NCERT Exemplar Class 9 Maths Solutions Chapter 12 Herons Formula

# NCERT Exemplar Class 9 Maths Solutions Chapter 12 Herons Formula

Edited By Ravindra Pindel | Updated on Aug 31, 2022 03:13 PM IST

NCERT exemplar Class 9 Maths solutions chapter 12 details Heron’s formula and related questions. The Heron's formula helps in finding the areas of triangles if side lengths are given. The NCERT exemplar Class 9 Maths chapter 12 solutions are highly accurate and elaborate in nature (prepared by experienced subject experts at Careers 360) and provides an outstanding approach to study NCERT Class 9 Maths. These NCERT exemplar Class 9 Maths chapter 10 solutions build a strong foundation of Heron’s Formula, also these NCERT exemplar Class 9 Maths solutions chapter 12 follow the syllabus recommended by the CBSE.

Also, read - NCERT Solutions for Class 9 Maths

## Exercise 12.1

Question:1

An isosceles right triangle has area $8\; cm^{^{2}}$ . The length of its hypotenuse is
(A) $\sqrt{32}\; cm$
(B) $\sqrt{16}\; cm$
(C) $\sqrt{48}\; cm$
(D) $\sqrt{24}\; cm$

An isosceles right triangle is given.
According to definition of right triangle, one angle should be 90o
According to definition of isosceles triangle any two sides equal.
i.e., AB = BC
Suppose equal sides of triangle be = x cm
[AB = BC = x]

Area of isosceles triangle = $\frac{1}{2}$ × base × height
$\Rightarrow$ 8 cm2 = $\frac{1}{2}$ × AB × BC
$\Rightarrow$ 8 × 2 = x × x [$\Theta$ AB = BC = x]
$\Rightarrow$ 16 =$x^{2}$
$\Rightarrow$ x = $\sqrt{16}$
$\Rightarrow$ x = 4 cm
So AB = BC = 4 cm
In $\Delta$ABC, using Pythagoras theorem
(AC)2 = (AB)2 + (BC)2
(AC)2 = (4)2 + (4)2
(AC)2 = 16 + 16
AC = $\sqrt{32}\; cm$
Hence hypotenuse of DABC is $\sqrt{32}\; cm$.
Hence option (A) is correct.

Question:2

The perimeter of an equilateral triangle is 60 m. The area is
(A) $10\sqrt{3}\; m^{2}$
(B) $15\sqrt{3}\; m^{2}$
(C) $20\sqrt{3}\; m^{2}$
(D) $100\sqrt{3}\; m^{2}$

[D]
Given perimeter of equilateral triangle = 60 m
Suppose the sides of equilateral triangle, AB = BC = CA = x m
We know that perimeter of equilateral triangle = 3 × side

60 = 3 × x
60 = 3x
$\frac{60}{3}\; = x$
x = 20 m
i.e., sides AB = BC = CA = 20 m
We know that
Area of equilateral triangle = $\frac{\sqrt{3}}{4}\; \times \; side^{2}$
$\frac{\sqrt{3}}{4}\; \times \; \left ( 20 \right )^{2}\; = \frac{\sqrt{3}}{4}\; \times \;20\times \; 20$
= $\sqrt{3}$ × 5 × 20 = $\sqrt{3}$ × 100 = $100\sqrt{3}\; m^{^{2}}$
Hence area of equilateral triangle is $100\sqrt{3}\; m^{^{2}}$.
Hence option (D) is correct.

