# NCERT Solutions Class 10 Maths Chapter 12

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

NCERT Solutions for Class 10 Mathematics Chapter 12 by Extramarks are made by subject matter experts to help students with their preparation for board exams. The solutions have in-depth answers to all the questions listed at the end of the NCERT Class 10 Mathematics Chapter 12. Whether students are looking for accurate answers to questions or want to cross check their answers, they can  bank  on the NCERT Solutions by Extramarks.

## Access NCERT Solutions for Class 10 Mathematics Chapter 12 – Areas related to circles

With adequate practice of the Class 10 Mathematics NCERT book questions, students can perform better in their board exams. This is where NCERT Solutions for Class 10 Mathematics Chapter 12 can help students with their preparation. The solutions have all the  answers explained in a simple and comprehensive manner so that students do not have difficulty in understanding the reasoning. Students can access these solutions on Extramarks website.

## NCERT Solutions for Class 10 Mathematics

Extramarks provides NCERT Solutions for Class 10 Mathematics for the following chapters:

• Chapter 1 – Real Numbers
• Chapter 2 – Polynomials
• Chapter 3 – Pair of Linear Equations in Two Variables
• Chapter 4 – Quadratic Equations
• Chapter 5 – Arithmetic Progressions
• Chapter 6 – Triangles
• Chapter 7 – Coordinate Geometry
• Chapter 8 – Introduction to Trigonometry
• Chapter 9 – Some Applications of Trigonometry
• Chapter 10 – Circles
• Chapter 11 – Constructions
• Chapter 13 – Surface Areas and Volumes
• Chapter 14 – Statistics
• Chapter 15 – Probability

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

We are all surrounded by circular objects. In some way or another, understanding the perimeter, segment and sectors helps in real life. Hence, Class 10 Mathematics NCERT Solutions Chapter 12 is a really helpful guide for all students irrespective of their level. Subject experts at Extramarks have not only answered every textbook question comprehensively but have also brought together important techniques on how to solve even the trickiest problems through step-by-step methods.

### Perimeter and Area of a Circle

Chapter 12 focuses on numerous problems related to the perimeter and area of a  circle. Although it might feel like Class 10 Mathematics Chapter 12 Areas Related to Circles is just another chapter that requires the student to memorize everything and apply the formula to solve a problem, it is not the case. Along with formula usage, solving the problems of this chapter demands accurate analytical skills. And NCERT Solutions Class 10 Mathematics Chapter 12 can help students to strengthen both their conceptual base as well as analytical skills.

### Area of Sector and Segment of a Circle

When a part of a circular region is enclosed by two radii and the corresponding art, it is called the sector of the circle. The part that is enclosed between the chord and arc is known as the segment of the circle. This segment is further classified into the major segment and minor segment, wherein the major segment takes up the larger area.

With the definitions just stated, it may seem like Chapter 12 Class 10 Mathematics can easily be learned even if students choose to leave it for the last minute. It is certainly not the case. In fact, it is a chapter filled with minor details that can’t be overlooked. . To avoid losing marks , we recommend practicing NCERT textbook questions and referring to NCERT Solutions for Class 10 Mathematics Chapter 12 by Extramarks.

The solutions cover the following exercises given in the chapter:

### Areas of Combinations of Plane Figures

Chapter 12 also explains how to solve the complicated problems related to Areas of Combinations of Plane Figures. This is the part that involves students engaging in both analytical and critical thinking skills. Sometimes the questions in Class 10 Board examinations are so tricky that they cannot be solved using merely formulae. . It requires students to think calmly and strategically. This section is mostly about calculating the area of diverse designs and patterns. Though the first step is mastering the formulae, the second step to get the right answer requires a lot of practice. And the best way to ace both is NCERT Solutions for Class 10 Mathematics Chapter 12.

Q.1 The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

Ans.

$\begin{array}{l}\text{It is given that the radii of first and second circles are}\\ \text{19 cm and 9 cm respectively.}\\ \text{Let the radius of the third circle is r.}\\ \text{According to question,}\\ \text{Circumference of the third circle}=\text{Sum of the circumferences}\\ \text{of the first and second circles}\\ \text{or 2}\mathrm{\pi }\text{r}=2\mathrm{\pi }×19+2\mathrm{\pi }×9=2\mathrm{\pi }\left(19+9\right)\\ \text{or 2}\mathrm{\pi }\text{r}=2×28×\mathrm{\pi }\\ \text{or r}=\frac{2×28×\mathrm{\pi }}{2×\mathrm{\pi }}=28\text{cm}\\ \text{Therefore, the radius of the circle which has circumference}\\ \text{equal to the sum of the circumferences of the given circles}\\ \text{is 28 cm.}\end{array}$

Q.2 The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.

Ans.

