NCERT Solutions for Class 9 Maths Chapter 12 - Herons Formula

NCERT Solutions for Class 9 Maths Chapter 12, Heron's Formula, covers the following two exercises:

Exercise 12.1: This exercise consists of six questions that are based on the concept of Heron's Formula. The questions cover topics such as finding the area of a triangle when the sides are given, finding the missing side when the area and two sides are given, and finding the height of a triangle when the area and the base are given. The exercise also includes word problems that require the application of Heron's Formula to find the area of triangles.

Exercise 12.2: This exercise consists of nine questions that are based on the real-life application of Heron's Formula. The questions cover topics such as finding the area of a field, the amount of fencing required to surround a triangular plot, the amount of carpet required to cover a triangular room, and the amount of paper required to cover a triangular cardboard. The questions require students to apply the formula to solve real-world problems.

Heron's Formula is a powerful tool for calculating the area of a triangle, and these exercises provide ample practice for students to understand the concept and apply it to various real-life situations.

Access NCERT Solutions Maths Chapter 12 - Heron’s formula

Exercise 12.1 

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 $180\ \text{cm}$, what will be the area of the signal board?

Ans:

Length of the side of traffic signal board $=a$ 

Perimeter of traffic signal board which is an equilateral triangle $=3\times a$

We know that,

$2s=$ Perimeter of the triangle, 

So, $2s=3a$

$\Rightarrow s=\dfrac{3}{2}a$

 Area of triangle can be evaluated by Heron’s formula:

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

Where,

$a$ , $b$ and $c$ are the sides of the triangle

\[s=\dfrac{a+b+c}{2}\]

Substituting $s=\dfrac{3}{2}a$ in Heron’s formula, we get:

Area of given triangle:

 $A=\sqrt{\dfrac{3}{2}a\left( \dfrac{3}{2}a-a \right)\left( \dfrac{3}{2}a-a \right)\left( \dfrac{3}{2}a-a \right)}$ 

$A=\dfrac{\sqrt{3}}{2}{{a}^{2}}\ \ ......\text{(1)}$

Perimeter of traffic signal board:

$P=180\ \text{cm}$

Hence, side of traffic signal board

$a=\left( \dfrac{180}{3} \right)$

$a=60\ \ ......\text{(2)}$

Substituting Equation (2) in Equation (1), we get:

Area of traffic signal board is $A=\dfrac{\sqrt{3}}{2}{{\left( 60\,cm \right)}^{2}}$

$\Rightarrow A=\left( \dfrac{3600}{4}\sqrt{3} \right)\,c{{m}^{2}}$

$\Rightarrow A=900\sqrt{3}\,c{{m}^{2}}$

Hence, the area of the signal board is $900\sqrt{3}\text{ c}{{\text{m}}^{2}}$.

2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are $122m,\,22m,$ and $120m$. The advertisements yield on earning of Rs. $5000\,per\,{{m}^{2}}$ per year. A company hired one of its walls for $3$ months. How much rent did it pay?

Ans: The length of the sides of the triangle are (say $a$, $b$ and $c$)

$a=122\ \text{m}$

$b=22\ \text{m}$

$c=120\ \text{m}$

Perimeter of triangle = sum of the length of all sides

Perimeter of triangle is:

$P=122+22+120$

$P=264\ \text{m}$

We know that,

$2s=$ Perimeter of the triangle, 

$2s=264\ \text{m}$

$s=132\ \text{m}$

Area of triangle can be evaluated by Heron’s formula:

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

Where,

$a$, $b$ and $c$ are the sides of the triangle

\[s=\dfrac{a+b+c}{2}\]

So, in this question,

$s=\dfrac{122+22+140}{2}$

$s=132\ \text{m}$

Substituting values of $s$ $a$, $b$, $c$ in Heron’s formula, we get:

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

$=\left[ \sqrt{132\left( 10 \right)\left( 110 \right)\left( 12 \right)} \right]\,{{m}^{2}}=1320\,{{m}^{2}}$ 

It is given that:

Rent of $1{{m}^{2}}$ area per year is:

