2 Marks Questions

Question 12 Marks

The length of the minute hand of a clock is $14 \ cm$. Find the area swept by the minute hand in $5$ minutes.

Answer

Here, $\theta$ $= 30^\circ$ and $r = 4 cm$
Area of sector OAPB =$\frac { \theta } { 360 } \times \pi r ^ { 2 }$
Let 'A' be the area of corresponding major sector.
Then, A = Area of sector $OAQB$
$\Rightarrow$A = Area of the circle - Area of the corresponding minor sector
$\Rightarrow A = \pi r ^ { 2 } - \frac { \theta } { 360 } \times \pi r ^ { 2 }$
$\Rightarrow A = \pi r ^ { 2 } \left( 1 - \frac { \theta } { 360 } \right)$
$\Rightarrow A = 3.14 \times 4 \times 4 \left( 1 - \frac { 30 } { 360 } \right) \mathrm { cm } ^ { 2 }$
$\Rightarrow A = 3.14 \times 4 \times 4 \times \frac { 11 } { 12 } \mathrm { cm } ^ { 2 }$$= \frac { 3.14 \times 44 } { 3 } \mathrm { cm } ^ { 2 }$$= 46.05 cm^2$

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Question 22 Marks

Area of a sector of angle p (in degrees) of a circle with radius R is

Answer

r = 12 cm, $\theta$ $= 120^o$
$\therefore$ Area of the corresponding sector of the circle = $\frac\theta{360^\circ}\mathrm{πr}^2\;=\;\frac{\displaystyle120}{\displaystyle360^\circ}\times3.14\;\times12\times12\;$= 150.72 $cm^2$
Area of $\triangle$AOB
Draw OM $\perp$ AB
In right triangle OMA and OMB,
OA = OB ....... Radii of the same circle
OM = OM ........ Common side
$\therefore$ $\triangle$OMA $\cong$ $\triangle$OMB ........RHS congruence criterion
$\therefore$ AM = BM ....... CPCT
$\Rightarrow$ AM = BM = $\frac 12$AB
and $\angle$AOM = $\angle$BOM [CPCT]
$\Rightarrow$ $\angle$AOM = $\angle$BOM = $\frac 12$$\angle$AOB
= $\frac 12$ $\times$ $120^o = 60^o$
$\therefore$ In right triangle OMA,
$cos60^o$ = $\frac {OM}{OA}$
$\Rightarrow$ $\frac 12$= $\frac {OM}{12}$
$\Rightarrow$ OM = 6 cm
$sin60^o$ = $\frac {AM}{OA}$
$\Rightarrow$ $\frac{\sqrt3}2$= $\frac {AM}{12}$
$\Rightarrow$ AM = 6$\sqrt3$ cm
$\Rightarrow$ 2AM = 12$\sqrt3$ cm
$\Rightarrow$ AB =12$\sqrt3$ cm
$\therefore$ Area of $\triangle$AOB =$\frac 12$ $\times$ AB $\times$ OM
= $\frac 12$ $\times$ 12$\sqrt3$ $\times$ 6 = 36$\sqrt3$ cm$^2$
= 36 $\times$ $1.73 cm^2 = 62.28 cm^2$
So, Area of the corresponding segment of the circle = Area of the correspoding sector of circle - Area of $\triangle$AOB
$= 150.72 - 62.28 = 88.44 cm^2$

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Question 32 Marks

To warm ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle $80^\circ$ to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (use $ \pi = 3.14 )$

Answer

$r = 15 cm, \theta = 60^o$
Area of the minor sector = $\frac\theta{360^\circ}\mathrm{πr}^2\;=\;\frac{\displaystyle60^\circ}{\displaystyle360^\circ}\times3.14\;\times15\times15$ = 117.75 cm$^2$
In $\triangle$AOB, draw OM $\perp$ AB
In right triangle OMA and OMB,
OA = OB .........Radii of the same circle
OM = OM .........Common side
$\therefore$ $\triangle$OMA $\cong$ $\triangle$OMB .........RHS congruence criterion
$\therefore$ AM = BM .......CPCT
$\Rightarrow$ AM = BM = $\frac 12$AB
$\angle$AOM = $\angle$BOM .......CPCT
$\Rightarrow$ $\angle$AOM = $\angle$BOM = $\frac 12$$\angle$AOB = $\frac 12$ $\times$ $60^o = 30^o$
$\therefore$ In right triangle OMA, $cos30^o$ = $\frac {OM}{OA}$
$\Rightarrow$ $\frac{\sqrt3}2$= $\frac {OM}{15}$
$\Rightarrow$ OM = $\frac{15\sqrt3}2$cm
sin$30^o$ = $\frac {AM}{OA}$
$\Rightarrow$ $\frac 12$= $\frac {AM}{15}$
$\Rightarrow$ AM = $\frac{15}2$cm
$\Rightarrow$ AB = 15 cm
$\therefore \text { Area of the corresponding minor segment of the circle }=\text { Area of minor sector }- \text { Area of } \triangle A O B$
$=117.75-97.3125=20.4375 cm^2$
and, area of the corresponding major segment of the circle $=\pi r ^2$ - area of the corresponding minor segment of the circle
$=3.14 \times 15 \times 15-20.4375$
$=706.5-20.4375=686.0625 cm^2$

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Question 42 Marks

A car has two wipers which do not overlap. Each wiper has a blade of length 25cm sweeping through an angle of $115^\circ$. Find the total area cleaned at each sweep of the blades.

