Exercise 7.1 - Chapter 7 Mensuration 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]
Updated By SaraNextGen
On April 24, 2024, 11:35 AM
Ex $7.1$ : Chapter 7 - Mensuration - 10th Maths Guide Samacheer Kalvi Solutions
Question 1.
The radius and height of a cylinder are in the ratio $5: 7$ and its curved surface area is 5500 sq.cm. Find its radius and height.
Solution:
$
\begin{aligned}
& \mathrm{r}=5 \mathrm{x} \\
& \mathrm{h}=7 \mathrm{x}
\end{aligned}
$
CSA of a cylinder $=2 \pi r h$
$
\begin{aligned}
& =2 x \frac{22}{7} \times 5 x \times 7 x=5500 \\
& 22 \emptyset x^2=550 \emptyset \\
& x^2=\frac{550}{22}=25 \\
& \therefore \quad x=5 \\
& \therefore \text { Radius }=5 \times 5=25 \mathrm{~cm} \\
& \text { height }=7 \times 5=35 \mathrm{~cm} \\
&
\end{aligned}
$
Question 2.
A solid iron cylinder has total surface area of 1848 sq.m. Its curved surface area is five - sixth of its total surface area. Find the radius and height of the iron cylinder.
The external radius and the length of a hollow
Solution:
$
\begin{aligned}
\text { C.S.A. } & =\frac{5}{6} \text { T.S.A } \\
2 \pi r(h+r) & =1848 \mathrm{~m}^2 \\
2 \pi r h+2 \pi r^2 & =1848 \mathrm{~m}^2 \\
\frac{5}{6} \times 1848+2 \pi r^2 & =1848 \\
1540+2 \pi r^2 & =1848 \\
2 \pi r^2 & =1848-1540 \\
& =308 \\
r^2 & =308 \times \frac{1}{2} \times \frac{7}{22} \\
r^2 & =49 \\
r & =7 m . \\
2 \pi r h & =\frac{5}{6} \times 1848 \\
2 \times \frac{22}{7} \times 7 \times h & =\frac{5}{6} \times 1848 \\
h & =35 \mathrm{~m} \\
\therefore \text { Radius } r=7 \mathrm{~m}, & \text { Height }=35 \mathrm{~m} .
\end{aligned}
$
Question 3.
The external radius and the length of a hollow wooden $\log$ are $16 \mathrm{~cm}$ and $13 \mathrm{~cm}$ respectively. If its thickness is $4 \mathrm{~cm}$ then find its T.S.A.
Solution:
$
\begin{aligned}
& \mathrm{R}=16 \mathrm{~cm} \\
& \mathrm{r}=\mathrm{R} \text { - thickness } \\
& \mathrm{r}=12 \mathrm{~cm} \\
& =16-4=12 \mathrm{~cm} \\
& \mathrm{~h}=13 \mathrm{~cm}
\end{aligned}
$
Total surface area of hollow cylinder $=2 \pi(R+r)(R-r+h)$ sq. units.
$
\begin{aligned}
& =2 x \frac{22}{7}(16+12)(16-12+13) \\
& =\frac{44}{\not 7}\left(\begin{array}{c}
4 \\
28
\end{array}\right)(17) \\
& \therefore \text { T.S.A }=2992 \mathrm{~cm}^2
\end{aligned}
$
Question 4.
A right angled triangle $\mathrm{PQR}$ where $\angle \mathrm{Q}=90^{\circ}$ is rotated about $\mathrm{QR}$ and $\mathrm{PQ}$. If $\mathrm{QR}=16 \mathrm{~cm}$ and $\mathrm{PR}=$ $20 \mathrm{~cm}$, compare the curved surface areas of the right circular cones so formed by the triangle.
Solution:
When it is rotated about $\mathrm{PQ}$ the C.S.A of the cone formed $=\pi \mathrm{rl}$.
.png)
When it is rotated about $\mathrm{QR} \mathrm{CSA}$ of the cone formed.
