Exercise 2.4 - Chapter 2 Measurements 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $2.4$
Question $1 .$
Two gates are fitted at the entrance of a library. To open the gates easily, a wheel is fixed at 6 feet distance from the wall to which the gate is fixed. If one of the gates is opened to $90^{\circ}$, find the distance moved by the wheel $(\pi=3.14)$.

Answer:
Let $\mathrm{A}$ be the position of the wall $\mathrm{AC}$ be the gate in initial position and $\mathrm{AB}$ be position when it is muved $90^{\circ}$.
Now the arc length $\mathrm{BC}$ gives the distance moved by the wheel.

Length of the arc $=\frac{\theta}{360^{\circ}} \times 2 \pi \mathrm{r}$ units $=\frac{90^{\circ}}{360^{\circ}} \times 2 \times 3.14 \times 6$ feets $=3.14 \times 3$ feets $=9.42$ feets
$\therefore$ Distance moved by the wheel $=9.42$ feets.

Question $2 .$
With his usual speed, if a person covers a circular track of radius $150 \mathrm{~m}$ in 9 minutes, find the distance that he covers in 3 minutes $(\pi=3.14)$.
Answer:
Radius of the circular track $=150 \mathrm{~m}$
Distance covers in 9 minutes $=$ Perimeter of the circle $=2 \times \pi \times r$ units
Distance covered in $9 \mathrm{~min}=2 \times 3.14 \times 150 \mathrm{~m}$
Distance covered in $1 \mathrm{~min}=\frac{2 \times 3.14 \times 150}{9} \mathrm{~m}$

Distance he covers in $3 \mathrm{~min}=314 \mathrm{~m}$
 

Question $3 .$
Find the area of the house drawing given in the figure.

Answer:
Area of the house $=$ Area of a square of side $6 \mathrm{~cm}$ + Area of a rectangle with $1=$ $8 \mathrm{~cm}, \mathrm{~b}=6 \mathrm{~cm}+$ Area of a $\Delta$ with $\mathrm{b}=6 \mathrm{~cm}$ and $\mathrm{h}=4 \mathrm{~cm}$ + Area of a parallelogram with $\mathrm{b}=8 \mathrm{~cm}, \mathrm{~h}=4 \mathrm{~cm}$
$
\begin{aligned}
&=(\text { side } \times \text { side })+(1 \times b)+\left(\frac{1}{2} \times \mathrm{b} \times \mathrm{h}\right)+\mathrm{bh} \mathrm{cm}^{2} \\
&=(6 \times 6)+(8 \times 6)+\left(\frac{1}{2} \times 6 \times 4\right)+(8 \times 4) \mathrm{cm}^{2} \\
&=36+48+12+32 \mathrm{~cm}^{2} \\
&=128 \mathrm{~cm}^{2} \\
&\text { Required Area }=128 \mathrm{~cm}^{2}
\end{aligned}$

Question $4 .$
Draw the top, front and side view of the following solid shapes

(i)

Question $5 .$
Guna has fixed a single door of width 3 feet in his room where as Nathan has fixed a double door, each of width $1 \frac{1}{2}$ feet in his room. From the closed position, if each of the single and double doors can open up to $120^{\circ}$, whose door takes a minimum area?
Answer:
Width of the door that Guna fixed $=3$ feet.
When the door is open the radius of the sector $=3$ feet

Angle covered $=120^{\circ}$
$\therefore$ Area required to open the door
$\begin{aligned}
&=\frac{120^{\circ}}{360^{\circ}} \times \pi r^{2}=\frac{120^{\circ}}{360^{\circ}} \times \pi \times 3 \times 3 \\
&=3 \pi \text { feet }^{2}
\end{aligned}$

(b) Width of the double doors that Nathan fixed $=1 \frac{1}{2}$ feet.
Angle described to open $=120^{\circ}$
Area required to open $=2 \times$ Area of the sector
$=2 \times \frac{120^{\circ}}{360^{\circ}} \times \pi \times \frac{3}{2} \times \frac{3}{2}$ feets $^{2}=\frac{3 \pi}{2}$ feet $^{2}{ }_{z}$ $=\frac{1}{2}(3 \pi)$ feet $^{2}$
$\therefore$ The double door requires the minimum area.
 

