In the figure above, find the value of y
28o
122o
150o
152o
Correct answer is B
ADB = 180o - 152o = 28o
28o + y + 30o = 180o
y = 180o - 58o
= 122o
40o
30o
25o
15o
Correct answer is D
PQM = 110o(Ext. < of a cyclic quad)
MPQ = 180o - (110 + 55)
= 180o - 165o
= 15o
7.3, 5.9
7.7, 12.5
12.5, 7.7
5.9, 7.3
Correct answer is C
\(\frac{PQ}{PN} = \frac{PM}{PR} = \frac{QM}{NR}\)
\(\frac{4.8}{12} = \frac{5}{PR}\)
PR = \(\frac{5 \times 12}{4.8} = \frac{50}{4}\)
= 12.5
\(\frac{PQ}{PN} = \frac{PM}{PT} = \frac{TM}{NT}\)
\(\frac{PT}{12} = \frac{5}{PR}\)
PT2 = 60
PT = \(\sqrt{60}\)
= 7.746
= 7.7
In the figure above, Find the value of x
130o
110o
100o
90o
Correct answer is A
No explanation has been provided for this answer.
Find the curved surface area of the frustrum in the figure
16\(\pi \sqrt{10}\)cm2
20\(\pi \sqrt{10}\)cm2
24\(\pi \sqrt{10}\)cm2
36\(\pi \sqrt{10}\)cm2
Correct answer is B
\(\frac{x}{4} = \frac{6 + x}{6}\)
6x = 4(6 + x) = 24 + 4x
x = 12cm
CSA = \(\pi RL - \pi rl\)
= \(\pi (6) \sqrt{{18^2} + 6^2} - \pi \times 4 \times \sqrt{{12^2} + 4^2}\)
= 6\(\pi \sqrt{360} - 4 \pi \sqrt{160}\)
= 36\(\pi \sqrt{10} - 16 \pi \sqrt{10}\)
= 20\(\pi \sqrt{10}\)cm2
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