Question

a chord 10cm long is drawn in a circle whose radius is √50cm find the area of the segments

Answer 1

Consider the centre of circle to be O, ends of chord as A, B and their mid-point as M.
We have, radius, r = 5√2 cm and chord length, l = 10 cm.
Now you can observe, ∆OAM & ∆OBM are isosceles & right angled at M.
Here, angle AOM = angle BOM = 45°, then area of ∆ AOB = 25 cm²
and the area of sector AOB = ¼(circle area) = (12.5)π cm², since angle AOB = 90°.
Then area of segments are 50(π-2)/4 and 50(3π+2)/4.

Answer 2

Answer:

The area of minor and major segments are 14.25sq.cm and 142.75sq.cm respectively.

Step-by-step explanation:

Given the length of the chord in a circle, $l = 10 cm$

The radius of the circle, $r = 5\sqrt{2} cm$

The ends of chord are joined at the through the radius at the centre of the circle.

Let AB be the chord, then OA and OB are the radius of the circle.

Thus, OAB represents an isosceles triangle.

Applying the Pythagoras theorem,

$OA^2+OB^2=AB^2$

$\implies \big(\sqrt{50} \big)^2+\big(\sqrt{50} \big)^2=\big(10 \big)^2\implies 50+50=100$

Proves that the given triangle is a right angled triangle.

And hence the angle at O is $\theta=90^{o}$.

Area of the sector is given by

$\dfrac{\theta}{360}\times \pi r^{2}$

$=\dfrac{90}{360}\times 3.14 \times \big(\sqrt{50} \big)^{2}$

$=\dfrac{1}{4}\times 3.14 \times 50=39.25cm^{2}$

Area of the right angled triangle AOB is given by

$\dfrac{1}{2}bh=\dfrac{1}{2}\times OA\times OB$

      $=\dfrac{1}{2}\times \sqrt{50} \times \sqrt{50}=\dfrac{1}{2}\times50=25cm^{2}$

A minor segment is the part of the circle between a chord and an arc.

So, area of the segment is measured by subtracting the area of the triangle from the area of the sector.

Thus, area of the minor segment is

$39.25-25=14.25cm^{2}$

Area of the major segment is measured by subtracting the area of the   minor segment from the area of the circle.

Area of the circle is given by

$\pi r^{2}=3.14\times \big(\sqrt{50} \big)^{2} =157cm^{2}$

Thus, the area of the major segment is

$157-14.25=142.75cm^{2}$

Therefore, the area of minor and major segments are 14.25sq.cm and 142.75sq.cm respectively.

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