### Video Transcript

A square of side length π is divided into four equal sections as shown. The two shaded sections are removed to leave a triangle. Part a) tick a box to show whether each statement is true or false. There are four statements in part a) and there is also a part b) that we will look at later.

The first statement says the following. Area of square equals two multiplied by area of final triangle. There are a couple of ways we could approach this. Firstly, we could work out the area of the square and the area of the triangle using our general formulas. The formula for the area of any square is the side length squared. In this case, as the side length of the square is π, then its area will be π squared.

The area of any triangle can be calculated by multiplying a half by the base by the perpendicular heights. The base of our triangle is length π. And the perpendicular height of the triangle is also length π. This means that the area of the triangle is equal to a half multiplied by π multiplied by π. π multiplied by π is equal to π squared. So, we have a half multiplied by π squared. This is equal to π squared over two, or a half of π squared.

The statement said that the area of the square was equal to two multiplied by the area of the triangle. Substituting in our areas gives us π squared is equal to two multiplied by π squared over two. The twos on the right-hand side of the equation will cancel. So, we are left with π squared is equal to π squared. This means that the first statement is true. The area of the square is equal to two multiplied by the area of the final triangle.

We could also have answered this first question by looking at the area of each part of the three shapes. As the area of the complete square is π squared, the area of each of the rectangles in the first shape would be π squared over two, or a half of π squared. In order to create the second shape, we split both of these rectangles in half. A half of a half is equal to a quarter.

Therefore, each of the triangles in the second shape has an area of π squared over four, or a quarter of π squared. Adding a quarter and a quarter gives us a half. Therefore, we can see that the area of the final triangle is π squared over two. As this is half of the area of the square, we can once again conclude that the area of the square is indeed two multiplied by the area of the final triangle.

The second statement says the following. Perimeter of square equals two multiplied by perimeter of final triangle. The perimeter of any shape is the distance around the outside. As the square has side length π, its perimeter will be equal to four π, as π plus π plus π plus π equals four π. The final triangle has side lengths π, π, and π. This means that it has perimeter π plus two π. This time, the statement asked us to consider whether the perimeter of the square was equal to two multiplied by the perimeter of the triangle.

Substituting in our expressions this time gives us four π is equal to two multiplied by π plus two π. We can expand the bracket by multiplying two by π and two by two π. This gives us two π plus four π. We can then subtract two π from both sides of the equation. This gives us two π is equal to four π. Finally, we can divide both sides of this equation by two. This gives us π is equal to two π. This cannot be correct as the length π is actually longer than the length π. We can, therefore, conclude that the statement, perimeter of square equals two multiplied by perimeter of final triangle, is false.

The third statement says the following. All angles in the final triangle are equal. All of the angles in any triangle are equal when the triangle is equilateral. This means that all three side lengths must also be equal. As the length π is not equal to the length of π, the final triangle is not equilateral. This means that the statement, all angles in the final triangle are equal, is false.

The final statement in part a) says the following. At least two sides of the final triangle are equal. We can see from the diagram that both of the sloping edges of the final triangle are of length π. This means that the triangle is isosceles, as two of the lengths are the same. Whilst the third side of the triangle is different in length, the statement is true. At least two of the sides of the final triangle are equal in length. In fact, we can go one stage further and say that exactly two sides of the final triangle are equal.

Part b) of the question says the following. The area of the square is 324 square centimetres. Work out the perimeter of the triangle.

We worked out in part a) that the area of the square was equal to π squared. Therefore, π squared is equal to 324 square centimetres. The opposite, or inverse, of squaring is square rooting. Therefore, we need to square root both sides of the equation to calculate π. The square root of π squared is equal to π. And on the right-hand side, we need to square root 324. This is equal to 18. Therefore, π is equal to 18 centimetres. The side length of the square is 18 centimetres. This means that the base of the final triangle is also 18 centimetres.

Letβs now consider one-half of the final triangle, as shown in pink. The pink triangle is right-angled with a base length of nine centimetres, as this is half of 18, a height of 18 centimetres, and a diagonal, or sloping, length π. In order to calculate the missing length of any right-angled triangle, we can use Pythagorasβ theorem. This states that π squared plus π squared is equal to π squared, where π is the longest side of the triangle known as the hypotenuse.

In our case, the hypotenuse is length π, and the two shorter sides have length nine and 18 centimetres. Substituting in these values gives us nine squared plus 18 squared is equal to π squared. Nine squared is equal to 81. 18 squared is equal to 324. We need to add these two numbers. This gives us 405.

Therefore, π squared is equal to 405. Finally, we need to square root both sides of this equation to calculate the length of π. As the square root of 405 is not a whole number, we will leave our answer as a surd for the time being. This will help us in terms of our accuracy. The length π is equal to the square root of 405.

We were asked to work out the perimeter of the triangle. The triangle has two sides of length π and one side of length π. Therefore, the perimeter is equal to π plus two π. π was equal to 18 centimetres. And π was equal to the square root of 405 centimetres. We can, therefore, say that the perimeter is equal to 18 plus two multiplied by the square root of 405.

Typing this into our calculator gives us an answer of 58.2492 and so on. In order to round this to one decimal place, we can look at the second digit after the decimal point. As this is a four, we will round down. The perimeter of the triangle, rounded to one decimal place, is 58.2 centimetres.