(a) ABCD is a trapezium with AB parallel to DC and... | WAEC SSCE - General Mathematics - 1996 (Essay)

Question 1 Report

(a) ABCD is a trapezium with AB parallel to DC and /AD/ = /AB/. If < BAD = 106°, find < BDC.

(b) The table below shows the distribution of 20 cards labelled A - E.

Card	A	B	C	D	E
Frequency	3	4	7	5	1

(i) If a card is selected at random from the pack, what is the probability that the card is E? (ii) If two cards are selected at random one after the other without replacement from the pack, what is the probability that one of the two cards is B?

Answer

(a) Finding \(\angle BDC\) in the trapezium

ABCD is a trapezium with \(AB \parallel DC\) and \(|AD| = |AB|\), and \(\angle BAD = 106^{\circ}\).

Since \(AB \parallel DC\) with \(AD\) as transversal, \(\angle BAD\) and \(\angle ADC\) are co-interior (allied) angles, so they add to \(180^{\circ}\):

\(\angle ADC = 180^{\circ} - 106^{\circ} = 74^{\circ}\).

Triangle ABD is isosceles because \(|AD| = |AB|\); the base angles are equal:

\(\angle ADB = \angle ABD = \dfrac{180^{\circ}-106^{\circ}}{2} = 37^{\circ}\).

Now \(\angle ADC = \angle ADB + \angle BDC\), so

\(\angle BDC = 74^{\circ} - 37^{\circ} = 37^{\circ}\).

(b) The cards

Card	A	B	C	D	E
Frequency	3	4	7	5	1

Total number of cards \(= 3+4+7+5+1 = 20\).

(i) Probability the card is E

\(P(E) = \dfrac{1}{20}\).

(ii) Probability that one of the two cards (drawn without replacement) is B

There are 4 B-cards and 16 non-B cards. It is easiest to find the probability of drawing no B, then subtract from 1.

\(P(\text{no B}) = \dfrac{16}{20}\times\dfrac{15}{19} = \dfrac{240}{380} = \dfrac{12}{19}\).

\(P(\text{at least one B}) = 1 - \dfrac{12}{19} = \dfrac{7}{19}\).

Read lesson note on Probability (WAEC)

Read lesson note on Inequalities (JAMB)

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