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NCERT Exemplar · Class 10 Mathematics Area Related to Circles

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EXERCISE 11.3 1–10 (part 4 of 7)

  1. Exercise 1

    Find the radius of a circle whose circumference is equal to the sum of the circumferences of two circles of radii 15\displaystyle 15 cm and 18\displaystyle 18 cm.

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    NCERT’s answer
    $\displaystyle 33$ cm
    \[2\pi R = 2\pi r_1 + 2\pi r_2 \] \[R = r_1+r_2 = 15+18 \] Answer: \(\displaystyle R = 33\) cm

  2. Exercise 2

    In Fig. 11.5\displaystyle 11.5, a square of diagonal 8\displaystyle 8 cm is inscribed in a circle. Find the area of the shaded region. NCERT_Question_Class10_Maths_Exemplar_Ch11_Ex11-3_Q2

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    \(\displaystyle (16 \pi-32) \mathrm{cm}^2\)
    NCERT_Solution_Class10_Maths_Exemplar_Ch11_Ex11-3_Q2 \[d = 2r \implies r = \dfrac{8}{2} = 4\text{ cm} \] \[\text{Area of circle} = \pi r^2 = 16\pi\text{ cm}^2 \] \[\text{Area of square} = \dfrac{d^2}{2} = \dfrac{8^2}{2} = 32\text{ cm}^2 \] \[\text{Shaded area} = 16\pi - 32 \text{ cm}^2 \] Answer: \(\displaystyle (16\pi - 32)\ \text{cm}^2\)

  3. Exercise 3

    Find the area of a sector of a circle of radius 28\displaystyle 28 cm and central angle 45\displaystyle 45°.

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    $\displaystyle 308$ \(\displaystyle cm^{2}\)
    NCERT_Solution_Class10_Maths_Exemplar_Ch11_Ex11-3_Q3 \[\text{Area of sector} = \dfrac{\theta}{360^\circ}\times\pi r^2 \] \[= \dfrac{45^\circ}{360^\circ}\times\dfrac{22}{7}\times28^2 \] \[= \dfrac18\times\dfrac{22}{7}\times784 = 308\text{ cm}^2 \] Answer: \(\displaystyle 308\ \text{cm}^2\)

  4. Exercise 4

    The wheel of a motor cycle is of radius 35\displaystyle 35 cm. How many revolutions per minute must the wheel make so as to keep a speed of 66\displaystyle 66 km/h?

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    NCERT’s answer
    500.
    \[\text{Circumference} = 2\pi r = 2\times\dfrac{22}{7}\times35 = 220\text{ cm} = 2.2\text{ m} \] \[\text{Speed} = 66\text{ km/h} = \dfrac{66000}{60} = 1100\text{ m/min} \] \[\text{Revolutions per minute} = \dfrac{1100}{2.2} = 500 \] Answer: \(\displaystyle 500\) revolutions per minute

  5. Exercise 5

    A cow is tied with a rope of length 14\displaystyle 14 m at the corner of a rectangular field of dimensions 20m × 16m. Find the area of the field in which the cow can graze.

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    NCERT’s answer
    $\displaystyle 154$ \(\displaystyle m^{2}\)
    NCERT_Solution_Class10_Maths_Exemplar_Ch11_Ex11-3_Q5 \[\text{Grazing area} = \dfrac{90^\circ}{360^\circ}\times\pi r^2 \quad \text{(quarter circle: rope shorter than each side)} \] \[= \dfrac14\times\dfrac{22}{7}\times14^2 = \dfrac14\times\dfrac{22}{7}\times196 = 154\text{ m}^2 \] Answer: \(\displaystyle 154\ \text{m}^2\)

  6. Exercise 6

    Find the area of the flower bed (with semi-circular ends) shown in Fig. 11.6. NCERT_Question_Class10_Maths_Exemplar_Ch11_Ex11-3_Q6

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    NCERT’s answer
    \(\displaystyle (380+25 \pi) \mathrm{cm}^2\)
    NCERT_Solution_Class10_Maths_Exemplar_Ch11_Ex11-3_Q6 \[r = \dfrac{10}{2} = 5\text{ cm} \] \[\text{Area of rectangle} = 38\times10 = 380\text{ cm}^2 \] \[\text{Two semicircular ends} = 2\times\dfrac12\pi r^2 = 25\pi\text{ cm}^2 \] \[\text{Total area} = 380 + 25\pi \text{ cm}^2 \] Answer: \(\displaystyle (380 + 25\pi)\ \text{cm}^2\)

