Explanation:
what is the problem to solve ?
I assume we need to calculate the area of the whole figure ?
in any case, we have a problem :
a rhombus is a tilted square, a special parallelogram, as it has 4 equal sides.
that would mean all 4 sides of that central rhombus are 5 in.
that would make both triangles equilateral triangles (all 3 sides are equally long : 5 in).
but in order for a right-angled triangle to have the Hypotenuse = 5 in and the height (= left leg) = 4 in, that must make the right leg
5² = 4² + leg²
25 = 16 + leg²
9 = leg²
leg = 3 in
and so, the baseline of the large triangle 2×3 = 6 in.
and not 5 in, which must be the top and base line of the rhombus.
so, the whole problem definition is wrong.
the only solution when accepting the given lengths, is that the triangle sides are NOT a straight extension of the rhombus sides.
the triangle side is the Hypotenuse of the smaller internal right-angled triangle
side² = 4² + (5/2)² = 16 + 6.25 = 22.25
side = 4.716990566... in
anyway, the area of each of the large triangles is
baseline×height / 2 = 5×4/2 = 10 in²
we have 2 triangles = 2×10 = 20 in²
the area of the rhombus is
baseline×height = 5×4 = 20 in²
please note that the area of a rhombus is also
diagonal1 × diagonal2 / 2
but that applies only, when we have the lengths of the diagonals. both approaches give the same result, of course.
so, the whole area is
20 + 20 = 40 in²