area of trapezium

Question

A trapezium has perpendicular diagonals and altitude 4. If one of the diagonals has length 5, find the area of the trapezium

A 12

B 50/3

C 25/2

D 40/5

I guess the answer is C. I divided the trapezium into two triangles and then it is 5(x+y)/2, in which x+y=length of the diagonal, which is 5, so it's 5(5)/2.

But what if it is a irregular trapezium? The diagonals will not be the same.

I didn't use the altitude so I think my answer is wrong. What should be the answer?

First, we need to know the nationality of this shape. Is it a British trapezium (US trapezoid), or a US trapezium (irregular quadrilateral). — Shaun Ault, May 04 '13 at 14:01

score 3 · Accepted Answer · answered May 04 '13 at 14:10

3

Trapezium

$AH=4$, $AC=5$ $\rightarrow$ $HC=3$.

$\displaystyle \frac{AB}{AO}=\frac{DC}{OC}=\frac{AC}{HC}=\frac{5}{3}$.

$\displaystyle AB+DC=(AO+OC)\cdot \frac{5}{3} = \frac{25}{3}$.

$\displaystyle S_{ABCD}=AH\cdot \frac{AB+DC}{2}=4\cdot \frac{25}{6} = \frac{50}{3}$.

answered May 04 '13 at 14:10

Oleg567

17,295

AB+DC=(AO+OC)⋅5/3 ??? How come? – zzz May 04 '13 at 14:14
i understand now... – zzz May 04 '13 at 14:41

area of trapezium

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