$\begin{array}{l}\text{It is given that the radii of first and second circles are}\\ \text{8 cm and 6 cm respectively.}\\ \text{Let the radius of the third circle is r.}\\ \text{According to question,}\\ \text{Area of the third circle}=\text{Sum of the areas of the}\\ \text{first and second circles}\\ \text{or}\mathrm{\pi }{\text{r}}^{2}=\mathrm{\pi }×{8}^{2}+\mathrm{\pi }×{6}^{2}=\mathrm{\pi }\left(64+36\right)\\ \text{or}\mathrm{\pi }{\text{r}}^{2}=100\mathrm{\pi }\\ \text{or r}=\sqrt{100}=10\text{cm}\\ \text{Therefore, the radius of the circle which has area}\\ \text{equal to the sum of the areas of the given circles}\\ \text{is 10 cm.}\end{array}$

Q.3 The following figure depicts an archery target marked with its five scoring areas from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions. Ans. $\begin{array}{l}\text{Radius of the region representing Gold score}=\mathrm{r}=\frac{21}{2}=10.5\text{cm}\\ \text{It is given that each band}\left(\text{except gold}\right)\text{is 10.5 cm wide.}\\ \text{Therefore,}\\ \text{Radius of the outer circle of the Red score}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{r}}_{1}=10.5+10.5=21\text{cm\hspace{0.17em}}\\ \text{Radius of the outer circle of the Blue score}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{r}}_{2}=21+10.5=31.5\text{cm}\\ \text{Radius of the outer circle of the Black score}\\ \text{}={\mathrm{r}}_{3}\\ \text{}=31.5+10.5\\ \text{}=42\text{cm}\\ \text{Radius of the outer circle of the White score}\\ \text{}={\mathrm{r}}_{4}\\ \text{}=42+10.5\\ \text{}=52.5\text{cm}\\ \text{Now,}\\ \text{Area of the Gold score}={\mathrm{\pi r}}^{2}\\ \text{}=\frac{22}{7}×\frac{21}{2}×\frac{21}{2}\\ \text{}=346.5{\text{cm}}^{2}\\ \text{Area of the Red score}={{\mathrm{\pi r}}_{1}}^{2}-\text{Area of the Gold score}\\ \text{}=\frac{22}{7}×{21}^{2}-346.5\\ \text{}=1039.5{\text{cm}}^{2}\\ \text{Area of the Blue score}={{\mathrm{\pi r}}_{2}}^{2}-{{\mathrm{\pi r}}_{1}}^{2}\\ \text{}=\frac{22}{7}×\left\{{\left(31.5\right)}^{2}-{21}^{2}\right\}\\ \text{}=1732.5{\text{cm}}^{2}\\ \text{Area of the Black score}={{\mathrm{\pi r}}_{3}}^{2}-{{\mathrm{\pi r}}_{2}}^{2}\\ \text{}=\frac{22}{7}×\left\{{\left(42\right)}^{2}-{\left(31.5\right)}^{2}\right\}\\ \text{}=2425.5{\text{cm}}^{2}\\ \text{Area of the White score}={{\mathrm{\pi r}}_{4}}^{2}-{{\mathrm{\pi r}}_{3}}^{2}\\ \text{}=\frac{22}{7}×\left\{{\left(52.5\right)}^{2}-{\left(42\right)}^{2}\right\}\\ \text{}=3118.5{\text{cm}}^{2}\end{array}$

Q.4 The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?

Ans.

$\begin{array}{l}\text{It is given that the wheels of a car are of diameter 80 cm each.}\\ \text{Therefore, circumference of each wheel}=\mathrm{\pi d}=\frac{22}{7}×80\\ \\ \text{Speed of the car}=66\text{km/h}=\frac{66×1000×100}{60\text{}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=110000\text{cm/minute}\\ \text{Distance covered by the car in 10 minutes}=11,00,000\text{cm}\\ \text{Number of revolutions made by each wheel to}\\ \text{cover}11,00,000\text{cm}=\frac{11,00,000\text{}}{\frac{22}{7}×80}=4375\end{array}$

Q.5 Tick the correct answer in the following and justify your choice: If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
(A) 2 units (B) π units (C) 4 units (D) 7 units

Ans.

$\begin{array}{l}\text{Let the radius of the circle be}\mathrm{r}.\\ \text{Given that,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Circumference of the circle}=\text{Area of the circle}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{\pi r}={\mathrm{\pi r}}^{2}\\ \text{or 2}=\mathrm{r}\\ \text{Therefore, radius of the circle is 2 units.}\end{array}$

Q.6 Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

Ans.

$\begin{array}{l}\text{It is given that radius of the circle is 6 cm and angle of the}\\ \text{sector is 60°.}\\ \text{Area of a sector of angle}\mathrm{\theta }=\frac{\mathrm{\theta }}{360\mathrm{°}}×{\mathrm{\pi r}}^{2}\\ \therefore \text{Area of a sector of angle}60\mathrm{°}=\frac{60\mathrm{°}}{360\mathrm{°}}×\frac{22}{7}×{6}^{2}=\frac{132}{7}{\text{cm}}^{2}\end{array}$

Q.7 Find the area of a quadrant of a circle whose circumference is 22 cm.

Ans.

$\begin{array}{l}\text{Let the radius of the circle be}\mathrm{r}.\\ \text{Given that,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Circumference of the circle}=\text{22 cm}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{\pi r}=\text{22 cm}\\ \text{or}\mathrm{r}=\frac{\text{22}}{2}×\frac{7}{22}=\frac{7}{2}\\ \text{Quadrant of a circle subtends an angle of 90° at the centre.}\\ \therefore \text{Area of a quadrant of the given circle}=\frac{90\mathrm{°}}{360\mathrm{°}}×\frac{22}{7}×{\left(\frac{7}{2}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{77}{8}{\text{cm}}^{2}\end{array}$

Q.8 The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Ans.

