# 9th Class Mathematics Mensuration Formulae

Formulae

Category : 9th Class

### Formulae 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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