NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles वृतों से संबंधित क्षेत्रफल
Important Questions for Class 10 Maths Chapter 11
Thank you for reading this post, don't forget to subscribe!Exercise 11.1
Question 1:
Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
6 cm त्रिज्या वाले एक वृत्त के एक त्रिज्यखंड का क्षेत्रफल ज्ञात कीजिए, जिसका कोण 60° है।
Solution
Given: r = 6 cm and \(\theta\) = 60°
Area of a sector = \(\frac{\theta}{360°} \times \pi r^2\)
\(\\\quad\) = \(\frac{60°}{360°} \times \frac{22}{7} \times 6^2\)
\(\\\quad\) = \(\frac{132}{7}\) cm2.
Question 2:
Find the area of a quadrant of a circle whose circumference is 22 cm.
एक वृत्त के चतुर्थांश का क्षेत्रफल ज्ञात कीजिए, जिसकी परिधि 22 cm है।
Solution
Circumference of a cirlce = 22 cm
\(\quad\quad 2\pi r\) = 22
\(\Rightarrow 2 \times \frac{22}{7} \times r\) = 22
\(\Rightarrow\) r = 22 \(\times \frac{7}{22} \times \frac{1}{2}\) = \(\frac{7}{2}\)
It should be noted that a quadrant of a circle is a sector which is making an angle of 90°.
\(\theta\) = 90°
Area of the quadrant = \(\frac{\theta}{360°} \times \pi r^2\)
\(\\\quad\) = \(\frac{90°}{360°} \times \frac{22}{7} \times (\frac{7}{2})^2\)
\(\\\quad\) = \(\frac{77}{8}\) cm2.
Question 3:
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
एक घड़ी की मिनट की सुई जिसकी लंबाई 14 cm है। इस सुई द्वारा 5 मिनट में रचित क्षेत्रफल ज्ञात कीजिए।
Solution
Length of minute hand = radius of the clock (circle)
radius, r = 14 cm
Angle swept by minute hand in 60 minutes = 360°
So, the angle swept by the minute hand in 5 minutes = 360° × \(\frac{5}{60}\) = 30°
\(\theta\) = 30°
Area of a sector = \(\frac{\theta}{360°} \times \pi r^2\)
\(\\\quad\) = \(\frac{30°}{360°} \times \frac{22}{7} \times 14^2\)
\(\\\quad\) = \(\frac{154}{3}\) cm2.
Question 4:
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (ii) major sector. (Use \(\pi\) = 3.14)
10 सेमी त्रिज्या वाले एक वृत्त की कोई जीवा केंद्र पर एक समकोण अंतरित करती है। निम्नलिखित के क्षेत्रफल ज्ञात कीजिए:
(i) संगत लघु वृत्तखण्ड (ii) संगत दीर्घ त्रिज्यखंड (\(\pi\) = 3.14 का प्रयोग कीजिए)।
Solution
r = 10 cm and \(\theta\) = 90°
Area of minor sector = \(\frac{\theta}{360°} \times \pi r^2\)
\(\\\quad\) = \(\frac{90°}{360°} \times 3.14 \times 10^2\)
\(\\\quad\) = 78.5 cm2.
And,
Area of \(\triangle\) AOB = \(\frac{1}{2} \times OB \times OA\) = \(\frac{1}{2} \times 10 \times 10\)
\(\\\quad\) = 50 cm2.