$The length of minute hand,r=14 cm In 60 minutes, a minute hand subtends an angle of 360° at the centre. So, in 5 minutes a minute hand subtends an angle of 360°×5 60° =30° ∴Area swept by the minute hand in 5 minutes = 30° 360° × 22 7 × ( 14 ) 2 = 154 3 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@420E@$

Q.9 A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment (ii) major sector. (Use π = 3.14)

Ans. $\begin{array}{l}\text{Let AB be the chord of the circle subtending 90° angle at}\\ \text{centre O of the circle.}\\ \left(\left(\mathrm{i}\right)\text{) Area of minor sector OACB}=\frac{90°}{360°}×{\mathrm{\pi r}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=\frac{1}{4}×3.14×10×10\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=78.5{\text{cm}}^{2}\\ \text{Area of}\mathrm{\Delta }\text{OAB}=\frac{1}{2}×\text{OA}×\text{OB}=\frac{1}{2}×10×10=50{\text{cm}}^{2}\\ \text{Area of minor segment ACB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{Area of minor sector OACB}-\text{Area of}\mathrm{\Delta }\text{OAB}\\ \text{}=78.5{\text{cm}}^{2}-50{\text{cm}}^{2}=28.5{\text{cm}}^{2}\\ \left(\left(\mathrm{ii}\right)\right)\text{Area of major sector OADB}=\frac{360°-90°}{360°}×{\mathrm{\pi r}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{270°}{360°}×3.14×10×10\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=235.5{\text{cm}}^{2}\end{array}$

Q.10 In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:

1. the length of the arc
2. area of the sector formed by the arc
3. area of the segment formed by the corresponding chord

Ans. $\begin{array}{l}\text{Radius of the given circle}=\mathrm{r}=21\text{cm}\\ \text{Angle subtended by the given arc}=60\mathrm{°}\\ \text{Length of an arc of a sector of angle}\mathrm{\theta }=\frac{\mathrm{\theta }}{360\mathrm{°}}×2\mathrm{\pi r}\\ \therefore \text{Length of arc ACB}=\frac{60\mathrm{°}}{360\mathrm{°}}×2×\frac{22}{7}×21\text{\hspace{0.17em}}=22\text{cm}\\ \text{Area of sector OACB}=\frac{60\mathrm{°}}{360\mathrm{°}}×\frac{22}{7}×21×21=231{\text{cm}}^{2}\\ \text{In}\mathrm{\Delta }\text{OAB,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{OAB}=\angle \text{OBA [As OA}=\text{OB]}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{OAB}+\angle \text{OBA}+\angle \text{AOB}=180\mathrm{°}\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\angle \text{OAB}+60\mathrm{°}=180\mathrm{°}\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{OAB}=\frac{180\mathrm{°}-60\mathrm{°}}{2}=\frac{120\mathrm{°}}{2}=60\mathrm{°}\\ \text{Therefore,}\mathrm{\Delta }\text{OAB is an equilateral triangle.}\\ \text{Area of}\mathrm{\Delta }\text{OAB}=\frac{\sqrt{3}}{4}×{\left(\text{side}\right)}^{2}=\frac{\sqrt{3}}{4}×{\left(\text{21}\right)}^{2}=\frac{441\sqrt{3}}{4}{\text{cm}}^{2}\\ \text{Area of segment ACB}=\text{Area of sector OACB}-\text{Area of}\mathrm{\Delta }\text{OAB}\\ \text{}=\left(231-\frac{441\sqrt{3}}{4}\right){\text{cm}}^{2}\end{array}$

Q.11

$A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. ( Use π=3.14 and 3 =1.73 ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@D1CC@$

Ans. $\begin{array}{l}\text{Radius of the given circle}=\mathrm{r}=15\text{cm}\\ \text{Angle subtended by the given chord}=60\mathrm{°}\\ \text{Area of sector OACB}=\frac{60\mathrm{°}}{360\mathrm{°}}×3.14×15×15=117.75{\text{cm}}^{2}\\ \text{In}\mathrm{\Delta }\text{OAB,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{OAB}=\angle \text{OBA [As OA}=\text{OB]}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{OAB}+\angle \text{OBA}+\angle \text{AOB}=180\mathrm{°}\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\angle \text{OAB}+60\mathrm{°}=180\mathrm{°}\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{OAB}=\frac{180\mathrm{°}-60\mathrm{°}}{2}=\frac{120\mathrm{°}}{2}=60\mathrm{°}\\ \text{Therefore,}\mathrm{\Delta }\text{OAB is an equilateral triangle.}\\ \text{Area of}\mathrm{\Delta }\text{OAB}=\frac{\sqrt{3}}{4}×{\left(\text{side}\right)}^{2}=\frac{\sqrt{3}}{4}×{\left(\text{15}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{225×1.73}{4}{\text{cm}}^{2}=97.3125{\text{cm}}^{2}\\ \text{Area of segment ACB}=\text{Area of sector OACB}-\text{Area of}\mathrm{\Delta }\text{OAB}\\ \text{}=117.75-97.3125=20.4375{\text{cm}}^{2}\\ \text{Area of major segment ADB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{Area of circle}-\text{Area of segment ACB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{\pi }{\left(15\right)}^{2}-20.4375\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3.14×225-20.4375=706.5-20.4375\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=686.0625{\text{cm}}^{2}\end{array}$

Q.12

$\begin{array}{l}\text{A chord of a circle of radius 12 cm subtends an}\\ \text{angle of 120° at the centre. Find the area of the}\\ \text{corresponding segment of the circle.}\\ \left(\mathbf{Use}\text{}\mathrm{\pi }=\mathbf{3}.\mathbf{14}\text{}\mathbf{and}\text{}\sqrt{3}=\mathbf{1}.\mathbf{73}\right)\end{array}$

