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# Geometry Exercise?

Shown are three semicircles and a square. Find the area of blue square.

### 2 Answers

- Pramod KumarLv 71 month ago
Refer to the figure below-

IG = (1/2 - a/2) = (1/2) ( 1 - a ) .................. (1)

Let's recollect Rectangular property of chords --

(If two chords intersect inside a circle, then the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord.)

DI = a + sqrt [ 1/4 - 1/4 ( 1 - a )² ] = (1/2) [ 2a + sqrt (2a - a²) ] .......... (2)

EI = DI - a = (1/2) [ 2a + sqrt (2a - a²) ] - a

Consider Triangle ODI :

1² = (a/2)² + (1/4) [ 2a + sqrt (2a - a²) ]²

=> 8 a⁴ - 4 a³ - 7 a² - 4 a + 4 = 0

From Wolfram Alpha --

a = 0.524956

Hence Required Area A = 0.2758 sqr units ................. Answer

- llafferLv 71 month ago
Let's try a few things. If we call point O the origin, then A is (-1, 0) and B is (1, 0).

Since AB = 2, AO and BO are both 1.

The centers of those circles (if the entire circle is drawn) would be:

(x - h)² + (y - h)² = r²

center is (-1/2, 0) and radius of 1/2 and center is (1/2, 0) and radius of 1/2, so:

(x + 1/2)² + y² = (1/2)² and (x - 1/2)² + y² = (1/2)²

(x + 1/2)² + y² = 1/4 and (x - 1/2)² + y² = 1/4

Now let's do the same with the large circle. Center is (0, 0) with a radius of 1:

(x - h)² + (y - h)² = r²

x² + y² = 1²

x² + y² = 1

Let's look at the OB semi-circle equation and the large semi-circle equation:

(x - 1/2)² + y² = 1/4 and x² + y² = 1

For the square to work, if we draw a lone parallel to DE and CF, perpendicular to AB, then call the points where they intersect EF and CD as P and Q, then the length:

CP and QF are the same, which is half the distance of CF. If we can find the distance of CF we can square it and get the area.

Let's solve both equations for y in terms of x (ignoring the minus side since there are no points with y < 0:

(x - 1/2)² + y² = 1/4 and x² + y² = 1

y² = 1/4 - (x +-1/2)² and y² = 1 - x²

y = √[1/4 - (x - 1/2)²] and y = √(1 - x²)

for a given x , we can find the values of y's for each and their distance would be the length of CF (I'll call this y) by subtracting the smaller from the larger:

y = √(1 - x²) - √[1/4 - (x - 1/2)²]

and we want the value x to be half the value of y:

x = y / 2

We now have a system of two equations and two unknowns. We want to solve for y, then square it to get the area. So substitute the expression of x in terms of y into the other equation and solve for y:

y = √(1 - x²) - √[1/4 - (x - 1/2)²]

y = √[1 - (y/2)²] - √[1/4 - (y/2 - 1/2)²]

y = √(1 - y²/4) - √[1/4 - (y - 1)/2)²]

y = √(1 - y²/4) - √[1/4 - (y - 1)²/4]

y = √(1 - y²/4) - √[(1 - (y - 1)²)/4]

y = √(1 - y²/4) - √[(1 - (y² - 2y + 1))/4]

y = √(1 - y²/4) - √[(1 - y² + 2y - 1)/4]

y = √(1 - y²/4) - √[(-y² + 2y)/4]

Let's get a common denominator in the first root, then we can pull out the 4's of the radical:

y = √(4/4 - y²/4) - √[(-y² + 2y)/4]

y = √[(4 - y²)/4] - √[(-y² + 2y)/4]

y = (1/2)√(4 - y²) - (1/2)√(-y² + 2y)

Let's multiply both sides by 2 to get rid of the fractions:

2y = √(4 - y²) - √(-y² + 2y)

Now let's square both sides to see what we get:

