9th Math Exercise 6.2 Federal Board Notes
9th Math Exercise 6.2 Federal Board Notes.
Q1 :- Find the length of arc and area of sector for given radius and central angle
(i)
$$r\;=\;5cm\;,\;\theta\;=\;\frac{\mathrm\pi}3radians\\l\;=\;?\;,\;A\;=\;?$$
$$l=r\;\theta\\=(5)\;(\frac{\mathrm\pi}3)\\=\;5.24cm$$
$$A=(\frac12)\;r^2\theta\\=\frac12{(5)}^2(\frac{\mathrm\pi}3)$$
$$=\frac{78.54}6\\=13.09\;cm^2$$
(ii):-
l=? A=?
r= 12cm
$$\theta=120^\circ\;=\;=\;120(\frac{\mathrm\pi}{180})rad$$
$$l=r\;\theta\\=(12)\;(\frac{2\mathrm\pi}3)=8\mathrm\pi\\=\;25.13cm$$
$$A=(\frac12)\;r^2\theta\\=\frac12{(12)}^2(\frac{2\mathrm\pi}3)=48\mathrm\pi\\=150.796\;cm^2$$
(iii)
r= 6 dm
l=?
A=?
$$\theta\;=\;60^\circ45’30”\\\theta\;=\;(\;60\;+\frac{45}{60}+\frac{30}{3600})^\circ\\(60\;+0.75+0.0083)$$
$$\theta\;=\;60.7583^\circ\\\theta\;=\;\frac{60.7583\mathrm\pi}{180}rad\\=\;60.7583(0.0175)rad$$
$$\theta\;=\;1.0633\;rad$$
$$l=r\;\theta\\=6(\;1.0633\;rad\;)\\=6.379\;dm$$
$$A=(\frac12)\;r^2\theta\\=\frac12{(6)}^2(\frac{60.7583\mathrm\pi}{180})\\=19.078\;dm^2$$
Q2 :-A central angle is a circle of radius 4 cm is 75 degree. Find the length of the intercepted arc and area bounded by the sector.
Radius = r=4cm,
Arc length = l ?
Area of sector=A=?
$$\theta\;=\;75^\circ$$
$$=\frac{75\mathrm\pi}{180}rad\\\frac{5\mathrm\pi}{12}\mathrm{rad}$$
$$l=r\;\theta\\l=4(\frac{5\mathrm\pi}{12})\\=\frac{5(3.1416)}3$$
$$=\frac{15.708}3cm\\=5.236\;cm\\=5.24\;cm$$
$$A=\frac12\;r^2\;\theta\\A=\frac12\;{(4)}^2\;(\;\frac{5\mathrm\pi}{12}\;)$$
$$=\frac12\;4\times4\times\;\frac{5\mathrm\pi}{12}\\=\;\frac{10\mathrm\pi}3$$
$$=\;\frac{10(3.1416)}3\\=10.472\;cm^2$$
Q3 :- Find radian measure of central angle of a circular sector of the following data.
(i) :-
l = 8cm , r = 4cm
$$l=r\;\theta\\8=4(\theta)\\\frac84=\theta\\\theta=2\;rad$$
(ii) :-
l = 8.5 m , r = 2.25 m
$$l=r\;\theta\\8.5=2.25(\theta)\\\frac{8.5}{2.25}=\theta\\\theta=3.78\;rad$$
(iii) :-
$$A=\;45\;cm^2,\;l=\;10.70\;cm$$
$$A=\frac12r^2\theta\\45=\frac12\;{(10.70)}^2\theta\\45\times2=114.49\;\theta$$
$$\frac{90}{114.49}=\;\theta\\\theta=\;0.786\;rad$$
(iv) :-
$$A=100\;cm^2\;,\;l=cm$$
$$\theta=?\\A=\frac12r^2\;\theta\;\;\;\;\;\;eq\;(i)$$
$$l=r\;\theta\\\frac l\theta\;=r$$
$$putting\;r\;=\;\frac l\theta in\;eq\;(i)\\A=\frac12.\;{(\frac l\theta)}^2.\theta$$
$$A=\frac12.\;\frac{l^2}{\theta^2}.\theta\\A=\frac12\;\frac{l^2}\theta$$
$$\theta\;=\frac{l^2}{2A}\\\theta\;=\frac{{(10)}^2}{2(100)}\\$$
$$=\frac12\\\theta=0.5\;rad$$
Q4 :- Find the radius of the circle.
(i) :-
$$l=\;4cm\;,\;\theta=\mathrm\pi\;\mathrm{rad}$$
$$l=r\;\theta\\4=r\;(\mathrm\pi)\\\frac4{\mathrm\pi}\;=\mathrm r$$
$$r=\frac4{3.1416}\\r=1.2732\;cm$$
(ii) :-
$$l=\;6\;m\;,\;\theta=15^\circ$$
$$l=\;r\;\theta\\6=r\;(15^\circ)\\6=\;r(\frac{15\mathrm\pi}{180})\\\frac{6\times180}{15\mathrm\pi}=r$$
$$r=\;\frac{6\times12}{\mathrm\pi}\\\frac{72}{3.1416}\\r=22.91\;m$$
(iii) :-
$$A=200\;cm^2\;,\;\theta=\frac{\mathrm\pi}4\;rad$$
$$A=\frac12\;r^2\;\theta\\200\;=\;\frac12r^2(\;\frac{\mathrm\pi}4)\\\frac{200\times2\times4}{\mathrm\pi}=r^2$$
$$r^2=509.29\\\sqrt{r^2\;}\;=\sqrt{509.29}$$
$$r=\;22.567\;cm$$
(iv) :– $$A=\;100\;dm^2,\;l=\;10\;dm$$
$$A=\frac12r^2\theta\\100=\frac12r\;(r\;\theta)\\100\times2=r.l$$
$$200=r\;(10)\;\\\frac{200}{10}=r\\=20\;dm$$
Q5 :- A 30 inch pendulum swings through an angle of 30 degree. Find the length of arc in inches through which the tip of the pendulum swings.