Ans. $\begin{array}{l}\text{Let AB is a chord of a circle of radius 12 cm which subtends an}\\ \text{angle of 120° at the centre}\mathrm{O}\text{.}\\ \text{Radius of the given circle}=\mathrm{r}=\text{OA}=\text{OB}=12\text{cm}\\ \text{Angle subtended by the given chord}=120\mathrm{°}\\ \text{Area of sector OACB}=\frac{120\mathrm{°}}{360\mathrm{°}}×3.14×12×12=150.72{\text{cm}}^{2}\\ \text{Let us draw a perpendicular OD on chord AB. It bisects the}\\ \text{chord AB.}\\ \therefore \text{AD}=\text{DB}\\ \text{In}\mathrm{\Delta }\text{ODA,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\text{OD}}{\mathrm{OA}}=\mathrm{cos}60\mathrm{°}\\ \mathrm{or}\text{}\frac{\mathrm{OD}}{12}=\frac{1}{2}\\ \mathrm{or}\text{OD}=6\text{cm}\\ \text{Also,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}sin60°}=\frac{\mathrm{AD}}{\mathrm{OA}}=\frac{\mathrm{AD}}{12}\\ \mathrm{or}\text{}\frac{\sqrt{3}}{2}=\frac{\mathrm{AD}}{12}\\ \mathrm{or}\text{}\mathrm{AD}=6\sqrt{3}\text{cm}\\ \mathrm{Now},\\ \mathrm{AB}=2\mathrm{AD}=2×6\sqrt{3}=12\sqrt{3}\text{cm}\\ \text{Area of}\mathrm{\Delta }\text{OAB}=\frac{1}{2}×\mathrm{AB}×\mathrm{OD}=\frac{1}{2}×12\sqrt{3}×6=62.28{\text{cm}}^{2}\\ \text{Area of segment ACB}=\text{Area of sector OACB}-\text{Area of}\mathrm{\Delta }\text{OAB}\\ \text{}=150.72-\text{}62.28=88.44{\text{cm}}^{2}\end{array}$

Q.13

$\begin{array}{l}\text{A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m}\\ \text{long rope}\left(\text{see the following figure}\right)\text{. Find}\\ \text{}\left(\text{i}\right)\text{the area of that part of the field in which the horse can graze.}\\ \text{}\left(\text{ii}\right)\text{the increase in the grazing area if the rope were 10 m long instead of 5 m.}\left(\mathbf{Use}\text{}\mathrm{\pi }=\mathbf{3}.\mathbf{14}\right)\end{array}$ Ans.

$\begin{array}{l}\left(\left(\mathrm{i}\right)\right)\text{The horse can graze a sector of 90° in a circle of radius 5 m.}\\ \text{Area that can be grazed by horse}\\ \text{}=\text{Area of sector of 90° in a circle of radius 5 m}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{90°}{360°}×3.14×25=19.625{\text{m}}^{2}\\ \left(\left(\mathrm{ii}\right)\right)\text{When length of rope is 10 m,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}area that can be grazed by horse}\\ \text{}=\frac{90°}{360°}×3.14×100=78.5{\text{m}}^{2}\\ \mathrm{Increase}\text{}\mathrm{in}\text{grazing area}=\left(78.5-19.625\right){\text{m}}^{2}=58.875{\text{m}}^{2}\end{array}$

Q.14 A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the following figure. Find:
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch. Ans. $\begin{array}{l}\text{Total length of wire required will be the length of 5 diameters}\\ \text{and the circumference of the brooch.}\\ \text{Radius of circle}=\frac{35}{2}\text{mm}\\ \text{Circumference of brooch}=2\mathrm{\pi r}=2×\frac{22}{7}×\frac{35}{2}=110\text{mm}\\ \text{Length of wire required}=110+5×35=110+175=285\text{mm}\\ \text{From the given figure, we observe that each of 10 sectors of the}\\ \text{circle subtends 36° at the centre of the circle.}\\ \text{Therefore, area of each sector}=\frac{36\mathrm{°}}{360\mathrm{°}}×{\mathrm{\pi r}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{10}×\frac{22}{7}×{\left(\frac{35}{2}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{385}{4}{\text{mm}}^{2}\end{array}$

Q.15 An umbrella has 8 ribs which are equally spaced (see the following figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella. Ans.

$\begin{array}{l}\mathrm{There}\text{are 8 ribs in the given umbrella. The arc between two}\\ \text{consecutive ribs subtends}\frac{360\mathrm{°}}{8}=45\mathrm{°}\text{at the centre of the}\\ \text{assumed flat circle.}\end{array}$ $\begin{array}{l}\text{Area between two consecutive ribs of circle}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{45\mathrm{°}}{360\mathrm{°}}×{\mathrm{\pi r}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{8}×\frac{22}{7}×{\left(45\right)}^{2}=\frac{22275}{28}{\text{cm}}^{2}\end{array}$

Q.16 A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

Ans.

$\begin{array}{l}\mathrm{It}\text{is obvious that each blade of wiper will sweep an area of}\\ \text{a sector of 115° in a circle of 25 cm radius.}\\ \\ \text{Area of a sector of 115° in a circle of 25 cm radius}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{115\mathrm{°}}{360\mathrm{°}}×{\mathrm{\pi r}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{115\mathrm{°}}{360\mathrm{°}}×\frac{22}{7}×{\left(25\right)}^{2}=\frac{158125}{252}{\text{cm}}^{2}\\ \therefore \text{Area swept by two blades}=2×\frac{158125}{252}{\text{cm}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{158125}{126}{\text{cm}}^{2}\end{array}$

Q.17 To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)

Ans.

