### Video Transcript

A cuboid is shown in the diagram. The base has a length of 12π₯π¦ centimeters and a width of four π¦ centimeters. The length of the diagonal π·πΊ is 15π₯π¦ centimeters. The surface area of the cuboid is 2448π¦ squared centimeters squared. The volume of the cuboid is ππ¦ cubed centimeters cubed, where π is an integer. Calculate the value of π.

This question is asking us about the volume of the cuboid. To find the volume of a cuboid, we multiply its length by its width by its
height. Weβve been given expressions for the length and the width of this cuboid in terms of
the letters π₯ and π¦. But we havenβt been told anything about the height of this cuboid.

So our first step is going to be to find an expression for the height in terms of π₯
and π¦. Letβs consider the right-angled triangle πΆπ·πΊ. As the base of the cuboid is a rectangle, its opposite sides are equal in length. So the side πΆπ· is equal to π΄π΅, which is 12π₯π¦ centimeters. We also know the length of the hypotenuse of this triangle: itβs 15π₯π¦
centimeters.

As we know two sides of a right-angled triangle and wish to calculate the third side,
we can apply Pythagorasβs theorem. Pythagorasβs theorem tells us that in a right-angled triangle, the sum of the squares
of the two shorter sides is equal to the square of the longest side, the
hypotenuse.

In our triangle, this gives πΆπΊ squared plus 12π₯π¦ squared is equal to 15π₯π¦
squared. Now, we can solve this equation to find an expression for πΆπΊ in terms of π₯ and
π¦. When we square a term in a bracket, we must remember to square each part. So 12π₯π¦ all squared is equal to 144 π₯ squared π¦ squared and 15π₯π¦ all squared is
equal to 225 π₯ squared π¦ squared.

We can then subtract 144 π₯ squared π¦ squared from each side, giving πΆπΊ squared is
equal to 225 π₯ squared π¦ squared minus 144 π₯ squared π¦ squared, which simplifies
to 81 π₯ squared π¦ squared. To find πΆπΊ, we then need to square root both sides of the equation, giving πΆπΊ is
equal to the square root of 81 π₯ squared π¦ squared.

To square root a product, we can find the product of the square roots. The square root of 81 is nine, the square root of π₯ squared is π₯, and the square
root of π¦ squared is π¦. So we have that πΆπΊ is equal to nine π₯ π¦. We now have an expression for the height of this cuboid, which we can substitute into
our formula for the volume.

We have 12π₯π¦ multiplied by four π¦ multiplied by nine π₯ π¦. 12 multiplied by four multiplied by nine is 432. π₯ multiplied by π₯ is π₯ squared and π¦ multiplied by π¦ multiplied by π¦ is π¦
cubed. So the volume is equal to 432 π₯ squared π¦ cubed. Now, remember weβre told in the question that this is equal to ππ¦ cubed.

Comparing these two expressions for the volume, we see that π is equal to 432 π₯
squared. And in order to find the value of π, we first need to find the value of π₯.

Now, the other piece of information that weβve been given in the question is that the
surface area of this cuboid is equal to 2448 π¦ squared centimeters squared.

The surface area of a cuboid is found by summing the areas of all six faces
together. And as this is a cuboid, the areas come in three pairs. We have the front and the back of the cuboid, the top and the base of the cuboid, and
the two vertical sides of the cuboid.

Letβs now find expressions for each of these areas, starting with the front and the
back of the cuboid. The front and the back are rectangles with measurements of 12π₯π¦ and nine π₯π¦
centimeters. So the area of the two rectangles is two multiplied by 12π₯π¦ multiplied by nine
π₯π¦. This expression simplifies to 216 π₯ squared π¦ squared.

Now, letβs consider the top and the base of the cuboid. These are rectangles with measurements of 12 π₯π¦ centimeters and four π¦
centimeters. The area of the two is, therefore, two multiplied by 12π₯π¦ multiplied by four π¦,
which simplifies to 96π₯π¦ squared.

Finally, we consider the vertical sides of the cuboid, which are rectangles with
measurements of four π¦ and nine π₯π¦ centimeters. Their areas are, therefore, two multiplied by four π¦ multiplied by nine π₯π¦, which
simplifies to 72 π₯π¦ squared.

So we now have our expression for the surface area of this cuboid. And in fact, the last two terms can be combined, giving 216 π₯ squared π¦ squared
plus 168 π₯π¦ squared. Now, remember weβre told that this is equal to 2448π¦ squared. So we can form an equation. We can also take a factor of π¦ squared from the right side of the equation. So we have 2448π¦ squared is equal to 216π₯ squared plus 168π₯ all multiplied by π¦
squared.

Comparing the coefficients of π¦ squared, we have that 2448 is equal to 216 π₯
squared plus 168 π₯, which gives us a quadratic equation in π₯. Here is that quadratic equation again: 216π₯ squared plus 168π₯ is equal to 2448. We can group all the terms on the same side of the equation by subtracting 2448 from
both sides.

Now, in fact, all of the coefficients in this equation are multiples of eight. So we can simplify by dividing through by eight, giving 27π₯ squared plus 21π₯ minus
306 is equal to zero. If you didnβt spot that the coefficients were all multiples of eight, you would spot
that they are even numbers. So you could halve them and then halve them again and then halve them a third time to
arrive at the same quadratic.

In fact, this quadratic can be simplified further as all of the coefficients are
multiples of three. So our fully simplified quadratic is nine π₯ squared plus seven π₯ minus 102 is equal
to zero.

Now, we want to solve this equation for π₯. And we have a variety of methods that we could try. We could factorize or we could complete the square or we could use the quadratic
formula. As the coefficients in this equation are quite large, itβs going to be most
straightforward to apply the quadratic formula.

Remember the quadratic formula tells us that the roots of the quadratic equation ππ₯
squared plus ππ₯ plus π is equal to zero are given by π₯ equals negative π plus
or minus the square root of π squared minus four ππ all over two π.

In our quadratic, π is equal to nine, π is equal to seven, and π is equal to
negative 102. So we have π₯ is equal to negative seven plus or minus the square root of seven
squared minus four multiplied by nine multiplied by negative 102 all divided by two
multiplied by nine.

Now, do be careful with the negatives in the bracket, but this simplifies to negative
seven plus or minus the square root of 3721 divided by 18. We can use a calculator to evaluate these two roots. And it gives exactly three or negative 3.7 recurring.

Now, while these are both roots to that quadratic equation, the value of π₯ needs to
make sense in the context of this question. π₯ doesnβt have to be an integer, but it does have to be a positive value. We know that π¦ must be a positive value as four π¦ is the width of the cuboid. So π₯ must also be positive in order to give a positive value when multiplied by 12π¦
for the length of the cuboid.

This tells us that the value of π₯ must be this positive value of three. Finally, we can substitute this value of π₯ into our expression for π. Remember π was equal to 432 π₯ squared. So we have 432 multiplied by three squared. Three squared is nine and 432 multiplied by nine gives 3888.

The value of π is 3888.