Great Problem From Test In India - How To Solve It

Thanks to Himanshu from India for sending me this problem! It was part of a math competition where you had about 1 minute to solve each problem (I admit I took much longer). Can you figure out the area of the folded rectangle?

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[r.d.w]

Once you have the 30cm2 from the congruent triangles, note that the original area (4X16) has had it's area which is greater than the 30 cut in half. Therefore 64-30 = 34, so 17cm2 was lost in the doubled up area, so 30+17 = 47cm2 without ever having to solve the area of the overlapping portion.

[NestorAbad]

Another way to calculate the distance AN is the following: the segments MN and AC must be perpendicular, as A falls on C when the rectangle is folded through MN (symmetry axis). So, as AC has a slope of -1/4, MN has slope 4, but as it rises 4 it must run 1 horizontally. This means that AN must be 1 cm longer than ND. So AN=8.5 cm and ND=7.5 cm.
Thanks for sharing!

[Someone-vjcz]

Presh: “Pythagorean theorem”

Me: has a heart attack

[timsullivan]

FASTEST solution: On the line BE (where B is beginning and E the end of the video, I placed cursor on point B, then moved along BE until reaching the point 03:56, revealing the answer to be 47 cm squared.

[JSSTyger]

I've been to Washington. The Pentagon is a lot bigger than 47cm²

[AA-]

I estimated 48cm^2 at first glance since the rectangle with area 4×16=64 was made up of 4 right angled triangles almost equal in area, so each triangle has around 16cm^2 and the overlapped shape has 3 of them so 16×3=48.

[craptify]

I estimated by taking the area of the rectangle of 64 and assumed we lose about 1/4 of it to "overlap". So I got 48 in my 5 second estimate. Might have worked well enough depending on the what the multiple choice options were...

[dilipgupta]

Yes I did it it but took 5 min
Thanks MYD, LOVE your videos

[foxyeipiccolicolpidigenio]

I considered the shape of the rectangle (00:18); ABMN=ABCD/2. Now look at the shape at 00:23. That shape is equal to ABMN+CDN. We already have ABMN (64/2=32). Now let's focus again on the rectangle: let ND=x and MC=16-x. Let's do some basic calculations:

Now we can find the area of CND: AREAcnd=(7.5)(4)/2=15. We now resume the formula previously written ABMN+CND=ABMND'=15+32=47.
I thought about sharing my resolution method as it saves us some calculations. It also seems to me a clearer way. Thank you so much for sharing the problem and good evening!

[KS-enhn]

I solved almost same question years ago at coaching classes. Feels good to see this question again. I did this by second method.

[wolframalpha]

Hey! I got the same answer in a different way :
When ABCD is divided along MN :
We are basically dividing ABCD into two equal halves!!(by figure),
To make ABCD half we need not always fold it from the center!!
So 64/2 = 32 which will be the areas of ABMN and DCMN.
Keep these results aside for a while now. Let's follow the normal way of assigning AN=x,
DN=16-X, BM=y, MC=16-y.
Now fold the paper along MC.
Solve for the unknown side AN of triangle ADN, If you look closely triangle ADN and ABM are congruent !! So they will have the same areas!! After calculating the area of triangle ADN subtract it from the area if ADMN = 32 we get 32-15 = 17, which is the area of triangle AMN! Now sum the areas up!! 15+15+17 = 47

[davecash]

All I see is a 3D wedge and I can’t unsee it

[johnchessant]

I found the general formula. If a is the short side and b is the long side, then the area is 3ab/4 - a^3/(4b).
For instance, for a = 4 and b = 16, the area is 3(4)(16)/4 - 4^3/(4(16)) = 48 - 1 = 47.

[DamienConcordel]

I took the axis of symmetry of the figure, which cuts MN at the dead center of the original rectangle (call it O).
I then took the projection of O on AN (call it P) and considered triangle APO.
we know all of its sides: OP = 2, AP = 8 and AO = √(2^2+8^2) = √68
Then I compared it with triangle AON. Both are right triangles and both share angle A, so they're congruent. Then it's just a matter of scaling the area of APO to the right amount.
The hypotenuse of APO becomes the long leg of AON. We know those two lengths, and the side length ratio to scale APO up to the size of AON is √(68) / 8.
The area of APO is 8, times 2 because we're considering only half of the overlap, times the scaling factor squared, or 17/16, and we get a total overlap area of 17 exactly.
Subtract from 64, and presto! 47 cm^2

[chirag]

Solved it in 2 minutes.Quite easy to understand.Only the calculation speed matters as with calculator and right concept u can do it in under a minute.Great stuff💯

[phildurre]

if you have just 1 minute, thats probably not how you do it. my method: rouph estimate is 3/4 of the original area, which is 48. Then you realise that the overlapping triangle is slightly larger than 1/4 because its base is more than 8cm. The result must be slightly less than 48. thats why you have 4 solutions to choose from, as you cant compute this in under 1 min.

[devondevon]

For me, the key was to find the area of triangle AMB, but to do so I needed the BM length as AB was given as 4. We know that the Length of AM and BM totals 16 and solving for x (using AB as 4) gives 8.5 and 7.5, so BM is 7.5 (as AM was the hypotenuse or diagonal )

[dinudeenudinesh]

I first time got the easy solution... I got second solution.. area of rect - area of AMN triangle.. 64-17=47 square units

[sanjayrai]

I did it the same way Presh did... and got it under 1 minute 30 sec.... great problem to test your mensuration skill... awesome vid and illustration.

[ghislainmaury]

I check the length from linear to folden rectangle :
Then BM = DN = D'N = a, and AN = MC = MC' = CN = b (redrawing graphic with linear and folden rectangle + understanding question + check similar length : ~55s)

BC = BM + MC => 16 = a+b => a = 16-b
with ABM and pythagore : AB² + BM² = AM² => 4² + a² = b² => 16 + (16-b)² = b² => 16 + 16² - 32b +b² = b² => b = 16(16+1)/32 (putting and solving that equation : ~25s)
=> b = 17/2 and a = 15/2

Then S = ABM + AND' + AMN = (4*15/2)/2 + (4*15/2)/2 + (4*17/2)/2 = 15*2+17 =47 (result nearly direct with 15/2 and 17/2 to be multiplied by 4/2 : ~5s)
OR S = ABCD - AMN = 4*16 - (4*17/2)/2 = 64-17 = 47

(Total time estimation it took me for it : 55+25+5 = ~85s , I was a little slow...)

In both case, the same UNIQUE calcul is necessary to have 'a' and 'b', so having the result by addition or substraction is the same if you has seen that both triangle ABM and AND' are the same and then AMN and DMN are the same too.


Question : Why didn't you show the 4 possible choices at the end of the video, just to see if with those choice, having an estimation should have been enough ?

With the graphic, a fast estimation is that a litle more than 1/4 of the rectangle is removed (1/4 of 64 = 16)
then S is a litle less that 64*3/4 = 48, so perhaps 47 if that answer is present...