Now, area of minor segment = area of the minor sector – the area of ΔAOB
= 78.5 – 50 = 28.5 cm2
(ii) Area of major sector = Area of the circle – Area of he minor sector
= (3.14×102)-78.5 = 235.5 cm2
Question 5:
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
(i) the length of the arc (ii) area of the sector formed by the arc
(iii) area of the segment formed by the corresponding chord
त्रिज्या 21 cm वाले वृत्त का एक चाप केंद्र पर का कोण अंतरित करता है। ज्ञात कीजिए:
(i) चाप की लंबाई (ii) चाप द्वारा बनाए गए त्रिज्यखंड का क्षेत्रफल
(iii) संगत जीवा द्वारा बनाए गए वृत्तखण्ड का क्षेत्रफल
Solution
r = 21 cm, \(\theta\) = 60°
(i) length of the arc = \(\frac{\theta}{360°} \times 2 \pi r\)
\(\\\quad\)= \(\frac{60°}{360°} \times 2 \frac{22}{7} \times 21\)
\(\\\quad\) = 22 cm
(ii) Area of sector = \(\frac{\theta}{360°} \times \pi r^2\)
\(\\\quad\) = \(\frac{60°}{360°} \times \frac{22}{7} \times 21^2\)
\(\\\quad\) = 231 cm2.
(iii) Area of segment = Area of sector – Area of \(\triangle\) ABC
Now, Draw OC \(\perp\) AB
\(\cos 30° = \frac{OC}{OA}\)
\(\Rightarrow \frac{\sqrt{3}}{2} = \frac{OC}{21}\)
\(\Rightarrow OC= \frac{21\sqrt{3}}{2}\)
And,
\(\sin 30° = \frac{AC}{OA}\)
\(\Rightarrow \frac{1}{2} = \frac{AC}{21}\)
\(\Rightarrow AC= \frac{21}{2}\)
AB = 2 \(\times \frac{21}{2}\) = 21 cm
Area of \(\triangle\) AOB = \(\frac{1}{2} \times AB \times OC\)
\(\\\quad\) = \(\frac{1}{2} \times 21 \times \frac{21\sqrt{3}}{2}\)
\(\\\quad\) = \(\frac{441\sqrt{3}}{4}\) cm2.
Now,
Area of segment = Area of sector – Area of \(\triangle\) AOB
\(\quad\quad\quad\quad\quad\quad\) = 231 – \(\frac{441\sqrt{3}}{4}\) cm2.
Question 6:
A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use \(\pi\) = 3.14 and \(\sqrt{3}\) = 1.73)
Solution
Area of segment = Area of sector – Area of \(\triangle\) ABC
Area of sector = \(\frac{\theta}{360°} \times \pi r^2\)
\(\\\quad\) = \(\frac{60°}{360°} \times 3.14 \times 15^2\)
\(\\\quad\) = 117.75 cm2.
Now, Draw OC \(\perp\) AB
\(\cos 30° = \frac{OC}{OA}\)
\(\Rightarrow \frac{\sqrt{3}}{2} = \frac{OC}{15}\)
\(\Rightarrow OC= \frac{15\sqrt{3}}{2}\)
And,
\(\sin 30° = \frac{AC}{OA}\)
\(\Rightarrow \frac{1}{2} = \frac{AC}{15}\)
\(\Rightarrow AC= \frac{15}{2}\)
AB = 2 \(\times \frac{15}{2}\) = 15 cm
Area of \(\triangle\) AOB = \(\frac{1}{2} \times AB \times OC\)
\(\\\quad\) = \(\frac{1}{2} \times 15 \times \frac{15\sqrt{3}}{2}\)
\(\\\quad\) = \(\frac{225\sqrt{3}}{4}\)
= \(\frac{225\times 1.73}{4}\) = 97.3125 cm2.
Now,
Area of minor segment = Area of sector – Area of \(\triangle\) AOB
\(\quad\quad\quad\quad\quad\quad\) = 117.75 – 97.3125 = 20.4375 cm2.
Area of major segment = Area of circle – Area of minor segment
= 3.14 \(\times 15^2\) – 20.4375
= 706.5 – 20.4375 = 686.0625 cm2.
