Exercise 7.1 - Chapter 7 Mensuration 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Ex $7.1$ : Chapter 7 - Mensuration - 9th Maths Guide Samacheer Kalvi Solutions
Question 1.
Using Heron's formula, find the area of a triangle whose sides are
(i) $10 \mathrm{~cm}, 24 \mathrm{~cm}, 26 \mathrm{~cm}$
(ii) $1.8 \mathrm{~m}, 8 \mathrm{~m}, 8.2 \mathrm{~m}$
Solution:
(i) sides : $10 \mathrm{~cm}, 24 \mathrm{~cm}, 26 \mathrm{~cm}$
Using Heron's formula
Area of the triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ sq. units
$\begin{aligned}
s &=\frac{a+b+c}{2}=\left(\frac{10+24+26}{2}\right) \mathrm{cm}=\frac{60}{2}=30 \mathrm{~cm} \\
\therefore \text { Area } &=\sqrt{30(30-10)(30-24)(30-26)} \\
&=\sqrt{30 \times 20 \times 6 \times 4}=\sqrt{600 \times 24}=\sqrt{14400}=120 \mathrm{sq} . \mathrm{cm}
\end{aligned}$
(ii) Sides : $1.8 \mathrm{~m}, 8 \mathrm{~m}, 8.2 \mathrm{~m}$
$\begin{aligned}
s &=\frac{1.8+8+8.2}{2}=\frac{18}{2}=9 \\
\therefore \text { Area } &=\sqrt{s(s-a)(s-b)(s-c)} \text { sq. units } \\
&=\sqrt{9(9-1.8)(9-8)(9-8.2})=\sqrt{9 \times 7.2 \times 1 \times 0.8} \\
&=\sqrt{51.84}=7.2 \text { sq.m }
\end{aligned}$

Question 2.
The sides of the triangular ground are $22 \mathrm{~m}, 120 \mathrm{~m}$ and $122 \mathrm{~m}$. Find the area and cost of leveling the ground at the rate of $₹ 20$ per $\mathrm{m}^{2}$.

Solution:
Sides : $22 \mathrm{~m}, 120 \mathrm{~m}, 122 \mathrm{~m}$
Using Heron's formula
$\begin{aligned} s &=\frac{22+120+122}{2}=\frac{264}{2}=132 \mathrm{~m} \\ \text { Area } &=\sqrt{s(s-a)(s-b)(s-c)} \\ &=\sqrt{132(132-22)(132-120)(132-122)} \\ &=\sqrt{132 \times 110 \times 12 \times 10} \\ &=\sqrt{1742400}=\sqrt{11 \times 11 \times 3 \times 3 \times 2 \times \times 2 \times 2 \times 2 \times 10 \times 10} \\ &=11 \times 3 \times 2 \times 2 \times 10=1320 \mathrm{~m}^{2} \end{aligned}$
$\therefore$ Cost of levelling $1320 \mathrm{~m}^{2}=1320 \times 20=₹ 26400$

Question $3 .$
The perimeter of a triangular plot is $600 \mathrm{~m}$. If the sides are in the ratio $5: 12: 13$, then find the area of the plot.
Solution:
$\mathrm{s}=600 \mathrm{~m}$
Side $s$ are in the ratio $5: 12: 13$
$\begin{aligned}
5 \mathrm{x}+12 \mathrm{x} &+13 \mathrm{x}=30 \mathrm{x} \\
s &=600 \Rightarrow \frac{30 x}{2}=600 \\
30 x &=1200 \\
x &=40
\end{aligned}$
$\therefore$ sides are 200
$\begin{aligned}
\therefore \text { Area } &=\sqrt{s(s-a)(s-b)(s-c)} \\
&=\sqrt{600(600-200)(600-480)(600-520)} \\
&=\sqrt{600 \times 400 \times 120 \times 80}=\sqrt{2304000000} \\
&=\sqrt{48 \times 48 \times 1000 \times 1000}=48 \times 1000=48000 \text { sq. m }
\end{aligned}$

Question 4.
Find the area of an equilateral triangle whose perimeter is $180 \mathrm{~cm}$.
Solution:

