Question

Find the area of the shaded region where ABC is a quadrant of radius 5cm and a semicircle is drawn with BC as diameter.

Answer 1

GIVEN:

• Radius (AB) = (AC) = 5 cm
• BC is diameter

TO FIND:

• What is the area of shaded region ?

SOLUTION:

Applying Pythagoras theorem in the right triangle ABC,we obtain

$\rm{\dashrightarrow (BC)^2 = (AB)^2 +(AC)^2 }$

$\rm{\dashrightarrow BC^2 = 5^2 + 5^2 }$

$\rm{\dashrightarrow BC^2 = 25 + 25}$

$\rm{\dashrightarrow BC^2 = 50 }$

$\rm{\dashrightarrow BC = \sqrt{50}}$

$\bf{\dashrightarrow BC = 5\sqrt{2} \: cm }$

Let A be the area of shaded region. Then,

$\rm{\dashrightarrow A = Area \: BXCYB }$

$\rm{\dashrightarrow A = Area \: BCYB - Area \: BCXC }$

$\rm{\dashrightarrow A = Area \: BCYB - ( Area \: BACXB - Area \: of \: \triangle BAC)}$

 ➠ ❮ A = (Area of semicircle with diameter BC) – [ Area of quadrant of a circle with AB as radius – Area of ⊿ ABC ❯ 

• Area of semicircle = 1/2 πr²
• Area of quadrant = 1/4 πr²
• Area of triangle = 1/2 base $\times$ height

$\rm{\dashrightarrow A = \Bigg[ \dfrac{1}{2} \Bigg\{ \dfrac{22}{7}\times (5 \sqrt{2})^2 \Bigg\} - \Bigg\{\dfrac{1}{4} \times \dfrac{22}{7} \times 5^2 - \dfrac{1}{2} \times 5 \times 5 \Bigg\} \Bigg]}$

$\rm{\dashrightarrow A = \Bigg\{ \dfrac{1}{2} \times \dfrac{22}{7} \times 25 \times 2 - \dfrac{1}{4} \times \dfrac{22}{7} \times 25 + \dfrac{1}{2} \times 25 \Bigg\}}$

$\rm{\dashrightarrow A = \dfrac{550}{7} - \dfrac{550}{28} + \dfrac{25}{2} }$

$\rm{\dashrightarrow A = 78.57-19.64 + 12.5 }$

$\bf{\dashrightarrow 71.43 \: cm^2 }$

❝ Hence, the area of shaded region is 71.43 cm² ❞

______________________

Answer 2

area of shaded region = 71.43cm^2