$R=Rs.\text{ 5000/}{{\text{m}}^{2}}$ 

So,

Rent of $1{{m}^{2}}$ area per month will be:

$R=Rs.\ \dfrac{5000}{12}/{{m}^{2}}$

Rent of $1320{{m}^{2}}$ area for $3$ months:

$R=\left( \dfrac{5000}{12}\times 3\times 1320 \right)/{{m}^{2}}$

$\Rightarrow R=Rs.\ 1650000$

Therefore, the total cost rent that company must pay is 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,

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

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

$A=\sqrt800 m^2$

$A=20\sqrt2 m^2$

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

Ans: Let the length of the third side of the triangle be $x$. 

Perimeter of the given triangle:

$P=42cm$

Let the sides of the triangle be $a$, $b$ and $c$.

$a=18cm$

$b=10cm$

$c=xcm$

Perimeter of the triangle = sum of all sides

$18+10+x=42$

$\Rightarrow 28+x=42$

$\Rightarrow x=14$

Area of triangle can be evaluated by Heron’s formula:

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

Where,

$a$, $b$ and $c$ are the sides of the triangle

\[s=\dfrac{a+b+c}{2}\]

$\Rightarrow s=\dfrac{18+10+14}{2}$

$\Rightarrow s=21cm$

Substituting values of  $s$ $a$, $b$, $c$ in Heron’s formula, we get:

Area of given triangle:

$A=\left[ \sqrt{21\left( 21-18 \right)\left( 21-10 \right)\left( 21-14 \right)} \right]$

$\Rightarrow A=\left[ \sqrt{21\left( 3 \right)\left( 11 \right)\left( 7 \right)} \right]$

$\Rightarrow A=21\sqrt{11}\text{ }c{{m}^{2}}$

Hence, the area of the given triangle is $21\sqrt{11}\text{ }c{{m}^{2}}$.

5. Sides of a triangle are in the ratio of $12:17:25$ and its perimeter is $540cm$. 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 $12x$, $17x$, and $25x$.

It is given that,

Perimeter of this triangle $=540cm$

Perimeter = sum of the length of all sides

$12x+17x+25x=540$

$\Rightarrow 54x=540$

$\Rightarrow x=10$

Sides of the triangle will be:

\[12\times 10=120cm\]

$17\times 10=170cm$

$25\times 10=250cm$

Area of triangle can be evaluated by Heron’s formula:

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

Where,

$a$, $b$ and $c$ are the sides of the triangle

\[s=\dfrac{a+b+c}{2}\]

$\Rightarrow s=\dfrac{120+170+250}{2}$

$\Rightarrow s=270cm$

Area of given triangle:

$A=\left[ \sqrt{270\left( 270-120 \right)\left( 270-170 \right)\left( 270-250 \right)} \right]$

$\Rightarrow A=\left[ \sqrt{270\left( 150 \right)\left( 100 \right)\left( 20 \right)} \right]$

$\Rightarrow A=9000c{{m}^{2}}$

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

6. An isosceles triangle has perimeter $30cm$ and each of the equal sides is $12cm$. Find the area of the triangle.

Ans: Let the third side of this triangle be $x$.

Measure of equal sides is $12cm$ as the given triangle is an isosceles triangle.

It is given that,

Perimeter of triangle, $P=30cm$

Perimeter of triangle = Sum of the sides 

$12+12+x=30$

$\Rightarrow x=6cm$

Area of triangle can be evaluated by Heron’s formula:

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

Where,

$a$, $b$ and $c$ are the sides of the triangle

\[s=\dfrac{a+b+c}{2}\]

$\Rightarrow s=\dfrac{12+12+6}{2}$

$\Rightarrow s=15cm$

Substituting values of $s$ $a$, $b$, $c$ in Heron’s formula we get:$A=\left[ \sqrt{15\left( 15-12 \right)\left( 15-12 \right)\left( 15-6 \right)} \right]$

$\Rightarrow A=\left[ \sqrt{15\left( 3 \right)\left( 3 \right)\left( 9 \right)} \right]$

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

Hence, the area of the given isosceles triangle is $9\sqrt{15}c{{m}^{2}}$.