Perimeter of an equilateral triangle $=180 \mathrm{~cm}$
$\therefore \text { One side }(a)=\frac{180}{3}=60 \mathrm{~m} \text {. }$
Area of an equilateral triangle $=\frac{\sqrt{3}}{4} a^{2}$ sq. units $=\frac{\sqrt{3}}{4} \times 60 \times 60$
$=900 \sqrt{3} \mathrm{~m}^{2}=900 \times 1.732=1558.8 \mathrm{~m}^{2}$

Question 5.
An advertisement board is in the form of an isosceles triangle with perimeter $36 \mathrm{~m}$ and each of the equal sides are $13 \mathrm{~m}$. Find the cost painting it at $₹ 17.50$ per square metre.
Solution:

Area of an isoceles triangle
$\begin{aligned}
h &=\sqrt{13^{2}-5^{2}}=\sqrt{169-25} \\
&=\sqrt{144}=12 \mathrm{~m}
\end{aligned}$
$\therefore$ Area of the triangular board
$\begin{aligned}
&=\frac{1}{2} \times b h \\
&=\frac{1}{2} \times 10 \times 12=60 \mathrm{~m}^{2}
\end{aligned}$
Cost of painting $1 \mathrm{~m}^{2}=₹ 17.50$
Cost of painting $60 \mathrm{~m}^{2}=60 \times 17.50=₹ 1050$

Question $6 .$
Find the area of the unshaded region.

Solution:
By the Pythagoras theorem
$\begin{aligned}
\mathrm{AB}^{2} &=\mathrm{AD}^{2}+\mathrm{DB}^{2} \\
&=12^{2}+16^{2}=144+256=400 \\
\mathrm{AB} &=20 \mathrm{~cm} . \\
s &=\frac{34+20+42}{2}=\frac{96}{2}=48 \\
\therefore \text { Area of the } \Delta \mathrm{ABC} &=\sqrt{s(s-a)(s-b)(s-c)} \\
&=\sqrt{48(48-34)(48-20)(48-42)} \\
&=\sqrt{48 \times 14 \times 28 \times 6}=\sqrt{112896} \\
&=\sqrt{336 \times 336}=336 \mathrm{sq} . \mathrm{cm} .
\end{aligned}$
Area of the triangle $\mathrm{ABD}$
$=\frac{1}{2} b h=\frac{1}{2} \times 12 \times 16=96 \mathrm{~cm}^{2}$
$\therefore$ Area of the unshaded region
$\begin{aligned}
&=\text { Area of } \Delta \mathrm{ABC}-\text { Area of } \Delta \mathrm{ABD} \\
&=336-96=240 \mathrm{~cm}^{2}
\end{aligned}$

 Question 7.
Find the area of a quadrilateral $\mathrm{ABCD}$ whose sides are $\mathrm{AB}=13 \mathrm{~cm}, \mathrm{BC}=12 \mathrm{~cm}, \mathrm{CD}=9 \mathrm{~cm}, \mathrm{AD}=14 \mathrm{~cm}$ and diagonal $\mathrm{BD}=15 \mathrm{~cm}$
Solution:

Area of the quadrilateral $\mathrm{A} \overline{\mathrm{B}} \overline{\mathrm{D}}$
$=$ Area of the $\triangle \mathrm{ABD}$
+ Area of the $\triangle \mathrm{BCD}$
Sides of the triangle $\mathrm{ABD}$ are $13 \mathrm{~cm}, 14 \mathrm{~cm}, 15 \mathrm{~cm}$.
$\begin{aligned}
s &=\frac{13+14+15}{2} \mathrm{~cm}=\frac{42}{2}=21 \mathrm{~cm} \\
\text { Area } &=\sqrt{s(s-a)(s-b)(s-c)} \\
&=\sqrt{21(21-13)(21-14)(21-15)} \\
&=\sqrt{21 \times 8 \times 7 \times 6}=\sqrt{7056}=84 \mathrm{~cm}^{2}
\end{aligned}$
Sides of the triangle
$B C D$ are $12 \mathrm{~cm}, 9 \mathrm{~cm}, 15 \mathrm{~cm}$
$\begin{aligned}
\therefore s &=\frac{12+9+15}{2}=\frac{36}{2}=18 \mathrm{~cm} \\
\text { Area } &=\sqrt{18(18-12)(18-9)(18-15)} \\
&=\sqrt{18 \times 6 \times 9 \times 3}=\sqrt{2916}=54 \mathrm{~cm}^{2}
\end{aligned}$
$\therefore$ Ärea of the quadrilateral
$=84 \mathrm{~cm}^{2}+54 \mathrm{~cm}^{2}=138 \mathrm{~cm}^{2}$