Question:3

The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then the area of the triangle is
(A) 1322 cm2
(B) 1311 cm2
(C) 1344 cm2
(D) 1392 cm2

[C]
Given; In DABC, a = 56 cm, b = 60 cm, c = 52 cm

(Semi perimeter) S = $\frac{a+b+c}{2}$
S = $\frac{56+60+52}{2}$
S = $\frac{168}{2}\; = \; 84\; cm$
Using Heron’s formula
Area of triangle $= \sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$
$= \sqrt{84\left ( 84-56 \right )\left ( 84-60 \right )\left ( 84-52 \right )}$
$= \sqrt{84\times 28\times 24\times 32}$
$= \sqrt{7\times 3\times 2\times 2\times 2\times 2\times 7\times 2\times 2\times 6\times 2\times 4\times 4}$
$= \sqrt{7\times 7\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 4\times 4\times 2}$
= 7 × 2 × 2 × 2 × 2 × 4 × 3 = 1344 $cm^{^{2}}$
Hence area of DABC is 1344 $cm^{^{2}}$.
Hence option (C) is correct.

Question:4

The area of an equilateral triangle with side 2$\sqrt{3}$ cm is
(A) 5.196 cm2
(B) 0.866 cm2
(C) 3.496 cm2
(D) 1.732 cm2

[A] Given side of equilateral triangle $= \; 2\sqrt{3\;}\; cm$
We know that area of equilateral triangle $= \; \frac{\sqrt{3}}{4}\; \times \left ( side \right )^{2}$
$= \; \frac{\sqrt{3}}{4}\; \times \left ( 2\sqrt{3} \right )^{2}$
$= \; \frac{\sqrt{3}}{4}\; \times\; 4\; \times \left ( \sqrt{3} \right )^{2}$ $\left [ \left ( \sqrt{3} \right )^{2}\; = \; \sqrt{3}\times \sqrt{3}\; = 3 \right ]$
$\frac{\sqrt{3}}{4}\; \times 4\times 3$
$\sqrt{3}\times 3$ = 1.732 × 3 [$\therefore \; \sqrt{3 }$ = 1.732]
= 5.196 $cm^{2}$
Hence area of equilateral triangle is 5.196 $cm^{2}$.
Hence option (A) is correct.

Question:5

The length of each side of an equilateral triangle having an area of $9\sqrt{3}\; cm^{2}$ is
(A) 8 cm
(B) 36 cm
(C) 4 cm
(D) 6 cm

[D]
Given area of equilateral triangle $9\sqrt{3}cm^{2}$
Let the side of equilateral triangle be = a cm
$So\; we\; know\; that\; area\; of\; triangle\; = \frac{\sqrt{3}}{4}\times \left ( side \right )^{2}$

Exercise 1.1
$9\sqrt{3}= \frac{\sqrt{3}}{4}\times \left ( side \right )^{2}$
$9\sqrt{3}= \frac{\sqrt{3}}{4}\times \left ( a \right )^{2}$ [$\therefore$ side = a]
$9\sqrt{3}\times 4\; = \; \sqrt{3}\times a^{2}$
$a^{2}= \; \frac{36\sqrt{3}}{\sqrt{3}}$
a2 = 36
a = $\sqrt{36}$
a = 6 cm
Hence, side of equilateral triangle is 6 cm.
Hence option (D) is correct.

Question:6

If the area of an equilateral triangle is $16\sqrt{3}cm^{2}$, then the perimeter of the triangle is:
(A) 48 cm
(B) 24 cm
(C) 12 cm
(D) 36 cm

[B]
Given area of equilateral triangle = $16\sqrt{3}cm^{2}$
Suppose the side of equilateral triangle be = a cm

We know that,
Area of equilateral triangle $= \frac{\sqrt{3}}{4}\left ( side \right )^{2}$
$16\sqrt{3}\; = \frac{\sqrt{3}}{4}\times \left ( a \right )^{2}$
$16\times 4\times \sqrt{3}= \sqrt{3}\times \left ( a \right )^{2}$
$\frac{64\times \sqrt{3}}{\sqrt{3}}= a^{2}$
a2 = 64
a = $\sqrt{64}$
a = 8 cm
Perimeter = 3a = 3(8) = 24 cm
Hence option (B) is correct.

