Find the area of the shaded region of this figure

Question

Find the area of the shaded region. (Each arcs of circles in the figure are assumed to be $\frac{1}{4}$ of a full circle)

This problem is classic...... – Brethlosze Aug 21 '17 at 16:57 — Brethlosze, Aug 21 '17 at 16:57

score 3 · Accepted Answer · answered Aug 21 '17 at 19:11

3

You need to find the area of 4 remaining parts and subtract them from the area of the square.

answered Aug 21 '17 at 19:11

Seyed

Brethlosze · Answer 2 · 2017-08-21T20:25:24.010

1

HINT: $$ A=4\int_{1/2}^{\sqrt 3/2}\sqrt{1-x^2}-1/2dx = 1-\sqrt 3-\frac \pi 3 = 0.31515 $$

edited Aug 21 '17 at 20:25

answered Aug 21 '17 at 17:08

Brethlosze

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Your answer is the best and should be the one that is accepted. The result is the same as that one and, oh, so much better! – Cye Waldman Aug 21 '17 at 20:08
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I'm not getting correct results with this one.link – Kitiara Aug 21 '17 at 20:13
Check the result. – Brethlosze Aug 21 '17 at 20:25
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@CyeWaldman, And oh! here is categorized for geometry :) – Seyed Aug 21 '17 at 21:55
The first time i faced this problem i didnt know algebra!. I just made it this way as vengeance... – Brethlosze Aug 21 '17 at 22:13
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You are assuming x = 1 in this form. – Kitiara Aug 21 '17 at 22:51
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@Kitiara I doesn't matter. The area must necessarily scale as $x^2$. – Cye Waldman Aug 21 '17 at 23:18
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@Seyed Thank you for pointing that out. By way apology, please accept my upvote. – Cye Waldman Aug 22 '17 at 15:30
@CyeWaldman, Thank you very much for your upvote Sir, and please believe me I was just joking, I put a smiley at the end of my comment. – Seyed Aug 22 '17 at 21:05
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@Seyed No offense taken. I was exhibiting my prejudice about geometry because the notation ABD, etc. drives me crazy. I cannot read a paper without constantly referring back to the figure to see what it means. That prejudice almost keep me from participating in this paper (http%3A%2F%2Fwww.ams.org%2Fnotices%2F201509%2Frnoti-p1036.pdf&usg=AFQjCNEvWsJAD1_j3IgvH5GcAhvGK_Jphg) because I had to read and understand several of Archimedes works in The Method. – Cye Waldman Aug 22 '17 at 23:23

Raffaele · Answer 3 · 2017-08-21T20:04:06.173

1

$FC=2x\sin 15°$

$\sin 15°=\sqrt{\dfrac{1-\cos 30°}{2}}=\sqrt{\dfrac{1-\frac{\sqrt 3}{2}}{2}}=\dfrac{1}{2}\,\dfrac{\sqrt{3}-1}{\sqrt{2}}$

$FC=2x\dfrac{\sqrt{3}-1}{2 \sqrt{2}}=x\dfrac{\sqrt{3}-1}{ \sqrt{2}}\\ Area_{FHGC}=FC^2=\left(x\dfrac{\sqrt{3}-1}{ \sqrt{2}}\right)^2=x^2(2-\sqrt 3)$

$area_{red}=\dfrac{1}{2} x^2 (t-\sin t)\\ area_{red}=\dfrac{1}{2}x^2\left(\dfrac{\pi}{6}-\dfrac{1}{2}\right) $

$Area=x^2\left[2-\sqrt 3+2\left(\dfrac{\pi}{6}-\dfrac{1}{2}\right)\right]\\ Area=x^2\left(1+\dfrac{\pi}{3}-\sqrt{3}\right)$

edited Aug 21 '17 at 20:04

answered Aug 21 '17 at 18:43

Raffaele

There is a mistake, i don't get the correct answer when i replace X with another number. – Kitiara Aug 21 '17 at 19:36
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Thank you. Now it works – Raffaele Aug 21 '17 at 19:58
Now it is correct, the second way of solving the question. Well done. – Kitiara Aug 21 '17 at 20:07

3 Answers3