What Is The Rectangle's Area?

Many students in the UK had difficulty with a similar problem on the GCSE exam. What's the rectangle's area, given the three circles are identical with radius r? Watch the video to learn how to solve this problem.

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[andeheri]

Finally something I actually solved in my head:)

[billy.]

I think this is a straight forward problem. No trick needed.

[ssccgl]

Hey This question was not so difficult 😂

[ArabianShark]

Easy problem, but it's nice to get a nice and easy one sometimes, just to bolster my own ego a little bit :)

[RonJohn]

0:20 Each side is length 4r, so the area is 16r^2. Let's see if I'm right.
0:55 Shows I must pay attention when they write *rectangle* not square.

[zippitzippit]

I think Presh wants to help us with our GCSE-s to be honest

[gedlangosz]

One of the hardest GCSE exam questions? mmm.

[manjulashah]

Thankyou sir helped me a bit . As my method was time consuming.

[-Stocking]

Wow you guys are SO GOOD AT THIS, why don't any of you have a Nobel Prize yet?

[japeking]

I draw one of your problems on my clipboard and then take it to my afternoon nap. This one was too straightforward so I'd solved it before I fell asleep and had to get back up and come see if I was right. Ruined my nap. Thanks a lot ;-)
I hope tomorrows choice is tougher ( but not too tough).

[MathsWithMelv]

Probably in the top 3 hardest GCSE problems so far! Not met a student that could do it!

[leejimmy]

I needed to solve a 5 circle version and this helped me immensely.

[ryanio]

Very succinct and clear explanation. Thanks Presh!

[daniellaclinton]

I feel silly for struggling with this q when I first saw it whilst preparing for my maths exam this tuesday so thank you for this explanation

[disco.jellyfish]

You can easily see, that the rectangles width is w=4r
Since these circles are as close as possible to each other the dots form a equilateral triangle.
The height of the Rectangle is 2r+2*d | 2*d is the vertical height difference between these two dots (right and bottom).
d=cos(pi/6)*r | (pi/6=30°)
So h=2r+2cos(pi/6)*r

This leads to the final result:
A=(4r)(2r+2cos(pi/6)*r)
A=(4r)(2r+2(sqrt(3)/2)r
A=(4r)(2r+sqrt(3)r)
A=(4r)r(2+sqrt(3))
A=4r²(2+sqrt(3))

Yeah... Same result.

[avhuf]

"hardest exams" ??? damn they must have low standards... so simple...

[randomusername]

YES! I finally figured one out without watching the video first!!

[alexhenson]

I have one for you!
You have a semicircle from the starting point with radius 4 and an angle θ, were 0<θ<π/2, starting from the y axis. You also have a line a1 parallel to the x axis were its points are dependent of the angle. after that you draw two lines, a2 and a3, from the line a1's points to the points A(-4, 0) and D(4, 0). What is the area of the shape made from the lines and the x axis and prove that its area is at its biggest when θ=π/6.


I hope this makes sense.

[browniboi]

This is easy if you've seen similar problems like it. But for us first-timers this was a challenging and interedting problem.

[lythd]

I think it is 4(2+2*sin60 deg)(r^2). Let me watch the video.

EDIT: Heyy I got it right! That wasn't that bad