Triangle to a hexagon

All of the triangles in the diagram have the same area. Therefore the total area is 4 times the area of the central triangle plus the area of the three squares $= 4 \times \left( \frac{3\times 4}{2} \right) + 4^2 + 3^2 +5^2=\boxed{74}$ .

Proof: Area of the left-upper triangle is $\frac{1}{2}ac \sin(180-B)$ .

Since $\sin(180-B)=\sin(B)=\frac{b}{c}$ , the area is $\frac{1}{2}ac \frac{b}{c}= \frac{1}{2}ab$ .

By the same logic, the area of the triangle on the right is $\frac{1}{2}bc \sin(180-A)= \frac{1}{2}bc \sin(A)= \frac{1}{2}bc \frac{a}{c}= \frac{1}{2}ab$ .

The area of the left-lower triangle is obviously just $\frac{1}{2}ab$ .

Thus, all three triangles have the same area of $\boxed{\frac{1}{2}ab}$ .

Draw another rectangle R that inscribes the hexagon and whose sides are parallel to the initial triangle catheti. This rectangle is $(4+3+4) \times (3+4+3)$ with an area of $110$ .

The area between R and the hexagon consists of 4 triangles. We calculate the areas.

• The top-left triangle is $\frac{6\times 4 }{2}=12$ ,
• the top-right triangle is $\frac {4\times 3}{2}=6$ ,
• the bottom-left triangle is $\frac{3\times 4}{2}=6$
• and the bottom-right triangle is $\frac{8\times 3}{2}=12$ .

Finally, subtracting triangles areas from R , the area of the hexagon is $110-12-6-6-12=74$ .

After finding the area of the rectangle surrounding the hexagon (11x10=110) Take away the 4 triangles in the corners (12,12,6,6) 110-12-12-6-6=74

It is possible to do this without any trigonometry.

We will solve for the general case, i.e. a triangle with side lengths $a, b, c$ . We will continue the side length of the largest square down through the smallest square (the $a \times a$ one), like so: The same thing can be done on the other side, through the $b \times b$ square. We can now add a perpendicular to that line from the corner of the $a \times a$ square. This creates two congruent triangles (one side and two angles are equal). Likewise, through the $b \times b$ square. The pairs of congruent triangles are shaded. Since the two triangles in each pair are congruent, and two triangles from different pairs share a height (the two in the original $a, b, c$ triangle), all the heights must be equal. The two unshaded triangles have a common base (it is the side length of the largest square). And because they both have the same base and the same height, their areas must be equal! It can easily be shown that the triangle at the bottom is identical to our original one.

Since the triangles all have the same area, the area of the entire figure will be $A=\frac{1}{2}ab \times 4+2(a^2+b^2)=2ab+2(a^2+b^2)=2(a^2+ab+b^2)$ .

In this case, we get $\boxed{74}$ .

If extend the segments of the squares of 3 and 4 we will have the next figure

The next is see the measure of some segments of withe and we will get a triangles rectangles also is so easy to calculate the areas and the sum

Red = Areas of triangles

I calculated the area of the two non right triangles using the length of two sides and the angle to find the length of the third side and then calculated the area from the length of all three sides I did use online calculators for this so in essence I think I cheated