Question $6 .$
In a rectangular field which measures $15 \mathrm{mx} 8 \mathrm{~m}$, cows are tied with a rope of length $3 \mathrm{~m}$ at four corners of the field and also at the centre. Find the area of the field where none of the cow can graze. $(\pi=3.14)$.
Answer:
Area of the field where none of the cow can graze $=$ Area of the rectangle $-[$ Area of 4 quadrant circles] - Area of a circle

Area of the rectangle $=1 \times$ b units $^{2}$
$=15 \times 8 m^{2}=120 m^{2}$
Area of 4 quadrant circles $=4 \times \frac{1}{4} \pi \mathrm{r}^{2}$ units
Radius ofthe circle $=3 \mathrm{~m}$
Area of 4 quadrant circles $=4 \times \frac{1}{4} \times 3.14 \times 3 \times 3=28.26 \mathrm{~m}^{2}$
Area of the circle at the middle $=\pi \mathrm{r}^{2}$ units
$=3.14 \times 3 \times 3 \mathrm{~m}^{2}=28.26 \mathrm{~m}^{2}$
$\therefore$ Area where none of the cows can graze
$=[120-28.26-28.26] \mathrm{m}^{2}=120-56.52 \mathrm{~m}^{2}=63.48 \mathrm{~m}^{2}$

Question $7 .$
Three identical coins, each of diameter $6 \mathrm{~cm}$ are placed as shown. Find the area of the shaded region between the coins. $(\pi=3.14)(\sqrt{3}=1.732)$

Answer:
Given diameter of the coins $=6 \mathrm{~cm}$
$\therefore$ Radius of the coins $=\frac{6}{2}=3 \mathrm{~cm}$
Area of the shaded region $=$ Area of equilateral triangle $-$ Area of 3 sectors of angle $60^{\circ}$

Area of the equilateral triangle $=\frac{\sqrt{3}}{4} \mathrm{a}^{2}$ units $^{2}=\frac{\sqrt{3}}{4} \times 6 \times 6 \mathrm{~cm}^{2}$ $=\frac{1.732}{4} \times 6 \times 6 \mathrm{~cm}^{2}=15.588 \mathrm{~cm}^{2}$
Area of 3 sectors $=3 \times \frac{\theta}{360^{\circ}} \times \pi r^{2}$ sq.units $=3 \times \frac{60^{\circ}}{360^{\circ}} \times 3.14 \times 3 \times 3 \mathrm{~cm}^{2}=1.458 \mathrm{~cm}^{2}$
$\therefore$ Area of the shaded region $=15.588-14.13 \mathrm{~cm}^{2}=1.458 \mathrm{~cm}^{2}$
Required area $=1.458 \mathrm{~cm}^{2}$ (approximately)
 

Question $8 .$
Using Euler's formula, find the unknowns.

Answer:
Euler's formula is given by $\mathrm{F}+\mathrm{V}-\mathrm{E}=2$
(i) $\mathrm{V}=6, \mathrm{E}=14$
By Euler's formula $=\mathrm{F}+6-14=2$
$\begin{aligned}
&\mathrm{F}=2+14-6 \\
&\mathrm{~F}=10
\end{aligned}$
(ii) $\mathrm{F}=8, \mathrm{E}=10$
By Euler's formula $=8+\mathrm{V}-10=2$
$\begin{aligned}
&\mathrm{V}=2-8+10 \\
&\mathrm{~V}=4
\end{aligned}$
(iii) $\mathrm{F}=20, \mathrm{~V}=10$
By Euler's formula $=20+10-\mathrm{E}=2$
$\begin{aligned}
&30-\mathrm{E}=2 \\
&\mathrm{E}=30-2 \\
&\mathrm{E}=28
\end{aligned}$