**NCERT Solutions class 10 exercise 12.3 of chapter 12-Areas related to circle**

**NCERT Solutions class 10 exercise 12.3 of chapter 12-Areas related to circle** will help the students of** 9 class** in the preparation of CBSE board exams. All** solutions** are the **solutions** of the unsolved questions of **class 9 NCERT** text book. Each **solution** is **solved** by a step by step method so,the students will not face any kind of problem. The** solutions** are created by an expert **math** teacher. You can also study here **NCERT solutions of maths and science,** sample papers,** solutions** of previous year question papers, tips for competitive exams and online jobs.

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**Pdf-NCERT solutions of class 10 maths chapter 12-Areas related to Circle**

**NCERT Solutions for Class 10 Mathsย ย Chapter 12-Areas related to circles**

**Exercise 12.1- Area Related to Circle**

**Exercise 12.2- Areas related to Circle**

**Exercise 12.3-Areas Related to the Circle**

**Solution of Latest Sample paper Class 10 maths for Term 1 2021 CBSE board**

**Q1.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.**

Ans. We are given PQ = 24 cm. PR = 7 cm

O is the centre of the circle given to us,so QR is the diameter of the circle

โด QPR = 90ยฐ ( angle made on the semicircle)

Applying the pythogorus theorem in ฮQPR

QRยฒ = PRยฒ + PQยฒ

= 24ยฒ + 7ยฒ = 576 + 49 = 625

QR = 25 cm

So, the diameter of the circle is = 25 cm

Radius of the circle is = 25/2 = 12.5 cm

Area of triangle

= 84

Area of the triangle = 84 sq.cm

Area of semicircle PQR

Area of shaded region = Area of the semicircle QPROQ – Area of triangle QPR

Area of the shaded region

**Q2. Find the area of the shaded region in the given figure . If radii of two concentric circles with centre O are 7 cm and 14 cm respectively and โ AOC =40ยฐ.**

Ans. We are given the radii of two concentric circles 7 cm and 14 cm subtending a

Area of the shaded region AFCDEB = Area of sector OAFC – Area of sector OBED

Area of the sector is given as

Area of sector OAFC

Area of sector OBED

Area of the shaded region AFCDEB

**Q3. Find the area of shaded region in the given figure ,if ABCD is a square of side 14 cm and APC and BPC are semicircles**

Ans. We are given side of square = 14 cm and semicircle APC and BPC

Area of shaded region = Area of the square ABCD – Area of two semicircles APC and BPC

Radii of two semicircles are = 14/2 = 7 cm

Area of the square ABCD = side ร side = 14 ร 14 = 196 cmยฒ

Area of two semicircles APC and BPC = 2รฯrยฒ/2 = (22/7)ร 7ร7= 22ร7 =154 cmยฒ

Area of shaded region = 196 – 154 = 42 cmยฒ

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**Q4. 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 OAB of side 12 cm as centre.**

Ans.

We are given the radius of arc = 6 cm

Side of square = 12 cm

Area of shaded region = Area of circle formed after extending the arc D to C – Area of sector DOC + Area of equilateral ฮAOB

Area of circle = ฯrยฒ = (22/7)ร6ร6=(792/7) cmยฒ

Area of sector DOC

Area of equilateral ฮAOB = [(Side)ยฒ/4] โ3

= [(12)ยฒ/4]โ3=(144/4)โ3= 36โ3 cmยฒ

Area of shaded region

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**Q5. From each corner of the 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 remaining portion of the square.[use ฯ = 22/7]**

Ans. We are given radii of 4 quadrant of circle created on the verteces of the square,and radii of the circle at the centre of the circle 1 cm and side of square 4 cm

Area of the shaded region = Area of the square – Area of 4 quadrant – Area of circle

= Side ร Side – 4ร(ฯrยฒ)/4- ฯrยฒ

= 4 ร 4 – (22/7)ร1ยฒ – (22/7)1ยฒ

= 16 – 22/7 – 22/7

=16 – 44/7

=(112-44)/7

=68/7

Hence the area of shaded region is (68/7) cmยฒ

**Q6.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 (shaded region)[use ฯ = 22/7]**

Ans.We are given radius of the circle 32 cm and three designs leaving an equilateral triangle(ABC) at the middle.