Exercise 12.2 

1. A park, in the shape of a quadrilateral $ABCD$, has $\angle C={{90}^{\circ }}$ , $AB=9m$, $BC=12m$, $CD=5m$ and $AD=8m$. How much area does it occupy?

Ans: It is given that:

$AB=9m$

$BC=12m$

$CD=5m$

$AD=8m$

$\angle C={{90}^{\circ }}$

Let us join $BD$. 

For $\Delta BCD$

We know that, in right angle triangle Pythagoras theorem is:

${{a}^{2}}+{{b}^{2}}={{c}^{2}}$

Where,

$a=$ perpendicular

$b=$ other side

$c=$ hypotenuse

In $\Delta BCD$, applying Pythagoras theorem,

$B{{D}^{2}}=B{{C}^{2}}+C{{D}^{2}}$

$\Rightarrow B{{D}^{2}}={{12}^{2}}+{{5}^{2}}$

\[\Rightarrow B{{D}^{2}}=169\]

\[\Rightarrow BD=\sqrt{169}\]

$\Rightarrow BD=13m$

Area of a right-angle triangle:

$A=\dfrac{1}{2}\times b\times h$

Where,

$b=$ base

$h=$ height

Area of $\Delta BCD$:

$A=\dfrac{1}{2}\times BC\times CD$

$\Rightarrow A=\dfrac{1}{2}\times 5\times 12$

$\Rightarrow A=30{{m}^{2}}$

For $\Delta ABD$

$AB=9m$

$AD=8m$

$BD=13m$

Area of triangle can be evaluated by Heron’s formula:

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

Where,

$a$, $b$ and $c$ are the sides of the triangle

\[s=\dfrac{a+b+c}{2}\]

$\Rightarrow s=\dfrac{9+8+13}{2}$

$\Rightarrow s=15cm$

Area of the $\Delta ABD$:

$A=\sqrt{15\left( 15-9 \right)\left( 15-8 \right)\left( 15-13 \right)}$

$\Rightarrow A\,=\sqrt{15\left( 6 \right)\left( 7 \right)\left( 2 \right)}$

$\Rightarrow A=6\sqrt{35}$

$A=\left( 6\times 5.916 \right){{m}^{2}}$

$A\,\,=35.496{{m}^{2}}$

Area of the park = Area of $\Delta ABD$ $+$ Area of $\Delta BCD$ 

${{A}_{total}}\,=35.496+30\ {{\text{m}}^{2}}$

\[{{A}_{total}}=65.496\ {{\text{m}}^{2}}\]

Hence, the area of the park is $65.496\text{ }{{m}^{2}}$.

2. Find the area of a quadrilateral $ABCD$ in which $AB=3cm$, $BC=4cm$, $CD=4cm$, $DA=5cm$ and $AC=5cm$.

Ans:

For $\Delta ABC$,

$AB=3cm$

$BC=4cm$

$AC=5cm$

We know that, in right angle triangle Pythagoras theorem is:

${{a}^{2}}+{{b}^{2}}={{c}^{2}}$

Where,

$a=$ perpendicular

$b=$ other side

$c=$ hypotenuse

Then,

$A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}$ 

$\Rightarrow {{\left( 5 \right)}^{2}}={{\left( 3 \right)}^{2}}+{{\left( 4 \right)}^{2}}$ 

Therefore, $\Delta ABC,$ is a right-angled triangle as it satisfies the Pythagoras theorem which is right-angled at point $B$.

Area of a right-angle triangle:

$A=\dfrac{1}{2}\times b\times h$

Where,

$b=$ base

$h=$ height

Area of $\Delta ABC$

$A=\dfrac{1}{2}\times AB\times BC$

$\Rightarrow A=\dfrac{1}{2}\times 3\times 4$

$\Rightarrow A=6c{{m}^{2}}$

For $\Delta ADC$

$AD=5cm$

$DC=4cm$

$AC=5cm$

Area of triangle can be evaluated by Heron’s formula:

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

Where,

$a$, $b$ and $c$ are the sides of the triangle

\[s=\dfrac{a+b+c}{2}\]

