Video: AQA GCSE Mathematics Higher Tier Pack 4 • Paper 1 • Question 9

Henrieta is solving a problem. Cube 1 has side lengths of 9 cm. Cube 2 has a surface area 3 times larger than that of cube 1. What is the surface area of cube 2? In order to solve the problem, Henrieta performs the following steps: Multiplying the side length of cube 1 by 3. Squaring the answer from the previous step. Multiplying this answer by 6 to find the surface area. Comment on Henrieta’s method.

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Video Transcript

Henrieta is solving a problem. Cube one has side lengths of nine centimeters. Cube two has a surface area three times larger than that of cube one. What is the surface area of cube two? In order to solve the problem, Henrieta performs the following steps: multiplying the side length of cube one by three, squaring the answer from the previous step, multiplying this answer by six to find the surface area. Comment on Henrieta’s method.

Let’s walk through what Henrieta did. She multiplied the side length of cube one by three. Nine times three equals 27. She squared the value from the previous step. 27 squared equals 729. She then multiplied that by six which equals 4374.

Henrieta is claiming that the surface area of cube two is 4374. We know that cube one has a side length of nine centimeters. If we want to find the surface area of cube one, we take its side length, square it, and then multiply by six: in this case, nine squared times six. And when we do that, we find that the surface area of cube one equals 486 centimeters squared.

We also know that cube two has a surface area three times larger than cube one. The surface area of cube two equals three times the surface area of cube one. We know the surface area of cube one is 486. Therefore, the surface area of cube two equals 1458 centimeters squared.

This means that Henrieta’s answer is not correct. Using her method, we found 4374 to be the surface area of cube two. But what went wrong? Why was her method incorrect?

Henrieta’s method is not correct. In her first step, she multiplied the side length of cube one by three. The surface area of cube two is three times larger, not the side length. The side length of cube two is actually the square root of three times larger than the side length of cube one. If we square the square root of three, we find the area of each face is three times larger.

Henrieta’s method could be corrected by multiplying the side length of cube one by the square root of three instead of by three.

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