Looking at a possible inconsistency in the areas of two right trapezoids forming a larger one

Question

Could anybody explain to me why the difference arises below?

• Area of right trapezoid for the total figure, i.e. $AEFD$, is $131978$
• For trapezoid $ABCD$ the area is $62196$
• For trapezoid $BEFC$ the area is $57630$

The sum of areas for $ABCD$ and $BEFC$ should be equal to that of $AEFD$, but clearly that's not the case here. Feel free to correct me if I'm wrong, but if not, where does the difference in values come from?

Answer 1

So, on the assumption that these are two trapezoids stacked on top of each other (with vertical bases and horizontal heights), your calculations -- yes, all three -- are correct.

The inconsistency lies in these not being trapezoids, or at length right trapezoids.

Consider: if we draw some blue lines to demarcate right triangles, and find the remaining leg of the triangles, we get this:

We have two triangles:

• Legs $565$ and $118$, and hypotenuse $570$
• Legs $292$ and $104$, and hypotenuse $310$

Apply the Pythagorean theorem, to "verify" each hypotenuse.

• The first would have a hypotenuse of about $577.19$
• The second would have one of about $309.9677$

Neither are quite equal to the claimed hypotenuses, so the only conclusion we can make is that the trapezoids are (just barely) not right trapezoids. The heights we're using are actually on sides that are just barely slanted.

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Addendum: Based on your previous question I would guess the error really arises from your calculation of the values of $292$ and $565$. I hypothesize that you approximated whatever values you obtained, e.g. rounded to the nearest whole number. That's not good enough a lot of the time in mathematics -- you'll have to use the exact value you get. For instance, in that question, you would get $6 \sqrt{2369} \approx 292.03$ -- it is not equal to $292$, but just very close. That barest difference, even if seemingly insignificant, is what caused the issue here.

Answer 2

I have subtracted length =43 from AD,BC to get two right angled triangles DGF and another smaller one after drawing a parallel line GF as shown. They should be similar if all is right.

Similarity by trigonometric ratio for bigger right triangle DGF

$$\tan \angle DGF= \dfrac{222}{292+565}\approx 0.2590 $$

and corresponding similarity ratio for smaller right triangle by proportion should be

$$\tan \angle DGF= \dfrac{118}{565}\approx 0.2088$$

which do not tally.

So which side would you like to alter to make the two ratios equal?

HINT:

It is a good practice to draw the trapeziums to scale. If you did that you would find C below line DF and next perhaps you would have helped yourself.

Anyways now push point C up by an amount $x$ , so $ ( 118 \to 118+x ) $ getting the same ratio so that D,C,F are in a line:

$$ \frac{118+x}{565}=\frac{222}{565+292}, \text{ I got }\; x= 28.3594 $$

Make a proportionate sketch on graph paper. Compute the three areas afresh, verify two smaller areas indeed sum up to the big trapezium area. Leave a comment ,if it succeeded or not.

Answer 3

Your figure simply has way too many lengths marked on, and they are not consistent.

If you knew that $ABCD$ and $BEFC$ are both trapezoids, and $ABE$ and $DCF$ are straight lines, then you would only need (say) the lengths $AB, BC, CD, DA$ and $BE$ to work out the others (extend $ABE$ and $DCF$ to meet at a point $G$, and use similar triangles). If you did this with the values you had for those five lengths, the answers you get for $CF$ and $EF$ wouldn't match your diagram. (As a simple check, for a trapezoid subdivided in this way into smaller trapezoids, you should always have $AB/DC=BE/CF$, which is not the case for your diagram.)

In your case, you also "know" that $ABCD$ is a right trapezoid. That means you need even fewer lengths to determine the whole diagram, since from three sides of a right trapezoid you can determine the other. Again, if you check this you will find that the four marked lengths of $ABCD$ are not consistent with it being a right trapezoid.