4 Marks Questions

Question 14 Marks

The volume of a right circular cone is $9856\text{ }c{{m}^{3}}$. If the diameter of the base if 28 cm, find:

Height of the cone
Slant height of the cone
Curved surface area of the cone.

Answer

Diameter of cone = 28 cm
$\therefore$ Radius of cone = 14 cm
Volume of cone = $9856\text{ }c{{m}^{3}}$

$\Rightarrow$ $\frac{1}{3}\pi {{r}^{2}}h$ = 9856
$\Rightarrow$ $\frac{1}{3}\times \frac{22}{7}\times 14\times 14\times h=9856$
$\Rightarrow$ $h=\frac{9856\times 3\times 7}{22\times 14\times 14}$ = 48 cm
Slant height of cone $\left( l \right)=\sqrt{{{r}^{2}}+{{h}^{2}}}$
=$\sqrt{{{\left( 14 \right)}^{2}}+{{\left( 48 \right)}^{2}}}$
=$\sqrt{196+2304}$
=$\sqrt{2500}$ = 50 cm
Curved surface area of cone = $\pi rl=\frac{22}{7}\times 14\times 50$=$2200\text{ }c{{m}^{2}}$

Question 24 Marks

What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. $\left( \text{Use }\pi \text{ = 3}\text{.14} \right)$

Answer

Height of the conical tent $\left( h \right)$ = 8 m and Radius of the conical tent $\left( r \right)$ = 6 m
Slant height of the tent $\left( l \right)=\sqrt{{{r}^{2}}+{{h}^{2}}}$
=$\sqrt{{{\left( 6 \right)}^{2}}+{{\left( 8 \right)}^{2}}}$
=$\sqrt{36+64}$
=$\sqrt{100}$
= 10 m
Area of tarpaulin = Curved surface area of tent = $\pi rl$ $=3.14\times 6\times 10=188.4\text{ }{{m}^{2}}$
Width of tarpaulin = 3 m
Let Length of tarpaulin = L
$\therefore $ Area of tarpaulin = $Length\times Breadth\text{ }=\text{ }L\times 3$ = 3L
Now According to question, 3L = 188.4
$\Rightarrow $ L = 188.4/3 = 62.8 m
The extra length of the material required for stitching margins and cutting is 20 cm = 0.2 m.
So the total length of tarpaulin bought is (62.8 + 0.2) m = 63 m

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