Areas Related to Circles
Category : 10th Class
Areas Related to Circles
Areas Related to Circles
10. Area of segment of a circle = Area of the corresponding sector - Area of the corresponding triangle.
Snap Test
(a) 70 cm (b) 50 cm
(c) 60 cm (d) 80 cm
(e) None of these
Ans. (a)
Explanation: Distance covered by the wheel in 1 revolution
\[=\,\,\,\frac{total\,dis\operatorname{tance}}{number\,of\,revolution}\,\,=\,\,\left( \frac{1100000}{5000} \right)\,cm\,=\,220\,cm\]
\[\therefore \] Circumference of the wheel \[=\text{ }220\text{ }cm\]
\[\Rightarrow ~~\,\,\,~2\pi r\,\,=\,\,220cm~~~\Rightarrow \text{ }\pi r\,\,=\,\,110\,cm\]
\[\Rightarrow ~~\,\,\,~r=\left( \frac{110\times 7}{22} \right)\,cm\,\,=\,\,35\,cm\]
Hence, the diameter of the wheel = 70 cm.
(a) \[8.625\,\,c{{m}^{2}}\]
(b) \[9.1\,\,c{{m}^{2}}\]
(c) \[9.625\,\,c{{m}^{2}}\]
(d) \[7.625\,\,c{{m}^{2}}\]
(e) None of these
Ans. (c)
Explanation: It is given that outer radius \[R\text{ }=\text{ }7\text{ }cm\] and inner radius \[r\,\,=\,\,3.5\,cm\].
Area of shaded region
\[=\text{ }(area\text{ }of\text{ }the\text{ }sector\text{ }with\text{ }radius\text{ }=\text{ }R\text{ }=\text{ }7\text{ }cm\text{ }and\text{ }central\text{ }angle\text{ }=\,\,\theta \,\,=\text{ }30{}^\circ )\]
\[-\text{ }(area\text{ }of\text{ }the\text{ }sector\text{ }with\text{ }radius\text{ }=\text{ }r\text{ }=\text{ }3.5\text{ }cm\text{ }and\text{ }central\text{ }angle\text{ }=\,\,\theta \,\,~=\text{ }30{}^\circ )\]
= \[\frac{\pi {{R}^{2}}\theta }{360}-\frac{\pi {{r}^{2}}\theta }{360}\,\,=\,\,\frac{\pi \theta }{360}({{R}^{2}}-{{r}^{2}})=\frac{\pi \theta }{360}(R+r)(R-r)\]
= \[\left[ \frac{22}{7}\times \frac{30}{360}\times (7+3.5)\times (7-3.5) \right]\,c{{m}^{2}}\]
= \[\left[ \frac{22}{7}\times \frac{30}{360}\times 10.5\,\,\times \,\,3.5 \right]\,c{{m}^{2}}\,=9.625\,c{{m}^{2}}\]
3. In the given figure, ABCPA is a quadrant of a circle of radius 14 cm. With AC as diameter, a semicircle is drawn. Find the area of the shaded region.
