Video: AQA GCSE Mathematics Foundation Tier Pack 2 β€’ Paper 3 β€’ Question 8

𝐴, 𝐡, and 𝐢 are three points plotted on a grid. a) Plot the point 𝐷 on the grid such that 𝐴𝐡𝐢𝐷 is a square. b) The point 𝐸 is the midpoint of 𝐴𝐡. Circle the two names for triangle 𝐡𝐢𝐸. [A] Right-angled [B] Scalene [C] Isosceles [D] Equilateral c) Circle the ratio which is equivalent to the β€œarea of triangle 𝐡𝐢𝐸 : area of square 𝐴𝐡𝐢𝐷”. [A] 1 : 2 [B] 1 : 3 [C] 1 : 4 [D] 1 : 8

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Video Transcript

𝐴, 𝐡, and 𝐢 are three points plotted on a grid. There are three parts to this question. A) Plot the point 𝐷 on the grid such that 𝐴𝐡𝐢𝐷 is a square. b) The point 𝐸 is the midpoint of 𝐴𝐡. Circle the two names for triangle 𝐡𝐢𝐸: right-angled, scalene, isosceles, equilateral. Part c) Circle the ratio which is equivalent to the area of triangle 𝐡𝐢𝐸 to the area of square 𝐴𝐡𝐢𝐷. Is it one to two, one to three, one to four, or one to eight?

The coordinates of the three points already on the grid are as follows. 𝐴 has coordinates seven, eight. 𝐡 is nine, three. And 𝐢 is four, one. We’re told that plotting point 𝐷 on the grid creates a square, 𝐴𝐡𝐢𝐷. As all four sides of a square are equal in length, the length of 𝐡𝐴 and 𝐡𝐢 will be equal to 𝐢𝐷 and 𝐷𝐴. Line 𝐴𝐡 will also be parallel to line 𝐢𝐷, and line 𝐡𝐢 will be parallel to line 𝐴𝐷.

Let’s first consider how we get from point 𝐡 to point 𝐴. To get from point 𝐡 to point 𝐴, we move five squares up and two squares to the left. This is because the difference in the 𝑦-coordinates is five, and the difference between the π‘₯-coordinates is two. As line 𝐡𝐴 is parallel to line 𝐢𝐷, we must also move five squares up and two squares to the left to get from 𝐢 to 𝐷. This means that point 𝐷 will have coordinates two, six.

Completing the square, we can now see that 𝐡𝐴 is parallel to 𝐢𝐷 and 𝐴𝐷 is parallel to 𝐡𝐢. All four sides of the shape 𝐴𝐡𝐢𝐷 are equal. Therefore, it is a square. The point 𝐷 on the grid such that 𝐴𝐡𝐢𝐷 is a square has coordinates two, six.

The second part of our question tells us that point 𝐸 is the midpoint of 𝐴𝐡. To work out where the midpoint of any two coordinates is, we add the two coordinates and divide by two. The π‘₯-coordinates of 𝐴 and 𝐡 are seven and nine. Seven plus nine is equal to 16. Dividing this by two gives us eight. This means that the π‘₯-coordinate of point 𝐸 is eight. The 𝑦-coordinates of 𝐴 and 𝐡 are eight and three. We add these and divide by two. Eight plus three is equal to 11. Dividing this by two gives us 5.5. This means that the 𝑦-coordinate of 𝐸 is 5.5. The point 𝐸, which is the midpoint of 𝐴𝐡, has coordinates eight, 5.5.

We were asked to consider the two names for triangle 𝐡𝐢𝐸 as shown in pink. As the angle at 𝐡 is the angle within the square, we know it is right-angled. Therefore, one name for triangle 𝐡𝐢𝐸 is right-angled. The other three types of triangles have the following properties. An equilateral triangle has three equal sides and three equal angles. An isosceles triangle has two equal sides and two equal angles. A scalene triangle has no equal sides and no equal angles.

As the initial shape 𝐴𝐡𝐢𝐷 was a square, we know that the length 𝐡𝐢 is equal to the length 𝐡𝐴. This means that the length 𝐡𝐸 is half of the length 𝐡𝐢. This means we can rule out equilateral triangle. The longest side of a right-angled triangle is known as the hypotenuse. This is longer than both of the other sides. Therefore, we can rule out isosceles. All three sides of the triangle 𝐡𝐢𝐸 are different lengths. Therefore, the triangle is scalene. The two names for triangle 𝐡𝐢𝐸 are right-angled and scalene.

The third part of the question asked us to consider the ratio of the area of the triangle 𝐡𝐢𝐸 to the area of the square 𝐴𝐡𝐢𝐷. Let’s first let the length of each side of the square be letter π‘₯. This means that the length 𝐡𝐸 will be a half π‘₯, as 𝐸 was the midpoint of 𝐴𝐡. The area of the square 𝐴𝐡𝐢𝐷 will therefore be equal to π‘₯ squared, as the area of any square is equal to one length squared.

The area of any triangle is equal to a half multiplied by the base multiplied by the height. In this case, the area of triangle 𝐡𝐢𝐸 is equal to a half multiplied by π‘₯ multiplied by a half π‘₯. One-half multiplied by one-half is one-quarter, and π‘₯ multiplied by π‘₯ is equal to π‘₯ squared. Therefore, the area of the triangle 𝐡𝐢𝐸 is equal to one-quarter π‘₯ squared. This means that the area of the triangle is a quarter of the area of the square.

We can show this on the grid by splitting the square into four identical triangles. Four triangles the same size as 𝐡𝐢𝐸 are the same area as the square 𝐴𝐡𝐢𝐷. This means that the ratio of the triangle to the square is one to four. The ratio that is equivalent to area of triangle 𝐡𝐢𝐸 to area of square 𝐴𝐡𝐢𝐷 is one to four.

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