Question:7

The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude
(A) $16\sqrt{5}cm$
(B) $10\sqrt{5}cm$
(C) $24\sqrt{5}cm$
(D) 28 $cm$

$Given\; a= \; 35cm,\; b= 54cm\; ,c= 61cm$

$S= \frac{a+b+c}{2}$
$S= \frac{35+54+61}{2}$
$\Rightarrow S= 75cm$
Using Heron's formula area of triangle $= \; \sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$
$= \; \sqrt{75\left ( 75-35 \right )\left ( 75-54 \right )\left ( 75-61 \right )}$
$= \sqrt{75\times 40\times 21\times 14}$
$= \sqrt{5\times 5\times 3\times 5\times 4\times 2\times 7\times 3\times 7\times 2}= \sqrt{5\times 5\times 5\times 3\times 3\times 2\times 2\times 2\times 2\times 7\times 7}$
$=$$5\times 3\times 2\times 7\times 2\sqrt{5}$
$= 420\sqrt{5}cm^{2}$
We know that area of $\bigtriangleup$ABC $= \frac{1}{2}\times base\times altitude$
$420\sqrt{5}= \frac{1}{2}\times 35\times AD$
$420\sqrt{5}\times 2= 35\times altitude$
$\Rightarrow \frac{420\sqrt{5}\times 2}{35}= \; altitude$
$\Rightarrow \frac{60\sqrt{5}\times 2}{5}= \; altitude$
$\Rightarrow 24\sqrt{5}= \; altitude$
$Hence,altitude\; is \; 24\sqrt{5}cm$
Hence option (C) is correct.

Question:8

The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is
(A) $\sqrt{15}cm^{2}$
(B) $\sqrt{\frac{15}{2}}cm^{2}$
(C) $2\sqrt{15}cm^{2}$
(D) $4\sqrt{15}cm^{2}$

[A]

$We\; know\; that,$$We\; know\; that,semi-perimeter$

$S= \frac{a+b+c}{2}$

$S= \frac{2+4+4}{2}$

$S= \frac{10}{2}= 2cm$

Using Heron’s formula area of $\Delta$ABC $= \sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$

$= \sqrt{5\left ( 5-2 \right )\left ( 5-4 \right )\left ( 5-4 \right )}$

$= \sqrt{5\times 3\times 1\times 1}$

$= \sqrt{15}cm^{2}$

Hence, area of given triangle is $\sqrt{15}cm^{2}$.

Hence option (A) is correct.

Question:9

The edges of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting it at the rate of 9 paise per $cm^{2}$ is
(A) Rs 2.00
(B) Rs 2.16
(C) Rs 2.48
(D) Rs 3.00

[B]
Solution.
To find the cost of painting, we have to find the area of the triangular board
Let the sides be denoted as, a = 6 cm, b = 8 cm, c = 10 cm
$semi-perimeter,S= \frac{a+b+c}{2}$
$S= \frac{6+8+10}{2}= \frac{24}{2}= 12cm$
We know that,
Using Heron’s formula, area of triangle $= \sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$
$= \sqrt{12\left ( 12-6 \right )\left ( 12-8 \right )\left ( 12-10\right )}$
$= \sqrt{12\times 6\times 4\times 2}$
$= \sqrt{6\times 2\times 6\times 2\times 2\times 2}$
$6\times 2\times 2= 24cm^{2}$
$Now\; cost\; of\; painting\; 1cm^{2}= 9\; paise= \frac{9}{100}Rs= 0.09Rs.$
$\left [ 1Rs= 100paise,1\; paise= \frac{1}{100}Rs \right ]$
$\therefore\; cost\; of\; painting\; 24cm^{2}= 24\times 0.09= 24\times \frac{9}{100}= Rs2.16$
Hence,cost of painting Rs. 2.16.
Hence option (B) is correct.

## Exercise 12.2

Question:1

The area of a triangle with base 4 cm and height 6 cm is 24 $cm^{2}$

[False]
Given, Base 4 cm and height 6 cm
We know that,
$Area\; of\; triangle\; = \frac{1}{2}\times base\times height$
$= \frac{1}{2}\times 6\times 4= 3\times 4= 12cm\; ^{2}$
Therefore the given statement is false.

