In the figure, PQ\\SR, ST\\, ST\\RQ, PS = 7cm, PT = 7cm, SR = 4cm. Find the ratio of the area QRST to the area of PQRS.

A. 56.77

B. 56.105

C. 28:105

D. 28:49

E. 56:49

In the figure, If PT is parallel to RS, PQ = PT, and angle SQT = 90^o, Find x

A. 35^o

B. 50^o

C. 55^o

D. 70^o

E. 80^o

In the figure, PS bisects angle QPR. Find the ratio SR:QR.

A. 1:2

B. 1:3

C. 1:4

D. 1:5

E. 1:6

FG is a given straight line and H is a fixed point. The construction marks shown in the diagram indicate the

A. perpendicular bisector of the line

B. perpendicular bisector of th eline JK

C. perpendicular bisector of the line HI

D. perpendicular from H to the line FG

In the figure, FGHJ is a circle of radius 3cm centre O. FOH, GOJ are perpendicular diameters. With G as centre of an arc of a circle is drawn to pass through F and H. Find the length of the perimeter of the lunar portion shaded.

A. 3\(\pi\)\(\sqrt{2 - 1}\)cm

B. \(\frac{9}{2}\)\(\pi\)cm

C. 3 \(\pi\)(1 + \(\frac{2}{2}\))

D. 3(1 + \(\frac{2}{2}\))

The angles marked in the figure are measured in degrees. Find x.

A. 45^o

B. 90^o

C. 60^o

D. 15^o

E. 30^o

In the figure, PS = SR and Rx is a tangent to the circle at R, < QRX is equal to 72^o. Find P

A. 20^o

B. 36^o

C. 72^o

D. 30^o

E. 15^o

If \(\theta\) = \(\frac{3}{2}\) and \(\theta\) is less than 90^o, calculate tan \(\frac{(90 - \theta)}{cos^2\theta}\)

A. \(\frac{1}{2}\)

B. \(\frac{4}{\sqrt{3}}\)

C. \(\sqrt{3}\)

D. 4\(\sqrt{3}\)

Find the area of the shades segments in the figure

A. \(\sqrt{3}

B. 4 \(\pi - \sqrt{3}\)

C. -\(\frac{2}{3} \pi\)

D. \(\frac{2\pi}{3}\) -3