Class 9 Maths Exercise 12.1 Solution

Given
Perimeter of Signal board \( =180 \mathrm{~cm} \)
Length of the side of the equilateral triangle \( =\mathrm{a} \)
The perimeter of the signal board \( = \) perimeter of the triangle
(Perimeter of Equilateral Triangle \( =3 \times \) Side \( ) \)
\(3 \mathrm{a}=180 \mathrm{~cm}\)
\(\mathrm{a}=60 \mathrm{~cm}\)
Therefore, each side of triangle \( =60 \mathrm{~cm} \).
Semi perimeter of the signal board \( (\mathrm{s})= \) \(\frac{ \text{Perimeter of board} }{ 2 }\)
\(=\frac{ 180 }{ 2 }= \) 90 cm
Using Heron's formula,
Area of the signal board \( =\sqrt{s(s-a)(s-b)(s-c)} \)
where, \( \mathrm{s}= \) semi perimeter of signal board \( =90 \mathrm{~cm} \mathrm{~a}=\mathrm{b}=\mathrm{c}=60 \mathrm{~cm} \)
Putting the value we
Area of signal board \( =\sqrt{90(90-60)(90-60)(90-60)} \)
Area of signal board \( =\sqrt{90 \times 30 \times 30 \times 30} \)
Area of signal board \( =100 \times 9 \sqrt{3} \)
Area of signal board \( =900 \sqrt{3} \mathrm{~cm}^{2} \)
Method 2:
we know that,
Area of an equilateral triangle \( =\frac{\sqrt{3}}{4} a^{2} \)
where \( \mathrm{a}= \) side of the equilateral triangle. putting the value of \( \mathrm{a}=60 \mathrm{~cm} \) we get,
Area \( =\frac{\sqrt{3}}{4}(60)^{2} \)
Area \( =\frac{\sqrt{3}}{4} \times 3600 \)
Area \( =900 \sqrt{3} \mathrm{~cm}^{2} \)

As shown in the figure the sides of the triangular side walls of flyover are \( 122 \mathrm{~m}, 22 \mathrm{~m} \), and 120 m .
Perimeter of the triangle \( =122+22+120=264 \mathrm{~m} \)
Semi perimeter of triangle(s) \( =\frac{264}{2} \)
\(=132 \mathrm{~m}\)
Now,
Using Heron's formula
Area of the advertisement \( =\sqrt{s(s-a)(s-b)(s-c)} \)
where, \( s= \) semi perimeter of triangle and \( a, b \), and \( c \) are the sides of the triangle.
\(=\sqrt{132(132-122)(132-22)(132-120)}\)
\(=\sqrt{132 \times 10 \times 110 \times 12}\)
\(=\sqrt{132 \times 132 \times 10 \times 10}\)
\(=132 \times 10 \mathrm{~m}^{2}\)
\( =1320 \mathrm{~m}^{2} \)
Rent of advertising per year \( = \) Rs. 5000 per \( \mathrm{m}^{2} \)
Rent of one wall for 1 year \( =1320 \times 5000 \)
Rent of one wall for 3 months \( =\frac{1320 \times 5000 \times 3}{12} \)
Rent of one wall for 3 months \( = \) Rs. 1650000

Given: The sides of the triangle are \( 15 \mathrm{~m}, 11 \mathrm{~m} \) and 6 m .
To find: Area painted in color
Explanation:
Perimeter of the triangular walls \( = \) sum of 3 sides
Perimeter \( =15+11+6=32 \mathrm{~m} \)
Semi perimeter of triangular walls \( (\mathrm{s})=\frac{32}{2} \)
\(=16 \mathrm{~m}\)
Using Heron's formula,
Area of the advertisement \( = \) Area of triangle
\( =\sqrt{s(s-a)(s-b)(s-c)} \)
where, \( \mathrm{s}= \) semi perimeter of the triangle \( \mathrm{a}, \mathrm{b} \) and c are the sides of the triangle Therefore for the given triangle,
\(\text {Area }=\sqrt{16(16-15)(16-11)(16-6)}\)
\(=\sqrt{16 \times 1 \times 5 \times 10}\)
\(=\sqrt{16 \times 5 \times 10}\)
\(=\sqrt{800} \mathrm{~m}^{2}\)
\( =\sqrt{(20 \times 20 \times 2}) \) (Taking 20 outside the square root, as a pair is present in square root, we get,)
\(=20 \sqrt{2} \mathrm{~m}^{2}\)
Thus the area of painted color is \( 20 \sqrt{2} \mathrm{~m}^{2} \)

