### Video Transcript

π΄πΆπ·πΉ is a rectangle. π΄π΅πΈπΉ is a square. The diagonal π΄π· intersects π΅πΈ
at πΊ. The area of triangle π΄π΅πΊ is five
centimetres squared. The area of triangle π΅πΆπΊ is 15
centimetres squared. Find the area of π΄πΆπ·πΉ.

Well, it says that π΄πΆπ·πΉ is a
rectangle. So we know that weβve got four
right angles, and opposite sides are equal in length. And π΄π΅πΈπΉ is a square. So again, weβve got four right
angles, and all of the sides are the same length. Also, because π΄π΅πΆ is a straight
line and π΄π΅πΈ is a right angle, we know that πΆπ΅πΊ is also a right angle.

Now we can also see that lines
π΄πΉ, π΅πΈ, and πΆπ· are parallel, as are lines π΄π΅πΆ and πΉπΈπ·. The question also tells us the area
of triangle π΄π΅πΊ is five square centimetres and the area of triangle π΅πΆπΊ is 15
square centimetres. So weβve 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 π΄π΅ the base of
triangle π΄π΅πΊ and π΅πΊ the perpendicular height, the area of triangle π΄π΅πΊ can
also be written as a half times the length π΄π΅ times the length of π΅πΊ. And of course, we know that thatβs
five square centimetres. And for triangle π΅πΆπΊ, weβll take
π΅πΆ as the base and again π΅πΊ as the perpendicular height. So weβve got a half times length
π΅πΆ times length π΅πΊ is equal to 15 square centimetres in this case.

Now if we look at these two, we can
see that theyβre both a half times something, and they both got π΅πΊ as the
perpendicular height. So the only thing that differs is
the length of the base, π΄π΅ and π΅πΆ. 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 π΅πΆ must be
three times bigger than π΄π΅ in order for that to work.

Now just to make our writing a
little bit more efficient, weβre going to define a new variable called π, which is
the length of π΄π΅. Now you donβt have to do this, but
it does make the writing little bit easier. So on our diagram, this distance
here is π, which means that this distance here between π΅ and πΆ is three times
that, three π.

Now remember, π΄π΅πΈπΉ was a
square, so π΄πΉ has got length π and πΉπΈ has also got length π. And because π΄πΆπ·πΉ is a
rectangle, that means that πΆπ· must have length π and πΈπ· must have length three
π as well. And the overall length of that
rectangle from πΉ to π· or π΄ to πΆ is π plus three π, which is four π.

Now letβs think about triangles
π΄π΅πΊ and π΄πΆπ·. Well, theyβre both right-angled
triangles. And this angle here is this angle
here and this angle here, so theyβve 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 π΄π΅πΊ
and π΄πΆπ· 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 π΄π΅πΊ are a quarter of the length of the corresponding sides in
π΄πΆπ· because length π΄π΅ is a quarter of the length of π΄πΆ. And this means that length π΅πΊ is
a quarter of the length of πΆπ·. And since we know that length πΆπ·
is equal to this letter π that we defined, we know that π΅πΊ has a length π over
four, or a quarter π.

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

Now itβs probably easier to think
of π as being π over one in this case, so weβve got three fractions multiplied
together. So on the numerator, Iβve got one
times π times π, which is π squared. And on the denominator, Iβve 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 squared is
equal to 40. And that means that π is equal to
the square root of 40. Well, it could be the positive or
negative version of that, mathematically speaking. But because weβre dealing with
length, weβre not interested in the negative version.

But before we worry about trying to
work out what the square root of 40 is, letβs think what weβre trying to
achieve. Weβre trying to find the area of
rectangle π΄πΆπ·πΉ. And to do that, weβre gonna do the
base times the height.

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

So Iβve 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.