} = 84 cm^{2}

Now, area of semicircle

∴ Area of the shaded portion

= 245.54 cm^{2} – 84 cm^{2} = 161.54 cm^{2}

Question 2.

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

Solution:

Radius of the outer circle (R) = 14 cm and θ = 40°

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.

Solution:

Side of the square = 14 cm

∴ Area of the square ABCD = 14 × 14 cm^{2}

= 196 cm^{2}

Similarly, area of semi-circle BPC = 77 cm^{2}

∴ Total area of two semi-circle = (77 + 77) cm^{2}

= 154 cm^{2}

Area of the shaded part

= Area of the square ABCD – Area of semi-circles

= (196 – 154) cm^{2} = 42 cm^{2}

Question 4.

Find the area of the shaded region in the 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 equilateral triangle OAB

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:

ABCD of the square ABCD = (4)^{2} cm^{2} = 16 cm^{2}.

Area of circel inside the square = πr^{2}

= π × (1)^{2} = π cm^{2}.

Question 6.

In a circular table cover of (he radius 32 cm, a design Is formed leaving an equilìtcral triangle ABC in the middle as shown in the figure. Find the area of the design (shaded region).

Solution:

Let O be the centre of a circular table and ABC be the equilateral triangle.

Then we draw OD ⊥ BC

In ∆OBD, we have: cos 60° = \(\frac{\mathrm{OD}}{\mathrm{OB}}\)

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 touch externally two of the remaining three circles. Find the area of the shaded region.

Solution:

Each side of the square ABCD = 14 cm

Question 8.

The given 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:

The distance around the track along the inner edge = Perimeter of GHI + Perimeter JKL + GL + IJ

= (π × 30 + π × 30 + 106 + 106)

= (60π + 212) m

= (60 × \(\frac{22}{7}\) + 212)m = \(\frac{2804}{7}\) m.

Similarly, area of another circular path

BCDJKLB = 1100 m^{2}

Area of rectangular path (ABLG) = 106 × 10

= 1060 m^{2}

Similarly, area of another rectangular path

IJDE = 1060 m^{2}

Total area of the track

= (1100 + 1100 + 1060 + 1060) m^{2}

= (2200 + 2120) m^{2} = 4320 m^{2}.

Question 9.

In the given 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:

O is the centre of the circle and OA = 7 cm

⇒ AB = 2OA = 2 × 7 cm = 14 cm

∵ AB and CD are perpendicular to each other.

⇒ OC ⊥ AB and OC = OA = 7 cm

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

= \(\frac{1}{2}\) × 14 cm × 7cm = 49 cm^{2}

Again, OD = OA = 7 cm

∴ Radius of the small circle = \(\frac{1}{2}\) (OD)

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

Solution:

Area of ∆ABC = 17320.5 cm^{2}

⇒ (side)^{2} = 40000

⇒ (side)^{2} = (200)^{2}

⇒ side = 200 cm

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

Since, each angle of an equilateral triangle is 60°,

∠A = ∠B = ∠C = 60°

Area of a sector having angle of sector (θ), as 60° and radius (r) = 100 cm = \(\frac{\theta}{360^{\circ}}\) × πr^{2}

Now, area of the shaded region = [Area of the equilateral triangle ABC] – [Area of 3 equal sectors]

= 17320.5 cm^{2} – 15700 cm^{2} = 1620.5 cm^{2}

Question 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:

∵ The circles touch each other.

∴ The side of the square ABCD

= 3 × diameter of a circle = 3 × (2 × radius of a circle) = 3 × (2 × 7) cm = 42 cm

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

= 1764 cm^{2}

Now, area of one circle

= πr^{2} = \(\frac{22}{7}\) × 7 × 7 cm^{2} = 154cm^{2}

∴ Total area of 9 circles = 9 × 154 cm^{2} = 1386 cm^{2}

∴ Area of the remaining portion of the handkerchief = (1764 – 1386) cm^{2} = 378 cm^{2}

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.

Solution:

(i) Here, radius (r) = 3.5 cm

Question 13.

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

Solution:

OABC is a square such that its side

OA = 20 cm

∴ OB^{2} = OA^{2} + AB^{2}

⇒ OB^{2} = 20^{2} + 20^{2} = 400 + 400 = 800

⇒ OB = \(\sqrt{800}=20 \sqrt{2}\)

⇒ Radius of the quadrant (r) = \(20 \sqrt{2}\) cm

Now, area of the quadrant OPBQ = \(\frac{1}{4}\) πr^{2}

= \(\frac{1}{4} \times \frac{314}{100}\) × 800 cm^{2} = 314 × 2 = 628 cm^{2}

Area of the square OABC = 20 × 20 cm^{2} = 400 cm^{2}

∴ Area of the shaded region

= 628 cm^{2} – 400 cm^{2} = 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). If ∠AOB = 30°, find the area of the shaded region.

Solution:

∵ Radius of bigger circle R = 21 cm and sector angle θ = 30°

∴ Area of the sector OAB

Question 15.

In the given 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:

Radius of the quadrant = 14 cm

Since, BC is a diameter in a semi-circle.

∴ ∠BAC = θ = 90°

Area of the semicircle ABPC

= [\(\frac{90^{\circ}}{360^{\circ}} \times \frac{22}{7}\) × 14 × 14] cm^{2}

= 22 × 7 cm^{2} = 154 cm^{2}

Area of right ∆ABC

= \(\frac{1}{2}\) × 14 × 14 cm^{2} = 98 cm^{2}

⇒ Area of segment BPC = 154 cm^{2} – 98 cm^{2}

= 56 cm^{2}

Now, area of the shaded region = [Area of semicircle BQC] – [Area of segment BRC]

= 154 cm^{2} – 56 cm^{2} = 98 cm^{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.

Solution:

Side of the square = 8 cm

Area of the square ABCD = 8 × 8 cm^{2} = 64 cm^{2}

Now, radius of the quadrant ADQB = 8 cm

∴ Area of the quadrant ADQB