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NCERT Exemplar · Class 10 Mathematics Area Related to Circles

60 questions · 60 still being checked

EXERCISE 11.4 1–10 (part 6 of 7)

  1. Exercise 1

    The area of a circular playground is 22176 m2\displaystyle 22176 \mathrm{~m}^2. Find the cost of fencing this ground at the rate of Rs 50\displaystyle 50 per metre.

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    NCERT’s answer
    Rs $\displaystyle 26400$
    \[\pi r^2 = 22176 \] \[r^2 = 22176\times\frac{7}{22} = 7056 \ \Rightarrow\ r = 84\text{ m} \] \[C = 2\pi r = 2\times\frac{22}{7}\times84 = 528\text{ m} \] \[\text{Cost} = 528\times50 = 26400 \] Answer: Rs $\displaystyle 26400$

  2. Exercise 2

    The diameters of front and rear wheels of a tractor are 80\displaystyle 80 cm and 2\displaystyle 2 m respectively. Find the number of revolutions that rear wheel will make in covering a distance in which the front wheel makes 1400\displaystyle 1400 revolutions.

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    NCERT’s answer
    $\displaystyle 560$
    \[d_1 = 80\text{ cm}, \quad d_2 = 200\text{ cm} \] \[\text{Distance covered by front wheel} = 1400\times\pi d_1 \] \[n = \frac{1400\times\pi d_1}{\pi d_2} = 1400\times\frac{80}{200} = 560 \] Answer: $\displaystyle 560$ revolutions

  3. Exercise 3

    Sides of a triangular field are 15\displaystyle 15 m, 16\displaystyle 16 m and 17\displaystyle 17 m. With the three corners of the field a cow, a buffalo and a horse are tied separately with ropes of length 7\displaystyle 7 m each to graze in the field. Find the area of the field which cannot be grazed by the three animals.

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    NCERT’s answer
    \(\displaystyle (24 \sqrt{21}-77) \mathrm{~m}^2\)
    Each animal grazes a sector of radius \(\displaystyle 7\text{ m}\), whose angle is the field's angle at that corner. NCERT_Solution_Class10_Maths_Exemplar_Ch11_Ex11-4_Q3 \[s = \frac{15+16+17}{2} = 24 \] \[\text{Area}(\triangle PQR) = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{24\times9\times8\times7} = 24\sqrt{21}\text{ m}^2 \] \[\angle P+\angle Q+\angle R = 180^\circ \quad \text{(angle sum property of a triangle)} \] \[\text{Grazed area} = \frac{\angle P+\angle Q+\angle R}{360^\circ}\pi r^2 = \frac{180^\circ}{360^\circ}\times\frac{22}{7}\times7^2 = 77\text{ m}^2 \] \[\text{Ungrazed area} = 24\sqrt{21}-77 \approx 109.98-77 = 32.98\text{ m}^2 \] Answer: \(\displaystyle (24\sqrt{21}-77)\text{ m}^2 \approx 32.98\text{ m}^2 \)

  4. Exercise 4

    Find the area of the segment of a circle of radius 12\displaystyle 12 cm whose corresponding sector has a central angle of 60\displaystyle 60° (Use π=3.14\displaystyle \pi=3.14).

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    NCERT’s answer
    \(\displaystyle (75.36-36 \sqrt{3}) \mathrm{~cm}^2\)
    Let \(\displaystyle O\) be the centre and \(\displaystyle AB\) the chord of the \(\displaystyle 60^\circ\) sector. NCERT_Solution_Class10_Maths_Exemplar_Ch11_Ex11-4_Q4 \[\text{Sector } OAB = \frac{60^\circ}{360^\circ}\pi r^2 = \frac16\times3.14\times12^2 = 75.36\text{ cm}^2 \] \[\triangle OAB \text{ is equilateral} \quad \text{(}OA = OB = 12\text{, included angle }60^\circ\text{)} \] \[\text{Area}(\triangle OAB) = \frac{\sqrt3}{4}(12)^2 = 36\sqrt3 \approx 62.35\text{ cm}^2 \] \[\text{Segment} = 75.36-36\sqrt3 \approx 75.36-62.35 = 13.01\text{ cm}^2 \] Answer: \(\displaystyle (75.36-36\sqrt3)\text{ cm}^2 \approx 13.01\text{ cm}^2\)

  5. Exercise 5

    A circular pond is 17.5\displaystyle 17.5 m is of diameter. It is surrounded by a 2\displaystyle 2 m wide path. Find the cost of constructing the path at the rate of Rs 25\displaystyle 25 per m2\displaystyle \mathrm{m}^2

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    NCERT’s answer
    Rs $\displaystyle 3061.50$
    Take \(\displaystyle \pi = 3.14\). \[r = \frac{17.5}{2} = 8.75\text{ m}, \quad R = r+2 = 10.75\text{ m} \] \[\text{Area of path} = \pi(R^2-r^2) = \pi(R-r)(R+r) = 3.14\times2\times19.5 \] \[= 3.14\times39 = 122.46\text{ m}^2 \] \[\text{Cost} = 122.46\times25 = 3061.50 \] Answer: Rs $\displaystyle 3061.50$