$\begin{array}{l}\mathrm{It}\text{is obvious that the lighthouse spreads light across a sector}\\ \text{of 80° in a circle of 16.5 km radius.}\\ \therefore \text{Required area}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{80\mathrm{°}}{360\mathrm{°}}×{\mathrm{\pi r}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{80\mathrm{°}}{360\mathrm{°}}×3.14×{\left(16.5\right)}^{2}=189.97{\text{km}}^{2}\end{array}$

Q.18

$\begin{array}{l}\text{A round table cover has six equal designs as shown}\\ \text{in the following figure. If the radius of the cover is}\\ \text{28 cm, find the cost of making the designs at the}\\ {\text{rate of ₹ 0.35 per cm}}^{\text{2}}\text{.}\left(\text{Use}\sqrt{\text{3}}=\text{1.7}\right)\end{array}$ Ans. $\begin{array}{l}\mathrm{It}\text{is obvious from the above figure that the designs are}\\ \text{segments of the circle.}\\ \text{Let us consider the segment APB. Chord AB is a side of}\\ \text{the hexagon. Each chord subtends}\frac{360°}{6}=60°\text{at the centre}\\ \text{of the circle.}\\ \text{In}\mathrm{\Delta }\text{OAB,}\\ \angle \text{OAB}=\angle \text{OBA [As OA}=\mathrm{OB}\right]\\ \angle \text{AOB}=60°\\ \mathrm{Also},\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{OAB}+\angle \text{OBA}+\angle \text{AOB}=180°\\ \mathrm{or}\text{2}\angle \text{OAB}=180°-60°=120°\\ \mathrm{or}\text{}\angle \text{OAB}=60°\\ \text{Therefore,}\mathrm{\Delta }\text{OAB is an equilateral triangle.}\\ \text{Area of}\mathrm{\Delta }\text{OAB}=\frac{\sqrt{3}}{4}×{\left(\text{side}\right)}^{2}=\frac{\sqrt{3}}{4}×{\left(\text{28}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=333.2{\text{cm}}^{2}\\ \therefore \text{Area of sector OAPB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{60°}{360°}×{\mathrm{\pi r}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{6}×\frac{22}{7}×{\left(28\right)}^{2}=\frac{1232}{3}{\text{cm}}^{2}\\ \text{Area of segment APB}=\text{Area of sector OAPB}-\text{Area of}\mathrm{\Delta }\text{OAB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\frac{1232}{3}-333.2\right){\text{cm}}^{2}\\ \therefore \text{Area of designs}=6×\left(\frac{1232}{3}-333.2\right){\text{cm}}^{2}=464.8{\text{cm}}^{2}\\ \text{Cost of making}464.8{\text{cm}}^{2}\text{design}=464.8×0.35=₹162.68\end{array}$

Q.19

$\begin{array}{l}\text{Tick the correct answer in the following:}\\ \text{Area of a sector of angle p}\left(\text{in degrees}\right)\text{of a circle}\\ \text{with radius R is}\\ \text{\hspace{0.17em}\hspace{0.17em}(A)\hspace{0.17em}\hspace{0.17em}}\frac{\text{P}}{180}×2\mathrm{\pi }\text{R \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(B)\hspace{0.17em}\hspace{0.17em}}\frac{\text{P}}{180}×\mathrm{\pi }{\text{R}}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}(C)}\frac{\text{P}}{360}×2\mathrm{\pi }\text{R \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(D)}\frac{\text{P}}{720}×2\mathrm{\pi }{\text{R}}^{2}\end{array}$

Ans.

$\begin{array}{l}\text{We know that area of a sector of angle}\mathrm{\theta }=\frac{\mathrm{\theta }}{360\mathrm{°}}×{\mathrm{\pi R}}^{2}\\ \therefore \text{Area of a sector of angle}\mathrm{P}=\frac{\mathrm{P}}{360\mathrm{°}}×{\mathrm{\pi R}}^{2}=\frac{\mathrm{P}}{360\mathrm{°}}×\frac{2}{2}{\mathrm{\pi R}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{P}}{720\mathrm{°}}×2{\mathrm{\pi R}}^{2}\\ \mathrm{Hence},\text{Option}\left(\text{D}\right)\text{is the correct answer.}\end{array}$

Q.20 Find the area of the shaded region in the following figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle. Ans.

$\begin{array}{l}\text{In the given figure, RQ is diameter. Therefore,}\angle \text{RPQ}=90\mathrm{°}\\ \text{By applying Pythagoras Theorem in}\mathrm{\Delta }\text{PQR, we get}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}QR}}^{2}={\mathrm{PQ}}^{2}+{\mathrm{PR}}^{2}={24}^{2}+{7}^{2}=625\\ \mathrm{or}\text{QR}=\sqrt{625}=25\text{}\\ \mathrm{Therefore},\text{radius of the circle}=\mathrm{OR}=\frac{\mathrm{QR}}{2}=\frac{25}{2}\\ \mathrm{Area}\text{of the semi-circle RPQOR}=\frac{1}{2}{\mathrm{\pi r}}^{2}=\frac{1}{2}×\frac{22}{7}×{\left(\frac{25}{2}\right)}^{2}=\frac{6875}{28}{\text{cm}}^{2}\\ \mathrm{Area}\text{of}\mathrm{\Delta }\text{PQR}=\frac{1}{2}×\mathrm{PQ}×\mathrm{PR}=\frac{1}{2}×24×7=84{\text{cm}}^{2}\\ \mathrm{Area}\text{of the shaded region}\\ \text{}=\mathrm{Area}\text{of semi-circle RPQOR}-\mathrm{Area}\text{of}\mathrm{\Delta }\text{PQR}\\ \text{\hspace{0.17em}}=\frac{6875}{28}-84=\frac{4523}{28}{\text{cm}}^{2}\end{array}$

