Question

See the attached figure!

In the given figure, O is the centre of the circle with:

$\tt{\leadsto AC = 24\:cm}$

$\tt{\leadsto AB = 7\:cm}$

$\tt{\leadsto \angle{BOD} = 90 \degree}$

Find the area of the shaded region.

Please don't post irrelevent answers. If you do so, I will report!

Please explain this problem clearly. Thank you!

​

Answer 1

Given :

Area of the shaded region = area of the circle - area of the triangle ABC - area of the quadrant COD

To find :

Here, in triangle ABC ∠CAB=90°

(angle in a semicircle)

Hence triangle ABC is right-angled at A

Let's Start :

Then applying Pythagoras theorem,

BC² = AC² + AB²

BC² = 24² + 7²

BC² = 625

BC = 25 cm

$\small\fbox\pink{ \sf \: Therefore diameter of circle = 25cm}$

$\sf \: i.e., Radius = \frac{25}{2} = 12.5 cm$

$\sf \: Then \: area \: of \: circle = πr² = 490.625 cm²$

$\small\fbox\red{ \sf Area of triangle ABC = 1/2 × Base × Height}$

$\sf= 84 cm²$

$\sf \: Area \: of \: quadrant = \frac{1}{4} × πr²$

Therefore area of shaded region,

= 122.65625 cm²

= 490.625 − 84 − 122.65625

= 283.96875 cm²

Answer 2

$\Large{\bf{\green{\mathfrak{\dag{\underline{\underline{Given:-}}}}}}}$

• AC = 24 cm, AB = 7 cm, $\angle{BOD}$ = 90°

$\Large{\bf{\orange{\mathfrak{\dag{\underline{\underline{To \: Find:-}}}}}}}$

• Area of Shaded region

$\Large{\bf{\red{\mathfrak{\dag{\underline{\underline{Solution:-}}}}}}}$

• Area of shaded region = (Area of semi circle - Area of ∆ABC) + Area of sector BOD

First,

• Area of ∆ABC =

We know that,

• Angle subtended by an semi circle is always 90°

Therefore,

• $\angle A$ = 90°

Now,

• ∆ABC is a right angled triangle.

• $\angle A$ = 90°

• AC = 24 cm

• AB = 7 cm

We know that,

$\boxed{\pink{\sf Area \: of \: a \: triangle \: = \: \dfrac{1}{2} \: * \: h \: * \: b}}$

Substituting the values,

• $\sf Area \: of \: a \: triangle \: = \: \dfrac{1}{2} \: * \: AB \: * \: AC$

• $\sf Area \: of \: a \: triangle \: = \: \dfrac{1}{2} \: * \: 7 \: * \: 24$

• $\sf Area \: of \: a \: triangle \: = \: \dfrac{1}{\cancel{2}} \: * \: 7 \: * \: \cancel{24}$

• $\sf Area \: of \: a \: triangle \: = \: 1 \: * \: 7 \: * \: 12$

• Area of ∆ABC = 84 cm².

Now,

• Since, ∆ABC is a right angled triangle.

• AB² + AC² = BC² (Phythagoras theorem)

• 7² + 24² = BC²

• 49 + 576 = BC²

• 625 = BC²

• BC = √625

• BC = 25 cm

Now,

• BC is diameter of the given circle.

We know that,

$\boxed{\pink{\sf Radius \: = \: \dfrac{Diameter}{2}}}$

Substituting the values,

• $\sf Radius \: = \: \dfrac{25}{2}$

• $\sf Radius \: = \: \dfrac{\cancel{25}}{\cancel{2}}$

• Radius = 12.5 cm

Now,

• Area of semi circle COBA =

We know that,

$\boxed{\pink{\sf Area \: of \: semi \: circle \: = \: \dfrac{1}{2} \: * \: \pi r^2}}$

Substituting the values,

• $\sf Area \: of \: semi \: circle \: = \: \dfrac{1}{2} \: * \: \dfrac{22}{7} \: * \: 12.5^2$

• $\sf Area \: of \: semi \: circle \: = \: \dfrac{1}{2} \: * \: \dfrac{22}{7} \: * \: 156.25$

• $\sf Area \: of \: semi \: circle \: = \: 0.5 \: * \: 3.14 \: * \: 156.25$

• Area of semi circle COBA = 245.3125 cm².

Now,

• Area of sector DOB =

We know that,

$\boxed{\pink{\sf Area \: of \: sector \: = \: \dfrac{\theta}{360^{\circ}} \: * \: \pi r^2}}$

Here,

• $\angle{DOB}$ = 90°

• Radius = 12.5 cm

Substituting the values,

• $\sf Area \: of \: sector \: = \: \dfrac{90^{\circ}}{360^{\circ}} \: * \: \dfrac{22}{7} \: * \: 12.5^2$

• $\sf Area \: of \: sector \: = \: \dfrac{\cancel{90^{\circ}}}{\cancel{360^{\circ}}} \: * \: \dfrac{22}{7} \: * \: 156.25$

• $\sf Area \: of \: sector \: = \: \dfrac{1}{4} \: * \: 3.14 \: * \: 156.25$

• $\sf Area \: of \: sector \: = \: 0.25 \: * \: 3.14 \: * \: 156.25$

• Area of sector DOB = 122.65625 cm².

Now,

• Area of Shaded region =

(Area of semi circle - Area of ∆ABC) + Area of sector BOD

Substituting the values,

• (245.3125 - 84) + 122.65625

• 161.3125 + 122.65625

• 283.96875

• $\sim$ 283.97 cm²

Therefore,

• Area of Shaded region = 283.97 cm².