Question

Find the area of shaded region​

Answer 1

Question

In the given figure, △ABC is right-angled at A. Semi circles are drawn on AB,AC and BC as diameters. It is given that AB=3 cm and AC=4 cm. Find the area of the shaded region in cm².

$\sf{Answer:→ \red{6cm²}}$

$\sf \green{{Step\ by\ step\ explaination}}$

Given that,

• ABC is a right angle triangle
• ∠A=90°
• AB = 3cm
• AC= 4cm

we need to find area of shaded region

Now In ∆ Abc And ∠A 90°

$\tt{By\ Pythagoras\ theorem \ = \red{h²=b²+ p²}}$

$\tt{→(BC)²=(AB)²+(AC)²}$

$\tt{→(BC)²=(3)²+(4)²}$

$\tt{→(BC)²=9+16}$

$\tt{→BC= \ sqrt{25}}$

$\tt{→BC= 5cm}$

• radius of semicircle AB= 3/2
• radius of semicircle AC=4/2= 2
• radius of semicircle BAC =5/2

$\tt{Area\ of\ shadded\ region \ = \red{Area\ of \ △ ABC)}+ \blue{(Area\ of\ semicircle\ AB)} + \red{(Area\ of\ semicircle\ AC)}-\pink{ (Area\ of\ semicircle\ BAC)}}$

$\tt{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \red{\dfrac{1}{2} (base \times height)}+ \pink{(\dfrac{1}{2} \pi r²)} + \blue{(\dfrac{1}{2} \pi r²)}-\pink{ (\dfrac{1}{2} \pi r²)}}$

$\tt{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \red{\dfrac{1}{2}( 3 \times 4)}+ \green{(\dfrac{1}{2} \pi (3/2)²)} + \green{(\dfrac{1}{2} \pi \times 2²)}-\pink{ (\dfrac{1}{2} \pi (5/2)²))}}$

$\tt{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \red{6}+ \dfrac{1}{2} \pi( \green{(\dfrac{3}{2})²} + \blue{4}-\pink{(\dfrac{5}{2})²}) \ \ \ \ taking \dfrac{1}{2} π\ comman}$

$\tt{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \red{6}+ \dfrac{1}{2} \pi \blue{(\dfrac{9}{4} + \blue{4}-\pink{\dfrac{25}{4})}}}$

$\tt{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \red{6}+ \dfrac{1}{2} \pi \dfrac{(9+16-25)}{4}}$

$\tt{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \red{6}+ \dfrac{1}{2} \pi ( \dfrac{0}{4})}$

$\tt{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \red{6}+ \dfrac{22 \times 0}{ 2 \times 7 \times 4}}$

$\tt{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \red{6}+ \dfrac{ 0}{ 56}}$

$\tt{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \red{6}+ 0}$

$\tt{Area\ of\ shadded\ region \ = \red{6cm²}}$