Processing math: 100%

WAEC Mathematics Past Questions & Answers - Page 161

801.

In the diagram P, Q, R, S are points on the circle RQS = 30^o. PRS = 50^o and PSQ = 20^o. What is the value of x^o + y^o?

260^o

130^o

100^o

80^o

View answer & explanation

Correct answer is B

Draw a line from P to Q

< PQS = < PRS (angle in the sam segment)

< PQS = 50^o

Also, < QSR = < QPR(angles in the segment)

< QPR = x^o

x + y + 5= = 180(angles in a triangle)

x + y = 180 - 50

x + y = 130^o

802.

In the diagram, PR is a diameter of the circle centre O. RS is a tangent at R and QPR = 58^o. Find < QRS

112^o

116^o

122^o

148^o

View answer & explanation

Correct answer is C

PRQ = 90 - 58 = 32^o(angle in a semi-circle)

Since PRS = 90^o(radius angular to tangent)

QRS = 90 + 32

= 122^o

803.

In the diagram, $\frac{PQ}{RS}$ , find x^o + y^o

360^o

300^o

270^o

180^o

View answer & explanation

Correct answer is D

PQR = QRS = Y(alt. amgles)

PQR + x = 180^o(angles on a straight line)

y + x = 180^o

804.

In the diagram, |QR| = 5cm, PQR = 60<sup>o</sup> and PSR = 45<sup>o</sup>. Find |PS|, leaving your answe in surd form.

4 $\sqrt{5}$ cm

3 $\sqrt{7}$ cm

4 $\sqrt{6}$ cm

5 $\sqrt{6}$ cm

View answer & explanation

Correct answer is D

tan 6^o = $\frac{|PR|}{|QR|}$

$\sqrt{3} = \frac{|PR|}{5}$

= |PR| = $5 \sqrt{3}$ cm

sin 45 = $\frac{|PR|}{|PS|}$

$\frac{1}{\sqrt{2}}$ = $\frac{5 \sqrt{3}}{|PS|}$

|PS| = $5 \sqrt{3}$ x $\sqrt{2}$

= 5 $\sqrt{6}$ cm

805.

In the diagram, PQUV, PQTU, QRTU and QRST are parallelograms. |UV| = 4.8cm and the perpendicular distance between PR and VS is 5cm. Calculate the area of quadrilateral PRSV

96cm²

72cm²

60cm²

24cm²

View answer & explanation

Correct answer is C

|VU| = |UT| = |TS|

|VS| = (4.8) x 3 = 14.4cm

|PQ| = |QR| = 4.8

|PR| = (4.8) x 2 = 9.6cm

Since quad. PRSV is a trapezium of height 5cm

Area of quad. PRSV = $\frac{1}{2}(a + b)h$

= $\frac{1}{2}(14.4 + 9.6) \times 5$

= $\frac{1}{2}(24) \times 5$

12 x 5 = 60cm²