A trapezium has vertices marked as P, Q, R and S (in that order anticlockwise). The side PQ is parallel to side SR.
Further, it is given that, PQ = 11 cm, QR = 4 cm, RS = 6 cm and SP = 3 cm.
What is the shortest distance between PQ and SR (in cm)?
The question asks for the shortest distance between the parallel sides PQ and SR of a trapezium PQRS. This distance is the height ($h$) of the trapezium relative to these parallel bases.
Given values are:
Let PQ be the longer base and SR be the shorter base. To find the height $h$, we can draw perpendiculars from the endpoints of the shorter base (S and R) to the longer base (PQ).
Let X and Y be the points on PQ such that SX is perpendicular to PQ and RY is perpendicular to PQ.
This forms:
The length of the base PQ can be expressed as $PX + XY + YQ$. Substituting the known values: $11 \text{ cm} = PX + 6 \text{ cm} + YQ$. This simplifies to $PX + YQ = 11 - 6 = 5 \text{ cm}$.
Using the Pythagorean theorem on the two right-angled triangles:
From Equation 1, $PX = \sqrt{9 - h^2}$. From Equation 2, $YQ = \sqrt{16 - h^2}$.
Substitute these expressions for PX and YQ into the equation $PX + YQ = 5$: $\sqrt{9 - h^2} + \sqrt{16 - h^2} = 5$.
This equation can be solved for $h$. An efficient method is to test the given options. Let's test $h = 2.4$ cm.
The sum $PX + YQ = 5$ cm matches the condition derived from the base length PQ. Therefore, the height $h = 2.4$ cm is correct.
The shortest distance between the parallel sides PQ and SR is the calculated height $h$.
The shortest distance is 2.4 cm.
In the given figure, $P, Q$, and $R$ are three points on a circle of radius 10 cm with $O$ as its center, $\overline{PQ} = \overline{RQ}$, and $\angle PQR = 45^\circ$. The figure is representative.
The area of the shaded region $PQRO$ is ______________ cm$^2$.
