Additional Questions - Chapter 7 Mensuration 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Questions
Exercise 7.1
Question 1.
Using Heron's formula, find the area of a triangle whose sides are $41 \mathrm{~m}, 15 \mathrm{~m}, 25 \mathrm{~m}$.
Solution:
$\begin{aligned}
& s=\frac{41+15+25}{2}=40.5 \\
\text { Area } &=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{40.5 \times 0.5 \times 25.5 \times 15.5} \\
=& \sqrt{\frac{405}{10} \times \frac{5}{10} \times \frac{255}{10} \times \frac{155}{10}}=\sqrt{\frac{5 \times 9 \times 9 \times 5 \times 5 \times 3 \times 17 \times 5 \times 31}{10 \times 10 \times 10 \times 10}} \\
=& \frac{1}{100} \times 5 \times 9 \times 5 \times \sqrt{3 \times 17 \times 31}=\frac{225}{100} \sqrt{1581}=\frac{9}{4} \sqrt{1581}
\end{aligned}$

Question $2 .$
Find the area of an equilateral triangle whose perimeter is $150 \mathrm{~m}$.
Solution:
$3 \mathrm{a}=150$
$a=\frac{150}{3}=50 \mathrm{~m}$
Area $=\frac{\sqrt{3}}{4} a^{2}=\frac{\sqrt{3}}{4} \times 50 \times 50=625 \sqrt{3}$

Question $3 .$
Find the area of a quadrilateral whose sides are $\mathrm{PQ}=15 \mathrm{~cm}, \mathrm{QR}=8 \mathrm{~cm}, \mathrm{RS}=25 \mathrm{~cm}, \mathrm{PS}=12$ $\mathrm{cm}$ and $\mathrm{LQ}=90^{\circ}$

Solution:
$\begin{aligned}
P R &=\sqrt{225+64}=\sqrt{289}=17 \mathrm{~cm} \\
s &=\frac{15+17+18}{2}=20 \mathrm{~cm}
\end{aligned}$
Area of $\Delta \mathrm{PQR}=\sqrt{20 \times 5 \times 12 \times 3}=\sqrt{5 \times 4 \times 5 \times 4 \times 3 \times 3}=60$
$\Delta$ PRS of Area $=\sqrt{27(27-17)(27-25)(27-12)}=\sqrt{27 \times 10 \times 2 \times 15}$
$=\sqrt{3 \times 3 \times 3 \times 5 \times 2 \times 2 \times 5 \times 3}=9 \times 5 \times 2=90$
Total area $=90+60=150 \mathrm{~cm}^{2}$
 

Exercise $7.2$
Question 1.
Find the TSA and LSA of a cuboid whose length, breadth and height are $10 \mathrm{~cm}, 12 \mathrm{~cm}$ and 14 cm respectively.
Solution:
TSA $=2(\mathrm{lb}+\mathrm{bh}+\mathrm{Ih})$
$=2(10 \times 12+12 \times 14+10 \times 14)=2(120+168+140)=856 \mathrm{~cm}^{2}$
LSA $=2(b h+I h)=2(12 \times 14+10 \times 14)=2(168+140)=2(308)=616 \mathrm{~cm}^{2}$
 

Question $2 .$
A cuboid has total surface area of $40 \mathrm{~m}^{2}$ and its lateral surface area is $26 \mathrm{~m}^{2}$. Find the area of its base.
Solution:
Area of its base $=\frac{\mathrm{TSA}-\mathrm{LSA}}{2}=\frac{40-26}{2} \mathrm{~m}^{2}=7 \mathrm{~m}^{2}$

Question $3 .$
Find the surface area of a cube whose edge is
(i) $27 \mathrm{~cm}$
(ii) $3 \mathrm{~cm}$
(iii) $6 \mathrm{~cm}$
(iv) $2.1 \mathrm{~cm}$
Solution:
(i) $\mathrm{TSA}=6 \mathrm{a}^{2}=6 \times 27^{2}=6 \times 729=4374 \mathrm{~cm}^{2}$
(ii) $\mathrm{TSA}=6 \mathrm{a}^{2}=6 \times 3^{2}=54 \mathrm{~cm}^{2}$
(iii) $\mathrm{TSA}=6 \mathrm{a}^{2}=6 \times 6^{2}=6 \times 36=216 \mathrm{~cm}^{2}$
(iv) $\mathrm{TSA}=6 \mathrm{a}^{2}=6 \times 2.1^{2}=6 \times 4.41=26.46 \mathrm{~cm}^{2}$

Exercise $7.3$
Question 1.
Find the volume of a cube whose surface area is a $96 \mathrm{~cm}^{2}$.
Solution:
$\begin{aligned}
&6 \mathrm{a}^{2}=96 \mathrm{~cm}^{2} \\
&a^{2}=\frac{96}{6} \\
&a=6
\end{aligned}$
Volume $=\mathrm{a}^{3}=6^{3}=36 \times 6=216 \mathrm{~cm}^{3}$

Question $2 .$
The volume of a cuboid is $440 \mathrm{~cm}^{3}$ and the area of its base is $88 \mathrm{~cm}^{2}$, find its height.
Solution:
$\begin{aligned}
&1 \times \mathrm{b} \times \mathrm{h}=440 \mathrm{~cm}^{3} \\
&h=\frac{440}{88}=5 \mathrm{~cm}
\end{aligned}$

Question $3 .$
How many 3 metre cubes can be cut from a cuboid measuring $18 \mathrm{~m} \times 12 \mathrm{~m} \times 9 \mathrm{~m}$ ?
Solution:

Question $4 .$
The outer dimensions of a closed wooden box are $10 \mathrm{~cm}$ by $8 \mathrm{~cm}$ by $7 \mathrm{~cm}$. Thickness of the wood is $1 \mathrm{~cm}^{3}$. Find the total cost of wood required to make box if $1 \mathrm{~cm}^{3}$ of wood costs ₹ $2.00 ?$
Solution:
$\text { Volume of wood }=12 \times 10 \times 9-10 \times 8 \times 7=1080-560=520 \mathrm{~cm}^{3} \times 2.00=₹ 1040$