Drawing AD median that is AD โฅ BC

The centre O of the circle is the centroid of the triangle

We know the cetroid of te triangle divides each median into 1 : 2 and bisects the sides of triagle ABC.

AO = (2/3) of AD

32 = (2/3) of AD

AD = 48 cm

BD = BC/2

Applying pythogorus theorem in ฮABD

ABยฒ = BDยฒ + ADยฒ

AB = BC(sides of equilateral triangle)

BCยฒ = (BC/2)ยฒ + ADยฒ

48ยฒ = BCยฒ – (BC/2)ยฒ

48ยฒ = (4BCยฒ -BCยฒ)/4 = 3BCยฒ/4

BCยฒ = (4ร 48ร 48)/3

BC = 2ร4ร4โ3= 32โ3 cm

Area of equilateral triangle ABC

= 16ร16ร3โ3

=768โ3 cmยฒ

Area of circle =ฯrยฒ = (22/7)ร32ร32 =(22528/7) cmยฒ

Hence the area of design = [(22528/7) -768โ3 ]cmยฒ

**Q7.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 circles toches externally two of the remaining three circles. Findย the area of shaded region.**

Ans. We are given the side of square = 14 cm

Radii of each quadrant of the circle is = 14/2 = 7 cm

The area of shaded region = Area of the square – Area of quadrants inside the square

= Side รSide – 4[1/4 (ฯrยฒ)]

= 14 ร 14 – (22/7) ร7ร7

= 196 -154 = 42

Hence area of the shaded region is 42 cmยฒ

**Q8. 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**

**[use ฯ = 22/7]**

Ans.

(i) The distance around the track along its inner edge = Length of rectangular part + Length of both semicircular parts

The distance around the track along its inner edge = 106ร2 + (2ฯr/2 )ร2

= 212 + 2ร(22/7) ร30

= 212 + 1320/7

= (1484 + 1320)/7

=2804/7 cm

(ii) The area of the track = Area of both rectangular paths + Area of both circular path

The area of the track = 2ร106 ร10 + 2รฯ(40ยฒ- 30ยฒ)/2

= 2120 + (22/7)( 1600 -900)

= 2120 + (22/7) 700

=2120 + 2200 =4320 mยฒ

**Q9.In the given figure ,AB and CD are two diameter of 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 shaded region.**

**[use ฯ = 22/7]**

Ans. The area of shaded region = Area of the segments formed by the chords DC and AC + Area of the circle of diameter BO (7 cm)

Area of the segments formed by the chords DC and AC = Area of the sector OAB – Area of the right ฮOBC

=2[ (ฯrยฒ/4) – (1/2) OC รOD]

=2 [(22/7) ร7ร7/4 – (1/2) ร7ร7]

=2[77/2 – 49/2]

=2 [28/2 ]= 28 cmยฒ

Area of the circle of diameter OB

= ฯrยฒ = (22/7) ร(7/2)(7/2) = 77/2 = 38.5 cmยฒ

The area of shaded region = 28 + 38.5 = 66.5 cmยฒ

**Q10.The area of an equilateral triangle ABC is 17320.5 cmยฒ, with each vertex of the triangle as centre ,a circle is drawn with a radius equal to half the length of the side of the triangle (see the given figure). Find the area of the shaded region**

**[use ฯ= 3.14 and โ3 = 1.73205]**

Ans.

Let the side of the equilateral triangle be a.

Area of equilateral triangle = 17320.5 cm2

aยฒโ3/4 = 17320.5

aยฒ =ย 17320.5ร4/โ3

aยฒ =ย 17320.5ร4/1.73205

a = 200

Length ofย side of the triangle is 200 cm

Radii of each circle is 200/2=100 cm

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