Q.21

$\begin{array}{l}\text{Find the area of the shaded region in the following}\\ \text{figure, if radii of the two concentric circles with}\\ \text{centre O are 7 cm and 14 cm respectively and}\\ \angle \text{AOC}=\text{40°.}\end{array}$ Ans. $\begin{array}{l}\mathrm{Radius}\text{of inner circle}=7\text{\hspace{0.17em}}\mathrm{cm}\\ \mathrm{Radius}\text{of outer circle}=14\text{\hspace{0.17em}}\mathrm{cm}\\ \mathrm{Area}\text{of the shaded region}\\ \text{}=\mathrm{Area}\text{of sector OAFC}-\mathrm{Area}\text{of sector OBED}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{40\mathrm{°}}{360\mathrm{°}}×\mathrm{\pi }{\left(14\right)}^{2}-\frac{40\mathrm{°}}{360\mathrm{°}}×\mathrm{\pi }{\left(7\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{9}×\frac{22}{7}×196-\frac{1}{9}×\frac{22}{7}×49\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{616}{9}-\frac{154}{9}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{462}{9}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{154}{3}{\text{cm}}^{2}\end{array}$

Q.22 Find the area of the shaded region in the following figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles. Ans. $\begin{array}{l}\text{Radius of each semicircle}=7\text{cm}\\ \mathrm{Area}\text{of each semicircle}=\frac{1}{2}{\mathrm{\pi r}}^{2}=\frac{1}{2}×\frac{22}{7}×7×7=77{\text{cm}}^{2}\\ \mathrm{Area}\text{of square ABCD}={\left(\mathrm{side}\right)}^{2}={14}^{2}=196{\text{cm}}^{2}\\ \mathrm{Area}\text{of shaded region}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{Area}\text{of square ABCD}-\mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{two}\text{semicircles}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=196-2×77=42{\text{cm}}^{2}\end{array}$

Q.23 Find the area of the shaded region in the following 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. Ans. $\begin{array}{l}\text{Area of the circle}={\mathrm{\pi r}}^{2}=\frac{22}{7}×{6}^{2}=\frac{22}{7}×36=\frac{792}{7}{\text{cm}}^{2}\\ \text{Area of the sector}=\frac{60\mathrm{°}}{360\mathrm{°}}×\frac{22}{7}×{6}^{2}=\frac{792}{42}{\text{cm}}^{2}=\frac{132}{7}{\text{cm}}^{2}\\ \text{Area of the given equilateral triangle}=\frac{\sqrt{3}}{4}{\left(\mathrm{side}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\sqrt{3}}{4}×{12}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=36\sqrt{3}{\text{cm}}^{2}\\ \text{Area of the shaded region}=\text{Area of the circle}+\text{Area of the triangle}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{Area of the sector}\\ \text{}=\frac{792}{7}+36\sqrt{3}-\frac{132}{7}=\left(\frac{660}{7}+36\sqrt{3}\right){\text{cm}}^{2}\end{array}$

Q.24 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 following figure. Find the area of the remaining portion of the square. Ans. $\begin{array}{l}\text{Area of the square ABCD}={\left(\mathrm{side}\right)}^{2}={4}^{2}=16{\text{cm}}^{2}\\ \text{Area of each sector}=\frac{90\mathrm{°}}{360\mathrm{°}}×\frac{22}{7}×{1}^{2}=\frac{11}{14}{\text{cm}}^{2}\\ \text{Area of the circle}={\mathrm{\pi r}}^{2}=\frac{22}{7}×{1}^{2}=\frac{22}{7}{\text{cm}}^{2}\\ \text{Area of the shaded region}=\text{Area of the square ABCD}\\ \text{}-\text{Area of the circle}\\ \text{}-4×\text{Area of the sector}\\ \text{}=16-\frac{22}{7}-4×\frac{11}{14}=\frac{68}{7}{\text{cm}}^{2}\end{array}$

Q.25 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 following figure. Find the area of the design (shaded region). Ans. $\begin{array}{l}\text{Radius of the circle}=32\text{cm}\\ \text{AD is the median of the triangle ABC.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AO}=\frac{2}{3}\mathrm{AD}=32\\ \text{or \hspace{0.17em}\hspace{0.17em}AD}=48\text{cm}\\ \text{In}\mathrm{\Delta }\text{\hspace{0.17em}ABD,}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB}}^{2}={\mathrm{AD}}^{2}+{\mathrm{BD}}^{2}={48}^{2}+{\left(\frac{\mathrm{AB}}{2}\right)}^{2}\\ \text{or \hspace{0.17em}\hspace{0.17em}AB}=32\sqrt{3}\text{cm}\\ \mathrm{Area}\text{of equilateral}\mathrm{\Delta }\text{\hspace{0.17em}ABC}=\frac{\sqrt{3}}{4}{\left(\mathrm{side}\right)}^{2}=\frac{\sqrt{3}}{4}{\left(32\sqrt{3}\right)}^{2}=768\sqrt{3}{\text{cm}}^{2}\\ \text{Area of the circle}={\mathrm{\pi r}}^{2}=\frac{22}{7}×{\left(32\right)}^{2}=\frac{22528}{7}{\text{cm}}^{2}\\ \text{Area of the design}=\text{Area of the circle}-\mathrm{Area}\text{of \hspace{0.17em}}\mathrm{\Delta }\text{\hspace{0.17em}ABC}\\ \text{}=\left(\frac{22528}{7}-768\sqrt{3}\right){\text{cm}}^{2}\end{array}$

Q.26 In the following 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. Ans.