$\Rightarrow s=\dfrac{5+4+5}{2}$

$\Rightarrow s=7cm$

Area of $\Delta ACD$:

$A=\sqrt{7\left( 7-5 \right)\left( 7-4 \right)\left( 7-5 \right)}$

$\Rightarrow A=\,\sqrt{7\left( 2 \right)\left( 2 \right)\left( 3 \right)}$

$A\,\,=2\sqrt{21}c{{m}^{2}}$

$\,\Rightarrow A\,=\left( 2\times 4.583 \right)c{{m}^{2}}$ 

$\Rightarrow A\,=9.166c{{m}^{2}}$

Area of $ABCD$= Area of $\Delta ABC$$+$ Area of $\Delta ACD$

$\Rightarrow {{A}_{total}}\,=\left( 6+9.166 \right)$ 

$\Rightarrow {{A}_{total}}=15.166c{{m}^{2}}$ 

Hence, the area of the given quadrilateral is $15.166c{{m}^{2}}$.

3. Radha made a picture of an aeroplane with colored papers as shown in the given figure. Find the total area of the paper used.

Ans: For triangle $\text{I}$ 

This triangle is an isosceles triangle with side length $5cm$, $5cm$ and $1cm$

Area of triangle can be evaluated by Heron’s formula:

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

Where,

$a$, $b$ and $c$ are the sides of the triangle

\[s=\dfrac{a+b+c}{2}\]

$\Rightarrow s=\dfrac{5+5+1}{2}$

$s=5.5cm$

Area of triangle $I$

$A=\sqrt{5.5\left( 5.5-5 \right)\left( 5.5-5 \right)\left( 5.5-1 \right)}\,$

$\Rightarrow A=\sqrt{5.5\left( 0.5 \right)\left( 0.5 \right)\left( 4.5 \right)}$

$\Rightarrow A=0.75\sqrt{11}c{{m}^{2}}$

$\Rightarrow A=\left( 0.75\times 3.317 \right)c{{m}^{2}}$

$\Rightarrow A=2.488c{{m}^{2}}$ 

For quadrilateral $\text{II}$ 

This quadrilateral is a rectangle. 

$l=6.5cm$

$b=1cm$

Area of rectangle, $A=l\times b$

$\Rightarrow A=\left( 6.5\times 1 \right)c{{m}^{2}}$ 

$\Rightarrow A=6.5c{{m}^{2}}$

For quadrilateral $\text{III}$  

This quadrilateral is a trapezium. 

Lengths of parallel sides of the given trapezium are $1cm$ and $2cm$.

From above figure, perpendicular height of parallelogram:

\[h=\sqrt{{{1}^{2}}-{{\left( 0.5 \right)}^{2}}}\,\]

\[\Rightarrow h=\sqrt{0.75}\]

$\Rightarrow h=0.866cm$

Area of trapezium

$A=\dfrac{1}{2}\times h\times \left( sum\text{ of the parallel sides} \right)$

Substituting values of unknown variables, we get:

$A=\dfrac{1}{2}\times 0.866\times \left( 2+1 \right)$

$\Rightarrow A=0.433\times 3$

$\Rightarrow A=1.299c{{m}^{2}}$

From given figure, it is obvious that 

Area of triangle $\left( \text{IV} \right)$ = Area of triangle in $\left( \text{V} \right)$ 

$A=\dfrac{1}{2}\times b\times h$

Where $b=1.5cm$

$h=6cm$

$\Rightarrow A=\left( \dfrac{1}{2}\times 1.5\times 6 \right)c{{m}^{2}}$ 

$\Rightarrow A=4.5c{{m}^{2}}$

Total area of the paper used

${{A}_{total}}=Ar\left( I \right)+Ar\left( II \right)+Ar\left( III \right)+Ar\left( IV \right)+Ar\left( V \right)$ 

$\Rightarrow {{A}_{total}}=~2.488+6.5+1.299+4.5+4.5$

$\Rightarrow {{A}_{total}}=19.287c{{m}^{2}}$

Hence, the total area of the used paper is $19.287c{{m}^{2}}$.