Question $8 .$
A park is in the shape of a quadrilateral. The sides of the park are $15 \mathrm{~m}, 20 \mathrm{~m}, 26 \mathrm{~m}$ and $17 \mathrm{~m}$ and the angle between the first two sides is a right angle. Find the area of the park.

Solution:
Area of the quadrilateral = Area of $\triangle \mathrm{ABD}+$ Area of $\triangle \mathrm{BCD}$ $\begin{aligned} \Delta \mathrm{ABD} \text { is right angled triangle } \\ \therefore \text { Area } &=\frac{1}{2} b h \\ &=\frac{1}{2} \times 20 \times 15 \mathrm{~m}^{2}=150 \mathrm{~m}^{2} \\ \text { In } \Delta \mathrm{ABD}, \mathrm{BD}^{2} &=\mathrm{AD}^{2}+\mathrm{AB}^{2} \\ &=15^{2}+20^{2}=225+400=625 \mathrm{~m}^{2} \\ \mathrm{BD} &=\sqrt{625}=25 \mathrm{~m} \\ \therefore \text { In } \Delta \mathrm{BCD}, s &=\frac{25+26+17}{2}=\frac{68}{2}=34 \\ \text { Area of } \Delta \mathrm{BCD} &=\sqrt{s(s-a)(s-b)(s-c)} \\ &=\sqrt{34(34-25)(34-26)(34-17)} \\ &=\sqrt{34 \times 9 \times 8 \times 17}=\sqrt{41616}=204 \mathrm{~m}^{2} \\ \therefore \text { Area of the quadrilateral } &=(150+204) \mathrm{m}^{2}=354 \mathrm{~m}^{2} \end{aligned}$

Question $9 .$
A land is in the shape of rhombus. The perimeter of the land is $160 \mathrm{~m}$ and one of the diagonal is $48 \mathrm{~m}$. Find the area of the land.
Solution:

Perimeter of the rhombus land $=160 \mathrm{~m}$
One of the diagonal $=48 \mathrm{~m}$
$\therefore$ Area of the land $=2 \times$ Area of the $\triangle \mathrm{ABC}$.
$s=\frac{40+40+48}{2}=\frac{128}{2}=64 \mathrm{~m}$
Area of $\Delta \mathrm{ABC}=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{64(64-40)(64-40)(64-48)}$
$=\sqrt{64 \times 24 \times 24 \times 16}=\sqrt{589824}$
$=768 \mathrm{~m}^{2}$
$\therefore$ Area of the land $=2 \times 768 \mathrm{~m}^{2}=1536 \mathrm{~m}^{2}$

Question $10 .$
The adjacent sides of a parallelogram measures $34 \mathrm{~m}, 20 \mathrm{~m}$ and the measure of the diagonal is $42 \mathrm{~m}$. Find the area of Parallelogram.

Solution:
Area of the parallelogram $=2 \times$ Area of the $\triangle \mathrm{ABC}$
$s=\frac{34+20+42}{2}=\frac{96}{2}=48 \mathrm{~m}$
Area of $\triangle \mathrm{ABC}=\sqrt{48(48-34)(48-20)(48-42)}$
$=\sqrt{48 \times 14 \times 28 \times 6}=\sqrt{112896}=336 \mathrm{~m}^{2}$
$\therefore \quad$ Area of the parallelogram $=2 \times 336 \mathrm{~m}^{2}=672 \mathrm{~m}^{2}$