Circles - I Practice Set
Complete 15 Questions with Explanations
1. Find the perimeter of a circular garden which occupies an area of 254 4/7 sqm.
Area = $254 \frac{4}{7} = \frac{1782}{7}$ sqm.
$\pi r^2 = \frac{1782}{7} \implies \frac{22}{7} \times r^2 = \frac{1782}{7} \implies r^2 = 81 \implies r = 9$ m.
Perimeter = $2\pi r = 2 \times \frac{22}{7} \times 9 = \frac{396}{7} = 56 \frac{4}{7}$ m.
$\pi r^2 = \frac{1782}{7} \implies \frac{22}{7} \times r^2 = \frac{1782}{7} \implies r^2 = 81 \implies r = 9$ m.
Perimeter = $2\pi r = 2 \times \frac{22}{7} \times 9 = \frac{396}{7} = 56 \frac{4}{7}$ m.
2. If the perimeter of a circle is equal to that of a square, then what is the ratio between their areas?
$2\pi r = 4a \implies a = \frac{\pi r}{2}$.
Ratio = $\frac{\pi r^2}{a^2} = \frac{\pi r^2}{(\pi r/2)^2} = \frac{4}{\pi} = \frac{4 \times 7}{22} = \frac{14}{11}$.
Ratio = $\frac{\pi r^2}{a^2} = \frac{\pi r^2}{(\pi r/2)^2} = \frac{4}{\pi} = \frac{4 \times 7}{22} = \frac{14}{11}$.
3. If the radius of one circle is Nine times the radius of another, how many times does the area of the greater circle contain the area of the smaller circle?
Let smaller radius be $r$, then larger radius $R = 9r$.
Area Ratio = $\frac{\pi R^2}{\pi r^2} = \frac{(9r)^2}{r^2} = 81$.
Area Ratio = $\frac{\pi R^2}{\pi r^2} = \frac{(9r)^2}{r^2} = 81$.
4. If the radius of the circle is reduced by (33 1/3)% then its circumference is reduced by:
Circumference ($C = 2\pi r$) is directly proportional to radius. If $r$ reduces by $X\%$, $C$ also reduces by $X\%$.
5. Two smaller parks of diameters 24 m and 32 m are to be replaced by a bigger circular park. What would be the radius of the new park?
$r_1 = 12, r_2 = 16$.
$\pi R^2 = \pi(12^2 + 16^2) = \pi(144 + 256) = 400\pi$.
$R = \sqrt{400} = 20$ m.
$\pi R^2 = \pi(12^2 + 16^2) = \pi(144 + 256) = 400\pi$.
$R = \sqrt{400} = 20$ m.
6. The area of a circle is 23 43/56 sq m. Find the length of an arc subtending 45 deg at the centre.
Area = $\frac{1331}{56} \implies \frac{22}{7}r^2 = \frac{1331}{56} \implies r^2 = \frac{1331 \times 7}{56 \times 22} = \frac{121}{16} \implies r = \frac{11}{4}$.
Arc Length = $\frac{\theta}{360} \times 2\pi r = \frac{45}{360} \times 2 \times \frac{22}{7} \times \frac{11}{4} = \frac{1}{8} \times \frac{121}{7} = \frac{121}{56} = 2 \frac{9}{56}$ m.
Arc Length = $\frac{\theta}{360} \times 2\pi r = \frac{45}{360} \times 2 \times \frac{22}{7} \times \frac{11}{4} = \frac{1}{8} \times \frac{121}{7} = \frac{121}{56} = 2 \frac{9}{56}$ m.
7. The length of common chord is 48 cm. Diameters are 52 cm and 60 cm. Distance between centres is:
Half chord = 24 cm. $R_1 = 26, R_2 = 30$.
$d_1 = \sqrt{26^2 - 24^2} = 10$. $d_2 = \sqrt{30^2 - 24^2} = 18$.
Distance = $10 + 18 = 28$ cm.
$d_1 = \sqrt{26^2 - 24^2} = 10$. $d_2 = \sqrt{30^2 - 24^2} = 18$.
Distance = $10 + 18 = 28$ cm.
8. An arc 17/3 cm long subtends 51 deg. What is the circumference?
$\text{Arc Length} = \frac{\theta}{360} \times \text{Circumference}$
$\frac{17}{3} = \frac{51}{360} \times C \implies C = \frac{17 \times 360}{3 \times 51} = \frac{360}{9} = 40$ cm.
$\frac{17}{3} = \frac{51}{360} \times C \implies C = \frac{17 \times 360}{3 \times 51} = \frac{360}{9} = 40$ cm.
9. Radii 15 cm and 10 cm touch externally. Length of common tangent AB is:
$AB = \sqrt{D^2 - (r_1-r_2)^2}$. Since touching externally, $D = 15+10 = 25$.
$AB = \sqrt{25^2 - 5^2} = \sqrt{600} = 10\sqrt{6}$ cm.
$AB = \sqrt{25^2 - 5^2} = \sqrt{600} = 10\sqrt{6}$ cm.
10. A chord is equal to its radius (13 cm). Angle subtended in major segment?
Chord = Radius forms an equilateral triangle at the center. Angle at center = 60°. Angle in major segment = Half of center angle = 30°.
11. Find area of sector: radius 13/2 cm, arc length 9/2 cm.
Area = $\frac{1}{2} \times \text{radius} \times \text{arc length} = \frac{1}{2} \times \frac{13}{2} \times \frac{9}{2} = \frac{117}{8} = 14 \frac{5}{8}$ sqcm.
12. Diameter is 117 cm less than circumference. Length of diameter?
$\pi D - D = 117 \implies D(\frac{22}{7} - 1) = 117 \implies D(\frac{15}{7}) = 117 \implies D = \frac{117 \times 7}{15} = \frac{819}{15} = 54 \frac{9}{15}$ cm.
13. Radius 63 cm. 12 revolutions in 8 seconds. Speed in km/hr?
Distance in 8 sec = $12 \times 2 \times \frac{22}{7} \times 63 = 4752$ cm = 47.52 m.
Speed = $\frac{47.52}{8} \times \frac{18}{5} = 5.94 \times 3.6 = 21.384$ km/hr.
Speed = $\frac{47.52}{8} \times \frac{18}{5} = 5.94 \times 3.6 = 21.384$ km/hr.
14. Area of circle inscribed in equilateral triangle of side 32 cm?
In-radius ($r$) of equilateral triangle = $\frac{a}{2\sqrt{3}} = \frac{32}{2\sqrt{3}} = \frac{16}{\sqrt{3}}$.
Area = $\pi r^2 = \pi (\frac{16}{\sqrt{3}})^2 = \frac{256}{3}\pi$ sqcm.
Area = $\pi r^2 = \pi (\frac{16}{\sqrt{3}})^2 = \frac{256}{3}\pi$ sqcm.
15. Circular disc area 1.96 pi m² rolls 5.28 km. Revolutions?
Area = $1.96\pi \implies \pi r^2 = 1.96\pi \implies r = 1.4$ m.
Circumference = $2\pi r = 2 \times \frac{22}{7} \times 1.4 = 8.8$ m.
Revolutions = $\frac{5280}{8.8} = 600$.
Circumference = $2\pi r = 2 \times \frac{22}{7} \times 1.4 = 8.8$ m.
Revolutions = $\frac{5280}{8.8} = 600$.
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