Equal Triangle Algebra

As the areas are same........................................ 1/2×(x+1)×h =1/2×5×h On solving we get x=4. As base of both the triangles are same, the hypotenuese of both triangles will be equal...... Therefore, (x+y)=(2y+2) Substitute x=4 and on simplification we get y=2.

If these triangles are in fact equal, then we can assume the bases and hypotenuses are equal to each other.

First find $x$ by setting the bases equal to each other. Subtract 4 from each side. $\begin{aligned} x+1&=5\\ x=4\\ \end{aligned}$ Then use your result from $x$ to solve for $y$ and set the hypotenuses equal to each other. Solve for $y$ by subtracting $y$ and 2 from each side. $\begin{aligned} 4+y&=2y+2\\ 4&=y+2\\ 2&=y\\ \end{aligned}$

Since the areas are equal, we have

$\dfrac{1}{2}(h)(x+1) = \dfrac{1}{2}(h)(5)$

$x+1=5$

$x=4$

We observed that the bases are also equal, so the triangles are congruent. So

$x+y = 2y+2$

$4+y=2y+2$

$4-2=2y-y$

$2 = y$

Note: In my solution, I assumed that the bases are not equal. But if we look further, we can see that the bases are equal.