NCERT Solutions for Class 9 Maths Chapter 10 Heron's Formula Ex 10.1

Exercise 10.1

1. A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is\[\mathbf{180}\text{ }\mathbf{cm}\] , what will be the area of the signal board?

Ans:

Side of traffic signal board$=a$ 

Perimeter of traffic signal board $=3\left( a \right)$ 

 By Heron’s formula, 

Area of triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$ 

Area of given triangle $=\sqrt{\frac{3}{2}\left( \frac{3}{2}a-a \right)\left( \frac{3}{2}a-a \right)\left( \frac{3}{2}a-a \right)}$ 

$=\sqrt{\frac{3}{2}a\left( \frac{a}{2} \right)\left( \frac{a}{2} \right)\left( \frac{a}{2} \right)}$ 

$=\frac{\sqrt{3}}{2}{{a}^{2}}$ 

Area of given triangle$=\frac{\sqrt{3}}{2}{{a}^{2}}$ …..(1)

Perimeter of traffic signal board $=180cm$ 

Side of traffic signal board $\left( a \right)=\left( \frac{180}{3} \right)cm=60cm$

Using equation (1), area of traffic signal $=\frac{\sqrt{3}}{2}\left( 60\,c{{m}^{2}} \right)$ 

$=\left( \frac{3600}{4}\sqrt{3} \right)c{{m}^{2}}=900\sqrt{3}c{{m}^{2}}$ 

2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are\[122\text{ }m,\text{ }22\text{ }m\] , and \[\mathbf{120}\text{ }\mathbf{m}\] (see the given figure). The advertisements yield an earning of \[\mathbf{Rs}.\text{ }\mathbf{5000}\text{ }\mathbf{per}\text{ }\mathbf{m2}\] per year. A company hired one of its walls for \[\mathbf{3}\] months. How much rent did it pay?

Flyover

Ans:

The sides of the triangle (i.e.,\[a,\text{ }b,\text{ }c\] ) are of \[122\text{ }m,\text{ }22\text{ }m,\] and \[120\text{ }m\] respectively.

Perimeter of triangle$=\left( 122+22+120 \right)m$ 

$2s=264m$ 

$s=132m$ 

By Heron’s formula,

Area of triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

Area of given triangle $=\sqrt{132\left( 132-122 \right)\left( 132-22 \right)\left( 132-120 \right)}{{m}^{2}}$ 

$=\sqrt{132\left( 10 \right)\left( 110 \right)\left( 12 \right)}$ 

Rent of $1\,{{m}^{2}}$ area per year$=Rs.5000$ 

Rent of $1\,{{m}^{2}}$ area per month$=Rs.\frac{5000}{12}$ 

Rent of \[1320\text{ }{{m}^{2}}\] area for \[3\]  months $=\frac{5000}{12}\left( 3 \right)\left( 1320 \right)=1650000$

Therefore, the company had to pay $Rs.1650000$  

3. There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN”. If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Ans: It is given that the sides of the wall are 15 m, 11 m and 6 m.

So, the semi perimeter of triangular wall (s) = (15+11+6)/2 m = 16 m

Using Heron’s formula,

$A = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} $ 

$A = \sqrt {16(16 - 15)(16 - 11)(16 - 6)} $

$A = \sqrt 8 00{m^2}$

$A = 20\sqrt 2 {m^2}$.

4. Find the area of a triangle two sides of which are \[\mathbf{18}\text{ }\mathbf{cm}\] and \[\mathbf{10}\text{ }\mathbf{cm}\] the perimeter is \[\mathbf{42}\text{ }\mathbf{cm}.\] 

Ans:

Let the third side of the triangle be\[x\] .

Perimeter of the given triangle \[=\text{ }42\text{ }cm\text{ }\]

\[18\text{ }cm+10\text{ }cm+x=42\] 

$x=14cm$ 

$s=\frac{perimeter}{2}=\frac{42}{2}=21cm$ 

By Heron’s formula, Area of triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

Area of given triangle $=\sqrt{21\left( 21-18 \right)\left( 21-10 \right)\left( 21-14 \right)}c{{m}^{2}}$ 

$=\sqrt{21\left( 3 \right)\left( 11 \right)\left( 7 \right)}$

$=21\sqrt{11}c{{m}^{2}}$ 

5. Sides of a triangle are in the ratio of \[\mathbf{12}:\mathbf{17}:\mathbf{25}\] and its perimeter is\[\mathbf{540}\text{ }\mathbf{cm}\] . Find its area.

Ans:

Let the common ratio between the sides of the given triangle be\[x\] . Therefore, the side of the triangle will be \[12x,\text{ }17x,\] and\[25x\] .

Perimeter of this triangle \[=540\text{ }cm\] 

\[\Rightarrow 12x+17x+25x=540\text{ }cm\] 

\[\Rightarrow 54x=540\text{ }cm\] 

\[\Rightarrow x=10cm\]

Sides of the triangle will be\[120\text{ }cm,\text{ }170\text{ }cm,\text{ }250\text{ }cm\].

$s=\frac{perimeter\,of\,triangle}{2}=\frac{540\,cm}{2}=270cm$ 

By Heron’s formula, Area of triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

Area of given triangle $=\sqrt{270\left( 270-120 \right)\left( 270-170 \right)\left( 270-250 \right)}c{{m}^{2}}$ 

$=\sqrt{270\left( 150 \right)\left( 100 \right)\left( 20 \right)}$

$=9000c{{m}^{2}}$ 

Therefore, the area of this triangle is $9000c{{m}^{2}}$.

6. An isosceles triangle has perimeter \[\mathbf{30}\text{ }\mathbf{cm}\] and each of the equal sides is\[\mathbf{12}\text{ }\mathbf{cm}\] . Find the area of the triangle.

Ans:

Let the third side of this triangle be\[x\] .

Perimeter of triangle \[=\text{ }30\text{ }cm\] 

\[\Rightarrow 12\text{ }cm+12\text{ }cm+x=30\text{ }cm\] 

\[\Rightarrow x=6\text{ }cm\] 

$s=\frac{perimeter\,of\,triangle}{2}=\frac{30\,cm}{2}=15cm$ 

By Heron’s formula, Area of triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

Area of given triangle $=\sqrt{15\left( 15-12 \right)\left( 15-12 \right)\left( \left( 15-6 \right) \right)}c{{m}^{2}}$ 

$=\sqrt{15\left( 3 \right)\left( 3 \right)\left( 9 \right)}$

$=9\sqrt{15}c{{m}^{2}}$

Conclusion:

NCERT Solutions for Class 9 Maths Ch 10 Ex 10.1 Heron’s Formula by Vedantu are vital for mastering the technique of finding the area of triangles when the lengths of all three sides are known. Heron's Formula is a practical tool that simplifies this process without needing to know the height of the triangle. Focus on accurately calculating the semi-perimeter and correctly applying the formula to find the area. By diligently working through Vedantu’s step-by-step solutions, you can ensure a thorough understanding and ability to tackle similar problems in your exams.

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