# Video: AQA GCSE Mathematics Higher Tier Pack 3 • Paper 3 • Question 22

Russ draws an isosceles triangle 𝐴𝑂𝐶, which intersects with a circle of center 𝑂. He then draws a point 𝐵, which is the midpoint of 𝐴𝐶. Finally, Russ draws another point 𝑃, which is at the intersection of 𝑂𝐵 and the circumference of the circle. 𝑃 is also the midpoint of 𝑂𝐵. 𝑂𝐴 = 𝑂𝐶. 𝑂𝑃 = 𝑃𝐵 = 20 mm. Angle 𝑂𝐴𝐶 = 54°. Russ says, “The area of the triangle is more than 15% larger than the shaded area.” Is Russ correct? You must show your working.

08:39

### Video Transcript

Russ draws an isosceles triangle 𝐴𝑂𝐶, which intersects with a circle of center 𝑂. He then draws a point 𝐵, which is the midpoint of 𝐴𝐶. Finally, Russ draws another point 𝑃, which is at the intersection of 𝑂𝐵 and the circumference of the circle. 𝑃 is also the midpoint of 𝑂𝐵. 𝑂𝐴 equals 𝑂𝐶; 𝑂𝑃 equals 𝑃𝐵, which equals 20 millimeters; and angle 𝑂𝐴𝐶 equals 54 degrees. Russ says, “The area of the triangle is more than 15 percent larger than the shaded area.” 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 𝑂𝐵, 20 millimeters plus 20 millimeters, which equals 40 millimeters. We’re not given the length of the base.

If we label the distance from 𝐴 to 𝐵 as 𝑥 and the distance from 𝐵 to 𝐶 as 𝑥, we can call the base of this isosceles triangle two 𝑥. We know that 𝐴𝐵 is equal to 𝐵𝐶 because we’re told that point 𝐵 is the midpoint of 𝐴𝐶. So we call the base of the triangle two 𝑥.

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

We can use what we know about right-angled triangles to solve for 𝑥. 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 we’re 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 𝑥 millimeters. Our goal is to solve for 𝑥. And that means we need to get 𝑥 out of the denominator. We do that by multiplying both sides of the equation by 𝑥.

On the left, we have 𝑥 times tan of 54 degrees. On the right, the 𝑥s 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 𝑥. And on the right, we have 40 divided by tan of 54 degrees. When we do that division, 𝑥 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 𝑥. We substitute what we know 𝑥 to be. Plugging that into our calculator gives us 1162.468045 continuing. We’re measuring in millimeters, and it’s an area. So our units will be millimeters squared. For now, let’s keep the area in this format.

I’m 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 𝜋𝑟 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, we’ll need to calculate the measure of angle 𝐴𝑂𝐶. Because we know that this is an isosceles triangle, we can say that angle 𝐴𝐶𝑂 equals 54 degrees. And that means 54 degrees plus 54 degrees plus angle 𝐴𝑂𝐶 must equal 180 degrees. They are the three angles that make up this isosceles triangle. 108 degrees plus angle 𝐴𝑂𝐶 equals 180 degrees. Subtract 108 degrees from both sides of the equation and we find that angle 𝐴𝑂𝐶 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 𝑃 falls on the circumference of the circle and 𝑂𝑃 equals 20 millimeters. 𝑂𝑃 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, we’re 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 we’ll 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.