geometry help?

Question

how do i get the area of a trapezoid with bases of 24 cm and 10 cm and with legs of 15 cm and 13 cm? please explain step by step.

Answer 1

no sweat, because you are not given a height, but two hypotenuses of 13 and 15. We can solve for height by visual inspection.

To make a trapezoid first draw two horizontal lines of 10 cm and 24 cm.

the top 10cm is the top base and bottom is 24 cm.

Notice that the bottom base must be a total of 24 cm. Therefore chop the bottom base into sections of 10 cm directly below the top base and have the two sides extend by (x) on one side and (14-x) on the other. Because 10 + x + (14 - x) = 24

Should look like:

...............10
........./..---------..\
.13../...|...........|...\15
...../.....|.h........|......\
.../------|---------|-------\
.......x.......10......(14-x)

now to solve for height h, notice there are two right triangles of height h.

hypotenuses 13 and 15, legs h, and x and (14-x)

c2 = a2 + b2
13^2 = x^2 + h^2
15^2 = (14-x)^2 + h^2

since heights are equal =>
13^2 - x^2 = h^2 = 15^2 - (14 - x)^2
169 - x2 = 225 - (196 - 28x +x2)
140 = 28 x

x = 5

now you have trapezoid that looks like this

...............10
........./..---------...\
.13../...|...........|...\.15
...../.....|.h........|......\
.../------|---------|-------\
.......5.......10........9


height is therefore 12 by pythagorean theorem
5^2 + 12^2 = 13^2
9^2 + 12^2 = 15^2

area of trapezoid
= .5(24+10)*12 = 204

This was an exercise of pythagorean theorem and algebra.

Answer 2

from the top of the trapezoid send to send two segments to the 24 cm base. This is h, the height.

The inside of the trapezoid is a rectangle, 10 by h. this leaves 24-10 or 14 cm do distribute to each base of the two triangles that are formed -- one on the left and one on the right.

say the one on the right has height of h and base of s. from the Pythogorean theorm s^2 +h^2=13^2

for the one one the left with have height of h and base of 14-s and by similar reasoning we have

(14-s)^2+h^2=15^2

we have two equations and two unknowns

h^2+s^2=169 (I)
h^2+(14-s)^2=225 (II)
expanding the second
h^2+s^2-28s+196=225

plugging (I) into (II) we have 169-28s+196=225

or s=5
therefore h^2+25=169 ==> h=12

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

Answer 3

Let α and β be the angles at the base of the trapezoid. Then
cosα = - (13^2 - 15^2 - (24 - 10)^2)/(2*15*(24 - 10))
cosα = - (169 - 225 - 196)/420
cosα = - (169 - 225 - 196)/420
cosα = 0.6
sinα = √(1 - 0.36) = 0.8
h = 0.8*15 = 12
A = 6(10 + 24)
A = 204 cm^2

Answer 4

I got 85 sq cm as the final answer.
step 1. draw the trap and label all the sides
step 2. draw in two altitudes forming two little triangles on the left and right sides of the trap
step 3. label this height h, and label one of the little bases of the triangle x. The other is 24 - 10 - x = 14 -x.
step 4. Use a^2 + b^2 = C^2 on the two little triangles to get two equations:
h^2 + x^2 = 13^2 and h^2 + (14-x)^2 = 15^2
step 5. Solve these two equations together by simplifying and subtracting them to get h = 5
step 6. A = (1/2) (5)(10 + 24) = 85

Answer 5

let x+b+y=B=>x+y=14
b=10;B=24
h=trapezoid high
15^2=x^2+h^2 (1)
13^2=y^2+h^2 (2)
(1)-(2)=> (15^2-13^2)=x^2-y^2
(15-13)(15+13)=(x+y)(x-y)
2*28=14*(x-y)=>x-y=4 and x+y=14=>x=9

h^2=15^2-9^2=>h^2=6*24=>h=12

A=(24+10)*12/2=204 cm^2