Let O be the centre of the circle and AB and CD are two parallel chords on the same side of the radius. OP is perpendicular to AB and OQ is perpendicular to CD. If AB = 10 cm, CD = 24 cm and PQ = 7 cm, then the diameter (in cm) of the circle is equal to:

The correct option is 26.

Let $r$ be the radius of the circle with centre $O$.
We are given two parallel chords $AB$ and $CD$ on the same side of the centre $O$ (the question mentions "same side of the radius", which implies they are on the same side of the centre/diameter parallel to them).
The lengths of the chords are:
$AB=10\text{cm}$
$CD=24\text{cm}$

We are given that:
$OP\perp AB$ and $OQ\perp CD$.
Since a perpendicular drawn from the centre of a circle to a chord bisects the chord, we have:
$AP=PB=\frac{AB}{2}=\frac{10}{2}=5\text{cm}$
$CQ=QD=\frac{CD}{2}=\frac{24}{2}=12\text{cm}$

Since the chords are parallel and on the same side of the centre, the points $O$, $Q$, and $P$ are collinear, lying on a line perpendicular to both chords.
Let $OQ=x\text{cm}$.
We are given that the distance between the two parallel chords is $PQ=7\text{cm}$.
Therefore, the distance $OP$ is:
$OP=OQ+PQ=x+7$

Now, let us apply the Pythagorean theorem to the right-angled triangles $\Delta OQC$ and $\Delta OPA$:
In $\Delta OQC$:
$O{C}^2=O{Q}^2+C{Q}^2$
$r^2=x^2+{12}^2$
$r^2=x^2+144$ (Equation 1)

In $\Delta OPA$:
$O{A}^2=O{P}^2+A{P}^2$
$r^2={(x+7)}^2+5^2$
$r^2=x^2+14x+49+25$
$r^2=x^2+14x+74$ (Equation 2)

Equating the expressions for $r^2$ from Equation 1 and Equation 2:
$x^2+144=x^2+14x+74$
Subtracting $x^2$ from both sides:
$144=14x+74$
$14x=144-74$
$14x=70$
$x=5\text{cm}$

Now, substitute the value of $x$ back into Equation 1 to find the radius $r$:
$r^2=5^2+144$
$r^2=25+144$
$r^2=169$
$r=\sqrt{169}=13\text{cm}$

Since the radius of the circle is $13\text{cm}$, the diameter is:
$d=2r=2\times 13=26\text{cm}$