$
\begin{aligned}
& =\pi r l . \\
\text { here } r & =\sqrt{l^2-h^2} \\
& =\sqrt{20^2-16^2} \\
& =\sqrt{400-256} \\
& =\sqrt{144}=12 \mathrm{~cm}
\end{aligned}
$
$
\begin{aligned}
\text { CSA } & =\pi r l=\frac{22}{7} \times 12 \times 20 \\
& =\frac{5280}{7} \\
& =754.28 \mathrm{~cm}^2 .
\end{aligned}
$
$
1005.71>754.28
$
$\therefore \mathrm{CSA}$ of the cone rotated about its $\mathrm{PQ}$ is larger than the CSA of the cone rotated about $\mathrm{QR}$.
Question 5.
4 persons live in a conical tent whose slant height is $19 \mathrm{~cm}$. If each person require $22 \mathrm{~cm}^2$ of the floor area, then find the height of the tent.
Solution:
Base area of the cone $=\pi r^2=$ sq units.
.png)
$
\begin{aligned}
\pi r^2 & =4 \times 22 \mathrm{~cm}^2 \\
r^2 & =88 \times \frac{7}{22} \\
& =\frac{616}{22}=28 \mathrm{~cm}^2
\end{aligned}
$
$
\begin{aligned}
l & =19 \mathrm{~cm} \\
h & =\sqrt{l^2-r^2} \\
& =\sqrt{19^2-28} \\
& =\sqrt{361-28} \\
& =\sqrt{333} \\
& =18.25 \mathrm{~cm} .
\end{aligned}
$
Question 6.
A girl wishes to prepare birthday caps in the form of right circular cones for her birthday party, using a sheet of paper whose area is $5720 \mathrm{~cm}^2$, how many caps can be made with radius $5 \mathrm{~cm}$ and height $12 \mathrm{~cm}$.
Solution:
$\begin{aligned}
r & =5 \mathrm{~cm} \\
h & =12 \mathrm{~cm} \\
& =l=\sqrt{r^2+h^2} \\
& =\sqrt{25+144} \\
& =\sqrt{169} \\
& =13 \mathrm{~cm}
\end{aligned}$
.png)
$
\begin{aligned}
\text { Kequired no. of caps } & =\frac{\text { Area of the paper }}{\text { Area of } 1 \text { cap }} \\
& =\frac{5720}{\pi r l}=\frac{5720}{\frac{22}{7} \times 5 \times 13} \\
& =\frac{5720}{\frac{1430}{7}}=572 \varnothing \times \frac{7}{143 \varnothing} \\
& =\frac{4004}{143}=28 \text { caps }
\end{aligned}
$
Question 7.
The ratio of the radii of two right circular cones of same height is $1: 3$. Find the ratio of their curved surface area when the height of each cone is 3 times the radius of the smaller cone.
Solution:
$\begin{aligned}
r_1: r_2=1: 3 & \Rightarrow 1 x: 3 x \\
h_1=3 x, h_2=3 x, l_1 & =\sqrt{r_1^2+h_1^2} \\
& =\sqrt{x^2+(3 x)^2} \\
& =\sqrt{x^2+9 x^2}=\sqrt{10 x^2} \\
& =\sqrt{10} x \\
l^2 & =\sqrt{r_2^2+h_2^2} \\
& =\sqrt{(3 x)^2+(3 x)^2} \\
& =\sqrt{9 x^2+9 x^2} \\
& =\sqrt{18 x^2}=3 \sqrt{2} x
\end{aligned}$
$
\begin{aligned}
\therefore \mathrm{CSA}_1: \mathrm{CSA}_2 & =\frac{\pi r_1 l_1}{\pi r_2 l_2}=\frac{\pi \times 1 x \times \sqrt{10} x}{\pi \times 3 x \times 3 \sqrt{2} x} \\
& =\frac{\pi \times 1 \not k \times \sqrt{10} \not k}{\pi \times 3 \not x \times 3 \sqrt{2} \not k} \\
& =\frac{\sqrt{10}}{9 \sqrt{2}}=\frac{\sqrt{2 \times 5}}{9 \sqrt{2}} \\
& =\frac{\sqrt{2} \times \sqrt{5}}{9 \sqrt{2}}=\frac{\sqrt{5}}{9} \\
& =\sqrt{5}: 9 .