4y² = (4 - y²) - 2√(4 - y²) * √(-y² + 2y) + (-y² + 2y)

Simplify the non-radical terms:

4y² = 4 - y² - 2√(4 - y²) * √(-y² + 2y) - y² + 2y

4y² = 4 - 2y² - 2√(4 - y²) * √(-y² + 2y) + 2y

6y² = 4 - 2√(4 - y²) * √(-y² + 2y) + 2y

We can now divide both sides by 2 to simplify:

3y² = 2 - √(4 - y²) * √(-y² + 2y) + y

Now we'll simplify the product of two roots as the root of the product and multiply out what's inside:

3y² = 2 - √[(4 - y²)(-y² + 2y)] + y

3y² = 2 - √(-4y² + 8y + y⁴ - 2y³) + y

3y² = 2 - √(y⁴ - 2y³ - 4y² + 8y) + y

Move all non-radical terms to the left side:

3y² - y - 2 = -√(y⁴ - 2y³ - 4y² + 8y)

Square both sides again:

(3y² - y - 2)² = y⁴ - 2y³ - 4y² + 8y

9y⁴ - 3y³ - 6y² - 3y³ + y² + 2y - 6y² + 2y + 4 = y⁴ - 2y³ - 4y² + 8y

9y⁴ - 6y³ - 11y² + 4y + 4 = y⁴ - 2y³ - 4y² + 8y

and put into standard form:

8y⁴ - 4y³ - 7y² - 4y + 4 = 0

Now we have a polynomial that we can solve for. This won't be rational so I'll use Netwton's method about 3 times to get an estimate that should be to about 3 or 4 DP of precision. I'll use 0.5 as a first guess:

f(y) = 8y⁴ - 4y³ - 7y² - 4y + 4

f'(y) = 32y³ - 12y² - 14y - 4

y₁ = y₀ - f(y₀) / f'(y₀)

y₁ = y₀ - (8y₀⁴ - 4y₀³ - 7y₀² - 4y₀ + 4) / (32y₀³ - 12y₀² - 14y₀ - 4)

y₁ = 0.5 - (8(0.5⁴) - 4(0.5³) - 7(0.5²) - 4(0.5) + 4) / (32(0.5³) - 12(0.5)² - 14(0.5) - 4)

y₁ = 0.5 - (8 * 0.0625 - 4 * 0.125 - 7 * 0.25 - 4 * 0.5 + 4) / (32 * 0.125 - 12 * 0.25 - 14 * 0.5 - 4)

y₁ = 0.5 - (0.5 - 0.5 - 1.75 - 2 + 4) / (4 - 3 - 7 - 4)

y₁ = 0.5 - 0.25 / (-10)

y₁ = 0.5 + 0.025

y₁ = 0.525

y₂ = y₁ - (8y₁⁴ - 4y₁³ - 7y₁² - 4y₁ + 4) / (32y₁³ - 12y₁² - 14y₁ - 4)

y₂ = 0.525 - (8(0.525)⁴ - 4(0.525)³ - 7(0.525)² - 4(0.525) + 4) / (32(0.525)³ - 12(0.525)² - 14(0.525) - 4)

y₂ = 0.525 - (8(0.075969140625) - 4(0.144703125) - 7(0.275625) - 4(0.525) + 4) / (32(0.144703125) - 12(0.275625) - 14(0.525) - 4)

y₂ = 0.525 - (0.607753125 - 0.5788125 - 1.929375 - 2.1 + 4) / (4.6305 - 3.3075 - 7.35 - 4)

y₂ = 0.525 - (-0.000434375) / (-10.027)

y₂ = 0.525 - 0.0000433205

y₂ = 0.5249566795

That's already good to 5DP, so let's round to that:

y₂ = 0.52496

That's the height of the square. Square that to get the area:

0.52496² = 0.2755830016

Round that to 4DP to get:

0.2756 unit²