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NCERT Exemplar · Class 10 Mathematics Area Related to Circles

60 questions · 60 still being checked

EXERCISE 11.3 11–16 (part 5 of 7)

  1. Exercise 11

    Find the area of the shaded region in Fig. 11.10\displaystyle 11.10, where arcs drawn with centres A, B, C and D intersect in pairs at mid-points P, Q, R and S of the sides AB, BC, CD and DA, respectively of a square ABCD (Use π=3.14\displaystyle \pi=3.14). NCERT_Question_Class10_Maths_Exemplar_Ch11_Ex11-3_Q11

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    30.$\displaystyle 96$ \(\displaystyle cm^{2}\)
    NCERT_Solution_Class10_Maths_Exemplar_Ch11_Ex11-3_Q11 \[\text{Area(square)} = 12^2 = 144\text{ cm}^2 \] \[\text{4 corner sectors, } r = 6\text{ cm}, \ \theta = 90^\circ \] \[4 \times \frac{90^\circ}{360^\circ}\pi r^2 = \pi r^2 = 3.14 \times 36 = 113.04\text{ cm}^2 \] \[\text{Shaded area} = 144 - 113.04 = 30.96\text{ cm}^2 \] Answer: \(\displaystyle 30.96\ \text{cm}^2\)

  2. Exercise 12

    In Fig. 11.11\displaystyle 11.11, arcs are drawn by taking vertices A, B and C of an equilateral triangle of side 10\displaystyle 10 cm. to intersect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of the shaded region (Use π=3.14\displaystyle \pi=3.14). NCERT_Question_Class10_Maths_Exemplar_Ch11_Ex11-3_Q12

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    39.$\displaystyle 25$ \(\displaystyle cm^{2}\)
    NCERT_Solution_Class10_Maths_Exemplar_Ch11_Ex11-3_Q12 Shaded region: three equal corner sectors. \[r = AF = \tfrac12 \times 10 = 5\ \text{cm} \] \[\theta = 60^\circ \quad \text{(angle of an equilateral triangle)} \] \[3 \times \frac{60^\circ}{360^\circ}\,\pi r^2 = \frac12 \pi r^2 = \frac12 \times 3.14 \times 25 \] \[= 39.25\ \text{cm}^2 \] Answer: \(\displaystyle 39.25\ \text{cm}^2\)

  3. Exercise 13

    In Fig. 11.12\displaystyle 11.12, arcs have been drawn with radii 14\displaystyle 14 cm each and with centres P, Q and R. Find the area of the shaded region. NCERT_Question_Class10_Maths_Exemplar_Ch11_Ex11-3_Q13

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    $\displaystyle 308$ \(\displaystyle cm^{2}\)
    Each shaded sector has radius \(\displaystyle r = 14 \) cm and angle equal to a vertex angle of \(\displaystyle \triangle PQR\). NCERT_Solution_Class10_Maths_Exemplar_Ch11_Ex11-3_Q13 \[\angle P + \angle Q + \angle R = 180^\circ \quad \text{(angle sum property of a triangle)} \] \[\text{Area} = \frac{\angle P}{360^\circ}\pi r^2 + \frac{\angle Q}{360^\circ}\pi r^2 + \frac{\angle R}{360^\circ}\pi r^2 = \frac{\angle P+\angle Q+\angle R}{360^\circ}\pi r^2 \] \[= \frac{180^\circ}{360^\circ}\times\frac{22}{7}\times 14^2 = \frac12\times\frac{22}{7}\times196 \] \[= 308 \text{ cm}^2 \] Answer: \(\displaystyle 308\ \text{cm}^2 \)

  4. Exercise 14

    A circular park is surrounded by a road 21\displaystyle 21 m wide. If the radius of the park is 105\displaystyle 105 m, find the area of the road.

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    $\displaystyle 15246$ \(\displaystyle m^{2}\)
    \[R_1 = 105 \text{ m}, \quad R_2 = 105 + 21 = 126 \text{ m} \] \[\text{Area of road} = \pi R_2^2 - \pi R_1^2 = \pi (R_2-R_1)(R_2+R_1) \] \[= \frac{22}{7}\times 21\times 231 \] \[= 66\times231 = 15246 \text{ m}^2 \] Answer: \(\displaystyle 15246\ \text{m}^2 \)

  5. Exercise 15

    In Fig. 11.13\displaystyle 11.13, arcs have been drawn of radius 21\displaystyle 21 cm each with vertices A, B, C and D of quadrilateral ABCD as centres. Find the area of the shaded region. NCERT_Question_Class10_Maths_Exemplar_Ch11_Ex11-3_Q15

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    $\displaystyle 1386$ \(\displaystyle cm^{2}\)
    Each shaded sector has radius \(\displaystyle r = 21 \) cm and angle equal to an interior angle of quadrilateral \(\displaystyle ABCD\). NCERT_Solution_Class10_Maths_Exemplar_Ch11_Ex11-3_Q15 \[\angle A+\angle B+\angle C+\angle D = 360^\circ \quad \text{(angle sum property of a quadrilateral)} \] \[\text{Area} = \frac{\angle A+\angle B+\angle C+\angle D}{360^\circ}\pi r^2 = \frac{360^\circ}{360^\circ}\times\frac{22}{7}\times21^2 \] \[= \frac{22}{7}\times441 = 1386 \text{ cm}^2 \] Answer: \(\displaystyle 1386\ \text{cm}^2 \)

  6. Exercise 16

    A piece of wire 20\displaystyle 20 cm long is bent into the form of an arc of a circle subtending an angle of 60\displaystyle 60° at its centre. Find the radius of the circle.

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    \(\displaystyle \frac{60}{\pi} \mathrm{~cm}\)
    \[\text{Arc length} = \frac{\theta}{360^\circ}\times2\pi r \] \[20 = \frac{60^\circ}{360^\circ}\times2\pi r = \frac{\pi r}{3} \] \[r = \frac{60}{\pi} = 60\times\frac{7}{22} = \frac{210}{11} \text{ cm} \] Answer: \(\displaystyle r = \dfrac{210}{11}\ \text{cm} \approx 19.1\ \text{cm} \)