  7. Exercise 7

    In Fig. 11.7\displaystyle 11.7, AB is a diameter of the circle, AC=6 cm\displaystyle \mathrm{AC}=6 \mathrm{~cm} and BC=8 cm\displaystyle \mathrm{BC}=8 \mathrm{~cm}. Find the area of the shaded region (Use π=3.14\displaystyle \pi=3.14). NCERT_Question_Class10_Maths_Exemplar_Ch11_Ex11-3_Q7

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    NCERT’s answer
    54.$\displaystyle 5$ \(\displaystyle cm^{2}\)
    NCERT_Solution_Class10_Maths_Exemplar_Ch11_Ex11-3_Q7 \[\angle ACB = 90^\circ \quad \text{(angle in a semicircle)} \] \[AB^2 = AC^2 + BC^2 = 6^2 + 8^2 = 100 \] \[AB = 10\text{ cm}, \quad r = 5\text{ cm} \] \[\text{Area of circle} = \pi r^2 = 3.14 \times 25 = 78.5\text{ cm}^2 \] \[\text{Area of } \triangle ABC = \tfrac12 \times AC \times BC = \tfrac12 \times 6 \times 8 = 24\text{ cm}^2 \] \[\text{Shaded area} = 78.5 - 24 = 54.5\text{ cm}^2 \] Answer: \(\displaystyle 54.5\ \text{cm}^2\)

  8. Exercise 8

    Find the area of the shaded field shown in Fig. 11.8. NCERT_Question_Class10_Maths_Exemplar_Ch11_Ex11-3_Q8

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    NCERT’s answer
    \(\displaystyle (32+2 \pi) \mathrm{m}^2\)
    \[\text{Rectangle} = 8 \times 4 = 32\ \text{m}^2 \] \[r = 6 - 4 = 2\ \text{m} \] \[\text{Semicircle} = \tfrac12 \pi r^2 = \tfrac12 \pi (2)^2 = 2\pi\ \text{m}^2 \] \[\text{Shaded area} = 32 + 2\pi\ \text{m}^2 \] Answer: \(\displaystyle (32 + 2\pi)\ \text{m}^2\)

  9. Exercise 9

    Find the area of the shaded region in Fig. 11.9. NCERT_Question_Class10_Maths_Exemplar_Ch11_Ex11-3_Q9

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    NCERT’s answer
    \(\displaystyle (248-4 \pi) \mathrm{m}^2\)
    \[\text{Outer rectangle} = 26 \times 12 = 312\ \text{m}^2 \] \[2r = 12 - 4 - 4 = 4\ \text{m} \quad\Rightarrow\quad r = 2\ \text{m} \] \[\text{Length of inner track} = 26 - 3 - 3 = 20\ \text{m} \] \[\text{Straight part} = 20 - 2r = 16\ \text{m} \] \[\text{Rectangle} = 16 \times 4 = 64\ \text{m}^2 \] \[2 \times \tfrac12 \pi r^2 = \pi (2)^2 = 4\pi\ \text{m}^2 \] \[\text{Unshaded area} = 64 + 4\pi\ \text{m}^2 \] \[\text{Shaded area} = 312 - (64 + 4\pi) = 248 - 4\pi\ \text{m}^2 \] Answer: \(\displaystyle (248 - 4\pi)\ \text{m}^2\)

  10. Exercise 10

    Find the area of the minor segment of a circle of radius 14\displaystyle 14 cm, when the angle of the corresponding sector is 60\displaystyle 60°.

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    NCERT’s answer
    \(\displaystyle \left(\frac{308}{3}-49 \sqrt{3}\right) \mathrm{cm}^2\)
    NCERT_Solution_Class10_Maths_Exemplar_Ch11_Ex11-3_Q10 \[\text{Sector } OAB = \frac{60^\circ}{360^\circ}\,\pi r^2 = \frac16 \times \frac{22}{7} \times 14^2 = \frac{308}{3}\ \text{cm}^2 \] \[\angle OAB = \angle OBA = \frac{180^\circ - 60^\circ}{2} = 60^\circ \quad \text{(} OA = OB \text{)} \] \[\triangle OAB \text{ is equilateral, side } 14\ \text{cm} \] \[\text{Area}(\triangle OAB) = \frac{\sqrt3}{4}(14)^2 = 49\sqrt3\ \text{cm}^2 \] \[\text{Minor segment} = \frac{308}{3} - 49\sqrt3\ \text{cm}^2 \approx 102.67 - 84.87 = 17.80\ \text{cm}^2 \] Answer: \(\displaystyle \left(\tfrac{308}{3} - 49\sqrt3\right)\ \text{cm}^2 \approx 17.80\ \text{cm}^2\)