To find: Area of the triangle
Given:
Two sides of the triangle are 18 cm and 10 cm (Given)
Perimeter of the triangle \( =42 \mathrm{~cm} \)
Perimeter of Triangle \( = \) Sum of sides of triangle
Let the third side of triangle be \( x \), So
\(x+18+10=42\)
Third side of triangle, \( x=42-(18+10) \)
\(=14 \mathrm{~cm}\)
Semi perimeter of triangle \( (\mathrm{s})=\frac{42}{2} \)
\(=21 \mathrm{~cm}\)
Using Heron's formula,
\(\text { Area of the triangle }=\sqrt{s(s-a)(s-b)(s-c)}\)
\(=\sqrt{21(21-18)(21-10)(21-14)}\)
\(=\sqrt{21 \times 3 \times 11 \times 7}\)
\(=\sqrt{7 \times 3 \times 7 \times 3 \times 11}\)
\(=21 \sqrt{11} \mathrm{~cm}^{2}\)

Let the common ratio of the sides of the triangle be \(x\) and sides are \( 12 x, 17 x\) and \(25 x\)
The perimeter of the triangle \( =540 \mathrm{~cm} \)
Sum of sides of the triangle \( =540 \mathrm{~cm} \)
\(12 x+17 x+25 x=540 \mathrm{~cm}\)
\(54 x=540\)
\(x=10\)
Sides of triangle:
\(12 x=12 \times 10=120 \mathrm{~cm}\)
\(17 x=17 \times 10=170 \mathrm{~cm}\)
\(25 x=25 \times 10=250 \mathrm{~cm}\)
Semi perimeter of triangle \( (\mathrm{s})=\frac{540}{2} \)
\(=270 \mathrm{~cm}\)
Now,
Using Heron's formula,
Area of the triangle \( =\sqrt{s(s-a)(s-b)(s-c)} \)
where, \( \mathrm{s}= \) semi-perimeter of the triangle and \( \mathrm{a}, \mathrm{b} \) and c are the sides of the triangle
\(=\sqrt{270(270-120)(270-170)(270-250)}\)
\(=\sqrt{270 \times 150 \times 100 \times 20}\)
\(=\sqrt{ (81000000)} \mathrm{cm}^{2}\)
\(=9000 \mathrm{~cm}^{2}\)
Area of triangle \( =9000 \mathrm{~c m}^{2} \)

Length of equal sides of the triangle \( =12 \mathrm{~cm} \)
Perimeter of the triangle \( =30 \mathrm{~cm} \)
Third side of triangle \( =30-(12+12) \)
\(=6 \mathrm{~cm}\)
Semi perimeter of triangle \( (\mathrm{s})=\frac{30}{2} \)
\( =15 \mathrm{~cm} \)
Now,
Using Heron's formula,
Area of the triangle \( =\sqrt{s(s-a)(s-b)(s-c)} \)
where, \( s= \) semi perimeter of the triangle
\( \mathrm{a}, \mathrm{b} \) and c are the sides of the triangle
\(=\sqrt{15(15-12)(15-12)(15-6)}\)
\(=\sqrt{15 \times 3 \times 3 \times 9}\)
\(=9 \sqrt{15} \mathrm{~cm}^{2}\)
Hence, the area of triangle is \( 9 \sqrt{15} \mathrm{~cm}^{2} \)