  6. Exercise 6

    In Fig. 11.17\displaystyle 11.17, ABCD is a trapezium with AB∥DC,AB=18 cm,DC=32 cm\displaystyle \mathrm{AB} \| \mathrm{DC}, \mathrm{AB}=18 \mathrm{~cm}, \mathrm{DC}=32 \mathrm{~cm} and distance between AB and DC=14 cm\displaystyle \mathrm{DC}=14 \mathrm{~cm}. If arcs of equal radii 7\displaystyle 7 cm with centres A, B, C and D have been drawn, then find the area of the shaded region of the figure. NCERT_Question_Class10_Maths_Exemplar_Ch11_Ex11-4_Q6

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    NCERT’s answer
    \(\displaystyle 196 \mathrm{~cm}^2\)
    Each corner carries a sector of radius \(\displaystyle 7\text{ cm}\), whose angle is the trapezium's angle at that vertex. NCERT_Solution_Class10_Maths_Exemplar_Ch11_Ex11-4_Q6 \[\text{Area(trapezium)} = \frac12(AB+DC)\times h = \frac12(18+32)\times14 = 350\text{ cm}^2 \] \[\angle A+\angle B+\angle C+\angle D = 360^\circ \quad \text{(angle sum property of a quadrilateral)} \] \[\text{Sum of 4 sectors} = \frac{\angle A+\angle B+\angle C+\angle D}{360^\circ}\pi r^2 = \frac{360^\circ}{360^\circ}\times\frac{22}{7}\times7^2 = 154\text{ cm}^2 \] \[\text{Shaded area} = 350-154 = 196\text{ cm}^2 \] Answer: \(\displaystyle 196\text{ cm}^2\)

  7. Exercise 7

    Three circles each of radius 3.5\displaystyle 3.5 cm are drawn in such a way that each of them touches the other two. Find the area enclosed between these circles.

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    NCERT’s answer
    \(\displaystyle 1.967 \mathrm{~cm}^2\) (approx)
    Centres of three mutually tangent circles of equal radius form an equilateral triangle of side 2r. NCERT_Solution_Class10_Maths_Exemplar_Ch11_Ex11-4_Q7 \[\text{Side of }\triangle X_1X_2X_3 = 2r = 7\text{ cm} \] \[\text{Area}(\triangle) = \frac{\sqrt3}{4}(7)^2 = \frac{49\sqrt3}{4} \approx 21.22\text{ cm}^2 \] \[\text{3 corner sectors, each }60^\circ\colon\quad 3\times\frac{60^\circ}{360^\circ}\pi r^2 = \frac12\times\frac{22}{7}\times(3.5)^2 = 19.25\text{ cm}^2 \] \[\text{Enclosed area} = 21.22-19.25 = 1.97\text{ cm}^2 \] Answer: \(\displaystyle \approx1.97\text{ cm}^2\)

  8. Exercise 8

    Find the area of the sector of a circle of radius 5\displaystyle 5 cm, if the corresponding arc length is 3.5\displaystyle 3.5 cm.

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    NCERT’s answer
    \(\displaystyle 8.7 \mathrm{~cm}^2\)
    \[\text{Area of sector} = \frac{1}{2} \times r \times l \] \[= \frac{1}{2} \times 5 \times 3.5 \] \[= 8.75 \text{ cm}^2 \] Answer: \(\displaystyle 8.75 \text{ cm}^2\)

  9. Exercise 9

    Four circular cardboard pieces of radii 7\displaystyle 7 cm are placed on a paper in such a way that each piece touches other two pieces. Find the area of the portion enclosed between these pieces.

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    NCERT’s answer
    \(\displaystyle 42 \mathrm{~cm}^2\)
    \[\text{Let centres be } A,B,C,D. \quad AB=BC=CD=DA=2r=14 \text{ cm} \quad \text{(circles touch)} \] NCERT_Solution_Class10_Maths_Exemplar_Ch11_Ex11-4_Q9 \[\text{Area of square } ABCD = (14)^2 = 196 \text{ cm}^2 \] \[\text{Area of 4 quarter circles} = 4\times\frac{90^\circ}{360^\circ}\times\pi r^2 = \pi r^2 \] \[= \frac{22}{7}\times 7^2 = 154 \text{ cm}^2 \] \[\text{Enclosed area} = 196 - 154 = 42 \text{ cm}^2 \] Answer: \(\displaystyle 42 \text{ cm}^2\)

  10. Exercise 10

    On a square cardboard sheet of area 784 cm2\displaystyle 784 \mathrm{~cm}^2, four congruent circular plates of maximum size are placed such that each circular plate touches the other two plates and each side of the square sheet is tangent to two circular plates. Find the area of the square sheet not covered by the circular plates.

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    NCERT’s answer
    \(\displaystyle 168 \mathrm{~cm}^2\)
    \[\text{Side of square sheet} = \sqrt{784} = 28 \text{ cm} \] \[2r = \frac{28}{2} = 14 \text{ cm} \quad \text{(two plates span a side)} \quad \Rightarrow \quad r = 7 \text{ cm} \] NCERT_Solution_Class10_Maths_Exemplar_Ch11_Ex11-4_Q10 \[\text{Area of 4 circles} = 4\pi r^2 = 4\times\frac{22}{7}\times 49 = 616 \text{ cm}^2 \] \[\text{Uncovered area} = 784 - 616 = 168 \text{ cm}^2 \] Answer: \(\displaystyle 168 \text{ cm}^2\)