4. A triangle and a parallelogram have the same base and the same area. If the sides of triangle are $26cm$, $28cm$ and $30cm$, and the parallelogram stands on the base $28cm$, find the height of the parallelogram.

Ans: Length of the sides of the triangle are $26cm$, $28cm$ and $30cm$.

Area of triangle can be evaluated by Heron’s formula:

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

Where,

$a$, $b$ and $c$ are the sides of the triangle

\[s=\dfrac{a+b+c}{2}\]

$\Rightarrow s=\dfrac{26+28+30}{2}$

$\Rightarrow s=42cm$

Area of given triangle

$A=\sqrt{42\left( 42-26 \right)\left( 42-28 \right)\left( 42-30 \right)}\text{ }$

$\Rightarrow A=\sqrt{42\left( 16 \right)\left( 14 \right)\left( 12 \right)}$

$\Rightarrow A=336c{{m}^{2}}$

Let the height of the parallelogram be $h$. 

It is given that,

Area of parallelogram = Area of triangle

Area of parallelogram: $A=b\times h$ 

Where, $b=28cm$

So, 

$\text{h}\times \text{28 cm}=336c{{m}^{2}}$

$\text{h}=12\,\text{cm}$

Therefore, the height of the parallelogram is $12cm$.

5. A rhombus shaped field has green grass for $18$ cows to graze. If each side of the rhombus is \[30m\] and its longer diagonal is $48m$, how much area of grass field will each cow be getting? 

Ans:

Let $ABCD$ be a rhombus-shaped field. 

For $\Delta BCD$

Let the sides of $\Delta BCD$ be $a$, $b$ and $c$

$a=30cm$

\[b=30cm\]

$c=48cm$

Area of triangle can be evaluated by Heron’s formula:

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

Where,

$a$, $b$ and $c$ are the sides of the triangle

\[s=\dfrac{a+b+c}{2}\]

$\Rightarrow s=\dfrac{30+30+48}{2}$

$\Rightarrow s=54cm$

Therefore, area of $\Delta BCD$

$A=\sqrt{54\left( 54-48 \right)\left( 54-30 \right)\left( 54-30 \right)}$

$\Rightarrow A=\sqrt{54\left( 6 \right)\left( 24 \right)\left( 24 \right)}$

$\Rightarrow A=432\text{ }c{{m}^{2}}$

Area of field:

${{A}_{total}}=2\times Ar\left( \Delta BCD \right)$

$\Rightarrow {{A}_{total}}=2\times 432$

$\Rightarrow {{A}_{total}}=864\text{ }c{{m}^{2}}$

Area for grazing for $1$ cow:

${{A}_{required}}=\dfrac{864}{18}$

$\Rightarrow {{A}_{required}}=48\text{ }{{\text{m}}^{2}}$

Hence, each cow will get $48\text{ }{{m}^{2}}$ area of grass field for grazing.

6. An umbrella is made by stitching $10$ triangular pieces of cloth of two different colors, each piece measuring $20cm$, \[50cm\] and $50cm$. How much cloth of each color is required for the umbrella?

Ans: Let’s the sides of a triangular piece of cloth be $a$, $b$ and $c$

$a=20cm$

\[b=50cm\]

$c=50cm$

Area of triangle can be evaluated by Heron’s formula:

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

Where,

$a$, $b$ and $c$ are the sides of the triangle

\[s=\dfrac{a+b+c}{2}\]

$\Rightarrow s=\dfrac{20+50+50}{2}$

$\Rightarrow s=60cm$

Area of each triangular piece

$A=\sqrt{60\left( 60-50 \right)\left( 60-50 \right)\left( 60-20 \right)}$

$\Rightarrow A=\sqrt{60\left( 10 \right)\left( 10 \right)\left( 40 \right)}$

$\Rightarrow A=200\sqrt{6}\text{ }c{{m}^{2}}$

Since there are $5$ triangular pieces made of two different colored cloths in which $3$ triangular pieces are of black color and the remaining $2$ are of white color.