Question 7:
A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use \(\pi\) = 3.14 and \(\sqrt{3}\) = 1.73)
Solution
Area of segment = Area of sector – Area of \(\triangle\) ABC
Area of sector = \(\frac{\theta}{360°} \times \pi r^2\)
\(\\\quad\) = \(\frac{120°}{360°} \times 3.14 \times 12^2\)
\(\\\quad\) = 150.72 cm2.
Now, Draw OD \(\perp\) AB
\(\cos 60° = \frac{OC}{OA}\)
\(\Rightarrow \frac{1}{2} = \frac{OC}{12}\)
\(\Rightarrow OD= \frac{12}{2}\) = 6 cm
And,
\(\sin 60° = \frac{AC}{OA}\)
\(\Rightarrow \frac{\sqrt{3}}{2} = \frac{AC}{12}\)
\(\Rightarrow AC= \frac{12\sqrt{3}}{2}\)
AB = 2 \(\times \frac{12\sqrt{3}}{2}\) = 12\(\sqrt{3}\) cm
Area of \(\triangle\) AOB = \(\frac{1}{2} \times AB \times OD\)
\(\\\quad\) = \(\frac{1}{2} \times 12\sqrt{3} \times 6\)
\(\\\quad\) = 36\(\sqrt{3}\)
= 36 (1.73) = 63.28 cm2.
Now,
Area of minor segment = Area of sector – Area of \(\triangle\) AOB
\(\quad\quad\quad\quad\quad\quad\) = 150.72 – 63.28 = 88.44 cm2.
Question 8:
A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see Fig. 11.8). Find
(i) the area of that part of the field in which the horse can graze.
(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use \(\pi\) = 3.14)
Solution
θ = 90° and r = 5 m
(i) Area of the field that horse can graze
\(\quad\) = Area of sector
\(\quad\) = \(\frac{\theta}{360°} \times \pi r^2\)
\(\\\quad\) = \(\frac{90°}{360°} \times 3.14 \times 5^2\)
\(\\\quad\) = 19.625 m2.
(ii) Length of the rope is increased from 5 m to 10 m
New redius, r = 10 m
Area grazed by horse
\(\quad\) = Area of sector
\(\quad\) = \(\frac{\theta}{360°} \times \pi r^2\)
\(\\\quad\) = \(\frac{90°}{360°} \times 3.14 \times 10^2\)
\(\\\quad\) = 78.5 m2.
\(\therefore\) Area increased = 78.5 – 19.625 = 58.875 m2.
Question 9:
A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. 11.9. Find :
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.
Solution
d = 35 m, r = \(\frac{35}{2}\)
(i) length of the silver wire required = Circumference of Circle
\(\quad\) = 2\(\pi r\) = 2 \(\times \frac{22}{7} \times \frac{35}{2}\)
\(\quad\) = 110 mm
Total length of the silver required = 110 + 5\(\times\)diameters
\(\quad\quad\quad\quad\quad\quad\quad\quad\) = 110 + 5\(\times\) 35 = 285 mm
(ii) Sector angle of each brooch, \(\theta\) = \(\frac{360°}{10}\) = 36°
the area of each sector of brooch = \(\frac{\theta}{360°} \times \pi r^2\)
\(\\\quad\) = \(\frac{36°}{360°} \times \frac{22}{7} \times (\frac{35}{2})^2\)
\(\\\quad\) = \(\frac{385}{5}\) mm2
Question 10:
An umbrella has 8 ribs which are equally spaced (see Fig. 11.10). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
Solution
Radius, r = 45 cm
Total number of ribs (n) = 8
Central angle formed by any two consecutive ribs = \(\frac{360°}{8}\) = 45°
Area between the two consecutive ribs = \(\frac{\theta}{360°} \times \pi r^2\)
\(\\\quad\) = \(\frac{45°}{360°} \times \frac{22}{7} \times 45^2\)
\(\\\quad\) = \(\frac{22275}{28}\) cm2
Question 11:
A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.