$\begin{array}{l}\text{Area of the square ABCD}={\left(\mathrm{side}\right)}^{2}={14}^{2}=196{\text{cm}}^{2}\\ \text{Distance between two centres of circles}=\text{AB}=14\text{cm}\\ \text{Radius of each circle}=\frac{\mathrm{AB}}{2}=\frac{14}{2}=7\text{cm}\\ \text{Area of each sector}=\frac{90\mathrm{°}}{360\mathrm{°}}×\frac{22}{7}×{7}^{2}=\frac{77}{2}{\text{cm}}^{2}\\ \text{Area of the shaded region}=\text{Area of the square ABCD}\\ \text{}-4×\text{Area of the sector}\\ \text{}=196-4×\frac{77}{2}=42{\text{cm}}^{2}\end{array}$

Q.27 The following 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:

1. the distance around the track along its inner edge
2. the area of the track.

Ans. $\begin{array}{l}\text{Distance around the track along its inner edge}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{AB}+\mathrm{arc}\text{BEC}+\mathrm{CD}+\mathrm{arc}\text{DFA}\\ \text{}=106+\frac{1}{2}×2\mathrm{\pi r}+106+\frac{1}{2}×2\mathrm{\pi r}=\frac{2804}{7}\text{m}\\ \text{Area of track}\\ \text{}=\mathrm{area}\text{}\mathrm{of}\text{rectangle GHIJ}-\mathrm{area}\text{}\mathrm{of}\text{rectangle ABCD}\\ \text{\hspace{0.17em}}+\text{area of semicircle HKI}-\text{area of semicircle BEC}\\ \text{}+\text{area of semicircle GLJ}-\text{area of semicircle AFD}\\ \text{}=106×80-106×60+\frac{1}{2}×\frac{22}{7}×{\left(40\right)}^{2}-\frac{1}{2}×\frac{22}{7}×{\left(30\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\frac{1}{2}×\frac{22}{7}×{\left(40\right)}^{2}-\frac{1}{2}×\frac{22}{7}×{\left(30\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4320{\text{m}}^{2}\end{array}$

Q.28 In the following 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. Ans. $\begin{array}{l}\text{Area of smaller circle}={\mathrm{\pi r}}^{2}=\frac{22}{7}×\frac{7}{2}×\frac{7}{2}=\frac{77}{2}{\text{cm}}^{2}\\ \text{Area of a quadrant of the bigger circle}\\ \text{=}\frac{1}{4}\mathrm{\pi }{\text{(7)}}^{2}\\ \text{=}\frac{1}{4}×\frac{22}{7}×{\text{(7)}}^{2}\\ =\frac{77}{2}{\text{cm}}^{2}\\ \mathrm{Area}\text{of triangle OBC =}\frac{1}{2}×7×7=\frac{49}{2}{\text{cm}}^{2}\\ \mathrm{Area}\text{of a shaded segment of the bigger circle}\\ \text{=Area of a quadrant}-\mathrm{Area}\text{of triangle OBC}\\ \text{}=\frac{77}{2}\text{}-\frac{49}{2}=\frac{28}{2}{\text{= 14 cm}}^{2}\\ \mathrm{Total}\text{area of the shaded region}\\ \text{=Area of smaller circle}+2\left(\mathrm{Area}\text{of a shaded segment}\right)\\ \text{=}\frac{77}{2}\text{+2}×14=\frac{133}{2}{\text{=66.5 cm}}^{2}\end{array}$

Q.29

$\begin{array}{l}{\text{The area of an equilateral triangle ABC is 17320.5 cm}}^{2}\text{.}\\ \text{With each vertex of the triangle as centre, a circle is}\\ \text{drawn with radius equal to half the length of the side}\\ \text{of the triangle}\left(\text{see the following figure}\right)\text{. Find the}\\ \text{area of the shaded region.}\left(\text{Use}\mathrm{\pi }=\text{3.14 and}\sqrt{\text{3}}=\text{1.73205}\right)\end{array}$ Ans. $\begin{array}{l}\mathrm{Let}\text{side of equilateral triangle be}\mathrm{a}\text{.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Area of equilateral triangle}=17320.5{\text{cm}}^{2}\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\sqrt{3}}{4}{\mathrm{a}}^{2}=17320.5\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}^{2}=\frac{4×17320.5}{\sqrt{3}}\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=200\text{cm}\\ \text{Each sector is of 60°.}\\ \text{So area of sector ADEF}=\frac{60\mathrm{°}}{360\mathrm{°}}{\mathrm{\pi r}}^{2}=\frac{1}{6}×3.14×{\left(100\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{15700}{3}\\ \text{Area of}\mathrm{shaded}\text{region}=\mathrm{Area}\text{of equilateral triangle}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-3×\text{area of each sector}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=17320.5-3×\frac{15700}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1620.5{\text{cm}}^{2}\end{array}$

Q.30 On a square handkerchief, nine circular designs each of radius 7 cm are made (see the following figure). Find the area of the remaining portion of the handkerchief. Ans.