So,

Area of black colored cloth required

${{A}_{black}}=200\sqrt{6}\times 3$

$\Rightarrow {{A}_{black}}=600\sqrt{6}\text{ }c{{m}^{2}}$

Area of white colored cloth required

${{A}_{white}}=200\sqrt{6}\times 2$

$\Rightarrow {{A}_{white}}=400\sqrt{6}\text{ c}{{\text{m}}^{2}}$

Hence, the area of black colored cloth and white colored cloth required is $600\sqrt{6}\text{ c}{{\text{m}}^{2}}$ and $400\sqrt{6}c{{m}^{2}}$ respectively.

7. A kite in the shape of a square with a diagonal $32cm$ and an isosceles triangle of base $8\ cm$ and sides $6\ cm$ each is to be made of three different shades as shown in the given figure. How much paper of each shade has been used in it?

Ans: It is given that the length of a diagonal of square is $32cm$.

Let a side of the square be $x\ \text{cm}$.

We know that in square both diagonals are perpendicular bisectors of each other.

So,

Base of the $Ist$ triangle will be of $32cm$ and its height will be $16cm$.

Now, 

Area of the triangle is:

$A=\dfrac{1}{2}\times b\times h$

$\Rightarrow A=\dfrac{1}{2}\times 32\times 16$

$\Rightarrow A=256\text{ }c{{m}^{2}}$

From the figure we can observe that the area of the $Ist$ triangle is equal to the area of the $I{{I}^{nd}}$ triangle by symmetry.

Area of $Ist$ shade = Area of $I{{I}^{nd}}$ shade

Hence, area of $I{{I}^{nd}}$ shade is:

$A=256\text{ }c{{m}^{2}}$

For $\text{II}{{\text{I}}^{\text{rd}}}$ triangle 

Let the sides of the triangle be $a$, $b$ and $c$.

$a=6cm$

$b=6cm$

$c=8cm$

Area of triangle can be evaluated by Heron’s formula:

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

Where,

$a$, $b$ and $c$ are the sides of the triangle

\[s=\dfrac{a+b+c}{2}\]

$\Rightarrow s=\dfrac{6+6+8}{2}$

$\Rightarrow s=10cm$

Area of $\text{II}{{\text{I}}^{\text{rd}}}$ triangle

$A=\sqrt{10\left( 10-6 \right)\left( 10-6 \right)\left( 10-8 \right)}\,\text{c}{{\text{m}}^{2}}$ 

\[\Rightarrow A=\sqrt{10\left( 4 \right)\left( 4 \right)\left( 2 \right)}\]

\[\Rightarrow A=\left( 4\times 2\sqrt{5} \right)\]

\[\Rightarrow A=8\sqrt{5}\]

$\Rightarrow A=17.92\text{ }c{{m}^{2}}$ 

Hence, the area of paper required for 1st shade, 2nd shade, and 3rd shade is $256\text{ }c{{m}^{2}}$, $256\text{ }c{{m}^{2}}$ and $17.92\text{ }c{{m}^{2}}$ respectively.

8. A floral design on a floor is made up of $16$ tiles which are triangular, the sides of the triangle being $9cm$, $28cm$ and $35cm$. Find the cost of polishing the tiles at the rate of $50\text{p per c}{{\text{m}}^{2}}.$ 

Ans: It can be observed that each tile is triangular.

Let’s consider a triangular tile having sides $a$, $b$ and $c$

$a=9cm$

$b=28cm$

$c=35cm$

Area of triangle can be evaluated by Heron’s formula:

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

Where,

$a$, $b$ and $c$ are the sides of the triangle

\[s=\dfrac{a+b+c}{2}\]

$\Rightarrow s=\dfrac{9+28+35}{2}$

$\Rightarrow s=36cm$

Area of each tile

$A=\sqrt{36\left( 36-35 \right)\left( 36-28 \right)\left( 36-9 \right)}\text{ c}{{\text{m}}^{2}}$ 

$\Rightarrow A=\sqrt{36\left( 1 \right)\left( 8 \right)\left( 27 \right)}$ 

$\Rightarrow A=36\sqrt{6}$ 

$\Rightarrow A=88.2\,c{{m}^{2}}$

It is given that there are $16$ triangular tiles

Thus,

Area of $16$ tiles

\[{{A}_{total}}=\left( 16\times 88.2 \right)\]

\[\Rightarrow {{A}_{total}}=1411.2\text{ }c{{m}^{2}}\]

Cost of polishing per $\text{c}{{\text{m}}^{2}}$ area 

$C=50p$

Cost of polishing $1411.2\,\text{c}{{\text{m}}^{2}}$ area

${{C}_{total}}=Rs.\left( 1411.2\times 0.50 \right)$

$\Rightarrow {{C}_{total}}=Rs.705.60$

Therefore, it will cost $Rs.\text{ }705.60$ while polishing all the tiles.

