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)?

Trapezium Height: Shortest Distance Between Parallel Sides

This problem requires us to determine the shortest distance between the parallel sides of a trapezium (also known as a trapezoid). In any trapezium, this shortest distance is defined as its height. We are provided with the lengths of all four sides of a specific trapezium, labeled PQRS, where the side PQ is given to be parallel to the side SR.

Trapezium Data and Goal

A trapezium is a four-sided polygon (quadrilateral) with at least one pair of parallel sides. In this question, PQ and SR are the parallel sides.

The given side lengths are:

- Side PQ (parallel to SR) = 11 cm
- Side QR = 4 cm
- Side RS (parallel to PQ) = 6 cm
- Side SP = 3 cm

Our objective is to find the perpendicular distance between the parallel sides PQ and SR, which represents the height of the trapezium.

Constructing Auxiliary Lines to Find Distance

To find the shortest distance (height) between the parallel sides PQ and SR, we can draw perpendicular lines from the vertices of the shorter parallel side (SR) to the longer parallel side (PQ).

- Draw a perpendicular line segment from S to PQ, and let the point where it meets PQ be T. So, ST is perpendicular to PQ.
- Draw another perpendicular line segment from R to PQ, and let the point where it meets PQ be U. So, RU is perpendicular to PQ.

This construction creates three distinct shapes within the trapezium PQRS:

- A rectangle STUR (since ST and RU are both perpendicular to PQ, they are parallel to each other, and SR is parallel to TU).
- Two right-angled triangles: ▵PST and ▵QRU.

From the properties of the rectangle STUR:

- ST = RU = h (this is the height we need to find).
- TU = SR = 6 cm.

Now, consider the side PQ. We know that PQ is composed of three segments: PT, TU, and UQ.

$$\text{PQ}=\text{PT}+\text{TU}+\text{UQ}$$

Substitute the given values:

$$11=\text{PT}+6+\text{UQ}$$

Rearrange the equation to find the sum of PT and UQ:

$$\text{PT}+\text{UQ}=11-6$$

$$\text{PT}+\text{UQ}=5\text{ cm}$$

Let's denote the length of PT as $x$. Then, the length of UQ will be $(5-x)$ cm.

Utilizing the Pythagorean Theorem for Height Calculation

We now have two right-angled triangles, ▵PST and ▵QRU, and we can apply the Pythagorean theorem ($a^2+b^2=c^2$) to each of them.Solving the System of Equations to Find the Shortest Distance

From the equations above, we have:

Equation 1 (from ▵PST):

$$9=x^2+h^2$$

We can express $h^2$ from this equation:

$$h^2=9-x^2\text{(Eq. A)}$$

Equation 2 (from ▵QRU):

$$16=(5-x{)}^2+h^2$$

Now, substitute the expression for $h^2$ from Eq. A into Equation 2:

$$16=(5-x{)}^2+(9-x^2)$$

Expand the term $(5-x{)}^2$. Recall the identity $(a-b{)}^2=a^2-2ab+b^2$:

$$(5-x{)}^2=5^2-2(5)(x)+x^2=25-10x+x^2$$

Substitute this expansion back into the equation:

$$16=(25-10x+x^2)+(9-x^2)$$

Combine the constant terms and notice that the $x^2$ terms cancel each other out:

$$16=25+9-10x+x^2-x^2$$

$$16=34-10x$$

Now, we solve for $x$:

$$10x=34-16$$

$$10x=18$$

$$x=\frac{18}{10}$$

$$x=1.8\text{ cm}$$

Finally, we substitute the value of $x$ back into Eq. A to find the value of $h^2$:

$$h^2=9-x^2$$

$$h^2=9-(1.8{)}^2$$

$$h^2=9-3.24$$

$$h^2=5.76$$

To find $h$, take the square root of $5.76$:

$$h=\sqrt{5.76}$$

$$h=2.4\text{ cm}$$

Final Result: Shortest Distance

The shortest distance between the parallel sides PQ and SR of the trapezium, which is its height, is found to be 2.40 cm.