$\begin{array}{l}\mathrm{Side}\text{of the square}=42\text{cm}\\ \therefore \text{Area of square}=1764{\text{cm}}^{2}\\ \mathrm{Area}\text{of each circle}=\frac{22}{7}×{7}^{2}=154{\text{cm}}^{2}\\ \mathrm{Area}\text{of 9 circles}=9×154=1386{\text{cm}}^{2}\\ \mathrm{Required}\text{area}=1764\text{}-1386=378{\text{cm}}^{2}\end{array}$

Q.31 In the following 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 Ans. $\begin{array}{l}\mathrm{Given}\text{that, r}=3.5\text{cm and OD}=2\text{cm}\\ \mathrm{Area}\text{of the quadrant OACB}=\frac{90\mathrm{°}}{360\mathrm{°}}×\frac{22}{7}×{\left(3.5\right)}^{2}=\frac{77}{8}{\text{cm}}^{2}\\ \mathrm{Area}\text{of}\mathrm{\Delta }\text{\hspace{0.17em}OBD}=\frac{1}{2}×\mathrm{OB}×\mathrm{OD}=\frac{1}{2}×3.5×2=3.5{\text{cm}}^{2}\\ \text{Area of}\mathrm{shaded}\text{region}=\mathrm{Area}\text{of the quadrant OACB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\mathrm{Area}\text{of}\mathrm{\Delta }\text{\hspace{0.17em}OBD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{77}{8}-3.5=\frac{49}{8}{\text{cm}}^{2}\end{array}$

Q.32 In the following 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) Ans. $\begin{array}{l}\mathrm{Given}\text{that, OA}=20\text{cm}\\ \text{Area of square OABC}={20}^{2}=400{\text{cm}}^{2}\\ \text{Radius of the quadrant\hspace{0.17em}\hspace{0.17em}OPBQ}=\mathrm{Diagonal}\text{of the square OABC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=20\sqrt{2}\text{cm}\\ \mathrm{Area}\text{of the quadrant OPBQ}=\frac{90°}{360°}×3.14×{\left(20\sqrt{2}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=628{\text{cm}}^{2}\\ \text{Area of}\mathrm{shaded}\text{region}=\mathrm{Area}\text{of the quadrant OPBQ}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{Area of square OABC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=628-400=228{\text{cm}}^{2}\end{array}$

Q.33 AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see the following figure). If ∠AOB = 30°, find the area of the shaded region Ans. $\begin{array}{l}\text{Area of}\mathrm{shaded}\text{region}=\mathrm{Area}\text{of the sector ABO}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\mathrm{Area}\text{of the sector CDO}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{30\mathrm{°}}{360\mathrm{°}}×\frac{22}{7}×{\left(21\right)}^{2}-\frac{30\mathrm{°}}{360\mathrm{°}}×\frac{22}{7}×{\left(7\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{308}{3}{\text{cm}}^{2}\end{array}$

Q.34 In the following 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. Ans.

$\begin{array}{l}\text{Area of}\mathrm{shaded}\text{region}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{Area}\text{of the semicircle}+\mathrm{Area}\text{}\mathrm{of}\text{}\mathrm{\Delta ABC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\mathrm{Area}\text{of the quadrant ABC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}×\frac{22}{7}×{\left(\frac{\sqrt{{14}^{2}+{14}^{2}}}{2}\right)}^{2}+\frac{1}{2}×{\left(14\right)}^{2}-\frac{90°}{360°}×\frac{22}{7}×{\left(14\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=154+98-154\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=98{\text{cm}}^{2}\end{array}$

Q.35 Calculate the area of the designed region in the following figure common between the two quadrants of circles of radius 8 cm each. Ans. $\begin{array}{l}\text{The designed area is the common region between two}\\ \text{equal sectors BAEC and DAFC.}\\ \mathrm{Area}\text{of the sector BAEC}=\mathrm{Area}\text{of the sector DAFC}\\ \text{}=\frac{90\mathrm{°}}{360\mathrm{°}}×\frac{22}{7}×{\left(8\right)}^{2}=\frac{352}{7}{\text{cm}}^{2}\\ \mathrm{Area}\text{of}\mathrm{\Delta }\text{ADC}=\mathrm{Area}\text{of}\mathrm{\Delta }\text{ABC}=\frac{1}{2}×8×8=32{\text{cm}}^{2}\\ \mathrm{Area}\text{of the designed portion}\\ \text{}=2×\left(\mathrm{Area}\text{of segment AEC}\right)\\ \text{}=2×\left(\mathrm{Area}\text{of sector BAEC}-\mathrm{Area}\text{of}\mathrm{\Delta }\text{ABC}\right)\\ \text{}=2×\left(\frac{352}{7}-32\right)=\frac{256}{7}{\text{cm}}^{2}\end{array}$

## 1. Why should I refer to NCERT Solutions Class 10 Mathematics Chapter 12?

A student must refer to NCERT Solutions Class 10 Mathematics Chapter 12 because of the below-stated reasons:

• Solutions are prepared by subject matter experts while ensuring that they adhere to CBSE guidelines.
• Use of simple language and step-by-step instructions make solutions easy to understand for students.
• Solutions can be used for last-minute revision to understand the concepts better and stay ahead of the competition.

## 2. How many exercises are there in NCERT Solutions for  Class 10 Mathematics Chapter 12?

There are a total of 3 exercises in NCERT Solutions for Class 10 Mathematics Chapter 12. The 1st exercise includes 5 questions. The 2nd exercise includes 14 questions, and the 3rd and final exercise includes 16 questions. These questions in every exercise are mostly based on the perimeter and area of a circle, segment and sector along with an area of a combination plane.