Mathematics Questions & Answers - Free Online Practice

1,491.

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

1,492.

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

1,493.

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

1,494.

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