Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

$Question Watch Video Show Examples In circle T, the length of UV=(5)/(3)pi and m/_UTV=100^(@). Find the area shaded below. Express your answer as a fraction times pi. Answer Attempt 1 out of 2 Sign out N Apr 19 7.26$

Question $\newline$ Watch Video $\newline$ Show Examples $\newline$ In circle $T$ , the length of $U V=\frac{5}{3} \pi$ and $\mathrm{m} \angle U T V=100^{\circ}$ . Find the area shaded below. Express your answer as a fraction times $\pi$ . $\newline$ Answer Attempt $1$ out of $2$ $\newline$ Sign out $\newline$ $\mathbb{N}$ $\newline$ Apr $19$ $\newline$ $7$ . $26$

View step-by-step help

Home

Math Problems

Grade 6

Evaluate variable expressions: word problems

Full solution

Q. Question

\newline

Watch Video

\newline

Show Examples

\newline

In circle

T

, the length of

U V=\frac{5}{3} \pi

and

\mathrm{m} \angle U T V=100^{\circ}

. Find the area shaded below. Express your answer as a fraction times

\pi

\newline

Answer Attempt

1

out of

2

\newline

Sign out

\newline

\mathbb{N}

\newline

Apr

19

\newline

7

26

Find Radius of Circle: First, we need to find the radius of the circle. Since UV is a chord and we have the measure of the central angle UTV, we can use the formula for the length of a chord: $2r \sin(\theta/2) = \text{chord length}$ , where $r$ is the radius and $\theta$ is the central angle.
Calculate Chord Length: Plug in the given values: $2r \sin(100^\circ/2) = (5/3)\pi$ .
Solve for Radius: Simplify the equation: $2r \sin(50^\circ) = (5/3)\pi$ .
Find Area of Sector: Now, solve for $r$ : $r = \frac{(5/3)\pi}{2 \sin(50^\circ)}$ .
Calculate Area of Triangle: Next, find the area of the sector formed by angle UTV, which is $\frac{\theta}{360} \times \pi r^2$ .
Subtract Triangle Area: Plug in the values for $\theta$ and $r$ : $\frac{100}{360} \times \pi \left(\frac{(5/3)\pi}{2 \sin(50^\circ)}\right)^2$ .
Correct Calculation: Simplify the expression to find the area of the sector: $\frac{100}{360} \times \pi \times \frac{(25/9)\pi^2}{4 \sin^2(50^\circ)}$ .
Correct Calculation: Simplify the expression to find the area of the sector: $\frac{100}{360} \times \pi \times \frac{(25/9)\pi^2}{4 \sin^2(50^\circ)}$ .Now, find the area of the triangle UTV using the formula $\frac{1}{2}ab\sin(C)$ , where $a$ and $b$ are the sides adjacent to angle $C$ .
Correct Calculation: Simplify the expression to find the area of the sector: $\frac{100}{360} \times \pi \times \frac{(25/9)\pi^2}{4 \sin^2(50^\circ)}$ .Now, find the area of the triangle UTV using the formula $\frac{1}{2}ab\sin(C)$ , where $a$ and $b$ are the sides adjacent to angle $C$ .Since the triangle is isosceles with the radius as its legs, $a = b = r$ , and $C = 100^\circ$ , the area of the triangle is $\frac{1}{2}r^2\sin(100^\circ)$ .
Correct Calculation: Simplify the expression to find the area of the sector: $\frac{100}{360} \times \pi \times \frac{(25/9)\pi^2}{4 \sin^2(50^\circ)}$ .Now, find the area of the triangle UTV using the formula $\frac{1}{2}ab\sin(C)$ , where $a$ and $b$ are the sides adjacent to angle $C$ .Since the triangle is isosceles with the radius as its legs, $a = b = r$ , and $C = 100^\circ$ , the area of the triangle is $\frac{1}{2}r^2\sin(100^\circ)$ .Plug in the value of $r$ : $\frac{1}{2}\left(\frac{(5/3)\pi}{2 \sin(50^\circ)}\right)^2\sin(100^\circ)$ .
Correct Calculation: Simplify the expression to find the area of the sector: $\frac{100}{360} \times \pi \times \frac{(25/9)\pi^2}{4 \sin^2(50^\circ)}$ .Now, find the area of the triangle UTV using the formula $\frac{1}{2}ab\sin(C)$ , where $a$ and $b$ are the sides adjacent to angle $C$ .Since the triangle is isosceles with the radius as its legs, $a = b = r$ , and $C = 100^\circ$ , the area of the triangle is $\frac{1}{2}r^2\sin(100^\circ)$ .Plug in the value of $r$ : $\frac{1}{2}\left(\frac{(5/3)\pi}{2 \sin(50^\circ)}\right)^2\sin(100^\circ)$ .Simplify the expression to find the area of the triangle: $\frac{1}{2}ab\sin(C)$ $0$ .
Correct Calculation: Simplify the expression to find the area of the sector: $\frac{100}{360} \times \pi \times \frac{(25/9)\pi^2}{4 \sin^2(50^\circ)}$ .Now, find the area of the triangle UTV using the formula $\frac{1}{2}ab\sin(C)$ , where $a$ and $b$ are the sides adjacent to angle $C$ .Since the triangle is isosceles with the radius as its legs, $a = b = r$ , and $C = 100^\circ$ , the area of the triangle is $\frac{1}{2}r^2\sin(100^\circ)$ .Plug in the value of $r$ : $\frac{1}{2}\left(\frac{(5/3)\pi}{2 \sin(50^\circ)}\right)^2\sin(100^\circ)$ .Simplify the expression to find the area of the triangle: $\frac{1}{2}ab\sin(C)$ $0$ .Finally, subtract the area of the triangle from the area of the sector to find the area of the shaded region: $\frac{1}{2}ab\sin(C)$ $1$ .
Correct Calculation: Simplify the expression to find the area of the sector: $\frac{100}{360} \times \pi \times \frac{(25/9)\pi^2}{4 \sin^2(50^\circ)}$ .Now, find the area of the triangle UTV using the formula $\frac{1}{2}ab\sin(C)$ , where $a$ and $b$ are the sides adjacent to angle $C$ .Since the triangle is isosceles with the radius as its legs, $a = b = r$ , and $C = 100^\circ$ , the area of the triangle is $\frac{1}{2}r^2\sin(100^\circ)$ .Plug in the value of $r$ : $\frac{1}{2}\left(\frac{(5/3)\pi}{2 \sin(50^\circ)}\right)^2\sin(100^\circ)$ .Simplify the expression to find the area of the triangle: $\frac{1}{2}ab\sin(C)$ $0$ .Finally, subtract the area of the triangle from the area of the sector to find the area of the shaded region: $\frac{1}{2}ab\sin(C)$ $1$ .Oops, I made a mistake in the calculation. I should have used the correct formula for the area of the triangle in a circle, which is $\frac{1}{2}ab\sin(C)$ $2$ , where $C$ is the central angle. Let's correct that.