## NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circles (Ex 12.3) Exercise 12.3

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NCERT Solutions for Class 10 Maths Chapter 12 - Exercise

## Access NCERT Solutions for Class 10 Mathematics Chapter 12 – Areas Related to Circles

Exercise 12.3

1. Find the area of the shaded region in the given figure, if PQ = 24 cm, PR = 7 cm and O is the center of the circle. $pi =\dfrac{22}{7}$

(Image will be uploaded soon)

Ans: From the given figure,

RQ is the diameter of the circle which implies that \[\angle RPQ = {90^ \circ }\]

Thus,

By applying Pythagoras theorem in \[PQR\],

\[ \Rightarrow R{P^2} + P{Q^2} = R{Q^2}\]

\[ \Rightarrow {(7)^2} + {(24)^2} = R{Q^2}\]

\[ \Rightarrow RQ = 25\]

Thus, Radius of circle, \[OR = \frac{{RQ}}{2} = \frac{{25}}{2}\]

We know that, RQ is the diameter of the circle, it divides the circle in two equal parts.

So, for area of shaded region,

Area of semicircle \[ = \frac{1}{2}\pi {r^2}\]

\[ = \frac{1}{2}\pi {\left( {\frac{{25}}{2}} \right)^2}\]

\[ = \frac{{6875}}{{28}}c{m^2}\]

Area of \[PQR\] \[ = \frac{1}{2} \times PQ \times PR\]

\[ = \frac{1}{2} \times 24 \times 7\]

\[ = 84c{m^2}\]

Area of shaded region = Area of semi - circle RPQOR − Area of \[PQR\]

\[ = \frac{{6875}}{{28}} - 84 = \frac{{4532}}{{28}}c{m^2}\]

Therefore, the area of the shaded region in the given figure is \[161.85 c{m^2}\].

2. Find the area of the shaded region in the given figure, if radii of the two concentric circles with center O are 7 cm and 14 cm respectively and \[\angle AOC = {40^ \circ }\]. $pi =\dfrac{22}{7}$

(Image will be uploaded soon)

Ans:

(Image will be uploaded soon)

Given that,

Radius of inner circle = 7 cm

Radius of outer circle = 14 cm

Angle subtended is \[{40^ \circ }\]

Now,

Area of shaded region = Area of sector OAFC − Area of sector OBED

\[ = \frac{{{{40}^ \circ }}}{{{{360}^ \circ }}} \times \pi {(14)^2} - \frac{{{{40}^ \circ }}}{{{{360}^ \circ }}} \times \pi {(7)^2}\]

\[ = \frac{{616}}{9} - \frac{{159}}{9}\]

\[ = \frac{{154}}{3}c{m^2}\]

Therefore, the area of the shaded region in the given figure is \[ = 51.33 c{m^2}\].

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. $pi =\dfrac{22}{7}$

(Image will be uploaded soon)

Ans:

(Image will be uploaded soon)

From the above figure it is evident that the radius of each semi-circle is 7 cm.

For area of shaded region,

So, Area of each semi-circle \[ = \frac{1}{2}\pi {r^2}\]

\[ = \frac{1}{2} \times \frac{{22}}{7} \times {(7)^2}\]

\[ = 77c{m^2}\]

Area of square ABCD \[ = {(side)^2} = {(14)^2} = 196c{m^2}\]

Area of the shaded region = Area of square ABCD − Area of semi-circle APD − Area of semi-circle BPC

\[ = 196 - 77 - 77 = 42c{m^2}\]

Therefore, the area of the shaded region in the given figure is \[42c{m^2}\].

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 center. $pi =\dfrac{22}{7}$

(Image will be uploaded soon)

Ans:

(Image will be uploaded soon)

Given that,

Radius of the circle is 6cm

We know that each interior angle of an equilateral triangle is of measure \[{60^ \circ }\].

