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Midterm

Mathematics 21a Study Guide - Midterm Guide: Cone


Department
Mathematics
Course Code
Mathematics 21a
Professor
math
Study Guide
Midterm

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Notes: Important formulas to be kept in mind:
Parameters of a Cone: Radius of the base (r ), Height of the cone (h ) and Slant
Height of a Cone (l )
Volume of a Cone = \frac{1}{3} \pi r^2 h
Curved Surface area of a Cone = \pi r l
Total Surface area of a Cone = \pi r^2+ \pi r l
Question 1: Find the volume of a cone whose slant height is 17 \ cm and radius
of base is 8 \ cm .
Answer:
Volume of a Cone = \frac{1}{3} \pi r^2 h
l = 17 \ cm, \ r = 8 \ cm
Therefore h = \sqrt{l^2 - r^2} = \sqrt{17^2-8^2} = 15 \ cm
Therefore Volume = \frac{1}{3} \times \frac{22}{7} \times (8)^2 \times (15) =
1005.71 \ cm^3
\\
Question 2: The curved surface area of a cone is 12320 \ cm^2 . If the radius of
its base is 56 \ cm , find its height.
Answer:
Curved surface area of the cone = 12320 \ cm^2
r = 56 \ cm
\pi r l = 12320 \Rightarrow l = \frac{12320}{\pi . 56} = 70 \ cm
Therefore h = \sqrt{l^2 - r^2} = \sqrt{70^2-56^2} = 1542 \ cm
\\
Question 3: The circumference of the base of a 12 \ m high conical tent is 66 \
m . Find the volume of the air contained in it.
Answer:
h = 12 \ m
Circumference of the base = 66 \ m
Therefore 2 \pi r = 66 \Rightarrow r = \frac{33}{\pi}
Volume of a Cone = \frac{1}{3} \pi r^2 h
= \frac{1}{3} \times \pi \times (\frac{33}{\pi})^2 \times (12) = 1386 \ m^3
\\
Question 4: The radius and the height of a right circular cone are in the ratio
5:12 and its volume is 2512 \ cm^3 . Find the radius and slant height of the
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cone. (Take \pi = 3.14 )
Answer:
Given: Radius and the height of a right circular cone are in the ratio 5:12
Let r = 5x and h = 12 x
Volume = 2512 \ cm^3
Therefore: \frac{1}{3} \pi r^2 h = 2512
\Rightarrow \frac{1}{3} \times 3.14 \times (5x)^2 \times (12x) = 2512
\Rightarrow x^3 = 8 \ or \ x = 2
Therefore Radius = 5 (2) = 10 \ cm and Height = 12(2) = 24 \ cm
\\
Question 5: Two right circular cones X and, Y are made. X having three times
the radius of Y and Y having half the volume of X . Calculate the ratio
between the heights of X and Y .
Answer:
Let the radius of cone Y is x . Therefore the radius of cone X = 3x .
Let the height of cone Y = h_y and the height of cone X = h_x
Given: V_y = \frac{1}{2} V_x
\Rightarrow \frac{1}{3} \pi (x)^2 (h_y) = \frac{1}{2} \{ \frac{1}{3} \pi (3x)^2
(h_x) \}
\Rightarrow \frac{h_x}{h_y} = \frac{x^2}{(3x)^2}
\Rightarrow \frac{h_x}{h_y} = \frac{2}{9}
\\
Question 6: The diameters of two cones are equal. If their slant heights are in
the ratio 5 : 4 , find the ratio of their curved surface areas.
Answer:
Let the radius of Cone 1 and Cone 2 = x
Let the slant height of the cones be l_1 and l_2 respectively.
Curved Surface Area of a cone = \pi r l
Therefore
\frac{Curved \ Surface \ Area \ of \ Cone \ 1}{Curved \ Surface \ Area \ of \
Cone \ 2} = \frac{\pi \times x \times l_1}{\pi \times x \times l_2} = \frac{5}
{4}
\\
Question 7: There are two cones. The curved surface area of one is twice that of
the other. The slant height of the latter is twice that of the former. Find the
ratio of their radii.
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