Video: Pack 1 β€’ Paper 1 β€’ Question 21

Pack 1 β€’ Paper 1 β€’ Question 21

06:14

Video Transcript

𝐴𝐢𝐷𝐹 is a rectangle. 𝐴𝐡𝐸𝐹 is a square. The diagonal 𝐴𝐷 intersects 𝐡𝐸 at 𝐺. The area of triangle 𝐴𝐡𝐺 is five centimetres squared. The area of triangle 𝐡𝐢𝐺 is 15 centimetres squared. Find the area of 𝐴𝐢𝐷𝐹.

Well, it says that 𝐴𝐢𝐷𝐹 is a rectangle. So we know that we’ve got four right angles, and opposite sides are equal in length. And 𝐴𝐡𝐸𝐹 is a square. So again, we’ve got four right angles, and all of the sides are the same length. Also, because 𝐴𝐡𝐢 is a straight line and 𝐴𝐡𝐸 is a right angle, we know that 𝐢𝐡𝐺 is also a right angle.

Now we can also see that lines 𝐴𝐹, 𝐡𝐸, and 𝐢𝐷 are parallel, as are lines 𝐴𝐡𝐢 and 𝐹𝐸𝐷. The question also tells us the area of triangle 𝐴𝐡𝐺 is five square centimetres and the area of triangle 𝐡𝐢𝐺 is 15 square centimetres. So we’ve been given the areas of some triangles.

Now if you remember, the area of a triangle is equal to a half times its base times its perpendicular height. So if we call 𝐴𝐡 the base of triangle 𝐴𝐡𝐺 and 𝐡𝐺 the perpendicular height, the area of triangle 𝐴𝐡𝐺 can also be written as a half times the length 𝐴𝐡 times the length of 𝐡𝐺. And of course, we know that that’s five square centimetres. And for triangle 𝐡𝐢𝐺, we’ll take 𝐡𝐢 as the base and again 𝐡𝐺 as the perpendicular height. So we’ve got a half times length 𝐡𝐢 times length 𝐡𝐺 is equal to 15 square centimetres in this case.

Now if we look at these two, we can see that they’re both a half times something, and they both got 𝐡𝐺 as the perpendicular height. So the only thing that differs is the length of the base, 𝐴𝐡 and 𝐡𝐢. And we can also see that the area of the second triangle is three times bigger than the area of the first triangle. So we can see that 𝐡𝐢 must be three times bigger than 𝐴𝐡 in order for that to work.

Now just to make our writing a little bit more efficient, we’re going to define a new variable called π‘Ž, which is the length of 𝐴𝐡. Now you don’t have to do this, but it does make the writing little bit easier. So on our diagram, this distance here is π‘Ž, which means that this distance here between 𝐡 and 𝐢 is three times that, three π‘Ž.

Now remember, 𝐴𝐡𝐸𝐹 was a square, so 𝐴𝐹 has got length π‘Ž and 𝐹𝐸 has also got length π‘Ž. And because 𝐴𝐢𝐷𝐹 is a rectangle, that means that 𝐢𝐷 must have length π‘Ž and 𝐸𝐷 must have length three π‘Ž as well. And the overall length of that rectangle from 𝐹 to 𝐷 or 𝐴 to 𝐢 is π‘Ž plus three π‘Ž, which is four π‘Ž.

Now let’s think about triangles 𝐴𝐡𝐺 and 𝐴𝐢𝐷. Well, they’re both right-angled triangles. And this angle here is this angle here and this angle here, so they’ve got that angle in common. Now angles in a triangle sum to 180 degrees. So that means that this angle must be the same as this angle, because the other two angles are the same. This means that triangles 𝐴𝐡𝐺 and 𝐴𝐢𝐷 are similar.

In similar triangles, all the sides are in the same ratios, and the sides of one triangle are a simple multiple of the sides of the other triangle. We can see that the corresponding sides in triangle 𝐴𝐡𝐺 are a quarter of the length of the corresponding sides in 𝐴𝐢𝐷 because length 𝐴𝐡 is a quarter of the length of 𝐴𝐢. And this means that length 𝐡𝐺 is a quarter of the length of 𝐢𝐷. And since we know that length 𝐢𝐷 is equal to this letter π‘Ž that we defined, we know that 𝐡𝐺 has a length π‘Ž over four, or a quarter π‘Ž.

Now we know the perpendicular height of triangle 𝐴𝐡𝐺. We can plug that into the formula for the area and find out the value of π‘Ž. And in our case, the area was five. So the area of five is equal to a half times the base length, π‘Ž, times the perpendicular height length, π‘Ž over four.

Now it’s probably easier to think of π‘Ž as being π‘Ž over one in this case, so we’ve got three fractions multiplied together. So on the numerator, I’ve got one times π‘Ž times π‘Ž, which is π‘Ž squared. And on the denominator, I’ve got two times one times four, which is eight.

Now I can multiply both sides of my equation by eight so that the eights cancel on the right-hand side. And that gives me π‘Ž squared is equal to 40. And that means that π‘Ž is equal to the square root of 40. Well, it could be the positive or negative version of that, mathematically speaking. But because we’re dealing with length, we’re not interested in the negative version.

But before we worry about trying to work out what the square root of 40 is, let’s think what we’re trying to achieve. We’re trying to find the area of rectangle 𝐴𝐢𝐷𝐹. And to do that, we’re gonna do the base times the height.

Now the base length here is four π‘Ž, and the height is π‘Ž. So that area, four π‘Ž times π‘Ž, is four π‘Ž squared. But we know that π‘Ž squared is equal to 40, so we can just plug that number in for π‘Ž squared here. And four times 40 is 160. So without forgetting the units, the area of rectangle 𝐴𝐢𝐷𝐹 is 160 square centimetres.

So I’ve ruled there was quite a lot to do in this question. We had to know the properties of rectangles and squares. We had to remember the formula for the area of a triangle. And we also had to recognise some similar triangles and use that information to work out the length of various sides and then use that to work out the area of the final rectangle.

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