Answer

Given, $r = 21 cm$ and $\theta = 60^o$
Length of arc = $=\;\frac\theta{360}2\mathrm{πr}\;=\frac{60}{360}\times2\times\frac{22}7\times21\;=\;22\;\mathrm{cm}$
Area of the sector = $=\;\frac\theta{360}\mathrm{πr}^2\;=\frac{60}{360}\times\frac{22}7\times21\times21\;=\;231\;\mathrm{cm}^2$
Area of segment formed by corresponding chord = Area of the sector - Area of $\triangle$OAB
= $\frac\theta{360}\mathrm{πr}^2$ - Area of $\triangle$OAB
⇒ Area of segment = 231 - Area of $\triangle$OAB ....... (1)
In right angled triangle OMA and OMB,
OM = OB [Radii of the same circle]
OM = OM [Common]
$\therefore$ $\triangle$OMA $\cong$ $\triangle$OMB [ RHS congruency]
$\therefore$ AM = BM [By CPCT]
$\therefore$ M is the mid-point of AB and $\angle$AOM = $\angle$BOM
$\Rightarrow$ $\angle$AOM = $\angle$BOM = $\frac{1}2$$\angle$AOB =$\frac{1}2$ × $60^o = 30^o$
Therefore, in right angled triangle OMA,
$cos30^o $= $\frac{OM}{OA}$
$\Rightarrow$ $\frac{\sqrt3}2$= $\frac{OM}{21}$
$\Rightarrow$ OM = $\frac{21\sqrt3}2$ cm
Also, $sin30^o$ = $\frac{AM}{OA}$
$\Rightarrow$ $\frac{1}2$= $\frac{AM}{21}$
$\Rightarrow$ AM = $\frac{21}2$ cm
$\therefore$ AB = 2 AM = 2 $\times$ $\frac{21}2$ = 21 cm
$\therefore$ Area of $\triangle$OAB = $\frac{1}2$ $\times$ AB $\times$ OM = $\frac{1}2$ $\times$ 21 $\times$ $\frac{21\sqrt3}2$ =$\frac{441\sqrt3}4$ cm$^2$
Using eq. (1),
Area of segment formed by corresponding chord = [231 - $\frac{441\sqrt3}4$] cm$^2$
= 40.05 cm$^2$

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Question 52 Marks

An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, Find the area between the two consecutive ribs of the umbrella.

Answer

Area of the segment AYB = Area of sector OAYB - Area of ∆ OAB

Now, area of the sector OAYB = $\frac{120}{360} \times \frac{22}{7} \times 21 \times 21 \mathrm{cm}^{2}=462 \mathrm{cm}^{2}$

For finding the area of ∆ OAB, draw OM ⊥ AB as shown in Fig.

Note that OA = OB. Therefore, by RHS congruence, ∆ AMO ≅ ∆ BMO.

So, M is the mid-point of AB and $\angle \mathrm{AOM}=\angle \mathrm{BOM}=\frac{1}{2} \times 120^{\circ}=60^{\circ}$

Let OM = x cm

So, from ∆ OMA, $\frac{\mathrm{OM}}{\mathrm{OA}}=\cos 60^{\circ}$

or, $\frac{x}{21}=\frac{1}{2} \quad\left(\cos 60^{\circ}=\frac{1}{2}\right)$

or, $x=\frac{21}{2}$

So, $\mathrm{OM}=\frac{21}{2} \mathrm{cm}$

Also, $\frac{\mathrm{AM}}{\mathrm{OA}}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}$

So, $\mathrm{AM}=\frac{21 \sqrt{3}}{2} \mathrm{cm}$

Therefore, $A B=2 A M=\frac{2 \times 21 \sqrt{3}}{2} \mathrm{cm}=21 \sqrt{3} \mathrm{cm}$

So, area of ∆ OAB $=\frac{1}{2} \mathrm{AB} \times \mathrm{OM}=\frac{1}{2} \times 21 \sqrt{3} \times \frac{21}{2} \mathrm{cm}^{2}$

$=\frac{441}{4} \sqrt{3} \mathrm{cm}^{2}$

Therefore, area of the segment AYB = $\left(462-\frac{441}{4} \sqrt{3}\right) \mathrm{cm}^{2}$ [From (1), (2) and (3)]

$=\frac{21}{4}(88-21 \sqrt{3}) \mathrm{cm}^{2}$

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Question 62 Marks

A chord of a circle of radius 12 cm subtends an angle of $120^{\circ}$ at the centre. find the area of the corresponding segment of the circle. [Use $\pi=3.14$ and $\sqrt{3}=1.73$ ]

Answer

Let us draw a perpendicular OV on chord ST . It will bisect the chord ST.
$SV=VT$
In $\triangle OVS$,
$\frac{O V}{O S}=\cos 60^{\circ} $
$\frac{O V}{12}=\frac{1}{2} $
$O V=6 cm $
$\frac{S V}{S O}=\sin 60^{\circ}=\frac{\sqrt{3}}{2} $
$\frac{S V}{12}=\frac{\sqrt{3}}{2} $
$S V=6 \sqrt{3} cm $
$ST=2 SV $
$=2 \times 6 \sqrt{3} $
$=12 \sqrt{3} cm$
Area of $\Delta OST =\frac{1}{2} \times S T \times O V$
$\frac{1}{2} \times 12 \sqrt{3} \times 6 $
$=36 \sqrt{3} $
$=36 \times 1.73 $
$=62.28 cm^2$
Area of sector OSUT $=\frac{120^{\circ}}{360^{\circ}} \times \pi(12)^2$
$=\frac{1}{3} \times 3.14 \times 144=150.72 cm^2$
Area of segment SUT $=$ Area of sector $\triangle$ SUT - Area of $\triangle OST$
$=150.72-62.28 $
$=88.44 cm^2$

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