Question

in the given figure o is the center of circle AOB is the diameter ac=12cm and ab=16cm find the area of shaded portion​

Answer 1

Answer:

Area of semicircle-area of triangle

Step-by-step explanation:

πr^2/2-1/2×b×h

Answer 2

Area of shaded region is 61.14 cm²

Step-by-step explanation:

Given: AC= 12 cm , AB= 16 cm  AB is diameter

Therefore  

∠ACB = 90° (angles in a semicircle are right angle)

∴ ΔACB  is a right angle triangle right angled at C

Using Pythagoras theorem:  The square of the hypotenuse of a right-angled triangle is equal to the sum of the squares on the other two sides.

$H^2= P^2+B^2$ where H is hypotenuse , P is perpendicular and B is base

$\Rightarrow AB^2=12^2+16^2\\\Rightarrow AB^2= 400\\\Rightarrow AB=\sqrt{400} =20$

Now as we know area of semi-circle is given by $area=\dfrac{\pi r^2}{2}$  where r is radius of circle and area of triangle is given by $area= \dfrac{1}{2} \times base\times height$

Now r= 10 cm ,base=12cm and height = 16cm

Area of shaded region = area of semicircle- area of triangle

$= \dfrac{\pi \times 10^2}{2} -\dfrac{1}{2} \times 12\times 16\\\\=\dfrac{22 \times 100}{7\times 2} - 6 \times 16\\\\=157.14-96=61.14cm^2$

Hence area of shaded region is 61.14 cm²

#Learn more

Find the area of the shaded region

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