CSEC Math Jan 2012 P2 Q3

CSEC Math Jan 2012 Paper 2 Question 3

CSEC Math Jan 2012 Paper 2 Question 3

Updated on April 12th, 2012: Added some video explanations

This question tests your knowledge of numbers, sets and geometric constructions.

In part (a) we are given a universal set \(U\) defined as

$$
U = \{ 51, 52, 53, 54, 55, 56, 57, 58, 59 \}.
$$

Now, if \(A\) and \(B\) are subsets of \(U\), such that

$$
\begin{eqnarray}
A &=& \{ \text{ odd numbers } \} \\
B &=& \{ \text{ prime numbers } \}.
\end{eqnarray}
$$

Then, the members of the sets \(A\) and \(B\) are

$$
\begin{eqnarray}
A &=& \{ 51, 53, 55, 57, 59 \} \\
B &=& \{ 53, 59 \}.
\end{eqnarray}
$$

Here is a Venn diagram that represents the sets \(A\), \(B\) and \(U\).

A Venn diagram to represent the sets A, B and U

A Venn diagram to represent the sets A, B and U

In part (b), we need to do some constructions using a pair of compasses, a ruler and a pencil.

Firstly, we are asked to construct a triangle \(CDE\) in which \(DE = 10 \text{ cm}\), \(DC = 8 \text{ cm}\) and \(\angle CDE = 45^\circ\).

Here is an image of the final construction along with the tools used:

A construction of triangle CDE

A construction of triangle CDE

And here are the steps to reproduce it.

  1. Draw line \(DE\) of length 10 cm.
  2. Construct angle \(CDE\) of \(45^\circ\), such that \(CD\) has length 8 cm.
    1. Using \(D\) as center and any suitable radius, draw an arc to cut at \(a\).
    2. Then with \(a\) as center and radius \(Da\), draw an arc to cut at \(b\).
    3. Next using \(Da\) again, with \(b\) as center, draw another arc to cut at \(c\).
    4. Now with \(b\) and \(c\) as centers and any convenient radius, draw arcs to intersect at \(d\).
    5. Draw a line from \(D\) through \(d\) intersecting the arc \(a\) at \(e\).
    6. Using \(a\) and \(e\) as centers, draw equal arcs to intersect at \(f\).
    7. Draw a line from \(D\) through \(f\).
    8. Draw line \(DC\) of length 8 cm on line \(Df\). Angle \(CDE\) is equal to \(45^\circ\).
  3. Draw line \(CE\) to complete the construction of triangle \(CDE\).

Secondly, we are asked to construct a line, \(CF\), perpendicular to \(DE\) such that \(F\) lies on \(DE\).

Here is an image of the final construction:

A construction of the line CF perpendicular to DE

A construction of the line CF perpendicular to DE

And here are the steps to reproduce it.

  1. Using \(C\) as center, draw an arc to cut \(DE\) at \(g\) and \(h\).
  2. Now with \(g\) and \(h\) as centers, draw arcs of equal radii to intersect at \(i\).
  3. Draw a line from \(C\) through \(i\) cutting \(DE\) at \(F\). \(CF\) is perpendicular to \(DE\) as required.

Finally, using a protractor I have measured that \(\angle DCE\) has a size of approximately \(83^\circ\).

A protractor being used to measure the size of angle DCE

A protractor being used to measure the size of angle DCE