Formulae
Category : 9th Class
Circumference and Area of a Circle
For a circle of radius r, we have
(i) Circumference of the circle \[=2\pi r\]
(ii) Area of the circle \[=\pi {{r}^{2}}\]
(iii) Area of the semicircle \[=\frac{1}{2}\pi {{r}^{2}}\]
(iv) Perimetre of the semicircle \[=\pi r+2r\]
Area of Ring
Let R & r be the outer and inner radii of ring. Here area of the ring \[=\pi ({{R}^{2}}-{{r}^{2}})\]
Length of Arc, Area of Sector and Segment
Let an arc AB makes an angle\[\theta <180{}^\circ \]at the center of a circle of radius r then we have:
(i) Length of the arc AB\[=\frac{2\pi r\theta }{360}\]
(ii) Area of the sector OACB \[=\frac{\pi {{r}^{2}}\theta }{360}\]
(iii) Perimetre of the sector\[\text{OACB}=\text{OA}+\text{OB}+\]length of\[\overset\frown{AB}+2r+\frac{2\pi \theta }{360}\]
(iv) Area of the minor segment ACBA = (Area of the sector OACB) - (area of\[\Delta OAB\])\[=\left( \frac{\pi {{r}^{2}}\theta }{360}-\frac{1}{2}{{r}^{2}}\sin \theta \right)\]
(v) Area of the major segment BDAB = (area of the circle) - (area of the minor segment ACBA)
A chord of a circle of radius 14 cm makes a right angle of the centre. Find the area of the major segments of the circle.
(a) \[\text{59}0\,\text{c}{{\text{m}}^{\text{2}}}\]
(b) \[\text{56}0\,\text{c}{{\text{m}}^{\text{2}}}\]
(c) \[\text{595}\,\text{c}{{\text{m}}^{\text{3}}}\]
(d) \[\text{995}\,\text{c}{{\text{m}}^{\text{2}}}\]
(e) None of these
Answer: (b)
Explanation
Let AB be the chord of a circle of centre 0 & radius = 14 cm
so that\[\angle \text{AOB}=\text{9}0{}^\circ \]
Area of the sector OACB
\[=\frac{\pi {{r}^{2}}\theta }{360}c{{m}^{2}}=\left( \frac{22}{7}\times 14\times 14\times \frac{90}{360} \right)c{{m}^{2}}=154c{{m}^{2}}\]
Area of \[\Delta \text{OAB}=\frac{1}{2}{{r}^{2}}\sin \theta \]
\[=\left( \frac{1}{2}\times 14\times 14\times \sin 90{}^\circ \right)=\text{98c}{{\text{m}}^{\text{2}}}=\] Area of the minor segment ACBA
= (area of the sector OACB) - (area of the \[\Delta \text{OAB}\])\[=(\text{154}-\text{98})\text{c}{{\text{m}}^{\text{2}}}=\text{56}\,\text{c}{{\text{m}}^{2}}\] Area of the major segment BDAB
= (area of the circle) - (area of the minor segment)
\[\text{=}\left\{ \left( \frac{22}{2}\times 14\times 14 \right)-56 \right\}\text{c}{{\text{m}}^{\text{2}}}=(\text{616}-\text{56})\text{c}{{\text{m}}^{\text{2}}}=\text{56}0\text{ c}{{\text{m}}^{\text{2}}}\]
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