(a) The mean of 1, 2, x, 11, y, 14, arranged in ascending order, is 8 and the median is 9. Find the values of x and y. (b) In the diagram, MN || PQ, |LM| = ...

Answer Details

(a) Find x and y.

The six numbers in ascending order are \(1,\;2,\;x,\;11,\;y,\;14\).

Use the mean. The mean of the six values is 8, so their sum is \(6\times 8=48\):

\[1+2+x+11+y+14=48\]

\[28+x+y=48\Rightarrow x+y=20\quad(1)\]

Use the median. For six values the median is the average of the 3rd and 4th values, which are \(x\) and \(11\):

\[\frac{x+11}{2}=9\Rightarrow x+11=18\Rightarrow x=7\]

Substitute into (1):

\[7+y=20\Rightarrow y=13\]

Check ascending order: \(1,2,7,11,13,14\) is valid. So \(x=7,\;y=13\).

(b) Area of quadrilateral MPQN.

Since \(MN\parallel PQ\), triangles \(LMN\) and \(LPQ\) are similar (equal angles at \(L\), and corresponding angles equal).

From the diagram, \(|LM|=3\text{ cm}\) and \(|MP|=1\text{ cm}\), so:

\[|LP|=|LM|+|MP|=3+1=4\text{ cm}\]

The ratio of corresponding sides is:

\[\frac{|LM|}{|LP|}=\frac{3}{4}\]

The ratio of areas of similar triangles is the square of the ratio of sides:

\[\frac{\text{Area }\Delta LMN}{\text{Area }\Delta LPQ}=\left(\frac{3}{4}\right)^{2}=\frac{9}{16}\]

Given \(\text{Area }\Delta LMN=18\text{ cm}^{2}\):

\[18=\frac{9}{16}\times\text{Area }\Delta LPQ\]

\[\text{Area }\Delta LPQ=18\times\frac{16}{9}=32\text{ cm}^{2}\]

The quadrilateral \(MPQN\) is the region between the two parallel lines:

\[\text{Area }MPQN=\text{Area }\Delta LPQ-\text{Area }\Delta LMN=32-18=14\text{ cm}^{2}\]