Ejeje J.
asked 08/05/25Find the area of the shaded region.
https://ibb.co/bjxZQDbF
6 Answers By Expert Tutors
Kara Z. answered 10/12/25
Master's in Chemical Engineering with 15+ Years Tutoring Experience
The easiest method for finding the area of the shaded region will be to subtract the area of the triangle from the area of the semicircle.
According to Thale's Theorem, any angle inscribed in a semicircle must be a right triangle with its right angle on the arc of the circle. (This is a specialized case of the Inscribed Angle Theorem.) Since we know that one angle of our triangle is 30°, we can use the Triangle Sum Theorem which states the the three interior angles of a triangle must add up to 180°. This means that the third angle of the triangle must measure 60° and that we must have a 30-60-90 triangle.
The diagram labels the longer leg of the triangle (opposite the 60° angle) as 9√3. According to the ratios of a 30-60-90 triangle, we can divide the long leg by √3 to get the length of the short leg (opposite the 30° angle). That gives us 9. The ratios also tell us that we can multiply the short leg of the triangle by 2 to get the length of the hypotenuse (opposite the 90° angle). That gives us 18.
So we know that the base of our triangle is 9, its height is 9√3, and the diameter of the circle is 18. Since the diameter is twice the length of the radius of a circle, our radius is 9.
Area of a triangle = 1/2bh = 1/2(9)(9√3) = 81√3/2
Area of semicircle = 1/2π r^2 = 1/2π(9)^2 = 81π/2
Shaded area = (semicircle area) - (triangle area)
= 81π/2 - 81√3/2
= 81/2 (π-√3)
≈ 57.0864
Shankar B. answered 09/14/25
Advanced Math experience at Engineering Math levels
Our first objective should be to find the radius R of the circle.
From the figure, note that the triangle has one side as the diameter of the circle.
Let the radius of the circle be = R. This would make the longest side of the triangle is = 2R.
Based on the Inscribed Angle Theorem (An inscribed angle is half the measure of the central angle that subtends the same arc), the inscribed angle (which is the angle of the triangle at the point where it touches the circle) will be half of the angle made by the arc opposite to it.
The opposite arc is a semi-circle and makes a 180-degree angle at the center. Therefore, the inscribed angle is 90 degrees, which makes the inscribed triangle a right-angled triangle with hypotenuse = 2R.
We know from trigonometry that
Cos (30o) = √3 / 2 --> (1)
From the diagram,
Cos (30o) = 9 √3 / 2 R --> (2) - since Cosine of an angle in a right-angled triangle is equal to the adjacent side divided by the hypotenuse.
Putting (1) and (2) together,
√3 / 2 = 9 √3 / 2 R
This would give us R = 9
Using Sin(30o) = 1/2, we can calculate the third side of the triangle to be = R = 9
Calculating the shaded area is really easy at this point.
The shaded area
= Area of the semi-circle minus the area of the triangle
= 1/2 * π * 92 - 1/2 * 9 * 9 √3
= 1/2 * 81 * (3.1416 - 1.7321)
= 40.5 * 1.4095
= 57.8868 square units
Note that with higher precision for π and √3, you will get a refined answer of 57.0864 square units.
In order to find the area of the shaded region, we would first need to find the area of the semicircle and the triangle, then subtract the area of the triangle from the area of the semicircle, and that would leave us with the area of the shaded region.
Step 1: find the area of the semicircle
Area of a circle is πr2 so the semicircle is half that or (1/2)πr2.
To find the radius, we have to solve for the hypotenuse of the triangle. This happens to be a right triangle because a triangle inscribed in a semicircle is always a right triangle. The right angle is touching the side of the circle, so the 3rd angle must be 60 degrees, since the angles inside of a triangle must add up to 180ο (30+60+90=180). We can use the special right triangle formula for solving the hypotenuse (which happens to be the diameter of the circle). For a 30-60-90 triangle the formula for the respective sides are x, x√3, and 2x. The long leg is opposite the 60ο angle, which is 9√3 = x√3. Therefore, x=9 (short leg), so the hypotenuse is 2x=2(9)=18, which is the length of the diameter. Half of the diameter is the radius, so the radius is equal to 9. The area of the semicircle is
(1/2)πr2 = (1/2)(3.14)(9)2 = 127.17
Step 2: Find the area of the triangle, which is (1/2)(base)(height). Since it's a right triangle, we can use the legs as the base and the height, since they are perpendicular to each other. We found these lengths in step 1, so we have (1/2)(9)(9√3) = 70.15
Step 3: Now we can subtract them to get the shaded region:
127.17 - 70.15 = 57.07 units squared
The triangle is a right triangle, and we know this because an angle inscribed in a semi-circle is a right angle. So this is a 30-60-90 triangle. We know the length of the longer leg, so we can use that to find the shorter leg, and to find the hypotenuse, which is also the diameter of the circle (because the point it goes through denotes the center).
The formula for a 30-60-90 triangle is s, s√3, 2s. Our longer leg is 9√3, so our shorter leg is 9 and our hypotenuse is 18.
For a right triangle, the two legs serve as the base and the height, which1 we need to find the area, A=1/2bh. For this triangle, that would be A=1/2(9)(9√3) or 40.5√3.
The semi-circle has an area formula is A=1/2(πr2). For this circle, that is A=1/2(π)(92) or 40.5π.
To find the shaded area, we need to subtract these areas, so 40.5π-40.5√3 ≈ 57.086
Aaron P. answered 08/10/25
I empower students to build confidence in mathematics
2x2-10(2x)+16=0
first change to a quadratic equation first
let 2x=y
y2-10y+16=0 now we have a quadratic equation
a* c=product
a=1, b=1* 16
sum=-10
now we find two numbers if multiplied together gives me 16 and if added together give -10
the numbers are -8,-2
continuation
y2 -8y-2y+16=0
y(y-8)-2(y-8)
(y-2)(y-8)=0
y-2=0 y=2
y-8=0 y=8
now we let 2x=y
y=2
y=8
when y=2
2x=21
x=1
y=8
2x=8
2x=23
x=3
Note that in indices if the bases are the same the powers are also similar
Aaron P.
request for more examples08/10/25
Doug C. answered 08/05/25
Math Tutor with Reputation to make difficult concepts understandable
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Cuatoi13 J.
Thank you for sharing26d