geometry problem help!! please help!?

Question

i have a trapezoid and the base1 is 10cm, base2 is 24 cm, the left side connecting the two bases is 13 cm and the right side is 15 cm. but i dont have the height...so i dont know how to find the area!!

(the trapezoid can be broken into a right triangle and a parralelgram, the side that i sed was 15 cm is the hypotnuse of the right triangle)

please help me asap....i have no idea how to do it without the height!!

Answer 1

On the right side we have h^2 + x^2 = 15^2
On the left side we have h^2 + (14-x)^2 =13^2
So 15^2 - x^2 = 13^2 - (14-x)^2 since they both = h^2
225 -x^2 = 169 -196 +28x -x^2
-x^2 + x^2 -28x = 169 - 196-225
-28x = - 252
x=9
So h^2 +9^2 =15^2
h^2=225-81=144
So h= 12
Now area = 12(24+10)/2 = 6* 34 = 204 cm^2

Answer 2

The second base is 14 cm longer than the first. So you have two triangles formed by dropping perpendiculars down from the top base to the bottom base.

Let the bases of these triangles be x and y. Then

x + y = 14
y = 14 - x

h = height of trapezoid

The two sides of the trapezoid are the hypotenuses of the triangles. So we have:

h² = 13² - x²
h² = 15² - y² = 15² - (14 - x)²

Setting them equal we have:

13² - x² = 15² - (14 - x)²
169 - x² = 225 - 196 + 28x - x²
169 = 29 + 28x
140 = 28x
5 = x

h² = 13² - x² = 13² - 5²
h² = 169 - 25 = 144
h = 12

So the area A, of the trapezoid is:

A = (1/2)(10 + 24)(12) = 204 cm²

Answer 3

Subtract base 1, 10, from base 2, 24, you get 14. Can you split the 14 up so that the 2 parts are the bases of right triangles with the same heights, one with a hypotenuse of 13 and one with 15? Yes. The 2 right triangles you should know best are the 3-4-5 (and its multiples) and the 5-12-13. Suppose the right triangle on the left is a 5-12-13. Then the one on the right would have a base of 9 (9+5=14), and it would be a 9-12-15. So the height of your trapezoid is 12, which makes its area 6(34) = 204 cm².

Answer 4

The only way that the figure can be broken into a right triangle and a parallelogram is if the 13 side is perpendicular to both bases. That makes the parallelogram a rectangle, so the height is 13. Now you can figure the bottom side of the right triangle as 14. But that does not give a consistent solution for the triangle -- Pythagoras is violated -- so the problem as stated has no solution.

Answer 5

From your description of a right triangle and a parallelogram, in order for this to work, the parallelogram would have to be a rectangle, and then the height would be 13. But this does not add up to a right triangle.
However, if there are two right triangles, attached to either side of the rectangle, that could work...
Looks like several people got it, but not by solving for the figure as you described. Instead they dissected it into a rectangle and two right triangles. Philo explained this on a more intutive level, several others showed the math, all arrived at a height of 12. I suggest you read over these replies and wade thru the math, doing it yourself, and you'll understand.
Next time you ask sometihng like this, Please try and post a picture somewhere in Photobucket or your 360 page, or at least describe it better. I wasted a lot of work trying to make it work with one triangle and a parallelgoram as you described, it just didn't work.

Answer 6

Who not solve for the height directly?

Let H be the height. By Pythagorean theorem, we have

√(13^2-H^2) + √(15^2-H^2) = 24-10 = 14......(1)

Solve (1) for H with a graphing calculator,
H = 12 cm

area = (1/2)(10+24)(12) = 204 cm^2

Answer 7

h^2 = ((a + b - c + d)(-a + b + c + d)(a - b - c + d)(a + b - c - d))/[4(a - c)^2]

a = 10
b = 13
c = 24
d = 15

h^2 = ((10 + 13 - 24 + 15)(-10 + 13 + 24 + 15)(10 - 13 - 24 + 15)(10 + 13 - 24 - 15))/(4(10 - 24)^2)

h^2 = (14 * 42 * -12 * -16)/(4(-14)^2)
h^2 = (112896)/(4 * 196)
h^2 = (112896)/(784)
h^2 = 144
h = 12

Area = h(a + c)/2
Area = (12(10 + 24))/2
Area = 6(34)
Area = 204cm^2

ANS : 204cm^2

Answer 8

D 15 in case you call "w" the width of the river in yards, a similar triangles supplies the ratio: 10 / 8 = w / 12 w = 10*12 / 8 = 15 the unusual difficulty on the diagram became the handwritten "no similarity" ... huh? the two triangles are comparable on the grounds that they each and each have an attractive attitude and a similar: attitude ABE = attitude CBD which potential the "third angles" must additionally be equivalent in comparable triangles, the ratios of area-lengths are equivalent