Mathematics Questions & Answers - Free Online Practice

1,601.

In the diagram. Find h

\(\frac{12}{7}\)cm

\(\frac{12}{7} \sqrt{6}\)cm

\(\frac{7}{12}\)cm

\(\frac{1}{2}\)cm

View answer & explanation

Correct answer is B

A_{\(\bigtriangleup\)} = \(\sqrt{S(S - a) (S - b)(S - c)}\) (Hero's Formula)

S = \(\frac{a + b + c}{2}\) = \(\frac{5 + 6 + 7}{2}\)

\(\frac{18}{2} = 9\)

A_{\(\bigtriangleup\)} \(\sqrt{9} \times 4 \times 3 \times 2\)

\(\sqrt{216} = 6 \sqrt{6}cm^3\)

A_{\(\bigtriangleup\)} = \(\frac{1}{2} \times 6 \times h\)

6\(\sqrt{6} = \frac{1}{2} \times 7 \times h\)

h = \(\frac{12}{h} \sqrt{6}\)

1,602.

In the diagram, PTS is a tangent to the circle TQR at T. Calculate < RTS

120^o

70^o

60^o

40^o

View answer & explanation

Correct answer is B

RTS = RQT (angle between a tangent and a chord at the point of contact is equal to the angle in the alternate segment) But R = Q + T = 180

RQT = 180^o - (50 + 60)

= 180^o - 110^o

= 70^o

Since RQT = RTS = 70^o

1,603.

In the diagram, O is the centre of the circle. If SOQ is a diameter and < PRS is 38^o, what is the value of < PSQ

148^o

104^o

80^o

52^o

View answer & explanation

Correct answer is D

< SRP = < SQP = 38^o (angle in the same segment of a circle are equal)

But < SPQ = 90^o (angle in a semicircle)

also < PSQ + < SQP + < SPQ = 180^o (angles in a triangle = 180^o)

< PSQ + 38^o + 90^o = 190^o

< PSQ = 128^o = 180^o

PSQ = 180^o - 128^o

PSQ = 52^o

1,604.

In the diagram, PQRS is a parallelogram. Find the value of < SQR

30^o

50^o

80^o

100^o

View answer & explanation

Correct answer is D

SQR + RQV + VQU = 18^o angle on a straight line SP is parallel to QR and PV is parallel to TR

< STP = < RQV = 30^o

But SQR + 30^o + 50^o = 180^o

SQR = 180 - 80

= 100^o