SOLUTION: Solve for the area of the composite figure. https://ibb.co/cXV52NVm If an isosceles trapezoid CDFG was added below, where the height was the same as the triangle above and the ba

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Question 1210163: Solve for the area of the composite figure.
https://ibb.co/cXV52NVm
If an isosceles trapezoid CDFG was added below, where the height was the same as the triangle above and the bases had a length of 12 and 24, then what is the area of the new composite figure?
https://ibb.co/jZWQ628k
Found 3 solutions by greenestamps, ArschlochGeometrie, mccravyedwin:
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


The figure APPEARS to be a kite plus a triangle. But there is nothing in either your written question or on the referenced figure that tells us so.

We don't even know whether the two segments that APPEAR to be perpendicular to each other in fact are perpendicular.

Without a description of the figure in your text or on the figure, we would only be guessing at the answer.

Re-post, defining the problem clearly.

------------------------------------------------------------------------

Tutor Edwin has solved the problem assuming that the part of the figure that looks like a kite is in fact a kite. Before the trapezoid was added, the area of the composite figure was 132.

When the trapezoid is added, Edwin uses a clever way of finding the area of the trapezoid by dividing it into 3 congruent triangles.

The area of the trapezoid is found more easily using the standard formula for the area of a trapezoid: height times average of the bases. In this problem that is



And then the total area of the new composite figure is 132+144 = 276.
Answer by ArschlochGeometrie(3)   (Show Source): You can put this solution on YOUR website!

Answer by mccravyedwin(407)   (Show Source): You can put this solution on YOUR website!

The figure on the site is not to scale.  Here it is drawn to scale.
I will assume a kite ABCD on the left, although I agree with 
greenestamps that nothing tells us that's the case.



Triangle DOA is a 45-45-90 triangle so. DO = AO = 6. Its area
is 
Triangle BOA is congruent to triangle DOA so it also has area 18.

You may recognize triangle DOC as a 6-8-10 right triangle and see
right off that OC=8.  If not, use the Pythagorean theorem,


Triangle DOC's area is 
Triangle BOC is congruent to triangle DOC so it also has area 24. 

So the area of kite ABCD is 18+18+24+24 = 84

Triangle BCE's height (the green line) is the same length as OC = 8.
so its area is 

So the kite's area plus triangle BCE's area is 48.

So the area of the composite figure is 84 + 48 = 132 square units.

-----------------------------
For the other problem, they have added on an isosceles trapezoid on the right,
I use different lettering, as they didn't put any lettering on the figures.



Let's draw in a couple of red lines and you will see what has been added:



Now you see that they have added on three more triangles all congruent to
triangle BCE.  We found triangle BCE's area to be 48, so we just add (3)(48)=144
to the area in the first problem, which was 132.

So the area of this new composite figure is 132+144 = 276 square units. 

Edwin
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