For area of shaded region,

Area of sector OCDE \[ = \frac{{{{60}^ \circ }}}{{{{360}^ \circ }}}\pi {r^2}\]

\[ = \frac{1}{6} \times \frac{{22}}{7} \times {(6)^2}\]

\[ = \frac{{132}}{7}c{m^2}\]

Area of \[OAB\] \[ = \frac{{\sqrt 3 }}{4}{(12)^2}\]

\[ = 36\sqrt 3 c{m^2}\]

Area of circle \[ = \pi {r^2}\]

\[ = \frac{{22}}{7} \times {(6)^2}\]

\[ = \frac{{792}}{7}c{m^2}\]

Area of shaded region = Area of ΔOAB + Area of circle − Area of sector OCDE

\[ = 36\sqrt 3 + \frac{{792}}{7} - \frac{{132}}{7}\]

\[ = \left( {36\sqrt 3 + \frac{{660}}{7}} \right)c{m^2}\]

Hence, the area of the shaded region in the given figure \[\left( {36\sqrt 3 + \frac{{660}}{7}} \right)c{m^2}\].

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. $pi =\dfrac{22}{7}$

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Ans:

(Image will be uploaded soon)

It is evident from the above figure that each quadrant is a sector of \[{90^ \circ }\] in a circle of 1 cm radius.

For area of shaded region,

Area of each quadrant \[ = \frac{{{{90}^ \circ }}}{{{{360}^ \circ }}}\pi {r^2}\]

\[ = \frac{1}{4} \times \frac{{22}}{7} \times {(1)^2}\]

\[ = \frac{{22}}{{28}}c{m^2}\]

Area of square \[ = {(side)^2}\]

\[ = {(4)^2}\]

\[ = 16c{m^2}\]

Area of circle \[ = \pi {r^2}\]

\[ = \pi {(1)^2}\] \[ = \frac{{22}}{7}c{m^2}\]

Area of the shaded region = Area of square − Area of circle – (4 × Area of quadrant)

\[ = 16 - \frac{{22}}{7} - \left( {4 \times \frac{{22}}{{28}}} \right)\]

\[ = 16 - \frac{{44}}{7}\]

\[ = \frac{{68}}{7}c{m^2}\]

Therefore, the area of the remaining portion of the square is \[\frac{{68}}{7}c{m^2}\].

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 the given figure. Find the area of the design (Shaded region). $pi =\dfrac{22}{7}$

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Ans:

(Image will be uploaded soon)

Given that,

Radius of circle = \[r\] = 32 cm

AD is the median of \[ABC\]

\[OA = \frac{2}{3}AD\]

\[AD = 48cm\]

In \[ABD\],

Using Pythagoras Theorem,

\[A{B^2} = A{D^2} + B{D^2}\]

\[A{B^2} = {(48)^2} + {\left( {\frac{{BC}}{2}} \right)^2}\]

\[A{B^2} = {(48)^2} + {\left( {\frac{{AB}}{2}} \right)^2}\]

\[ \Rightarrow \frac{{3A{B^2}}}{4} = {(48)^2}\]

\[ \Rightarrow AB = \frac{{48 \times 2}}{{\sqrt 3 }} = \frac{{96}}{{\sqrt 3 }}\]

\[ = 32\sqrt 3 \] cm

For area of design,

Area of equilateral triangle \[ = \frac{{\sqrt 3 }}{4} \times {(32)^2} \times 3\]

\[ = 96 \times 8 \times \sqrt 3 \]

\[ = 768\sqrt 3 c{m^2}\]

Area of circle \[ = \pi {r^2}\]

\[ = \frac{{22}}{7} \times {(32)^2}\]

\[ = \frac{{22528}}{7}c{m^2}\]

Area of design = Area of circle − Area of \[ABC\]

\[ = \left( {\frac{{22528}}{7} - 768\sqrt 3 } \right)c{m^2}\]

Hence, the area of the design (Shaded region) is \[\left( {\frac{{22528}}{7} - 768\sqrt 3 } \right)c{m^2}\].

7. In the given figure, ABCD is a square of side 14 cm. With centers 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. $pi =\dfrac{22}{7}$

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Ans:

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It is evident that,

Area of each of the 4 sectors is equal to each other

Sector of \[{90^ \circ }\] in a circle of 7 cm radius.

