Find the area of the shaded region.

Question

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Answer 1

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 (30^o) = √3 / 2 --> (1)

From the diagram,

Cos (30^o) = 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(30^o) = 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.

Answer 2

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 180^ο (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 60^ο 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

Answer 3

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

Answer 4

2^x2-10(2^x)+16=0

first change to a quadratic equation first

let 2^x=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 2^x=y

y=2

y=8

when y=2

2^x=2¹

^x=1

^y=8

2^x=8

2^x=2³

x=3

Note that in indices if the bases are the same the powers are also similar

Answer 5