Diagonals at right angle

In the trapezoid ABCD, this is given:
AB=12cm
CD=4cm
And diagonals crossed under a right angle. What is the area of this trapezoid ABCD?

Correct answer:

S =  64 cm²

Step-by-step explanation:

$$\begin{gathered} a=12\text{ cm } \\ c=4\text{ cm } \\ a:c=h_1:h_2 \\ h=h_1+h_2 \\ {h}_1=a\mathrm{/}2=12\mathrm{/}2=6\text{ cm } \\ {h}_2=c\mathrm{/}2=4\mathrm{/}2=2\text{ cm } \\ h=h_1+h_2=6+2=8\text{ cm } \\ S={\displaystyle \frac{a+c}{2}}\cdot h={\displaystyle \frac{12+4}{2}}\cdot 8=64{\text{cm}}^2 \end{gathered}$$

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Showing 1 comment:

Math student

I don't think the solution is correct.  The height is the radius if and only if the trapezoid is symmetrical.  But it doesn't have to be.  In fact, take a line segment of length 12 and start a 20 degree angle from one side and 70 from the other (for the diagonals). Intersect them and keep going until there is 4 long parallel line. Then increase the 20 degree angle - the height will increase and the area too.

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