### Video Transcript

A net of a shape is made from squares and rectangles. Now this question has got four parts in it. Letβs look at part a) first. a) Shade an area of the given diagram, which can be represented by the expression ππ.

Now weβre told that this net is made from squares and rectangles. And the fact that itβs a net means that it can be folded up to make a three-dimensional object. Because the shapes in the middle are rectangles, we know that these two lengths must be the same and this one, and this one, and this one. Because the other two shapes are squares, these four lengths must all be the same. They must all be length π. And because itβs a net, this side here must fold onto this side here, and they must be the same length. And similarly, this side here folds onto this side here. So they must both be length π.

Because we got a rectangle, this must also be length π. And because that shape is a square. All these other lengths must be π as well. Lastly, because these two sides must be the same length to fold onto each other, this length here must also be π. And because that bottom shape is are rectangle, this last length here must also be π as well.

So now we worked out what all the lengths are, letβs think about what that shape would be if we folded the net up into a three-dimensional shape. It will be a cuboid. It would have two ends, which were squares, π by π. And it would have four sides, which are rectangles, π by π.

Okay, letβs answer part a): Shade an area of the given diagram, which can be represented by the expression ππ.

Well, the expression ππ means π times π. And π times π could represent the area of a rectangle with side lengths π and π. And in our diagram, there are four options for rectangles which have side lengths of π and π. Iβm gonna shade this one, but I could easily have shaded this one or this one or this one.

Letβs make a note of that down here and move on to part b).

b) Shade an area of the given diagram, which can be represented by the expression π times π plus π. Well, this could represent the area of a rectangle with side lengths π and π plus π. Looking at that shape, this is length π and weβve got π here and π here. So if we add those two together, weβve got π plus π. So the area of this rectangle here is π times π plus π. Now, in a similar way, this length is π plus π and this length is π. So we could easily have shaded this area instead.

So letβs make a note of our answer to part b) and move on to part c).

Shade an area of the given diagram, which can be represented by the expression two π squared.

Well, two π squared can be expressed two times π times π. And we can think of π times π as representing the area of a square with sides of length π. So weβve got to find two squares with sides of length π on our diagram and shade them both in. Well, remembering all the side lengths that we worked out earlier, thereβs one square of side length π and thereβs another. So if we add those two areas together, weβve got π squared plus π squared. Thatβs two π squared.

Now remember, in part a), we had four options of which rectangle we shaded in. In part b), we had two options of which rectangle to shade in. But in part c), we must shade both of those squares together because their areas added together make two π squared. Okay, letβs move on to part d).

d) Write an expression for the surface area of the cuboid that the net makes.

Well, we saw earlier that the net folds up to make a cuboid like this. The two ends are squares π by π and the four sides are rectangles π by π. To get the total surface area then, we need to add up four times the area of a pink rectangle and add to that two times the area of an orange square. Well, the pink rectangle has side lengths of π and π. So the area of one of those is gonna be π times π. So the area of four of them is gonna be four times π times π. And the orange square has sides of length π.

So the area of one orange square is gonna be π times π. The area of two orange squares is gonna be two times π times π. And a slightly neater way of writing that is instead of four times π times π, weβll write four ππ. And instead of two times π times π, weβll write two π squared. So for part d), the total surface area of the net is equal to four ππ plus two π squared. Although, obviously, if you wrote two π squared plus four ππ the other way around, that would still be a correct answer.