For the area of shaded region,

Area of each sector \[ = \frac{{{{90}^ \circ }}}{{{{360}^ \circ }}}\pi {(7)^2}\]

\[ = \frac{1}{4} \times \frac{{22}}{7} \times {\left( 7 \right)^2}\]

\[ = \frac{{77}}{2}c{m^2}\]

Area of square ABCD \[ = {(side)^2}\]

\[ = {(14)^2} = 196c{m^2}\]

Area of shaded portion = Area of square ABCD – (4 × Area of each sector)

\[ = 196 - \left( {4 \times \frac{{77}}{2}} \right)\]

\[ = 42c{m^2}\]

Therefore, the area of shaded portion is \[42c{m^2}\].

8. The given figure depicts a racing track whose left and right ends are semicircular.

(Image will be uploaded soon)

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, $pi =\dfrac{22}{7}$

find

Ans:

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(i) The distance around the track along its inner edge

Ans: For inner edge,

Radius = \[r\]= 30cm

Distance around the track along its inner edge = AB + arc BEC + CD + arc DFA

\[ = 106 + \left( {\frac{1}{2} \times 2\pi r} \right) + 106 + \left( {\frac{1}{2} \times 2\pi r} \right)\]

\[ = 212 + \left( {\frac{1}{2} \times 2 \times \frac{{22}}{7} \times 30} \right) + \left( {\frac{1}{2} \times 2 \times \frac{{22}}{7} \times 30} \right)\]

\[ = 212 + \left( {2 \times \frac{{22}}{7} \times 30} \right)\]

\[ = \frac{{2804}}{7}m\]

Hence, distance around the track along its inner edge is \[\frac{{2804}}{7}m\].

(ii) The area of the track

Ans: Radius of inner edge = 30cm

Radius of outer edge = 40cm

Area of the track = (Area of GHIJ − Area of ABCD) + (Area of semi-circle HKI – Area of semi- circle BEC) + (Area of semi-circle GLJ − Area of semicircle AFD)

\[ = \left( {106 \times 80} \right) - \left( {106 \times 60} \right) + \left( {\frac{1}{2} \times \frac{{22}}{7} \times {{(40)}^2}} \right) - \left( {\frac{1}{2} \times \frac{{22}}{7} \times {{(30)}^2}} \right) + \left( {\frac{1}{2} \times \frac{{22}}{7} \times {{(40)}^2}} \right) - \left( {\frac{1}{2} \times \frac{{22}}{7} \times {{(30)}^2}} \right)\]

\[ = 106\left( {80 - 60} \right) + \left( {\frac{{22}}{7} \times {{\left( {40} \right)}^2}} \right) - \left( {\frac{{22}}{7} \times {{\left( {30} \right)}^2}} \right)\]

\[ = 2120 + \frac{{22}}{7}\left[ {{{\left( {40} \right)}^2} - {{\left( {30} \right)}^2}} \right]\]

\[ = 2120 + 2200\]

\[ = 4320{m^2}\]

Therefore, the area of the track is \[4320{m^2}\].

9. In the given figure, AB and CD are two diameters of a circle (with center 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.

$pi =\dfrac{22}{7}$

(Image will be uploaded soon)

Ans:

(Image will be uploaded soon)

Given that,

Radius of larger circle \[ = {r_1}\] = 7 cm

Radius of smaller circle \[ = {r_2}\] = \[\frac{7}{2}\] cm

For area of shaded region,

Area of smaller circle \[ = \pi {r_2}^2\]

\[ = \frac{{22}}{7} \times {\left( {\frac{7}{2}} \right)^2}\]

\[ = \frac{{77}}{2}c{m^2}\]

Area of semi-circle AECFB of larger circle \[ = \frac{1}{2}\pi {r_2}^2\]

\[ = \frac{1}{2} \times \frac{{22}}{7} \times {\left( 7 \right)^2}\]

\[ = \frac{{77}}{2}c{m^2}\]

Area of \[ABC\]\[ = \frac{1}{2} \times AB \times OC\]

\[ = \frac{1}{2} \times 14 \times 7\]

\[ = 49c{m^2}\]

Area of the shaded region = Area of smaller circle + Area of semi-circle AECFB − Area of \[ABC\]

\[ = \frac{{77}}{2} + 77 - 49\]

\[ = 28 + 38.5 = 66.5c{m^2}\]

Therefore, the area of shaded region is \[66.5c{m^2}\].

