Question

Q.1: The cost of fencing a circular field at the rate of Rs. 24 per metre is Rs. 5280. The field is to be ploughed at the rate of Rs. 0.50 per m². Find the cost of ploughing the field (Take π = 22/7).

Q.2: The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?

Q.3: Find the area of the sector of a circle with radius 4 cm and of angle 30°. Also, find the area of the corresponding major sector (Use π = 3.14)​

Answer 1

Solution : 1

$\bf \: \underline{Given} :$

$\sf \: Cost \: of \: fencing \: circular \: field = Rs. \: 5280$

$\sf \: Rate \: of \: ploughing \: per \: metre^{2} = Rs. \: 0.50$

$\bf \: \underline{To \: find} :$

$\sf \: We \: have \: to \: find \: the \: cost \: of \: ploughing \: the \: field.$

$\bf \: \underline{\underline{Solution} }:$

$\sf \: First \: of \: all \: we \: have \: find \: the \: circumference \: of \: the \: circular \: field.$

$\sf \: Circumference = \dfrac{Total \: cost}{Cost \: per \: metre}$

$\sf \: Circumference = \dfrac{5280}{24}$

$\sf \: Circumference = 220$

$\sf \: \underline{Therefore},$

$\sf \: \implies \: 2 \pi r = 220$

$\sf \:\implies \: 2 \times \dfrac{22}{7} \times r = 220$

$\sf \: \implies \:\dfrac{44}{7} \times r = 220$

$\sf \: \implies \:r = 220 \times \dfrac{44}{7}$

$\sf \: \implies \:r = 35$

$\sf \: \underline{ So, radius \: of \: the \: circular \: field = 35 \: metre. }$

$\bf \: Now,$

$\sf \: We \: will \: find \: the \: area.$

$\sf \: Area \: of \: circle = \pi r^{2}$

$\sf \: \implies \: \dfrac{22}{7} \times 35^{2}$

$\sf \: \implies \: \dfrac{22 \times 1225}{7}$

$\sf \: \implies \: 3850m^{2}$

$\sf \: Cost \: of \: ploughing \: per \: m ^{2} = Rs. \: 0.50$

$\sf \: So, cost \: of \: ploughing \: 3850 \: m^{2} = 3850 \times 0.05$

$\underline{ \boxed{ \pink{\sf \: Hence, C ost \: of \: ploughing \: 3850 \: m^{2} = Rs. \: 1925}}}$

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Solution : 2

$\bf \: \underline{Given} :$

$\sf \: Diameter \: of \: wheel \: of \: a \: car = 80 \: cm$

$\sf \: The \: speed \: of \: the \: car = 60 \: km \: per \: hrs.$

$\small \sf \: Time \: at \: which \: the \: car \: can \: cover \: a \: distance \: at \: a \: speed \: of \: 66 \: km \ hrs = 10 \: minutes.$

$\bf \: \underline{To \: find} :$

$\sf \: We \: have \: to \: find \: the \: no. \: of \: complete \: revolution \: that \: a \: car \: can \: cover \: in \: 10 \: days.$

$\bf \: \underline{\underline{Solution} }:$

$\sf \: First \: of \: all \: we \: will \: calculate \: the \: distance \: traveled \: by \: car \: in \: 10 \: minutes.$

$\sf \: Before \: that, we \: have \: to \: convert \: the \: minutes \: into \: hrs.$

$\sf \: \implies \: 1 \: minute \: = \: \dfrac{1}{60} \: hours$

$\sf \: \implies \: 10 \: minute \: = \: \dfrac{1}{60} \: hours$

$\sf \: \implies \: \: \dfrac{1}{6} \: hours$

$\sf \: \underline{ As \: we \: know \: that},$

$\sf \: Distance = Speed \times Time$

$\sf \: Distance = 66 \times \dfrac{1}{6}$

$\sf \: Distance = 11 \: km$

$\sf \: Now, we \: will \: convert \: the \: distance \: coved \: by \: car \: from \: km \: to \: cm.$

$\sf \: \implies \: 11 \: km \: = 11 \times 100000$

$\sf \: \implies \: 11 \: km \: = 1100000 \: cm$

$\sf \: Now, \: we \: will \: calculate \: the \: circumference.$

$\sf \: For \: that \: we \: need \: to \: find \: the \: radius.$

$\sf \: Radius = \dfrac{Diameter}{2}$

$\sf \: Radius = \dfrac{80}{2}$

$\sf \: Radius = 40$

$\sf \: Now, it's \: time \: to \: find \: the \: circumference.$

$\sf \: Circumference = 2 \pi r$

$\sf \: Circumference = 2 \times \dfrac{22}{7} \times 40$

$\sf \: Circumference = \dfrac{422}{7} \times 40$

$\sf \: Circumference = \dfrac{1760}{7} \: cm \: or \: 251.43 \: cm$

$\bf \: Now,$

$\sf \: No. \: of \: revolution \times Circumference = Distance \: traveled \: in \: 10 \: minutes$

$\sf \: No. \: of \: revolution \times \dfrac{1760}{7} = 11$

$\sf \: No. \: of \: revolution = 4375$

$\underline{ \boxed{ \pink{ \sf \: So, \: the \: number \: of \: revolution \: in \: 10 \: minutes = 4375}}}$

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Answer 2

Answer:

Question 1 :

Solution

Rate at which the field is fenced = Rs 24 per metre

Cost of fencing = Rs 5280

Rate of ploughing the field Rs 0.50 per sqm.

$perimeter = \frac{cost \: of \: fencing}{rate}$

$\implies \: \frac{5280}{24} = 220m$

The perimeter is 220 m

Now , the radius of the circular field ,

$2\pi \: r = 220 \\ \implies \: r \: = \frac{220 \times 7}{44} = 35m$

Cost of ploughing the field = area × rate

$\implies \: \frac{22}{7} \times 35 \times 35 \times 0.50 =rs \: 1925$

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Question 2 :

Solution

Diameter = 80 cm = 0.8 m

time taken to travel = 10 m = 1/6 hr

speed of car = 66km/hr

$distance \: = speed \times time \\ \implies \: 66 \times \frac{1}{6} = 11km = 11000m$

Distance travelled in one revolution = πd

$\frac{22}{7} \times \frac{8}{35} = \frac{88}{35} m$

$number \: of \: revolution \: = \frac{total \: distance}{distance \:travelled \: in \: one \: revolution}$

$\implies \: 11000 \div \frac{88}{35} \\ \\ \implies \: \: 11000 \times \frac{35}{88} = 4375$

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Question 3 :

Formula for finding area of sector =

$\frac{ \theta}{360} \times \pi {r}^{2}$

$\implies \: \frac{30}{360} \times 3.14 \times 4 \times 4$

$= 4.19 {m}^{2}$

Area of major sector

$\pi {r}^{2} - area \: of \: minor \: sector$

$\implies \: 3.14 \times 4 \times 4 - 4.19$

$\implies \: 46.05 {m}^{2}$