9. A field is in the shape of a trapezium whose parallel sides are $25m$ and $10m$. The non-parallel sides are $14m$ and $13m$. Find the area of the field.

Ans:

Draw a line $BE$ parallel to $AD$ and draw a perpendicular $BF$ on $CD$.

It can be observed that $ABED$ is a parallelogram.

In parallelogram $ABED$

$\text{BE}=\text{AD}=13\,\text{m}$ 

$\text{ED=AB=10 m}$ 

$\text{EC}=25-\text{ED}$

$\Rightarrow EC=15\,\text{m}$

For $\Delta BEC$

$BE=13m$

$EC=15m$

$BC=14m$

Area of triangle can be evaluated by Heron’s formula:

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

Where,

$a$, $b$ and $c$ are the sides of the triangle

\[s=\dfrac{a+b+c}{2}\]

$\Rightarrow s=\dfrac{13+15+14}{2}$

$\Rightarrow s=21m$

Area of triangle $\Delta BEC$

$A=\sqrt{21\left( 21-13 \right)\left( 21-14 \right)\left( 21-15 \right)}\text{ }$ 

$A=\sqrt{21\left( 8 \right)\left( 7 \right)\left( 6 \right)}$ 

$A=84{{m}^{2}}$

Also,

Area of $\Delta BEC=\dfrac{1}{2}\times \text{CE }\!\!\times\!\!\text{ BF}$

$\Rightarrow 84=\dfrac{1}{2}\times 15\times BF$

$\Rightarrow BF=\dfrac{84\times 2}{15}$

$\Rightarrow BF=11.2m$

Now, $ABED$ is a parallelogram

So, Area of $ABED$: $A=\text{BF }\!\!\times\!\!\text{ DE}$

$\Rightarrow A=112{{m}^{2}}$

Area of the field is:

${{A}_{total}}=Ar\left( \Delta BEC \right)+Ar\left( ABED \right)$

$\Rightarrow {{A}_{total}}=84+112$

$\Rightarrow {{A}_{total}}=196{{m}^{2}}$

Hence, the area of the field is $196{{m}^{2}}$.

NCERT Solutions for Class 9 Maths Chapter 12 Herons Formula - PDF Download

The fact that these NCERT Solution Class 9 Chapter 12 PDFs can be obtained for free from the internet is their best feature. The student only needs to read through the notes and solutions in their leisure time to brush up on their topics once they have downloaded the Heron's formula Class 9 Solutions. Students can read the text, look at these NCERT Maths solutions Class 9 Chapter 12 before contacting the professionals if they still have questions. Before any test, students can refer to the solutions while on the go with the free Heron's Formula Class 9 PDF.

You can opt for Chapter 12 - Heron’s Formula NCERT Solutions for Class 9 Maths PDF for Upcoming Exams and also You can Find the Solutions of All the Maths Chapters below.

NCERT Solutions for Class 9 Maths

• Chapter 1 - Number System

• Chapter 2 - Polynomials 

• Chapter 3 - Coordinate Geometry

• Chapter 4 - Linear Equations in Two Variables

• Chapter 5 - Introductions to Euclid's Geometry

• Chapter 6 - Lines and Angles

• Chapter 7 - Triangles

• Chapter 8 - Quadrilaterals

• Chapter 9 - Areas of Parallelogram and Triangles

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Heron's formula

• Chapter 13 - Surface area and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

A Brief Overview of Class 9 Maths Chapter 12 - Heron's Formula

12.1 Introduction

Chapter 12 Class 9 concepts you have learned in the previous chapters will be recalled a bit in this section. Revising the concepts of the properties of different shapes and various closed figures helps build the base to understand Heron's formula, which is then used to determine the area of a triangle.

