### Video Transcript

Russ draws an isosceles triangle
π΄ππΆ, which intersects with a circle of center π. He then draws a point π΅, which is
the midpoint of π΄πΆ. Finally, Russ draws another point
π, which is at the intersection of ππ΅ and the circumference of the circle. π is also the midpoint of
ππ΅. ππ΄ equals ππΆ; ππ equals ππ΅,
which equals 20 millimeters; and angle ππ΄πΆ equals 54 degrees. Russ says, βThe area of the
triangle is more than 15 percent larger than the shaded area.β Is Russ correct? You must show your working.

Russ is making a claim about the
area of the triangle compared to the area of the shaded region. Before we can say if Russ is
correct, we need to know the area of the triangle and the area of the shaded
region.

Starting with the area of a
triangle, the area of any triangle equals one-half the height times the base. The height of this triangle is the
length ππ΅, 20 millimeters plus 20 millimeters, which equals 40 millimeters. Weβre not given the length of the
base.

If we label the distance from π΄ to
π΅ as π₯ and the distance from π΅ to πΆ as π₯, we can call the base of this
isosceles triangle two π₯. We know that π΄π΅ is equal to π΅πΆ
because weβre told that point π΅ is the midpoint of π΄πΆ. So we call the base of the triangle
two π₯.

Now the question is how do we find
this π₯ value? Well, in an isosceles triangle, if
a line segment goes from the vertex angle to the base β for us, that would be line
segment ππ΅ β then the line segment is perpendicular to the base. It forms a right angle. This means that triangle π΄π΅π is
a right-angled triangle.

We can use what we know about
right-angled triangles to solve for π₯. We sometimes use the memory device
SOHCAHTOA to help us remember the relationships inside right-angled triangles, the
sine, cosine, and tangent, from angle 54 degrees.

We know the side length opposite
that angle. And weβre interested in the side
length adjacent to our angle. The relationship opposite over
adjacent is the tangent ratio. Tan of 54 degrees equals opposite
over adjacent, 40 millimeters over π₯ millimeters. Our goal is to solve for π₯. And that means we need to get π₯
out of the denominator. We do that by multiplying both
sides of the equation by π₯.

On the left, we have π₯ times tan
of 54 degrees. On the right, the π₯s cancel out,
leaving us with 40. And then we divide both sides of
the equation by tan of 54 degrees. On the left, the tangents cancel
out, leaving us with π₯. And on the right, we have 40
divided by tan of 54 degrees. When we do that division, π₯ is
equal to 29.0617 continuing.

Going back to the original equation
for area of a triangle, if we multiply one-half times two, they cancel. One-half times two equals one. And that means the area of this
triangle will be equal to 40 times π₯. We substitute what we know π₯ to
be. Plugging that into our calculator
gives us 1162.468045 continuing. Weβre measuring in millimeters, and
itβs an area. So our units will be millimeters
squared. For now, letβs keep the area in
this format.

Iβm going to clear some space. But remember, part of this question
is showing your work. And that means everything up until
this point is also part of the answer.

And now for the shaded area, the
shaded area is a portion of a circle. We start with the formula for the
area of a circle. Area equals ππ squared. And we multiply that formula by the
fraction created by taking the shaded angle and putting that over 360 degrees.

In order to find the measure of
this shaded angle, weβll need to calculate the measure of angle π΄ππΆ. Because we know that this is an
isosceles triangle, we can say that angle π΄πΆπ equals 54 degrees. And that means 54 degrees plus 54
degrees plus angle π΄ππΆ must equal 180 degrees. They are the three angles that make
up this isosceles triangle. 108 degrees plus angle π΄ππΆ
equals 180 degrees. Subtract 108 degrees from both
sides of the equation and we find that angle π΄ππΆ equals 72 degrees. And that means the interior angle
of the shaded region would be equal to 360 degrees minus 72 degrees, 288
degrees.

Now we need the radius of the
circle. The radius is 20 millimeters. We know this because point π falls
on the circumference of the circle and ππ equals 20 millimeters. ππ is a radius of this
circle. Back to our formula, 288 over 360
can be reduced to four-fifths. 20 squared equals 400. And bring down π. Four-fifths times 400 equals
320. And again, bring down π. The area of the shaded region
equals 320π millimeters squared.

Now that we have these two areas,
weβre ready to consider what Russ has said. We need to consider the area of the
triangle over the shaded area, which would look like this. From there, we need to divide the
area of the triangle by 320π. When we do that, we get 1.156328
continuing.

This number tells us that if we
knew the area of the shaded region, we could multiply it by 1.156328 continuing. And that would give us the area of
the triangle. This confirms that the triangle is
larger than the shaded area. To find out how much larger, we
need to subtract one from 1.156328 continuing, which gives us 0.156328
continuing.

The final step we need to do is
take this decimal value and turn it into a percent. We do that by multiplying the
decimal by 100. The percentage is 15.6328
continuing percent. If we round to two decimal places,
we look to the third decimal place, to the deciding digit, which is the two. And that means weβll round
down.

Our percentage is then 15.63
percent. The triangle is 15.63 percent
larger than the shaded area. Russ said that the triangle is more
than 15 percent larger. 15.63 is larger than 15. So Russ is correct.