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Area Of Circle Sector And Segment | Area Of Circle Sector And Segment Exercise 16A

Question 12

A chord 10 cm long is drawn in a circle whose radius is 5√2 cm. Find the areas of both segments. (Take pi = 3.14)

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Radius of circle = 5√2 cm

Chord = 10 cm

To find: Areas of both the segments

Now,

Area of minor segment = (area of sector OACBO) – (area of ∆OAB)

= 39.25 - 25 = 14.25 cm^2

Area of the major segment = (Area of circle) - (Area of minor segment)

\begin{array}{l} =\left(\frac{22}{7} \times 5 \sqrt{2} \times 5 \sqrt{2}-14.25\right) \mathrm{cm}^{2} \\ =\left(\frac{1100}{7}-14.25\right) \mathrm{cm}^{2}=(157-14.25) \mathrm{cm}^{2} \\ =142.75 \mathrm{cm}^{2} \end{array}

Therefore, Area of the major segment is 142.75 cm^2.

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Question 12

A chord 10 cm long is drawn in a circle whose radius is 5√2 cm. Find the areas of both segments. (Take pi = 3.14)

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

Radius of circle = 5√2 cm

Chord = 10 cm

To find: Areas of both the segments

Now,

Area of minor segment = (area of sector OACBO) – (area of ∆OAB)

= 39.25 - 25 = 14.25 cm^2

Area of the major segment = (Area of circle) - (Area of minor segment)

\begin{array}{l} =\left(\frac{22}{7} \times 5 \sqrt{2} \times 5 \sqrt{2}-14.25\right) \mathrm{cm}^{2} \\ =\left(\frac{1100}{7}-14.25\right) \mathrm{cm}^{2}=(157-14.25) \mathrm{cm}^{2} \\ =142.75 \mathrm{cm}^{2} \end{array}

Therefore, Area of the major segment is 142.75 cm^2.

Our top 5% students will be awarded a special scholarship to Lido.

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