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The height of a trapezoid is $10$ cm. The lengths of the two diagonals of the trapezoid are $30$ cm and $50$ cm. Calculate the area of the trapezoid.

On the homework I solved this using $${D_1D_2\over 2}$$ and my teacher marked me wrong. So I don't know what I did wrong. Please help. I know I can only use the formula if the diagonals are $90$ degrees. But how do I check that ?

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  • $\begingroup$ that is the formula of the area of a rhombus. $\endgroup$ Dec 20, 2012 at 5:02

6 Answers 6

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The area will be $\frac12\cdot 10\cdot (y+x+y+z)=5(x+2y+z)$

Now, $(y+z)^2+10^2=50^2$ and $(x+y)^2+10^2=30^2$

$(y+z)=\sqrt{50^2-10^2} CM=20\sqrt6 CM$

$(x+y)=\sqrt{30^2-10^2} CM=20\sqrt2 CM$

SO, the area will be $5(20\sqrt2(\sqrt3+1)) CM^2$

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To help you think about the solution, draw two parallel lines 10 cm apart. The bottom will contain the B1 and the top will contain the B2. Draw the two diagonals at the proper lengths from the base to the top line making sure they cross. Now imagine that you slide one of the diagonals along the bottom and top lines. Notice that (B2+B2)/2 does not change (i.e one stretches and one shrinks). So no matter where the diagonals cross you have the same area. Now slide them apart until the intersection point reaches the top line (i.e. B2 = 0). Now you have triangle with the same area as the trapezoid. You know the height and two sides of the triangle, a little geometry and you can compute the base and your're there.

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Gluing two copies of the trapezoid together with one rotated by 180 degrees can yield a parallelogram. One diagonal from each trapezoid cuts the parallelogram into two triangles, each of which has an altitude of 10 and sides adjacent to that altitude's vertex of lengths 30 and 50.

So the area is the same as that of a triangle with two sides of 30 and 50, and an altitude between them of 10. Is this enough of a help?

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  • $\begingroup$ Trying to understand what you said... Is the answer 193.1 cm? $\endgroup$
    – NGPP1
    Dec 20, 2012 at 0:35
  • $\begingroup$ This is the right answer, but I don't think your gluing construction works. $\endgroup$
    – Peter Shor
    Dec 20, 2012 at 0:36
  • $\begingroup$ I am still confused $\endgroup$
    – NGPP1
    Dec 20, 2012 at 0:42
  • $\begingroup$ @PeterShor, in my method, the area is around $386.37 cm^2$ $\endgroup$ Dec 20, 2012 at 5:03
  • $\begingroup$ @PeterShor I have added imagery that demonstrates the gluing construction works. I leave the labeling of edges and confirmation to you. The trapezoid's area is the same as that of the triangle at the end of the process: a triangle with altitude 10 and two surrounding sides of length 30 and 50. $\endgroup$
    – 2'5 9'2
    Dec 20, 2012 at 10:05
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Since you can drag base 1 left or right and not change the area then if you drag the smaller base until it forms right angles with base 2 you get a trapezoid made up of one square and one triangle. So we know base 1 = size of smaller base = one side of the square. We know that hypotenuse = size of smaller diagonal. Since the corner is a right angle we can calculate height as the square root of the (sum of smaller diagonal squared minus smaller base squared) then from there we know area of trapezoid = midsegment length times height so average the 2 known bases and multiply by the calculated height and you have the area of a trapezoid given only the length of the bases and the length of the diagonals.

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The theorem you need is that

the area of a trapezium is half that of the bounding parallelogram.

Consider a trapezium $ABCD.$ Its bounding parallelogram is defined as follows. Look at the diagonal $AC,$ then from $C$ draw a line parallel to the other diagonal $BD$ to meet $AB$ extended at $B'.$ Similarly, draw a line from $A$ to meet $CD$ extended at $D'.$ Then by construction $AB'CD'$ is a parallelogram and is referred to as the bounding parallelogram of the trapezium $ABCD.$ I shall leave it to you to convince yourself that the bounding parallelogram is uniquely defined (up to congruence, that is). You only need repeat the above construction for the other diagonal and prove congruence, or use corresponding transformations.

Now, having explained the term bounding parallelogram, let us prove the above theorem. First I shall use the notation $[XYZ],$ for example, to mean the area of the simple polygon $XYZ.$ Then we have that $$[ABDD']+[BB'CD]=[AB'CD'].$$ But $[ABCD]=[ABD]+[BCD].$ However, $[ABDD']=2[ABD]$ and $[BB'CD]=2[BCD],$ from which the result follows.

Now applying it to your case gives the area of trapezium $ABCD$ as $[AD'C],$ with $|AC|=30$ and $|BD|=|AD'|=50.$ Also, let $\angle BDC=\angle AD'C=\phi$ and $\angle ACD=\theta.$ If $M$ is the point where the altitude from $A$ intersects $D'C,$ then we have $$|D'C|=|D'M|+|MC|=|D'A|\cos \phi + |AC|\cos \theta=50 \cos \phi + 30 \cos \theta.$$ Also, $$10=|AM|=|D'A|\sin \phi = |AC|\sin \theta=50 \sin \phi = 30 \sin \theta,$$ giving $\sin \phi = 1/5$ and $\sin \theta = 1/3,$ so that $\cos \phi = \sqrt{24}/5$ and $\cos \theta = \sqrt{8}/3,$ and finally $|D'C|=10(\sqrt{24} + \sqrt{8}),$ so that $$[ABCD]=\frac 1 2\cdot 10\cdot 10(2\sqrt{6} + 2\sqrt{2})=100(\sqrt{6} + \sqrt{2}).$$

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the formula $\frac{d_1\cdot d_2}2$ is the area of the special quadrilateral called rhombus but not the trapezium..

If you want to find the area of the trapezium the like the above mentioned using the Pythogoras theorem. Find the sides and form an equation and with that solve the linear equation then you can solve the area of the trapezium.. finally you will get

area$=\frac{h(a+b)}2$

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