Question

Two circles of radius r units - one centrally positioned on the chord and another as the incircle of the triangle - are positioned in the semicircle of radius 13 units.

Determine the area of the triangle in units squared.​

Answer 1

Answer:

$\LARGE{ \underline{\underline{ \pink{ \bf{Required \: answer:}}}}}$

★Consider the ∆aob,

ob = 0.5 units

ab = 1 unit

$⇒ oa = \sqrt{1² - 0.5²} = \frac{ \sqrt{3} }{2} umits$

$Total \: area \: of \: triangles \: aob \: and \: boc \: = \frac{ \sqrt{3} }{4} </p><p></p><p>$

$Area \: of \: sector \: ab = \frac{1}{2} r²∠abc = \frac{\pi}{3} </p><p></p><p></p><p></p><p>$

★Area of intersection =2(Area of sector − Area of triangles)

$= ( \frac{\pi}{3} - \frac{ \sqrt{3} }{4} )$

$= \frac{4\pi - 3 \sqrt{3} }{6} sq.square$

Explanation:-

Answer 2

Answer:

Answer:

\LARGE{ \underline{\underline{ \pink{ \bf{Required \: answer:}}}}}

Requiredanswer:

★Consider the ∆aob,

ob = 0.5 units

ab = 1 unit

⇒ oa = \

$sqrt{1² - 0.5²} = \frac{ \sqrt{3} }{2} umits⇒oa= </p><p>1²−0.5²$

=

2

3

umits

$Total \: area \: of \: triangles \: aob \: and \: boc \: = \frac{ \sqrt{3} }{4} < /p > < p > < /p > < p >Totalareaoftrianglesaobandboc=$

4

3

</p><p></p><p>

Area \: of \: sector \: ab = \frac{1}{2} r²∠abc =

$\frac{\pi}{3} < /p > < p > < /p > < p > < /p > < p > < /p > < p >Areaofsectorab=$

2

1

r²∠abc=

3

π

</p><p></p><p></p><p></p><p>

★Area of intersection =2(Area of sector − Area of triangles)

= ( \frac{\pi}{3} - \frac{ \sqrt{3} }{4} )=(

3

π

−

4

3

)

$= \frac{4\pi - 3 \sqrt{3} }{6} sq.square= </p><p>6</p><p>4π−3 </p><p>3$

sq.square

Explanation:-