perpendicular bisector of the two lines
angle bisector of the two lines
bisector of the two lines
line parallel to the two lines
Correct answer is B
The required locus is angle bisector of the two lines
If the heights of two circular cylinder are in the ratio 2 : 3 and their volumes?
27 : 32
27 : 23
23 : 32
27 : 23
Correct answer is A
\(\frac{h_1}{h_2}\) = \(\frac{2}{3}\)
h2 = \(\frac{2h_1}{3}\)
\(\frac{r_1}{r_2}\) = \(\frac{9}{8}\)
r2 = \(\frac{9r_1}{8}\)
v1 = \(\pi\)(\(\frac{9r_1}{8}\))2(\(\frac{2h_1}{3}\))
= \(\pi\)r1 2h1 x \(\frac{27}{32}\)
v = \(\frac{\pi r_1 2h_1 \times \frac{27}{32}}{\pi r_1 2h_1}\) = \(\frac{27}{32}\)
v2 : v1 = 27 : 32
153\(\pi\)cm3
207\(\pi\)cm3
15 300\(\pi\)cm3
20 700\(\pi\)cm3
Correct answer is D
Volume of a cylinder = πr\(^2\)h
First convert 3m to cm by multiplying by 100
Volume of External cylinder = π \times 13\(^2\) \times 300
Volume of Internal cylinder = π \times 10\(^2\) \times 300
Hence; Volume of External cylinder - Volume of Internal cylinder
Total volume (v) = π (169 - 100) \times 300
V = π \times 69 \times 300
V = 20700πcm\(^3\)
If Cos \(\theta\) = \(\frac{12}{13}\). Find \(\theta\) + cos2\(\theta\)
\(\frac{169}{25}\)
\(\frac{25}{169}\)
\(\frac{169}{144}\)
\(\frac{144}{169}\)
Correct answer is A
Cos \(\theta\) = \(\frac{12}{13}\)
x2 + 122 = 132
x2 = 169- 144 = 25
x = 25
= 5
Hence, tan\(\theta\) = \(\frac{5}{12}\) and cos\(\theta\) = \(\frac{12}{13}\)
If cos2\(\theta\) = 1 + \(\frac{1}{tan^2\theta}\)
= 1 + \(\frac{1}{\frac{(5)^2}{12}}\)
= 1 + \(\frac{1}{\frac{25}{144}}\)
= 1 + \(\frac{144}{25}\)
= \(\frac{25 + 144}{25}\)
= \(\frac{169}{25}\)
Each of the interior angles of a regular polygon is 140°. How many sides has the polygon?
9
8
7
5
Correct answer is A
For a regular polygon of n sides
n = \(\frac{360}{\text{Exterior angle}}\)
Exterior < = 180° - 140°
= 40°
n = \(\frac{360}{40}\)
= 9 sides