Video: AQA GCSE Mathematics Higher Tier Pack 5 β€’ Paper 1 β€’ Question 12

𝐴𝐡𝐺𝐻 is a rectangle. 𝐡𝐢𝐷𝐺 is a rectangle. 𝐺𝐷𝐸𝐹 is a square. These three shapes join together to make a shape, 𝑅. Show that the total area of 𝑅 is 4π‘₯Β² + 8π‘₯ + 5.

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Video Transcript

𝐴𝐡𝐺𝐻 is a rectangle. 𝐡𝐢𝐷𝐺 is a rectangle. 𝐺𝐷𝐸𝐹 is a square. These three shapes join together to make a shape, 𝑅. Show that the total area of 𝑅 is four π‘₯ squared plus eight π‘₯ plus five.

Our first step is to work out the dimensions of each of the three shapes. 𝐴𝐡𝐺𝐻 is a rectangle. It has length π‘₯ plus two centimeters and width four centimeters. 𝐡𝐢𝐷𝐺 is also a rectangle. It has length two π‘₯ minus one centimeters and width four centimeters. This is because the lengths of 𝐡𝐢 and 𝐹𝐸 are equal. Likewise, 𝐴𝐻 is equal to 𝐢𝐷. Finally, 𝐺𝐷𝐸𝐹 is a square. All four lengths of a square are equal. In this case, they are two π‘₯ minus one centimeters.

We need to work out an expression for the area of each of these three shapes and then add them together to make the area of shape 𝑅. The area of a rectangle is calculated by multiplying the length by the width this means that the area of 𝐴𝐡𝐺𝐻 is equal to four multiplied by π‘₯ plus two. In the same way, the area of the rectangle 𝐡𝐢𝐷𝐺 is equal to four multiplied by two π‘₯ minus one.

The area of a square can be calculated by squaring its length. In this case, we have two π‘₯ minus one all squared. The total area of 𝑅 is therefore equal to two π‘₯ minus one all squared plus four multiplied by π‘₯ plus two plus four multiplied by two π‘₯ minus one.

Our next step is to multiply out or expand the brackets. Expanding a single bracket involves multiplying the number outside the bracket by each of the terms inside the bracket. If we look at the first rectangle, we need to multiply four by π‘₯ and four by two. Four multiplied by π‘₯ is equal to four π‘₯ and four multiplied by two is equal to eight. Therefore, the area of rectangle 𝐴𝐡𝐺𝐻 is four π‘₯ plus eight.

We can use the same method to expand four multiplied by two π‘₯ minus one. Four multiplied by two π‘₯ is equal to eight π‘₯ and four multiplied by negative one is equal to negative four. This means that the area of the rectangle 𝐡𝐢𝐷𝐺 is eight π‘₯ minus four.

Finally, we need to expand two π‘₯ minus one all squared. Squaring an expression means multiplying it by itself. In this case, we need to multiply two π‘₯ minus one by two π‘₯ minus one. We will look at two methods of expanding double brackets: the FOIL method and the grid method.

The FOIL method involves multiplying the First terms, the Outside terms, the Inside terms, and the Last terms. Multiplying the first terms two π‘₯ and two π‘₯ gives us four π‘₯ squared. Multiplying the outside terms two π‘₯ and negative one gives us negative two π‘₯. Multiplying the inside terms also gives us negative two π‘₯. Finally, multiplying the last terms gives us positive one. Negative one multiplied by negative one is equal to positive one. Remember that when you multiply two negative numbers, you get a positive answer.

We can now group or collect the like terms: negative two π‘₯ minus two π‘₯. This is equal to negative four π‘₯. We can therefore see that two π‘₯ minus one all squared or two π‘₯ minus one multiplied by two π‘₯ minus one is equal to four π‘₯ squared minus four π‘₯ plus one.

An alternative method to expand these brackets is the grid method. Once again, we need to multiply each term in the first bracket by each term in the second bracket. Two π‘₯ multiplied by two π‘₯ is four π‘₯ squared. Negative one multiplied by two π‘₯ is equal to negative two π‘₯. Two π‘₯ multiplied by negative one is also equal to negative two π‘₯. Negative one multiplied by negative one is equal to positive one. Once again, we have two negative numbers multiplying to give a positive answer. Grouping the two negative two π‘₯s gives us negative four π‘₯. Once again, we have proved that the area of the square is equal to four π‘₯ squared minus four π‘₯ plus one.

The total area of 𝑅 is therefore equal to four π‘₯ squared minus four π‘₯ plus one plus four π‘₯ plus eight plus eight π‘₯ minus four. We now need to collect the like terms. We can group negative four π‘₯, positive four π‘₯, and positive eight π‘₯ and we can also group positive one, positive eight, and negative four. There is only one term with π‘₯ squared. Therefore, our first term is four π‘₯ squared. Negative four π‘₯ plus four π‘₯ is equal to zero. Adding eight π‘₯ to this gives us positive eight π‘₯. One plus eight is equal to nine. Subtracting four gives us five.

This means that the total area of 𝑅 is four π‘₯ squared plus eight π‘₯ plus five as required in the question.

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