The chapter starts with recalling how to calculate the perimeter of various figures and shapes using some examples. The various ways a triangle's area is calculated are also discussed. The easy formulae recalling how to calculate the area of a triangle have been discussed before the chapter begins to teach how to apply Class 9 Heron's Formula.

Heron's formula will teach students the advanced methods of calculating the area of any irregular triangle, where one can use the methods of calculating the area of a triangle discussed previously.

12.2 Area of a Triangle - by Heron’s Formula

The subsection starts with a simple description of Heron's formula. The students are taught its derivation and its use. Simple examples are used to explain the derivation of the formula. Each term linked to derive the formula has been mentioned so that the students find it easy to derive it and find the value of the terms and put them in Heron's formula to arrive at the right answer.

The formula includes the terms that represent the sides and the perimeters. With the sides of the triangle given, the student will be able to find the semi-perimeters of the triangle. The sides are not easily measurable, and the students must apply Heron’s formula.

12.3 Applications of Heron’s Formula in Finding the Areas of a Quadrilateral

This Chapter Maths Class 9 section discusses the various ways in which the area of a quadrilateral can be derived using Heron's formula. The student's knowledge of shapes and figures is first tested. Then Heron's formula will be applied, letting the students find out the various parameters to calculate the area of a quadrilateral. The scalene quadrilateral is divided into two parts, forming two right-angled triangles. This lets the students calculate the area as per the formula. The students have to use their knowledge to find the area of a scalene triangle. Finding the dimension of the sides is the challenge, which is explained in detail in this section.

12.4 Summary

Once the students study Chapter 12 Maths Class 9 Heron's formula, they are well versed with the concepts. They have figured out the Heron's formula, and they now know how it can be used to calculate the area of a quadrilateral or the area of a scalene triangle. They will learn how Heron's formula can be used to calculate the area of a quadrilateral or the area of a scalene triangle. The end summarises Heron's formula and the terms used in it. The students need to be able to recollect the terms of the formula fast, which will help them to remember the ways of using the formula and arrive at the solution to the problem.

NCERT Solutions Class 9 Maths Chapter 12 Exercises

Key Takeaways of NCERT Solutions Class 9 Maths Chapter 12 Heron’s Formula

The key takeaways from the Maths Chapter 12 Class 9 Solutions are as follows:

• Chapter 12 Maths Class 9 Solutions explain the chapter in great detail that lays the foundation of future topics in Mathematics

• The expert teachers have prepared the Heron’s formula Class 9 Questions with Solutions with varied approaches that solve all the confusion that the students may have about this topic

• The NCERT Class 9 Maths Chapter 12 Solutions PDF file can be downloaded freely and referred to when the student wishes to.
  
  

Extra Questions for Practice

1. Find the area of the triangular plot whose sides are in the ratio of 3:5:7 and the perimeter is 300.

2. The perimeter of the rhombus sheet is 40 cm with one diagonal of 12 cm and is painted on both sides at the rate of Rs. 8  per m2. Find the cost of painting.

3. A field in the form of a parallelogram has sides of 40 m and 46 m, and one of its diagonals is 100 m long. Find the area of the field.
   
   

Vedantu’s NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula will help students quickly cover all the chapter's important concepts. Students are advised to use these solutions to strengthen their understanding and score well in their Class 9 Maths exam.

Conclusion 

NCERT Solutions for Class 9 Maths Chapter 12 - Heron's Formula provides an indispensable resource for students diving into the world of geometry and trigonometry. This chapter equips learners with a powerful tool, Heron's Formula, for calculating the area of triangles with ease. The solutions offer step-by-step explanations, guiding students through complex calculations and practical applications. By mastering this chapter, students not only gain a deeper understanding of geometry but also sharpen their problem-solving skills. Heron's Formula is a valuable addition to their mathematical toolkit, setting a solid foundation for future studies in mathematics and various real-life scenarios where the calculation of triangle areas is essential.