$$l\;=\;?\;,\;r\;=\;30\;inch\\\theta\;=\;30^\circ$$
$$=\frac{30\mathrm\pi}{180}\;rad\\=\frac{\mathrm\pi}6\;rad$$
$$l=\;r\;\theta\\=\;(30)\;(\frac{\mathrm\pi}6)$$
$$=\;5\;(\;3.1416)\\=15.71\;inch$$
Q6 :- A motorcycle is travelling on a curve along a highway. The curve is an arc of a circle with radius of 10 km. If the motorcycle speed is 42 km/h, what is the angle in degrees through which the motorcycle will true in 21 minutes.
r= 10 km
speed = 42 km/h
$$=\;\frac{42}{60\;}\;km/min$$
Distance after 21 minutes
= Arc length
$$=\;\frac{42}{60\;}\;\times\;21\\\theta\;=\;?$$
$$l=r\;\theta\;\\\frac{42}{60}\;\times\;21=10(\theta)$$
$$\\\frac{882}{60\times10}\;=\theta$$
$$\theta\;=\;1.47\;rad\\\theta\;=\;1.47\;\times\;\frac{180^\circ}{\mathrm\pi}\\\theta=\;\frac{264.6}{3.1416}$$
$$\theta\;=\;84.22^\circ$$
Q7 :- Find the perimeter and area of the half circle in the following figure by using the formulae.
$$l=r\;\theta\;,\;A=\;\frac12\;r^2\;\theta$$
$$d=12\;cm\\r=\;\frac d2\;=\;\frac{12}2\;\\r=\;6\;cm\\\theta=\mathrm\pi\;\;\mathrm{rad}$$
$$l\;=\;r\;\theta\\=\;6\;(\mathrm\pi)\\=\;18.85\;\mathrm{cm}$$
$$perimeter\;=\;r+d\;\\=\;18.85\;+\;2\\=\;30.85\;cm$$
$$A=\frac12\;r^2\;\theta\\=\;\frac12\;{(6)}^2\;(\mathrm\pi)\\$$
$$=\frac12\;\times36\;(\;3.1416\;)\\=\;18\;(\;3.1416\;)\\=\;56.55\;\mathrm{cm}^2\\$$
Q8 :- Find the circle measure of the angle between the hour and minute land of a circle.
1 :- 90′ clock
$$1\;\mathrm{Rev}\;=\;360^\circ\\\mathrm{Angle}\;\mathrm{between}\;\mathrm{two}\;\mathrm{consective}\;\mathrm{points}\;\\=\;\frac{360^\circ}{12}\\$$
$$=\;30^\circ\\$$
$$\mathrm{Angle}\;\mathrm{between}\;9\;\mathrm{and}\;12\\=\;30\;\times\;30^\circ\\=\;90^\circ\\$$
2 :- 2:30
$$\mathrm{Formula}\;:-\;\\\vert\;30\mathrm H\;-\;\frac{11}2\;\mathrm M\vert\;\\\vert\;30\mathrm H\;-\;\frac{11}2\;\mathrm M\;\vert\\$$
$$=\;\vert\;30\times2-\frac{11}2\;\times\;30\;\vert\\=\;\vert\;60\;-\;165\;\vert\\$$
$$=\;\vert-\;105\;\vert\\=\;105^\circ\\$$
$$105\times\frac{\mathrm\pi}{180\;}\;\mathrm{rad}\\=\;\frac{7\mathrm\pi}{\;12}\;\mathrm{rad}\\$$
$$\mathrm{Hence}\;\mathrm{angle}\;\mathrm{at}\;2:30\;\mathrm{will}\;\mathrm{be}\;\frac{7\mathrm\pi}{12}\;\mathrm{rad}\\$$
3 :- 6:45
$$\;\vert\;30\mathrm H\;-\;\frac{11}2\;\mathrm M\vert\;\\\vert\;30\times6\;-\;\frac{11}2\;\times45\;\vert\\$$
$$=\;\vert\;180\;-\;247.5\;\vert\;\\=\;\vert\;-\;67.5\;\vert\\=\;67.5^\circ\\$$
$$67.5\times\frac{\mathrm\pi}{180\;}\;rad\\=\;\frac{675\mathrm\pi}{1800}\;rad\\$$
$$=\;\frac{3\mathrm\pi}8\;rad\\$$
$$Angle\;at\;6:45\;is\;\\67.5^\circ\;or\;\frac{3\mathrm\pi}8\;rad\\$$
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