Can this be solved this without sin and cos? My daughter got this homework problem, and I can't seem to solve it. What makes it stranger is they haven't done geometry yet: like no sin and cos stuff. They did do area of triangles, but it seems like this is too advanced.  I am wondering if maybe I am missing some simple trick or something like that.
Here is the diagram:

The instruction are to calculate s1-s2.
Any ideas?  I know if you assume they are right angles you can use cos and sin, but I feel like something simpler should be involved.
 A: While this is probably not the way the problem was intended to be solved, we can do it with almost no calculation.
The area you are trying to compute is the area of the yellow trapezoid in the figure below, which is what is left of the $S_1$ triangle after we cut a copy of the $S_2$ triangle from the upper right corner.

By similar triangles, the original $S_2$ plus the blue trapezoid have the same total area as the copy of $S_2$ plus the yellow trapezoid,
so the blue and yellow trapezoids have equal areas.
But the two trapezoids together fill a rectangle $2\,\mathrm{cm}$ high and $10\,\mathrm{cm}$ wide, with an area of $20\,\mathrm{cm}^2.$
Since the trapezoids are equal, each has half the area of the rectangle,
that is, $10\,\mathrm{cm}^2.$
A: Obviously it is all about similar triangles.
You have 3 triangles similar here. S₁, S₂, and the big one of area (6+8)×10/2 = 70 cm².
S₁ dimension is obviously 8/14 × the one of the big one.
So its area is (8/14)² = (4/7)² = 16/49 the area of the big one.
S₂ dimension is (6/8) = 3/4 the dimension of S₁. So its area is 9/16 the area of S₁. So 16/49×9/16 = 9/46 the area of the big one.
So S₁-S₂ area is (16-9)/49 = 1/7 times the area of the big one.
So answer is 1/7×70 = 10 cm².
A: HINT 1: $S_{1}$ and $S_{2}$ are similar triangles, so they will have the same ratios for their side each side lengths.

HINT 2: Let $x$ be the width of $S_{1}$. Then $10-x$ is the width of $S_{2}$. Create a ratio (since $S_{1}$ is similar to $S_{1}$) using $x$, $10-x$, $8$, and $6$.
A: Comment expanded to answer per request.
Let $T$ be the area of trapezoid under the shaded area $S_1$. Notice
$$S_1 - S_2 = (S_1 + T) - (S_2 + T)$$
and $S_1 + T$ is the area of a right triangle with height $(8+6)$cm and base $10$cm while $S_2 + T$ is the area of a rectangle with width $10$cm and height $6$cm. We have
$$S_1 - S_2 = \frac12(8+6)(10)  - (10)(6)
= 70 - 60 = 10{\rm cm}^2$$
A: This answer is really based off of @chrslg's answer: https://math.stackexchange.com/a/4637428/1149060
I understand it like this.  All the triangles are similar so the ratio of length to area is constant.
The area of the big triangle is 70.  The length of the triangle is 14.
The s1 triangle has a length of 8.  That is 8/14 the length of the large triangle. Since the ratio of length to area is constant, it will be be 8/14 of the area of the large triangle or (8/14) * 70 = 40.
Same idea for the s2 triangle (6/14)*70 = 30
Therefore, the different is 40 - 30 = 10.  (10 cm2)
A: Let $S_3$ be the trapezoid's area. The area of the big triangle is $S= \frac{(14)(10)}{2} = 70 \text{cm}^2$. The area of the rectangle is $R=(6)(10) = 60 \text{cm}^2$. From the figure, we have two equations
$$
\begin{align}
S_1 + S_3 &= 70,\\
S_2 + S_3 &= 60.
\end{align}
$$
It is easy now to find $S_1-S_2$.