Solution
r = 25 cm, \(\theta\) = 115°
Area of sector = \(\frac{\theta}{360°} \times \pi r^2\)
\(\\\quad\) = \(\frac{115°}{360°} \times \frac{22}{7} \times 25^2\)
\(\\\quad\) = \(\frac{158125}{252}\) cm2
Total area cleaned at each sweep of the blades = 2 \(\times \frac{158125}{252}\) = \(\frac{158125}{126}\) cm2
Question 12:
To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use\(\pi\) = 3.14)
जहाजों को समुद्र में जलस्तर के नीचे स्थित चट्टानों की चेतावनी देने के लिए, एक लाइट हाउस 80°कोण वाले एक त्रिज्यखंड में 16.5 km की दूरी तक लाल रंग का प्रकाश फैलता है। समुद्र के उस भाग का क्षेत्रफल ज्ञात कीजिए जिसमें जहाजों को चेतावनी दी जा सके। (\(\pi\) = 3.14 का प्रयोग कीजिए)
Solution
r = 16.5 km, \(\theta\) = 80°
Area of the sea = \(\frac{\theta}{360°} \times \pi r^2\)
\(\\\quad\) = \(\frac{80°}{360°} \times 3.14 \times (16.5)^2\)
\(\\\quad\) = 189.97 km2
Question 13:
A round table cover has six equal designs as shown in Fig. 11.11. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per cm2. (Use \(\sqrt{3}\) = 1.7)
एक गोले मेजपोश पर छः समान डिजाइन बने हुए हैं जैसाकि आकृति 11.11 में दर्शाया गया है। यदि मेजपोश की त्रिज्या 28 cm है, तो ₹ 0.35 प्रति वर्ग सेंटीमीटर की दर से इन डिजाइनों को बनाने की लागत ज्ञात कीजिए। (\(\sqrt{3}\) = 1.7 का प्रयोग कीजिए)
Solution
6 equal designs = 6 \(\times\) area of one segment
here, \(\theta\) = 60°, r = 28 cm
Cost of making design = ₹ 0.35 per cm2
ΔAOB is an equilateral triangle.
So, its area will be \(\frac{\sqrt{3}}{4}\)×a2 sq.unit
Area of one segment = Area of sector – Area of \(\triangle\) AOB
\(\quad\quad\quad\quad\quad\) = \(\frac{\theta}{360°} \times \pi r^2\) – \(\frac{\sqrt{3}}{4}\)×a2
\(\quad\quad\quad\quad\quad\) = \(\frac{60°}{360°} \times \frac{22}{7}\times 28^2\) – \(\frac{\sqrt{3}}{4}\)× 282
\(\quad\quad\quad\quad\quad\) = 410.67 – 333.2 = 77.47 cm2
Area of 6 design = 6 \(\times\) 77.47 = 464.82 cm2
So, total cost of making design = 464.82 \(\times\) ₹ 0.35 = ₹ 162.687
Question 14:
Tick the correct answer in the following :
Area of a sector of angle p (in degrees) of a circle with radius R is
निम्नलिखित में सही उत्तर चुनिए:
त्रिज्या R वाले वृत्त के उस त्रिज्यखंड का क्षेत्रफल जिसका कोण p° है, निम्नलिखित है:
(A) \(\frac{p}{180} \times 2 \pi R \quad\) (B) \(\frac{p}{180} \times \pi R^2\)
(C) \(\frac{p}{360} \times 2 \pi R \quad\) (D) \(\frac{p}{720} \times 2 \pi R^2\)
Solution
Area of sector = \(\frac{\theta}{360°}\times \pi R^2\)
\(\theta\) = p
So, area of sector = \(\frac{p}{360°}\times \pi R^2\)
Multiplying and dividing by 2
= \(\frac{2}{2} \times \frac{p}{360}\times \pi R^2\)
= \(\frac{p}{720} \times 2 \pi R^2\) (D)
NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles वृतों से संबंधित क्षेत्रफल
