260o
130o
100o
80o
Correct answer is B
Draw a line from P to Q
< PQS = < PRS (angle in the sam segment)
< PQS = 50o
Also, < QSR = < QPR(angles in the segment)
< QPR = xo
x + y + 5= = 180(angles in a triangle)
x + y = 180 - 50
x + y = 130o
112o
116o
122o
148o
Correct answer is C
PRQ = 90 - 58 = 32o(angle in a semi-circle)
Since PRS = 90o(radius angular to tangent)
QRS = 90 + 32
= 122o
In the diagram, \(\frac{PQ}{RS}\), find xo + yo
360o
300o
270o
180o
Correct answer is D
PQR = QRS = Y(alt. amgles)
PQR + x = 180o(angles on a straight line)
y + x = 180o
4\(\sqrt{5}\)cm
3\(\sqrt{7}\)cm
4\(\sqrt{6}\)cm
5\(\sqrt{6}\)cm
Correct answer is D
tan 6o = \(\frac{|PR|}{|QR|}\)
\(\sqrt{3} = \frac{|PR|}{5}\)
= |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
96cm2
72cm2
60cm2
24cm2
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
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