Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is one of our many quizzes on Circle sectors and segments. Click on this to access more such competitive quizzes on Geometry.

Circle sectors and segments Quiz 2

mathematics

 A perfect circular clock is mounted on the wall. Ram wonders what the area of the segment is when the hour hand strikes the time of 3’ o clock. The length of hour hand is 4 cm.

 Area of the segment of the circle is to be found. Area of the segment of the circle is $A=\dfrac{1}{2}\times {{r}^{2}}\left[ \dfrac{\pi }{180}(\theta -\sin \theta ) \right]$  where the angle and radius are the variables.

The angle formed when the clock strikes 3o'clock is ${{90}^{\circ }}$. So, area of the segment of the circle is, $ \\ A=\dfrac{1}{2}\times {{r}^{2}}\left[ \dfrac{\pi }{180}(\theta -\sin \theta ) \right] \\ A=\dfrac{1}{2}\times {{4}^{2}}\left[ \dfrac{\pi }{180}(90-\sin 90) \right] \\ A=8\left[ \dfrac{\pi }{2}-\dfrac{\pi }{180} \right]=\dfrac{178\pi }{45}c{{m}^{2}}$ 

 Sarita wants to make a rangoli, for the measurements she makes a square with size 8 cm and a design as shown. Find the area of the shaded part  $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/9838d5fd-78c4-4fcd-9ed5-1ffb70557d3b6511139643405166377.png"/>

 Area of the segment of the circle is $A=\dfrac{1}{2}\times {{r}^{2}}\left[ \dfrac{\pi }{180}(\theta -\sin \theta ) \right]$  where the angle and radius are the variables. Area of shaded region $= 2\times$ Area of segment AC.

As seen from the diagram construction is made $ \\ $   <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/02d9d079-d63d-4e87-a18e-7802ee85c6c04568938342321868846.png"/>  $ \\ $ Area of shaded region $= 2\times$ Area of segment AC $ \\ $ Considering the quadrant ACD $ \\ $ Area of quadrant ACD $= \dfrac{\pi {{r}^{2}}}{4}=\dfrac{1}{4}\times 3.14\times 8\times 8=50.24c{{m}^{2}} $ $ \\ $ Area of $\Delta$ ACD $=\dfrac{1}{2}\times 8\times 8=32c{{m}^{2}}$ $ \\ $ Area of segment AC $=$ Area of quadrant ACD $-$ Area of $\Delta$ ACD = $50.24 - 32=18.24$ $ \\ $ Area of the region $= 18.24\times 2=36.48c{{m}^{2}}$ 

 Find the perimeter by the segment covered by two concentric circles with radius $20cm$ and $10cm$ with angle as ${{30}^{\circ }}$.

 Use the formula of perimeter to find the sum of all sides with two sides as 10cmother two sides as the arcs of two circles.

<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/9d368151-d8d7-4949-bf7f-2240acdc9076209566783971679946.png"/> $ \\ $ $l=2\pi r\times \dfrac{\theta }{{{360}^{\circ }}} \\ l=2\pi \times 20\times \dfrac{{{30}^{\circ }}}{{{360}^{\circ }}}=10.46 \\ l=2\pi \times 10\times \dfrac{{{30}^{\circ }}}{{{360}^{\circ }}}=5.233 \\ $ Since the perimeter includes two more sides as $(20-10)=10cm$  $ \\ $ Perimeter sum of all sides $= (10.46+5.233+10+10)=35.693$ 

 The standard measurement of shot describes the radius to be $20m$ and the central angle to be ${{34.92}^{\circ }}$. What can you say about the area of the field covered?

 The area sector is to be found. The area sector is given as $\left(\frac{θ}{360^{\circ}}\right)\times \pi \times r^2$. Here the angle and radius are the unknown

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/fc65c4b6-d93f-48a6-80e9-bb800730908c3979848641847185649.png"/> $ \\ $ The area of a sector is given as $ \\ A=\dfrac{\theta }{360}\times \pi \times {{r}^{2}} \\ A=\dfrac{{{34.92}^{\circ }}}{{{360}^{\circ }}}\times \pi \times {{20}^{2}}=121.832{{m}^{2}}$ 

 A designer wants to make a new surgical mask. She designs the template with 2 concentric circles with radii as 21cm and 7 cm. For cutting the template she uses an angle of ${{\left( 30 \right)}^{\circ }}$. If the total area of cloth she has is 1 sq. m. Find how many masks she can make.   

 Find area of the individual sectors and subtract them or use the formula - The area between a divided sectors is given as $\dfrac{\theta }{360}\times \pi \times ({{R}^{2}}-{{r}^{2}})$

Let $R$ and $r$ denote the radius of the sector such that, $R=21cm$ and $r=7cm$  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/2ec796f4-6d73-4145-8832-3395b11e7d506524585101972452821.png"/> $ \\ $ The area of a sector is given as $ \\ A=\dfrac{\theta }{360}\times \pi \times {{r}^{2}} \\ $ Here area of the region for mask $= \dfrac{\theta }{360}\times \pi \times ({{R}^{2}}-{{r}^{2}})$ $ \\ $ Substituting values, $\dfrac{30}{360}\times \pi \times ({{21}^{2}}-{{7}^{2}})=\dfrac{231}{2}-\dfrac{77}{6}=102.67c{{m}^{2}} \\ $ Since the area of cloth requires is, $\dfrac{1000}{102.67}=9.73\approx 9$ 