Trapezoid ABCD, where AB||CD and AB=1,BC=2,CD=4, and AD=3...?

Question

What's the area?
000A--1--B
00/000000\
0300000002
/000000000\
D-----4------C

I just put that pseudo-diagram in case I messed up somehow describing it.

Two people have said 4.7, but neither has said how they got it, which is what I'm looking for.

Answer 1

I would presume you know Hero's formula for calculating area of a triangle whose 3 sides are known :
Area of Triangle =Sqr root (s x (s-a) x (s-b) x (s-c)) where
s=( a+b+c)/2
so you draw a line parallel to AD from B meeting Cd at E.
This is a parallelogram as opposite sides are parallel and equal.
therefore DE = 1, therefore EC = 3 .
Therefore in Triangle BEC,
BE=3, EC=3,BC=2 and, using the above formula ,
Area= 2 x Sqr root of 2.
Using the other area formula of
Area = 0.5 x base x height
base = 3,
we get height of BEC = 4/3 Sqr root 2
and this is equal to the height of the trapezium
so using formula for area of trapezium
Area = 0.5 x (sum of parallel sides) x height
area = 4.713333 units

Answer 2

jazideol has the concepts dead on. He got the correct value for the height of the trapezoid. He used the correct formula for the area of the trapezoid. His calculator must be broken because he got the wromg number for the final calculation of the area.

Here's another way to do it.

Construct perpendiculars from both A and B down to the base CD. Let's label these segments AE and BF These each will be the height of the trapezoid. These also form two right triangles with a rectangle between them.

Let's say each of these heights, AE and BF measures h units in length. Since EF is 1 unit long there are only three more units of line length left to make the bases of the two triangles. Let the base of triangle BCF, CF=x. Then the base of triangle AED, ED=3-x.

We can restate the Pythagorean theorem as "The square of one leg is equal the the square of the hypotenuse minus the square of the second leg." Therefore, in triangle BCF, h^2=2^2-x^2, or h^2=4-x^2. In triangle AED, h^2=3^2-(3-x)^2 or h^2=9-(9-6x+x^2)

Both expressions of x are equal to h^2 so:

4-x^2= 9-(9-6x+x^2)
4-x^2=9-9+6x-x^2

The (-x^2) on each side cancel out and the 9-9 on the right is zero, so this simplifies to 4=6x or x=2/3.

Now we know CF=2/3 and BC=2 , so now we can find the height, h. h^2=2^2-(2/3)^2
h^2=4-4/9
h^2=36/9 - 4/9
h^2=32/9
h=sqrt(32/9)
h=sqrt32/3
h=(4/3)sqrt2

Area of the trapezoid equals (1/2 of the sum of the bases) times height.

A = (1/2)*(1+4)*(4/3)sqrt2

This calculates to be A=4.714 units approximately

Answer 3

The area equals 10*sqrt(2)/3 units or about 4.714 units, but the proof is rather difficult. Try to find the height of the trapezoid, it is equal to 4*sqrt(2)/3

Answer 4

enable segments AC and BD meet at P. in view that AB || CD, Triangles ABP and CDP are resembling one yet another with a facet ratio of (2/5). we can locate 2 numbers x and y such that ----- AP = 2x and BP = 2y. Then CP = 5x and DP = 5y back enable perspective ADP = ? and perspective BCP = ? back evaluate Rt. Triangles ADP and BCP Tan ? = AP/DP = ( 2x/5y ) and Tan ? = BP/CP = ( 2y/5x ) it can be shown that ? CPD = ? CQD + ? QCP + ? QDP => ninety ° = 40 5 ° + ? + ? ………… => ? + ? = 40 5 ° therefore Tan (? + ? ) = a million = => [2x/5y+ 2y/5x ] divided through [a million- 2x/5y 2y/5x ] => (10 x^2+10 y^2)/(25 xy (a million- 4/25 )) = (10 ( x^2+ y^2))/(21 x y) => xy = (10/21) ( x^2+ y^2 )……………………….. be conscious back in Triangle ABP : AB2 = AP2 + BP2 = 4 ( x 2 + y 2) => ( x 2 + y 2) = 4 and x y = 40/21 ultimately in view that portion of ABCD = (a million/2) BD.CD = (a million/2) [ BP + PD ] [ AP + CP ] = (a million/2) (7y.7x) => (40 9/2) (40/21) => one hundred forty/3 …………………… answer ......................... . Sir Duke : Kindly seem into this answer. I will be happy in case you could kindly make comments on it. .

Answer 5

area of the triangle ADC is 1/2 * 4 * 3 = 6
and area of traingle ABC is 1/2 * 1 * 2 = 1
so area of triangle is 6 + 1 = 7 units

Answer 6

4.71405