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Find the area of a trapezoid with bases 10 and 24 and legs o

## Find the area of a trapezoid with bases 10 and 24 and legs o

[From: Mathematics] [author: ] [Date: 06-15] [Hit: ]
Find the area of a trapezoid with bases 10 and 24 and legs of 13 and 15. Thats my problem.?Find the area of a trapezoid with bases 10 and 24 and legs of 13 and 15.......

Find the area of a trapezoid with bases 10 and 24 and legs of 13 and 15. Thats my problem.?
Find the area of a trapezoid with bases 10 and 24 and legs of 13 and 15.
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answers:
Philomel say: Do you was the surface area or the volume? If it's the surface area do you want to include the base?
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Puzzling say: Drop a couple of perpendiculars from the top base as shown below. They will form a pair of right triangles.

Just by inspection and knowledge of my Pythagorean Triples, I noticed that the left (red) triangle would work as a 5-12-13 triangle and the right (blue) triangle as a 9-12-15 (3:4:5 ratio) triangle. So the height is 12 units and it's easy to calculate the area from there.

However, let's assume you didn't notice that and wanted to figure it out yourself. First we know the bottom base is 24 units long so:
a + b + 10 = 24
a + b = 14

We can also use the Pythagorean Theorem to write equations for the sides of the two triangles:
a² + h² = 13²
b² + h² = 15²

We have three equations and three unknowns and you just need a little algebra to solve for h.
a + b = 14 . . . . . [eq. 1]
a² + h² = 13² . . . [eq. 2]
b² + h² = 15² . . . [eq. 3]

Let's start by solving for 'a' in the first equation:
a = 14 - b

Substitute that into the equation for the red triangle [eq. 2]:
(14 - b)² + h² = 169

Let's solve the bottom equation [eq. 3] in terms of h²
h² = 225 - b²

Substitute that in for h² so our only remaining variable is b:
(14 - b)² + 225 - b² = 169

Expand:
14² - 28b + b² - b² = 169 - 225

The b² terms cancel out:
196 - 28b = -56

Get like terms together:
-28b = -252

Change the sign on both sides:
28b = 252

Divide both sides by 28:
b = 252/28
b = 9

From there we can find h:
h² = 225 - 9²
h² = 225 - 81
h² = 144
h = √144 (we can ignore the negative square root since these are positive lengths)
h = 12

Now you simply use the formula for the area of a trapezoid which is the average of the bases times the height.
A = ½(b₁ + b₂)h
A = ½(10 + 24) * 12
A = ½(34) * 12
A = 17 * 12
A = 204 sq. units

Alternatively, you could use the area of the two triangles and the rectangle:
A = ½(5 * 12) + 10(12) + ½(9 * 12)
A = 30 + 120 + 54
A = 204 sq. units
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Φ² = Φ+1 say: The sum of infinite diminishing parallel sides (limiting to a triangle where the extensions of the two non-parallel opposite sides) is 24 + 10 + 25/6 + ... = 24(12/7)
So the other sides of this large triangle are 13(12/7) and 15(12/7).
Using Heron's formula, the area of the triangle with sides 24, 156/7, 180/7 is 1728/7
The area of the trapezoid must then be (1728/7)(1 - (5/12)²) which is 204 square units.
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say: could you explain or put a photo my friend?
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Pope say: Use a sketch to gather your thoughts. Let ABCD be the trapezoid.

AB = 24
BC = 13
CD = 10
DA = 15
AB || DC

You have both bases, so the height would settle the question.

Construct point E on AB, such that DE || CB. That makes BCDE a parallelogram.
BE = CD = 10.
DE = CB = 13
AE = AB - EB = 14

You now have all sides of ∆AED. The altitude from D is equal to the height of the trapezoid.
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Kairaa say: The area is: 204
your height would be 12
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