Find the area of the region ABCDEFA shawn in the figure, given that ABDE is a square of side 10 cm. BCD is a semicircle with BD as diameter, EF = 8 cm, AF = 6 cm and ∠AFE = 90° (π = 3.14)


115.25 cm2
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A bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter of the wheel.

Let the radius of the wheel be r cm. Distance covered by the wheel in one revolution


equals space fraction numerator Distance space moved over denominator Number space of space revolutions end fraction
equals space 11 over 5000 space km
equals space open parentheses 11 over 5000 straight x 1000 straight x 100 close parentheses space cm
equals space 220 space cm

Now, the circumference of the wheel
= 220 cm

rightwards double arrow space space space space space 2 πr space space equals space 220 space space cm
rightwards double arrow space space space space space space space πd space equals space 220 space cm
rightwards double arrow space space space space space space space space straight d space equals space fraction numerator 220 space straight x space 7 over denominator 22 end fraction equals space 70 space cm
Hence, the diameter of the wheel is 70 cm.
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Three horses are tethered with 7 cm long ropes at the three comers of a triangular field having sides 20 m, 34 m and 42 m. Find the area of the plot which can be grazed by the horses. Also, find the area of the point which remains ungrazed.


Clearly, let
∠A = ө1, ∠B = ө2 ∠C = ө3


Clearly, let∠A = ө1, ∠B = ө2 ∠C = ө3
We have,r = 7 m,a = 34

We have,
r = 7 m,
a = 34 m, b = 42 m, c= 20 m
Now, area which can be grazed by the horses = Sum of the areas of three sectors with central angles ө1, ө2 and ө3 each with radius (r) = 7 m


Clearly, let∠A = ө1, ∠B = ө2 ∠C = ө3
We have,r = 7 m,a = 34

Now, we can find out area of the triangle by using Heron's formula.
Here, we have


Clearly, let∠A = ө1, ∠B = ө2 ∠C = ө3
We have,r = 7 m,a = 34

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OAPB is a sector of a circle of radius 3.5 cm with the centre at O and ∠AOB = 120°. Find the length of OAPBO.

It is given that,


It is given that,= (3.5 + 3.5 + 14.7) cm = 21.7 cm

It is given that,= (3.5 + 3.5 + 14.7) cm = 21.7 cm
= (3.5 + 3.5 + 14.7) cm = 21.7 cm
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The minute hand of a clock is 12 cm long. Find the area of the face of the clock described by minute hand in 35 minutes.

Fig. 12.58

Angle described by the minute hand in 60 minutes = 360°.
Angle described by the minute hand in 35 minutes

         equals 360 over 60 space straight x space 35 space equals space 210 degree
Thus, we have,

             straight theta = 210degree, and r = 12 cm.
Therefore, required area

              equals space πr squared space straight x space straight theta over 360
equals space open square brackets 22 over 7 straight x space 12 space straight x space 12 space straight x 210 over 360 close square brackets space cm squared
equals space 264 space cm squared

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