SOLUTION: In trapezoid ABCD segments AB and CD are parallel. Point P is the intersection of diagonals AC and BD. The area of APAB is 16 square units, and the area of ▲PCD is 25 square unit
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-> SOLUTION: In trapezoid ABCD segments AB and CD are parallel. Point P is the intersection of diagonals AC and BD. The area of APAB is 16 square units, and the area of ▲PCD is 25 square unit
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Question 1203372: In trapezoid ABCD segments AB and CD are parallel. Point P is the intersection of diagonals AC and BD. The area of APAB is 16 square units, and the area of ▲PCD is 25 square units. What is the area of trapezoid ABCD?
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There's a strange typo when you wrote "The area of APAB is 16 square units".
I have a feeling it should read "The area of triangle PAB is 16 square units".
The copy/paste system probably confused the triangle symbol for the letter A. But it doesn't really explain why the triangle symbol shows up next to "PCD".
For many visual learners, such as myself, it helps to draw out a diagram. I often use GeoGebra.
Claim: Triangles PAB and PCD are similar.
How can we prove this claim? Well we just need two pairs of congruent angles.
One pair is angle APB = angle CPD, which are vertical angles.
Vertical angles are always congruent.
Then another pair is that
angle ABP = angle CDP
these are alternate interior angles. Such angles are congruent when we have parallel lines.
We will then use the angle angle (AA) similarity theorem to conclude that triangle PAB is similar to triangle PCD.
The order of the lettering is important so we know how the corresponding angles pair up.
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We are given
area(PAB) = 16
area(PCD) = 25
The ratio of those areas is 16:25
The square root of each part gives 4:5
This operation is valid because we have shown triangle PAB is similar to triangle PCD.
If two linear pieces are in ratio m:n, then their areas are in ratio m^2:n^2. This applies to similar figures only.
So if PB = 4 for instance, then its corresponding paired side PD would be 5 units long.
The thing is that we don't know how long PB is
Let's call it x.
PD must then be (5/4)x units long so PB and PD are in ratio 4:5
This bit of scratch work might help show why that is.
PB : PD
x : (5/4)x
4x : 5x
4 : 5
I multiplied both sides by 4 to get rid of the fraction.
Rotate the diagram so that segment BP is completely horizontal.
Triangles PAB and PAD have base PB and PD respectively.
Both triangles PAB and PAD have the same height h.
Whatever triangle PAB's area is, we multiply by 5/4 to get the area of triangle PAD.
But we were given PAB to be 16 square units.
area(PAD) = (5/4)*area(PAB)
area(PAD) = (5/4)*16
area(PAD) = 20
Follow a similar set of steps to conclude that
area(PCB) = (5/4)*area(PAB)
area(PCB) = (5/4)*16
area(PCB) = 20
We have all the pieces we need to find the area of the trapezoid
area(ABCD) = area(PAB) + area(PCD) + area(PCB) + area(PAD)
area(ABCD) = 16 + 25 + 20 + 20
area(ABCD) = 81
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