(a) \[94\text{ }c{{m}^{2}}~\]
(b) \[92\text{ }c{{m}^{2}}\]
(c) \[95\text{ }c{{m}^{2}}~\]
(d) \[98\text{ }c{{m}^{2}}\]
(e) None of these
Ans. (d)
Explanation: In right Angled \[\Delta \,ABC\]:
\[AC2\text{ }=\text{ }AB2\text{ }+\text{ }BC2\text{ }=\text{ }142\text{ }+\text{ }142\text{ }=\text{ }196\text{ }+\text{ }196\text{ }=\text{ }392\]
\[\Rightarrow \,\,\,\,AC=\sqrt{392}\,\,=\,\,14\sqrt{2}\,cm\]
Area of the shaded region
= Area of semicircle ACQA - Area of segment ACPA
= (Area of semicircle with diameter AC) - (Area of quadrant ABCPA – Area \[\Delta \,ABC\])
= (Area of semicircle with radius \[{{r}_{1}}\,=\,\,\frac{1}{2}\,AC\,\,\,=\,\,AC\,\,\,=\,\,7\sqrt{2}~cm)\,\,-\,\,\left( Area\text{ }of\text{ }quadrant\text{ }with\text{ }radius\text{ }{{r}_{2}}=\text{ }14\text{ }cm \right)\text{ }+\]
\[+\text{ }\left( Area\text{ }of\text{ }triangle\text{ }with\text{ }b\text{ }=\text{ }14\text{ }cm,\text{ }h\text{ }=\text{ }14\text{ }cm \right)\]
\[=\,\,\,\,\,\left( \frac{1}{2}\times \pi {{r}_{1}}^{2} \right)-\left( \frac{1}{4}\times \pi {{r}_{2}}^{2} \right)+\left( \frac{1}{2}\times b\times h \right)\]
= \[\left[ \left\{ \frac{1}{2}\times \frac{22}{7}\times {{(7\sqrt{2})}^{2}} \right\}-\left\{ \frac{1}{4}\times \frac{22}{7}\times 14\times 14 \right\}+\left\{ \frac{1}{2}\times 14\times 14 \right\} \right]\,c{{m}^{2}}\]
= \[\left( 154\text{ }-\text{ }154\text{ }+\text{ }98 \right)\text{ }c{{m}^{2}}=\text{ }98\text{ }c{{m}^{2}}\].
(a) 24 cm (b) 44 cm
(c) 45 cm (d) 30 cm
(e) None of these
Ans. (b)
Explanation: Radius of the circle \[r\text{ }=\text{ }28\text{ }cm\].
Length of the wire = circumference of the circle = \[2\pi r=\left( 2\times \frac{22}{7}\times 28 \right)\,cm\,\,\,=\,\,176\,\,cm\]
\[\therefore \] \[Perimeter\text{ }of\text{ }the\text{ }square\text{ }=\text{ }length\text{ }of\text{ }the\text{ }wire\text{ }=\text{ }176\text{ }cm\]
\[4a\text{ }=\text{ }176\text{ }cm\text{ }\left[ where\text{ }a\text{ }=\text{ }side\text{ }of\text{ }the\text{ }square \right]\]
\[a\,\,=\,\,\left( \frac{176}{4} \right)\,cm\,\,=\,\,44\,cm.\]
Hence, the length of the side of the square \[=\text{ }44\text{ }cm\].
(a) \[1246\text{ }{{m}^{2}}\]
(b) \[2618\text{ }{{m}^{2}}\]
(c) \[3264\text{ }{{m}^{2}}\]
(d) \[4460\text{ }{{m}^{2}}\]
(e) None of these
Ans. (b)
Explanation: \[Circumference\text{ }of\text{ }the\text{ }circular\text{ }park\text{ }=\text{ }352\text{ }m\]
\[\Rightarrow \,\,2\pi r\,=\,\,352\,m\,\,\,\Rightarrow \,\,\,r\,\,=\,\,\left( \frac{352}{2\pi } \right)\,m\,\,=\,\,\left( \frac{352\times 7}{2\times 22} \right)m\,\,=\,\,56\,m.\]
\[\therefore \] \[Inner\text{ }radius\text{ }\left( park\text{ }alone \right)\text{ }r\text{ }=\text{ }56\text{ }m.\]
\[Outer\text{ }radius\text{ }\left( including\text{ }road \right)\text{ }R\text{ }=\text{ }\left( r+7 \right)\text{ }m\text{ }=\text{ }63\text{ }m.\]
So, area of the road = \[\pi {{R}^{2}}\,\,-\,\,\pi {{r}^{2}}\]
\[=\,\,\pi ({{R}^{2}}\,\,-\,\,{{r}^{2}})\,\,=\,\,\left[ \frac{22}{7}\times ({{63}^{2}}-{{56}^{2}}) \right]\,{{m}^{2}}\]
\[=\,\,\,\,\left[ \frac{22}{7}\times (63+56)\,\,\times \,\,(63-56) \right]\,{{m}^{2}}\]
\[=\,\,\,\left( \frac{22}{7}\times 119\times 7 \right)\,{{m}^{2}}\,\,=\,\,2618\,{{m}^{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec