Following up on comments of @Mauro ALLEGRANZA regarding this interesting question, I believe Euclid does not prove the formula for the area of a triangle, nor does Elements VI, 23 show that the area of a parallelogram is equal to base times height.
"The ratio of the area of a triangle to the area of a parallelogram is equal to one half the ratio of their bases times the ratio of their heights," and "the area of a triangle is 1/2 base times height," are not equivalent propositions, nor does the second follow from the first in light of I, 41, as OP seems to claim in his second paragraph and fourth and sixth comments.
When Euclid proves in VI, 23 that the areas of equiangular parallelograms are in the ratio compounded of the ratios of their sides, he is speaking and thinking not arithmetically but geometrically. Allowing for incommensurability, he does not suppose the sides are in the ratio of a number to a number, and hence that multiplication of ratios is even possible. Compounding of ratios, as his proof makes clear, is a purely geometric idea and operation, working with lines and areas as continuous, unmeasured, possibly incommensurable, magnitudes.
Euclid is considering the ratio of the areas of two parallelograms, and he is not viewing those areas, or the sides that contain them, as subject to numerical measurement. And even where magnitudes, e.g. lengths or areas, prove to be equal, equality in the geometric parts of Elements is evidenced not by numerical measure but rather by spatial coincidence, either as a whole, as in I, 4, or when broken into parts, as in I, 35.
Although I, 41 of course proves that the area of a triangle is half that of a parallelogram of the same base and height, it jars with Euclid's entire approach to say either that the area of a triangle is equal to half its base times its height, or that the area of a parallelogram is equal to base times height. It is true that he did not have the concept of real numbers at his disposal, but that is only part of the reason. For Euclid geometry is not dependent on arithmetic. In fact, geometry can go where arithmetic cannot follow. Not until the seventeenth century does Descartes (Geometry) boast of overcoming what he calls the "scruple" the ancients had against introducing the terms of arithmetic into geometry. But even Isaac Newton, fifty years later, prefers to formulate and prove the propositions of Principia as geometric proportions rather than algebraic equations. And Dedekind of course (Continuity and Irrational Numbers), as zealous for the independence of arithmetic as Euclid was for that of geometry, sought to provide a foundation for continuity in the domain of numbers without appealing, as was usual before him, to geometric intuition.
I think it is fair to say, that Euclid proved that the ratio of the area of a triangle to that of a parallelogram is one-half the ratio compounded of the ratios of their bases and heights, or, with some stretching of the term, one-half the ratio of their bases times the ratio of their heights. But it does not follow from this, that in Euclid's view a triangle's area equals half its base times its height, or that a parallelogram's area equals base times height. Not that these formulas are to be rejected, of course, but they do not arise from, or comport well with, the principles and point of view of Elements.
HINT
Exactly the same idea. Split in two parts using the height, and in the half-triangle you have the hypotenuse of $17$ and one of the legs is $k/2$.
- By the Pythagorean theorem, height $h$ satisfies $h^2 + (k/2)^2 = 17^2$, can you find $h(k)$?
- Now the area of the big triangle is $k \cdot h(k) /2$, but you already know this is $120$, can you solve for $k$?
Remark It's obvious one of the answers will be $k=16$ because then the triangles are identical. Are there other values?
Update
You have $$k = \sqrt{4\left(17^2 - h^2\right)} = 2\sqrt{17^2 - h^2},$$ hence the final equation is
$$
120 = k(h) \cdot h /2
= \frac{h}{2} \cdot 2\sqrt{17^2 - h^2}
= h \sqrt{17^2 - h^2}
$$
To solve this, square both sides to get
$$
120^2 = h^2 \left(17^2 - h^2\right)
$$
and let $z = h^2$ to get a quadratic in $z$.
Best Answer
As in the comments, the answer must be (a).
However, to prove this if the answers had not been given, drop a perpendicular, $CE$, of height $h$ from $C$ onto $BQ$ where $BE=x$.
By similar triangles, $\frac{10}{h}=\frac{100}{100-x}$ and $\frac{40}{h}=\frac{100}{x}$.
Then $x=20,h=8$ and so the required area is $\frac{1}{2}\times50\times h=200$.
N.B. A neat general result for this set up is that the reciprocal of $h$ is the sum of the reciprocals of the two given heights i.e.
$$\frac{1}{h}=\frac{1}{10}+\frac{1}{40}.$$
Proof
Let the given heights be $L,M$ and the base $D$. By similar triangles: $$\frac{L}{h}=\frac{D}{D-x}\text { and }\frac{M}{h}=\frac{D}{x}.$$ Then $$\frac{D-x}{hD}=\frac{1}{L}\text { and }\frac{x}{hD}=\frac{1}{M}.$$ Now add the two LH terms: $$\frac{D-x}{hD}\text +\frac{x}{hD}=\frac{D}{hD}=\frac{1}{h}$$ and so $$\frac{1}{h}=\frac{1}{L}+\frac{1}{M}.$$