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

π΄, π΅, and πΆ are three points plotted on a grid. a) Plot the point π· on the grid such that π΄π΅πΆπ· is a square. b) The point πΈ is the midpoint of π΄π΅. Circle the two names for triangle π΅πΆπΈ. [A] Right-angled [B] Scalene [C] Isosceles [D] Equilateral c) Circle the ratio which is equivalent to the βarea of triangle π΅πΆπΈ : area of square π΄π΅πΆπ·β. [A] 1 : 2 [B] 1 : 3 [C] 1 : 4 [D] 1 : 8

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Video Transcript

π΄, π΅, and πΆ are three points plotted on a grid. There are three parts to this question. A) Plot the point π· on the grid such that π΄π΅πΆπ· is a square. b) The point πΈ is the midpoint of π΄π΅. Circle the two names for triangle π΅πΆπΈ: right-angled, scalene, isosceles, equilateral. Part c) Circle the ratio which is equivalent to the area of triangle π΅πΆπΈ to the area of square π΄π΅πΆπ·. 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. π΄ has coordinates seven, eight. π΅ is nine, three. And πΆ is four, one. Weβre told that plotting point π· on the grid creates a square, π΄π΅πΆπ·. As all four sides of a square are equal in length, the length of π΅π΄ and π΅πΆ will be equal to πΆπ· and π·π΄. Line π΄π΅ will also be parallel to line πΆπ·, and line π΅πΆ will be parallel to line π΄π·.

Letβs first consider how we get from point π΅ to point π΄. To get from point π΅ to point π΄, we move five squares up and two squares to the left. This is because the difference in the π¦-coordinates is five, and the difference between the π₯-coordinates is two. As line π΅π΄ is parallel to line πΆπ·, we must also move five squares up and two squares to the left to get from πΆ to π·. This means that point π· will have coordinates two, six.

Completing the square, we can now see that π΅π΄ is parallel to πΆπ· and π΄π· is parallel to π΅πΆ. All four sides of the shape π΄π΅πΆπ· are equal. Therefore, it is a square. The point π· on the grid such that π΄π΅πΆπ· is a square has coordinates two, six.

The second part of our question tells us that point πΈ is the midpoint of π΄π΅. To work out where the midpoint of any two coordinates is, we add the two coordinates and divide by two. The π₯-coordinates of π΄ and π΅ 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 πΈ is eight. The π¦-coordinates of π΄ and π΅ 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 π¦-coordinate of πΈ is 5.5. The point πΈ, which is the midpoint of π΄π΅, has coordinates eight, 5.5.

We were asked to consider the two names for triangle π΅πΆπΈ as shown in pink. As the angle at π΅ is the angle within the square, we know it is right-angled. Therefore, one name for triangle π΅πΆπΈ 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 π΄π΅πΆπ· was a square, we know that the length π΅πΆ is equal to the length π΅π΄. This means that the length π΅πΈ is half of the length π΅πΆ. 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 π΅πΆπΈ are different lengths. Therefore, the triangle is scalene. The two names for triangle π΅πΆπΈ are right-angled and scalene.

The third part of the question asked us to consider the ratio of the area of the triangle π΅πΆπΈ to the area of the square π΄π΅πΆπ·. Letβs first let the length of each side of the square be letter π₯. This means that the length π΅πΈ will be a half π₯, as πΈ was the midpoint of π΄π΅. The area of the square π΄π΅πΆπ· 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 π΅πΆπΈ 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 π΅πΆπΈ 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 π΅πΆπΈ are the same area as the square π΄π΅πΆπ·. This means that the ratio of the triangle to the square is one to four. The ratio that is equivalent to area of triangle π΅πΆπΈ to area of square π΄π΅πΆπ· is one to four.