\(\frac{20}{√3}\)m
\(\frac{40}{√3}\)m
20√3m
40√3m
Correct answer is D
\(\frac{40}{x}\) = tan 30°
x = \(\frac{40}{tan 30}\)
= \(\frac{40}{1\sqrt{3}}\)
= 40√3m
3√10
3√5
√26
√13
Correct answer is D
2x - y .....(i)
x + y.....(ii)
from (i) y = 2x - 4
from (ii) y = -x + 2
2x - 4 = -x + 2
x = 2
y = -x + 2
= -2 + 2
= 0
\(x_1\) = 2
\(y_1\) = 0
\(x_2\) = 4
\(y_2\) = 3
Hence, dist. = \(\sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}\)
= \(\sqrt{(3 - 0)^2 + (4 - 2)^2}\)
= \(\sqrt{3^2 + 2^2}\)
= \(\sqrt{13}\)
12cm
10cm
8cm
6cm
Correct answer is D
p from LM = \(\sqrt{10^2 - 8^2}\)
= \(\sqrt{36}\) = 6cm
x - y = 1
x + y = 1
x + y = 5
x - y = 5
Correct answer is C
m = tan 135° = -tan 45° = -1
\(\frac{y - y_1}{x - x_1}\) = m
\(\frac{y - 3}{x - 2}\) = -1
= y - 3 = -(x - 2)
= -x + 2
x + y = 5
A cone with the sector angle of 45° is cut out of a circle of radius r of the cone.
\(\frac{r}{16}\) cm
\(\frac{r}{6}\) cm
\(\frac{r}{8}\) cm
\(\frac{r}{2}\) cm
Correct answer is C
The formula for the base radius of a cone formed from the sector of a circle = \(\frac{r \theta}{360°}\)
= \(\frac{r \times 45°}{360°}\)
= \(\frac{r}{8} cm\)