\end{aligned}
$
$\therefore$ The ratio of their curved surface areas
$
\Rightarrow \sqrt{5}: 9
$
Question 8.
The radius of a sphere increases by $25 \%$. Find the percentage increase in its surface area.
Solution:
Surface area of sphere $\mathrm{A}=4 \pi \mathrm{r}^2$
New radius $=r^{\prime}=1.25 \mathrm{r}$,
$[\because \mathrm{r}+0.25 \mathrm{r}](25 \%=0.25)$
New surface area $=\mathrm{A}^{\prime}=4 \pi\left(\mathrm{r}^{\prime}\right)^2$
$
\begin{aligned}
& =4 \pi(1.25 \mathrm{r})^2 \\
& 4 \pi r^2[1.5625] \\
& \% \text { increase in } \mathrm{S} . \mathrm{A}=\frac{\text { New } \mathrm{SA}-\text { old } \mathrm{SA}}{\text { Old surface Area }} \times 100
\end{aligned}
$
$
\begin{aligned}
& =\frac{\mathrm{A}^{\prime}-\mathrm{A}}{\mathrm{A}} \times 100 \\
& =\left(\frac{1.5625 \mathrm{~A}-\mathrm{A}}{\mathrm{A}}\right) \times 100 \\
& =\frac{0.5625 \mathrm{~A}}{\mathrm{~A}} \times 100 \\
& =56.25 \%
\end{aligned}
$
Question 9.
The internal and external diameters of a hollow hemispherical vessel are $20 \mathrm{~cm}$ and $28 \mathrm{~cm}$ respectively. Find the cost to paint the vessel all over at $\square 0.14$ per $\mathrm{cm}^2$.
Solution:
External diameter $\mathrm{D}=28 \mathrm{~cm}$
Internal diameter $\mathrm{d}=20 \mathrm{~cm}$
$
\therefore \mathrm{R}=\frac{28}{2}=14 \mathrm{~cm}, r=\frac{20}{2}=10 \mathrm{~cm} \text {. }
$
T.S.A of the hemispherical vessel $=\pi\left(3 \mathrm{R}^2+r^2\right)$
.png)
$
\begin{aligned}
& =\frac{22}{7}(3 \times 196+100) \\
& =\frac{22}{7} \times 688 \\
& =\frac{15136}{7}=2162.28 \mathrm{~cm}^2
\end{aligned}
$
Cost of painting@₹ $0.14$ per $\mathrm{cm}^2$
$
\begin{aligned}
& =2162.28 \times 0.14 \\
& =₹ 302.72
\end{aligned}
$
Question 10.
The frustum shaped outer portion of the table lamp has to be painted including the top part. Find the total cost of painting the lamp if the cost of painting 1 sq.cm is $\square 2$.
.png)
Solution:
Here given that $\mathrm{R}=12 \mathrm{~m}$
$
\begin{aligned}
& \mathrm{r}=6 \mathrm{~m} \\
& \mathrm{~h}=8 \mathrm{~m}
\end{aligned}
$
$
\begin{aligned}
l & =\sqrt{h^2+(R-r)^2} \\
& =\sqrt{8^2+6^2}=\sqrt{64+36} \\
& =\sqrt{100}=10 \mathrm{~m}
\end{aligned}
$
$\therefore$ CSA of the frustum
$
\begin{aligned}
& =\pi(\mathrm{R}+r) l \\
& =\frac{22}{7}(12+6) 10 \\
& =\frac{220}{7} \times 18 \\
& =\frac{3960}{7}=565.71 \mathrm{~m}^2
\end{aligned}
$
Area of the top part $=\pi r^2$
$
\begin{aligned}
& =\frac{22}{7} \times 6 \times 6 \\
& =\frac{792}{7}=113.14 \mathrm{~m}^2
\end{aligned}
$
$\therefore$ The total area to be painted
$
\begin{aligned}
& =565.71+113.14 \\
& =678.85 \mathrm{~m}^2
\end{aligned}
$
$\therefore$ The cost of painting $₹ 2$ per $\mathrm{m}^2$
$
=₹ 1357.72
$
Also Read : Exercise-8.1-Chapter-8-Statistics-10th-Maths-Guide-Samacheer-Kalvi-Solutions
