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NCERT Exemplar · Class 10 Mathematics Area Related to Circles

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EXERCISE 11.4 11–20 (part 7 of 7)

  1. Exercise 11

    Floor of a room is of dimensions 5\displaystyle 5 m × 4\displaystyle 4 m and it is covered with circular tiles of diameters 50\displaystyle 50 cm each as shown in Fig. 11.18. Find the area of floor that remains uncovered with tiles. (Use π=3.14\displaystyle \pi=3.14) NCERT_Question_Class10_Maths_Exemplar_Ch11_Ex11-4_Q11

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    \(\displaystyle 4.3 \mathrm{~m}^2\)
    \[\text{Tiles along 5 m} = \frac{5}{0.5}=10, \quad \text{along 4 m} = \frac{4}{0.5}=8 \quad \Rightarrow \quad \text{Number of tiles} = 80 \] \[\text{Area of floor} = 5\times 4 = 20 \text{ m}^2 \] \[\text{Area of one tile} = \pi r^2 = 3.14\times(0.25)^2 = 0.19625 \text{ m}^2 \] \[\text{Area covered} = 80\times 0.19625 = 15.7 \text{ m}^2 \] \[\text{Uncovered area} = 20-15.7 = 4.3 \text{ m}^2 \] Answer: \(\displaystyle 4.3 \text{ m}^2\)

  2. Exercise 12

    All the vertices of a rhombus lie on a circle. Find the area of the rhombus, if area of the circle is 1256 cm2\displaystyle 1256 \mathrm{~cm}^2. (Use π=3.14\displaystyle \pi=3.14).

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    \(\displaystyle 800 \mathrm{~cm}^2\)
    \[\angle A + \angle C = 180^\circ \quad \text{(opposite angles of cyclic quadrilateral } ABCD\text{)} \] \[\angle A = \angle C \quad \text{(opposite angles of a rhombus)} \] \[\angle A = 90^\circ \;\Rightarrow\; ABCD \text{ is a square} \] NCERT_Solution_Class10_Maths_Exemplar_Ch11_Ex11-4_Q12 \[\pi r^2 = 1256 \quad \Rightarrow \quad r^2 = \frac{1256}{3.14}=400 \quad \Rightarrow \quad r=20 \text{ cm} \] \[AC = BD = 2r = 40 \text{ cm} \quad \text{(diagonal is a diameter)} \] \[\text{Area of square} = \frac{1}{2} \times AC \times BD = \frac{1}{2}(40)(40)=800 \text{ cm}^2 \] Answer: \(\displaystyle 800 \text{ cm}^2\)

  3. Exercise 13

    An archery target has three regions formed by three concentric circles as shown in Fig. 11.19. If the diameters of the concentric circles are in the ratio 1:2:3\displaystyle 1: 2: 3, then find the ratio of the areas of three regions. NCERT_Question_Class10_Maths_Exemplar_Ch11_Ex11-4_Q13

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    $\displaystyle 1$ : $\displaystyle 3$ : $\displaystyle 5$
    \[\text{Let the radii be } r, 2r, 3r \quad \text{(diameters in ratio } 1:2:3\text{)} \] \[A_1 = \pi r^2 \] \[A_2 = \pi(2r)^2-\pi r^2 = 3\pi r^2 \] \[A_3 = \pi(3r)^2-\pi(2r)^2 = 5\pi r^2 \] \[A_1:A_2:A_3 = 1:3:5 \] Answer: \(\displaystyle 1:3:5\)

  4. Exercise 14

    The length of the minute hand of a clock is 5\displaystyle 5 cm. Find the area swept by the minute hand during the time period 6\displaystyle 6 : 05\displaystyle 05 a m and 6\displaystyle 6 : 40\displaystyle 40 a m.

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    \(\displaystyle 45 \frac{5}{6} \mathrm{~cm}^2\)
    \[\text{Time elapsed} = 6{:}40 - 6{:}05 = 35 \text{ min} \] \[\theta = 35\times 6^\circ = 210^\circ \quad \text{(minute hand turns } 6^\circ\text{/min)} \] \[\text{Area swept} = \frac{\theta}{360^\circ}\times \pi r^2 = \frac{210}{360}\times\frac{22}{7}\times 5^2 \] \[= \frac{7}{12}\times\frac{22}{7}\times 25 = \frac{275}{6} \text{ cm}^2 \] Answer: \(\displaystyle \dfrac{275}{6} \text{ cm}^2 \approx 45.83 \text{ cm}^2\)

  5. Exercise 15

    Area of a sector of central angle 200\displaystyle 200° of a circle is 770 cm2\displaystyle 770 \mathrm{~cm}^2. Find the length of the corresponding arc of this sector.

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    NCERT’s answer
    \(\displaystyle 73 \frac{1}{3} \mathrm{~cm}\)
    \[\text{Area} = \frac{\theta}{360^\circ}\pi r^2 \] \[770 = \frac{200}{360}\times\frac{22}{7}\times r^2 \] \[r^2 = 441 \quad\Rightarrow\quad r = 21\text{ cm} \] \[\text{Arc length} = \frac{\theta}{360^\circ}\times 2\pi r = \frac{200}{360}\times 2\times\frac{22}{7}\times 21 \] \[= \frac{220}{3}\text{ cm} \] Answer: \(\displaystyle \dfrac{220}{3}\text{ cm} \approx 73.3\text{ cm} \)

  6. Exercise 16

    The central angles of two sectors of circles of radii 7\displaystyle 7 cm and 21\displaystyle 21 cm are respectively 120\displaystyle 120° and 40\displaystyle 40°. Find the areas of the two sectors as well as the lengths of the corresponding arcs. What do you observe?