10. The area of an equilateral triangle ABC is \[17320.5c{m^2}\]. With each vertex of the triangle as center, a circle is drawn with radius equal to half the length of the side of the triangle (See the given figure). Find the area of shaded region. [\mathbf{Use}\text{ }\pi =3.14 \text{ }\mathbf{and}\text{ }\sqrt{3}=1.73205]

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Ans:

Let the side of the equilateral triangle be a.

Given that,

Area of equilateral triangle \[ = 17320.5c{m^2}\]

\[ \Rightarrow \frac{{\sqrt 3 }}{4}{(a)^2} = 17320.5\]

\[ \Rightarrow \frac{{1.73205}}{4}{a^2} = 17320.5\]

\[ \Rightarrow {a^2} = 4 \times 10000\]

\[ \Rightarrow a = 200cm\]

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It is evident from the figure that, each sector is of measure \[{60^ \circ }\]

Area of sector ADEF \[ = \frac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times \pi \times {r^2}\]

\[ = \frac{1}{6} \times \pi \times {\left( {100} \right)^2}\]

\[ = \frac{{3.14 \times 10000}}{6}\]

\[ = \frac{{15700}}{3}c{m^2}\]

Area of shaded region = Area of equilateral triangle – (3 × Area of each sector)

\[ = 17320.5 - 3 \times \frac{{15700}}{3}\]

\[ = 17320.5 - 15700\]

\[ = 1620.5c{m^2}\]

Therefore, area of given shaded region is \[1620.5c{m^2}\].

11. On a square handkerchief, nine circular designs each of radius 7 cm are made (see the given figure). Find the area of the remaining portion of the handkerchief. $pi =\dfrac{22}{7}$

(Image will be uploaded soon)

Ans:

(Image will be uploaded soon)

It is evident from the above figure, that the side of the square is 42 cm.

So, for area of the remaining portion of handkerchief,

Area of square \[ = {(side)^2} = {(42)^2}\]

\[ = 1764c{m^2}\]

Area of each circle \[ = \pi {r^2}\]\[ = \frac{{22}}{7} \times {(7)^2}\]

\[ = 154c{m^2}\]

Area of 9 circles \[ = 9 \times 154\]

\[ = 1386c{m^2}\]

Area of the remaining portion of the handkerchief \[ = 1764 - 1386\]

\[ = 378c{m^2}\]

Therefore, the area of the remaining portion of the handkerchief is \[ = 378c{m^2}\].

12. In the given figure, OACB is a quadrant of circle with center O and radius 3.5 cm. If OD = 2 cm, find the area of the $pi =\dfrac{22}{7}$

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Ans:

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Given that radius is 3.5cm.

Quadrant OACB

Ans:

Since OACB is a quadrant, the angle at O is \[{90^ \circ }\].

Area of quadrant OACB \[ = \frac{{{{90}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}\]

\[ = \frac{1}{4} \times \frac{{22}}{7} \times {\left( {3.5} \right)^2}\]

\[ = \frac{1}{4} \times \frac{{22}}{7} \times {\left( {\frac{7}{2}} \right)^2}\]

\[ = \frac{{77}}{8}c{m^2}\]

Therefore, the area of quadrant OACB is \[\frac{{77}}{8}c{m^2}\].

Shaded region

Ans:

For the area of the shaded region,

Area of \[OBD\] \[ = \frac{1}{2} \times OB \times OD\]

\[ = \frac{1}{2} \times 3.5 \times 2\]

\[ = \frac{7}{2}c{m^2}\]

Area of the shaded region = Area of quadrant OACB − Area of \[OBD\]

\[ = \frac{{77}}{8} - \frac{7}{2}\]

\[ = \frac{{49}}{8}c{m^2}\].

Hence, the area of the shaded region in the given figure is \[\frac{{49}}{8}c{m^2}\].

