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Examples 11 to 15 of Mensuration

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(11) Find the ratio of  the areas of  the incircle and circumcircle  of  a square.

Soln. Let the side of  the square be 2x.

Then, its diagonal  =  2\sqrt2 x  ⇒   R  = \sqrt2  x

&    Radius  of  incircle   =   x    =  r

∴  \frac{Area\ of\ \ incircle}{Area\ of\ \ circumcircle} =  \frac{\pi r^2}{\pi R^2}

=    \frac{x^2}{\left({\sqrt2}^x\right)^2}   =  \frac{1}{2}

(12) In a circle of  radius 56 cm, an arc subtends an angle of  45o at the centre. Find the length of  the arc and the area of  the sector.

Soln. Length of  the arc  =  \frac{\pi\ r\ \theta}{180}  =  \frac{22\times56\times45}{7\times180}  =  44 cm.

Area of  the sector  =  \frac{\pi\ r^2\ \theta}{360}  =  \frac{22\times56\times56\times45}{7\times360} =  1232 cm2.

(13) Find the area of  the sector of  a circle whose radius is 21 cm and  angle of  sector is  60o.

Soln. Radius of  the circle ( r )  =  21 cm.  Central angle  (\theta )  =  600.

Area of  sector  = \frac{\pi r^2\theta}{360^o}

=  \frac{22}{7} \times 21 \times 21 \times \frac{60}{360}

=  231 cm2.

(14) Find the area of  shaded region.

Soln. Area of  equilateral triangle

=  \frac{\sqrt3}{4}x2

=  \frac{\sqrt3}{4} \times 10 \times 10

=  25\sqrt3

Area of  circle =  r2

=  \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}

=  \frac{77}{2}cm2.

∴  Area of  shaded region

=  \left (25\sqrt3 - \frac{77}{2} \right )  cm2.

(15) The diameter of  the wheel of  a bus is 140 cm. How many revolutions per minute must the wheel make in order to keep a speed of  132 km per hour ?

Soln. Here, Diameter  =  140 cm    \large \Rightarrow  Radius  =  70 cm

Distance covered in 1 revolution

=   2\pi r

=  \frac{2\times22\times70}{7}

=   440 cm

In 1 hr., distance covered  =  132  km.

min.,                                 distance covered is

In  60                                        13200000 cm

1   ”            ”              ”            ( ? )

\frac{1\times13200000}{60}   =  220000

i.e.  in 1 min., distance covered  =  220000 cm

\therefore Number of  revolutions =  \frac{Distance\ covered\ in\ 1\ min.}{Distance\ covered\ in\ 1\ revolution}

=  \frac{220000}{440}

=   500

 

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