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    NCERT’s answer
    Areas: \(\displaystyle \frac{154}{3} \mathrm{~cm}^2, 154 \mathrm{~cm}^2\); Arc lengths: \(\displaystyle \frac{44}{3} \mathrm{~cm}\); Arc lengths of two sectors of two different circles may be equal, but their area need not be equal.
    \[\text{Area}_1=\frac{120}{360}\pi(7)^2=\frac{1}{3}\times\frac{22}{7}\times49=\frac{154}{3}\text{ cm}^2 \] \[\text{Arc}_1=\frac{120}{360}\times2\pi(7)=\frac{44}{3}\text{ cm} \] \[\text{Area}_2=\frac{40}{360}\pi(21)^2=\frac{1}{9}\times\frac{22}{7}\times441=154\text{ cm}^2 \] \[\text{Arc}_2=\frac{40}{360}\times2\pi(21)=\frac{44}{3}\text{ cm} \] The two arcs are equal, \(\displaystyle \tfrac{44}{3}\text{ cm} \) each, though the sector areas differ. Answer: Area\(\displaystyle _1=\dfrac{154}{3}\text{ cm}^2\approx51.3\text{ cm}^2\), Area\(\displaystyle _2=154\text{ cm}^2\); both arcs \(\displaystyle =\dfrac{44}{3}\text{ cm}\approx14.7\text{ cm}\)

  7. Exercise 17

    Find the area of the shaded region given in Fig. 11.20. NCERT_Question_Class10_Maths_Exemplar_Ch11_Ex11-4_Q17

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    \(\displaystyle (180-8 \pi) \mathrm{~cm}^2\)
    NCERT_Solution_Class10_Maths_Exemplar_Ch11_Ex11-4_Q17 Each semicircle has diameter \(\displaystyle s\), the side of \(\displaystyle PQRS\), so \(\displaystyle r=\tfrac{s}{2}\) and its tip is \(\displaystyle \tfrac{s}{2}+r\) from the centre. \[\frac{s}{2}+r=s=\frac{14}{2}-3=4\text{ cm} \] \[r=\frac{s}{2}=2\text{ cm} \] \[\text{Design}=s^2+4\left(\tfrac12\pi r^2\right)=16+8\pi \] \[\text{Shaded}=14^2-(16+8\pi)=180-8\pi \] \[=180-\frac{176}{7}=\frac{1084}{7}\text{ cm}^2 \] Answer: \(\displaystyle (180-8\pi)\text{ cm}^2=\dfrac{1084}{7}\text{ cm}^2\approx154.9\text{ cm}^2 \)

  8. Exercise 18

    Find the number of revolutions made by a circular wheel of area 1.54 m2\displaystyle 1.54 \mathrm{~m}^2 in rolling a distance of 176\displaystyle 176 m.

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    NCERT’s answer
    $\displaystyle 40$
    \[\text{Area}=\pi r^2=1.54\text{ m}^2 \] \[r^2=1.54\times\frac{7}{22}=0.49 \quad\Rightarrow\quad r=0.7\text{ m} \] \[\text{Circumference}=2\pi r=2\times\frac{22}{7}\times0.7=4.4\text{ m} \] \[\text{Revolutions}=\frac{176}{4.4}=40 \] Answer: $\displaystyle 40$ revolutions

  9. Exercise 19

    Find the difference of the areas of two segments of a circle formed by a chord of length 5\displaystyle 5 cm subtending an angle of 90\displaystyle 90° at the centre.

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    \(\displaystyle \left(\frac{25 \pi}{4}+\frac{25}{2}\right) \mathrm{~cm}^2\)
    NCERT_Solution_Class10_Maths_Exemplar_Ch11_Ex11-4_Q19 \[AB^2=OA^2+OB^2=2r^2 \quad(\angle AOB=90^\circ) \] \[25=2r^2 \quad\Rightarrow\quad r^2=\frac{25}{2} \] \[\text{Minor segment}=\tfrac14\pi r^2-\tfrac12 r^2=\frac{25\pi}{8}-\frac{25}{4} \] \[\text{Difference}=(\pi r^2-\text{minor})-\text{minor}=\pi r^2-2(\text{minor}) \] \[=\frac{25\pi}{2}-\frac{25\pi}{4}+\frac{25}{2}=\frac{25\pi}{4}+\frac{25}{2} \] \[=\frac{25}{4}\times\frac{22}{7}+\frac{25}{2}=\frac{225}{7}\text{ cm}^2 \] Answer: \(\displaystyle \left(\dfrac{25\pi}{4}+\dfrac{25}{2}\right)\text{ cm}^2=\dfrac{225}{7}\text{ cm}^2\approx32.1\text{ cm}^2 \)

  10. Exercise 20

    Find the difference of the areas of a sector of angle 120\displaystyle 120° and its corresponding major sector of a circle of radius 21\displaystyle 21 cm.

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    \(\displaystyle 462 \mathrm{~cm}^2\)
    \[\text{Circle area}=\pi r^2=\frac{22}{7}\times21^2=1386\text{ cm}^2 \] \[\text{Minor sector}(120^\circ)=\frac{120}{360}\times1386=462\text{ cm}^2 \] \[\text{Major sector}(240^\circ)=1386-462=924\text{ cm}^2 \] \[\text{Difference}=924-462=462\text{ cm}^2 \] Answer: $\displaystyle 462$ cm\(\displaystyle ^2\)