13. In the given figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. [\mathbf{Use}\text{ }\pi =3.14]

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Ans:

(Image will be uploaded soon)

For radius

In \[OAB\],

Using Pythagoras Theorem,

\[O{B^2} = O{A^2} + O{B^2}\]

\[ = {(20)^2} + {(20)^2}\]

\[ \Rightarrow OB = 20\sqrt 2 \]

Radius of circle \[ = r = \]\[20\sqrt 2 \] cm

For the area of shaded region,

Area of quadrant OPBQ \[ = \frac{{{{90}^ \circ }}}{{{{360}^ \circ }}} \times 3.14 \times {\left( {20\sqrt 2 } \right)^2}\]

\[ = \frac{1}{4} \times 3.14 \times 800\]

\[ = 628c{m^2}\]

Area of square OABC \[ = {\left( {side} \right)^2}\]

\[ = {(20)^2}\]

\[ = 400c{m^2}\]

Area of shaded region = Area of quadrant OPBQ − Area of square OABC

\[ = \left( {628 - 400} \right)c{m^2}\]

\[ = 228c{m^2}\]

Therefore, the area of the shaded region is \[228c{m^2}\].

14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and center O (see the given figure). If \[\angle AOB = {30^ \circ }\], find the area of the shaded region. $pi =\dfrac{22}{7}$

(Image will be uploaded soon)

Ans:

(Image will be uploaded soon)

Given that,

Radius for sector OAEB = 21cm

Radius for sector OCFD = 7cm

Angle subtended is \[{30^ \circ }\]

Area of the shaded region = Area of sector OAEB − Area of sector OCFD

\[ = \left[ {\frac{{{{30}^ \circ }}}{{{{360}^ \circ }}} \times \pi \times {{\left( {21} \right)}^2}} \right] - \left[ {\frac{{{{30}^ \circ }}}{{{{360}^ \circ }}} \times \pi \times {{\left( 7 \right)}^2}} \right]\]

\[ = \frac{1}{{12}}\pi \left[ {\left( {21 - 7} \right)\left( {21 + 7} \right)} \right]\]

\[ = \frac{{22 \times 14 \times 28}}{{12 \times 7}} = \frac{{22 \times 14 \times 28}}{{12 \times 7}}\]

\[ = \frac{{308}}{3}c{m^2}\]

Therefore, the area of shaded region is \[\frac{{308}}{3}c{m^2}\].

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. $pi =\dfrac{22}{7}$

(Image will be uploaded soon)

Ans:

(Image will be uploaded soon)

Given that,

Radius of circle = \[r = \] 14cm

As ABC is a quadrant of the circle, \[\angle BAC\] will be of measure \[{90^ \circ }\].

In \[ABC\],

Using Pythagoras Theorem,

\[ \Rightarrow B{C^2} = A{C^2} + A{B^2}\]

\[ = {(14)^2} + {(14)^2}\]

\[ \Rightarrow BC = 14\sqrt 2 \]

Radius of semi-circle drawn on BC = \[{r_1}\] \[ = \frac{{14\sqrt 2 }}{2} = 7\sqrt 2 cm\]

Now, for area of the shaded region,

Area of \[ABC\] \[ = \frac{1}{2} \times AB \times AC\]

\[ = \frac{1}{2} \times {(14)^2}\]

\[ = 98c{m^2}\]

Area of sector ABDC \[ = \frac{{{{90}^ \circ }}}{{{{360}^ \circ }}} \times \pi {r^2}\]

\[ = \frac{1}{4} \times \frac{{22}}{7} \times {(14)^2}\]

\[ = 154c{m^2}\]

Area of semi-circle drawn on BC \[ = \frac{1}{2} \times \pi \times {r_1}^2\]

\[ = \frac{1}{2} \times \frac{{22}}{7} \times {\left( {7\sqrt 2 } \right)^2}\]

\[ = 154c{m^2}\]

Area of shaded region = Area of semi-circle on BC – (Area of sector ABDC – Area of ∆ABC)

\[ = 154 - \left( {154 - 98} \right)\]

\[ = 98c{m^2}\]

Hence, the area of the shaded region of the given figure is \[98c{m^2}\].

16. Calculate the area of the designed region in the given figure common between the two quadrants of circles of radius 8 cm each. $pi =\dfrac{22}{7}$

(Image will be uploaded soon)

Ans:

(Image will be uploaded soon)

Given that ,

Radius of each circle is 8cm

The designed area is the common region between two sectors BAEC and DAFC.

