jugoslav

Great solution, Tiggsy!!!

In rectangle ABCD, I is the midpoint of AD and E and F trisect BC.  BD intersects EI and FI and G and H.  Find the ratio of the area of triangle GHI to the area of rectangle ABCD.

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Since a square is also a rectangle, we can use it to make life easier.

Let the square side be 3 units.

ID = 1.5         EF = 1           ∠EIF = 2* tan^-1[(EF/2) / AB] = 18.92464442º

∠ADB = 45º           ∠EID = 90 + (∠EIF / 2) = 99.4623222º           ∠FID = 90 - (∠EIF / 2) = 80.5376778º

Now you have enough information to calculate the area of triangles GID and HID.

But if you're lazy like me, you can visit this website:      https://www.triangle-calculator.com/

The top is 1, and the bottom is 3

This one's very tough.

Line segments ED and E'D' are congruent. The distance between them is 1 unit.

So, we have (ED')² = (E'D')² + (EE')²

Radius        r = (ED') / 2

( I hope it's clear enough.) 

I'll come back if I get the right answer. 

And the answer is     a = 6 units  

The ratio is 1pi : 4pi  or 1/4

I think that your answer is not correct.

According to your calculations is as follows:

4 * tan∠FBC = 6.315789474

6.315789474 / tan∠EBD = 9.972299169     ( 4 + a )

AX and CZ are not congruent!!!

We need to look for the correct answer.

a = 6 units  

The height of a trapezoid is 8 cm. The lengths of the two diagonals of the trapezoid are 10 cm and 17 cm. Calculate the area of the trapezoid.

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Since the type of trapezoid is not specified, we can use a right trapezoid.

AB = sqrt(BD² - AD²)

CD = sqrt(AC² - AD²)

Area       A = AD [(AB + CD) / 2]  ==>  A = 84 cm²

A region is bounded by semicircular arcs constructed on the side of a square whose sides measure 2 / pi, as shown. What is the perimeter of this region?

Perimeter      P = 2[ pi(2 / pi)] = 4