3/4 divided by 12 as a fraction

We know that the octagon is regular which means all sides are equal.
since all sides are equal, then all sides are equal 10cm.
if all sides are equal then all angles within it are equal.
since the total angle in an octagon is 1080°, the sum of each angle within the octagon is 135°.

Please note that the shaded region comprises a rectangle /ADEH/ and a scalene triangle /ACD/.

So to get the area of the entire region, we have to solve for the area of the Scalene Triangle /ACD/ and add that to the area of the rectangle /ADEH/

Step 2 - Solving for /ACD/

The formula for the area of a Scalene Triangle is given as:

A = $\sqrt{S(S-a)(S-b)(S-c) square units}$

This formula assumes that we have all the sides. But we don't yet.

However, we know the side /CD/ is 10cm. Recall that side /CD/ is one of the sides of the octagon ABCDEFGH.

This is not enough. To get sides /AC/ and /AD/ of Δ ACD, we have to turn to another triangle - Triangle ABC. Fortunately, ΔABC is an Isosceles triangle.

Step 3 - Solving for side AC.

Since all the angles in the octagon are equal, ∠ABC = 135°.

Recall that the total angle in a triangle is 180°. Since Δ ABC is an Isosceles triangle, sides /AB/ and /BC/ are equal.

Recall that the Base angles of an isosceles triangle is always equal. That is ∠BCA and ∠BAC are equal. To get that we say:

180° - 135° = 45° [This is the sum total of ∠BCA and ∠BAC. Each angle therefore equals

45°/2 = 22.5°

Now that we know all the angles of Δ ABC and two sides /AB/ and /BC/, let's try to solve for /AC/ which is one of the sides of Δ ACD.

According to the Sine rule,

$\frac{Sin 135}{/AC/}$ = $\frac{Sin 22.5}{/AB/}$ = $\frac{Sin 22.5}{/BC/}$

Since we know side /BC/, let's go with the first two parts of the equation.

That gives us $\frac{0.7071}{/AC/} = \frac{0.3827}{10}$

Cross multiplying the above, we get

/AC/ = $\frac{7.0711}{0.3827}$

Side /AC/ = 18.48cm.

Returning to our Scalene Triangle, we now have /AC/ and /CD/.

To get /AD/ we can also use the Sine rule since we can now derive the angles in Δ ABC.

From the Octagon the total angle inside /HAB/ is 135°. We know that ∠HAB comprises ∠CAB which is 22.5°, ∠HAD which is 90°. Therefore, ∠DAC = 135° - (22.5+90)

∠DAC = 22.5°

Using the same deductive principle, we can obtain all the other angles within Δ ACD, with ∠CDA = 45° and ∠112.5°.

Now that we have two sides of ΔACD and all its angles, let's solve for side /AD/ using the Sine rule.

$\frac{Sin 112.5}{/AD/}$ = $\frac{Sin 45}{18.48}$