For the area of the designed region,

Area of sector \[ = \frac{{{{90}^ \circ }}}{{{{360}^ \circ }}} \times \frac{{22}}{7} \times {\left( 8 \right)^2}\]

\[ = \frac{1}{4} \times \frac{{22}}{7} \times 64\]

\[ = \frac{{22 \times 16}}{7}\]

\[ = \frac{{352}}{7}c{m^2}\]

Area of \[BAC\] \[ = \frac{1}{2} \times BA \times BC\]

\[ = \frac{1}{2} \times {(8)^2}\]

\[ = 32c{m^2}\]

Area of the designed portion = 2 × (Area of segment AEC) = 2 × (Area of sector BAEC − Area of \[BAC\])

\[ = 2 \times \left( {\frac{{352}}{7} - 32} \right)\]

\[ = \frac{{2 \times 128}}{7}\]

\[ = \frac{{256}}{7}c{m^2}\]

Hence, the area of the designed region is \[\frac{{256}}{7}c{m^2}\].

## NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circles Exercise 12.3

Opting for the NCERT solutions for Ex 12.3 Class 10 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 12.3 Class 10 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 10 students who are thorough with all the concepts from the Subject Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 10 Maths Chapter 12 Exercise 12.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 10 Maths Chapter 12 Exercise 12.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 10 Maths Chapter 12 Exercise 12.3 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

### Important points

1. You will be asked to determine the area of shaded regions in this exercise. To find the area of the shaded region, we will normally subtract the area of a smaller inner shape from the area of a larger outer shape. If any of the shapes is a composite shape, we must subdivide it into shapes for which we have area formulas. You’ll get a clear idea about this while solving exercise solutions.

2. Steps in finding the Area of the Shaded Region

First, Calculate the area of the inner shape.

Then calculate the area of the outer shape.

So, Area of shaded region = Area of outer shape – Area of the inner shape

There are a variety of combinations of shapes to solve areas of shaded regions that we will discuss in Exercise 12.3.

### NCERT Solutions for Class 10 Maths Chapter 12 Exercises

Chapter 12 Areas Related to Circle in PDF Format | |

5 Questions and Solutions | |

14 Questions and Solutions |

## FAQs on NCERT Solutions for Class 10 Maths Chapter 12 - Exercise

**1. Which are the best questions to solve from Exercise 12.3 of Class 10 Maths?**

All the questions have their own importance. The simpler ones are crucial for one or two mark questions and the advanced ones are generally asked for three or more marks. Hence. all the questions must be given equal importance for a better knowledge of all the contents of the chapter and to do well in the tests as well. Refer to the official website of Vedantu for more chapter related resources.

**2. Is Exercise 12.3 of Class 10 Maths important from an exam point of view?**

Yes, indeed it is very crucial for your examinations. Practice all of its contents, the easy ones as well as the tougher ones as each question will only help you understand better and enhance your problem-solving abilities and you will also get a clearer picture of what different types of problems can be set in your examinations.

**3. What is meant by Area Of A Circle Class 10?**

The Area Of A Circle is equal to 3.14*r. The length of an arc of a sector of a circle with radius r and angle measured in degrees is equal to theta/360 *2*pie*r. The area of a circle segment is the area of the corresponding sector minus the area of the associated triangle. Visit the page NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.3 on the official website of Vedantu or download the Vedantu app for more problems and explanations regarding the same at free of cost.

**4. What is meant by a Tangent?**

In mathematics, a circle's "Tangent" is a line that crosses the circle just once. It's termed the point of contact since the tangent and circle share a common point G. Only one tangent may be found at a circle's centre point. Practice the multiple problems related to this to grasp the idea better and also to score good marks in your tests.

**5. What does a Quadrant mean?**

A Quadrant is one-quarter of a circle. Each of the four regions is a quadrant when a circle is equally split into four parts by two perpendicular lines. In fact, anything divided into four equal pieces may be characterized as a quadrant, such as a public park quadrant. The four halves of the coordinate plane system are referred to as quadrants.