Area of Trapezium? Easy, or is it?

Drop perpendiculars from $B$ and $A$ to intersect $CD$ at $E$ and $F$ , respectively. Let $CD = a$ , $FD = b$ , and $BE = h$ . Since $BA = EF = 10$ , so $24 = CD = CE+EF+FD = a+10+b,$ $(1)\ a+b = 14.$ By Pythagorean theorem and using (1), we have $BC^2-CE^2 = BE^2 = AF^2 = AD^2-FD^2,$ $13^2-a^2 = h^2 = 15^2-b^2,$ $b^2-a^2 = 15^2-13^2,$ $(b-a)(a+b) = (15-13)(15+13)$ $14(b-a) = 56,$ $(2)\ b-a = 4.$ Adding and subtracting (1) and (2), we have $2b = 18$ and $2a = 10,$ or $b = 9$ and $a = 5$ . Thus $BC^2-CE^2 = BE^2,$ $13^2-a^2 = h^2,$ $h = 12.$ And it follows that the area of the trapezoid is $\frac{(AB+CD)\times BE}{2} = \frac{(10+24)\times 12}{2} = 204.$

Drop altitudes from B to E and A to F, both on CD. Length is h.

This creates two right triangles, BCE and ADF.

Let CE = x and DF = y

13^2 = x^2 + h^2 and 15^2 = y^2 + h^2 and x + y = 14 ===> y = 14 - x

Substituting for y and solving each equation for h^2 we get:

169 - x^2 = 225 - x^2 + 28x - 196

Solving: 140 = 28x =====> x = 5

5^2 + h^2 = 13^2 = 169

h^2 = 144 ====> h = 12

Area = 0.5 (b1 +b2) h = .5 (10+24) 12 = 204 in^2

Since the quadrilateral is a trapezoid, the area is equal to (a+b)/2 times h, where a and b are the parallel sides and h the distance between them. To find the value of h, one can use the formula h=sqr((-a+b+c+d)(a-b+c+d) (a-b+c-d) (a-b-c+d))/2 times abs(b-a). In this case it reduces to sqr((42) times (14) times (-16) times (-12))/28 or 336/28 which equals 12. Multiply by (24+10)/2 or 17, and you get the answer of 204.

The figure is a trapezoid with area = (h/2)(10 + 24), where h is the height equal to the perpendiculars drawn from B to CD and A to CD.. If the perpendicular from B meets CD at point P, let PC = x. Then at the other end of CD if the perpendicular from A meets CD at Q, QD = 14 -x. Using the Pythagorean theorem twice, we have the equations: x^2 + h^2 = 169 and h^2 + (14 - x)^2 = 225. These are easily solved for x = 5, h= 12, and the area = (12/2)(10 + 24) = 204. Ed Gray

Divide the trapezium into 2 triangles by drawing the line segment BD, call this distance x. By alternating angles, angle ABD = angle BDC. Call these angles theta. Apply the cosine rule to the triangles ABD and BDC creating simultaneous equations in terms of x and theta. These solve to give x = root 505 and cos theta = 19/root505. Use sin^2+cos^2=1 to obtain sin theta. Apply 1/2 abSinC to each triangle and sum to give required result.