Three Squares Form A Triangle

We want to find $\angle A + \angle B + \angle C$ .

Let angles $A', B', C'$ denotes the of the triangle $ABC$ as shown below.

We know that the sum of interior angles of a triangle is $180^\circ$ .

Thus, we have $\angle A'+ \angle B'+ \angle C' = 180^\circ$ .

From the diagram, we have the sum of two right angles (one from each of the two squares) and plus the two angles $A$ and $A'$ produces a full circle (360 degrees), so

$90^\circ + \angle A + 90^\circ + \angle A' = 360^\circ \qquad (1)$

Similarly, we have

$90^\circ + \angle B + 90^\circ + \angle B' = 360^\circ \qquad (2) \\ 90^\circ + \angle C + 90^\circ + \angle C' = 360^\circ \qquad (3)$

Adding these three equations, $(1), (2), (3)$ gives us

$\begin{aligned} ( \angle A + \angle B + \angle C) + (\angle A'+ \angle B'+ \angle C') + 6 \times 90^\circ &=& 3\times 360^\circ \\ ( \angle A + \angle B + \angle C) + 180^\circ + 540^\circ &=& 1080^\circ \\ \angle A + \angle B + \angle C &=& 1080^\circ - 540^\circ - 180^\circ \\ &=&360^\circ \end{aligned}$

Hence our answer is $\boxed{360^\circ}$ .

Although it's not strictly rigorous, it's quite simple to see that, as the sides of the squares are parallel, if you were to join up the angles (i.e. squish the squares to line thickness) then they would span one full rotation or 360 degrees.

This is just the Exterior Angle Theorem. The sum of the exterior angles is always 360 degrees.

Simple logic. Each corner of the triangle shares 2 90 degree angles. If you assume the triangle is 60 degrees on average per corner and the total of the four areas is 360 degrees, angle A is 120 degrees (360 -90 -90 -60) x 3 = 360 total.

Draw a figure or diagram them label it correctly. Then the rest is a piece of ice cream.

You can definitely take the squares away but leaving lines to show you the angles. Just join them together and you get 360 degrees.

Imagine the smaller triangle's three angles are x, y and Z. Then A=360-{(2×90)+x} B=360-{(2×90)+y} C=360-{(2×90)+z} _ _ _ _ __ A+B+C=1080-(540+x+y+z) [*We know the sum of three angles triangle (X+Y+Z=180⁰)] Thus, The answer is equals to 360⁰.

If you squish the interior triangle so point A touches the line CD then angle at A is zero, C and D are 180 each. The pink and/or blue squares (which could be rectangles) would change size to accommodate the squishing.

Construct a line parallel to CB that passes through A; Construct a line parallel to the other two sides of square CB that passes through A. Angles about a point = 360°.