If the diagram is the graph of y = x2, the shaded area is
64 square units
\(\frac{126}{4}\) square units
32 square units
\(\frac{64}{3}\) square units
Correct answer is D
A = \(\int^{x = 4}_{x = 0}\) ydx = \(\int^{4}_{0}\) x2dx = [\(\frac{x^3}{3} + c]^{4}_{0}\)
= \(\frac{64}{3}\) square units
32cm
26cm
20cm
16cm
Correct answer is B
\(\bigtriangleup\)OEN = \(\bigtriangleup\)OGN
OE = OG = r(radii)
(r - 8)2 + 122 = r2
r2 - 16r + 64 + 144 = r2
16r = 64 + 144 = r2
16r = 64 + 144 = 208
\(\frac{208}{16}\) = 13cm
FH = 2FO = 2OH = 2R
= 2 x 13cm
= 26cm
In the diagram, If < RPS = 50o, < RPQ = 30o and pq = QR, Find the value of < prs
50o
60o
70o
80o
Correct answer is C
< QPR = < PRQ = 30o (PQ = QR)
< SPQ + < QRS = 180o (supplementary)
80 + < QRS = 180o = < QRS = 180o - 80o = 100o
< QRP + < PRS = 100o = 30o + < PRS = 100o
= < PRS = 100o - 30o = 70o
7
8
9
12
Correct answer is A
degree of excellent in the pie-chart = 360o - (90o + 120o + 80o) = 360o - 290o = 70o
= 360o = 36 students
70 = \(\frac{70^o}{360^o}\) x 36 = 7
find the length XZ in the triangle
\(\sqrt{7m}\)
\(\sqrt{6m}\)
\(\sqrt{5m}\)
\(\sqrt{3m}\)
Correct answer is A
xz2 = xy2 + yx2 - 2(xy) (yz) cos 120o
= 22 + 12 - 2(2) (1) cos 1202
= 4 + 1 - 4x - cos 60 = 5 - 4x - \(\frac{1}{2}\)
5 + 2 = 7
xz = \(\sqrt{7}\)m
JAMB Subjects
Aptitude Tests