Answer :
see the attached figure to better understand the problem
we know that
The inscribed angle in a circle measures half of the arc it comprises.
in this problem
the inscribed angle= ∠ACB
and the arc it comprises measures 180°
then
the ∠ACB=180°/2-------> ∠ACB=90°
applying the Pythagorean theorem
AC²+CB²=AB²-------> AB²=24²+7²-------> AB²=625------> AB=25 cm
the diameter of circle is AB
radius=25/2--------> r=12.5 cm
[the area of a half circle]=pi*r²/2------> pi*12.5²/2--------> 245.44 cm²
[area of triangle ABC]=AC*CB/2--------> 24*7/2-------> 84 cm²
[the area of the shaded region]=[the area of a half circle]-[area of triangle ABC]
[the area of the shaded region]=245.44-84-------> 161.44 cm²
the answer is
the area of the shaded region is 161.44 cm²
we know that
The inscribed angle in a circle measures half of the arc it comprises.
in this problem
the inscribed angle= ∠ACB
and the arc it comprises measures 180°
then
the ∠ACB=180°/2-------> ∠ACB=90°
applying the Pythagorean theorem
AC²+CB²=AB²-------> AB²=24²+7²-------> AB²=625------> AB=25 cm
the diameter of circle is AB
radius=25/2--------> r=12.5 cm
[the area of a half circle]=pi*r²/2------> pi*12.5²/2--------> 245.44 cm²
[area of triangle ABC]=AC*CB/2--------> 24*7/2-------> 84 cm²
[the area of the shaded region]=[the area of a half circle]-[area of triangle ABC]
[the area of the shaded region]=245.44-84-------> 161.44 cm²
the answer is
the area of the shaded region is 161.44 cm²
The area of shaded region is [tex]\boxed{161.54{\text{ c}}{{\text{m}}^2}}.[/tex]
Further explanation:
The formula for area of semi-circle can be expressed as follows,
[tex]\boxed{{\text{Area of semi - circle}} = \frac{1}{2} \times \pi {r^2}}[/tex]
Here, [tex]r[/tex] is the radius of the circle.
Given:
The length of side AC is [tex]24{\text{ cm}}.[/tex]
The length of side AB is [tex]7{\text{ cm}}.[/tex]
Explanation:
In [tex]\Delta {\text{ABC}}[/tex] use the Pythagoras theorem to obtain the length of side BC.
[tex]\begin{aligned}{\text{B}}{{\text{C}}^{\text{2}}} &= {\text{A}}{{\text{C}}^{\text{2}}} + {\text{A}}{{\text{B}}^{\text{2}}}\\{\text{B}}{{\text{C}}^{\text{2}}} &= {7^2} + {24^2}\\{\text{B}}{{\text{C}}^{\text{2}}} &= 49 + 576\\{\text{B}}{{\text{C}}^{\text{2}}}&= 625\\{\text{BC}}&= 25\\\end{aligned}[/tex]
The diameter of the circle is [tex]25{\text{ cm}}.[/tex]
The radius of the circle is [tex]\dfrac{{25}}{2}{\text{ cm}}.[/tex]
The area of the semi-circle can be obtained as follows,
[tex]\begin{aligned}{\text{Area of semicircle}}&= \frac{1}{2} \times \frac{{22}}{7} \times {\left( {\frac{{25}}{2}} \right)^2}\\&= \frac{{13750}}{{56}}\\&= 245.54{\text{ c}}{{\text{m}}^2}\\\end{aligned}[/tex]
The area of triangle can be obtained as follows,
[tex]\begin{aligned}{\text{Area of triangle}}&= \frac{1}{2} \times 7 \times 24\\&=12 \times 7\\&= 84{\text{ c}}{{\text{m}}^2}\\\end{aligned}[/tex]
Area of shaded region can be obtained as follows,
[tex]\begin{aligned}{\text{Area of shaded region}} &= {\text{Area of semicircle}} - {\text{Area of triangle}}\\&= 245.54 - 84\\&= 161.54{\text{ c}}{{\text{m}}^2}\\\end{aligned}[/tex]
The area of shaded region is [tex]\boxed{161.54{\text{ c}}{{\text{m}}^2}}.[/tex]
Learn more:
- Learn more about inverse of the functionhttps://brainly.com/question/1632445.
- Learn more about equation of circle brainly.com/question/1506955.
- Learn more about range and domain of the function https://brainly.com/question/3412497
Answer details:
Grade: High School
Subject: Mathematics
Chapter: Circles
Keywords: area of shaded region, center, AB=7cm, AC=24 cm, figure, radius of circle, arc length, radian, central angle, intercepted, circle, circumference, sector of a circle, minor sector, major sector, segment, angle.