Question:2

[True]
Given, AB = AC = 4 cm and $\angle$A = $90^{\circ}$

.We know that
$Area\; of\; triangle= \frac{1}{2}\times base\times height$
$= \frac{1}{2}\times 4\times\; 4= 4\times 2= 8cm^{2}$
Hence, area of $\Delta$ is 8 $cm^{2}$ is True.

Question:3

The area of the isosceles triangle is $\frac{5}{4}\sqrt{11}cm^{2}$, if the perimeter is 11 cm and the base is 5 cm.

[True]
Given
$Area\; of\; isosceles\; triangle= \frac{5}{4}\sqrt{11}cm^{2}$
$Perimeter\; of\; triangle= 11cm$

We know that
$Perimeter= x+x+5 \; \; \; \; \; \; \; \; \; \; \left [ \therefore Let\; equal\; sides\; of\; triangle= x \right ]$
$11cm-5cm= 2x$
$\Rightarrow 6cm= 2x\Rightarrow x= \frac{6}{2}cm$
$\Rightarrow x=3cm$
Using Pythagoras theorem
$\left ( AB \right )^{2}=\left ( AD \right )^{2}+\left ( BD \right )^{2}$
$\left ( 3 \right )^{2}=\left ( AD \right )^{2}+\left ( \frac{5}{2} \right )^{2}$
$9-\frac{25}{4}=\left ( AD \right )^{2}$
$\Rightarrow AD= \sqrt{9-\frac{25}{4}}$
$\Rightarrow AD= \sqrt{\frac{36-25}{4}}$
$AD= \sqrt{\frac{11}{4}}=\frac{\sqrt{11}}{2}CM\; \; \; \; \; \; \; \left [ Q\sqrt{4} =\sqrt{2\times 2}=2\right ]$
$Area\; of\; isosceles\; triangle=\frac{1}{2}\times base\times height$
$= \frac{1}{2}\times 5\times\frac{\sqrt{11}}{2}=\frac{5\sqrt{11}}{4}cm^{2}$
Therefore the given statement is true.

Question:4

The area of the equilateral triangle is$20\sqrt{3}cm^{2}$ whose each side is 8 cm.

[False]
$Area\; of\; equilateral\; triangle =\frac{\sqrt{3}}{4}side^{2}$

$Required \; area =\frac{\sqrt{3}}{4}\times \left ( 8 \right )^{2}$
$=\frac{\sqrt{3}}{4}\times 8\times 8\; =16\sqrt{3}cm^{2}$
Therefore the given statement is false.

Question:5

If the side of a rhombus is 10 cm and one diagonal is 16 cm, the area of the rhombus is 96 $cm^{2}$.

[True]
$Side\; of\; rhombus = 10 cm$
$One\; Diagonal = 16 cm$

We know that diagonal of rhombus interest each other at right angle
$So\; In\; right\; \Delta AOB$
$Using\; Pythagoras\; theorem$
$(AB)^{2} = (OA)^{2} + (OB)^{2}$
$(10)^{2} = (8)^{2} + (OB)^{2}$
$100 = 64 + (OB)^{2}$
$100 -64=\left ( OB \right )^{2}$
$(OB)^{2} = 36$
$OB=\sqrt{36} \; \; \; \; \; \; \; \left [ Q\sqrt{36}=\sqrt{6\times 6}=6 \right ]$
$OB = 6 cm$
$\therefore BD = 2\; \times \; OB$
$BD = 2 \times 6 = 12 cm$
$Now, Area\; of\; rhombus = \frac{1}{2}\; \times product\; of \; its\; diagonals$
$= \frac{1}{2} \times BD \times AC$
$=\frac{1}{2} \times 12 \times 16$
$= 6 \times 16 = 96 cm^{2}$
Hence, area of rhombus is 96 $cm^{2}$
Therefore the given statement is true.

Question:6

The base and the corresponding altitude of a parallelogram are 10 cm and 3.5 cm, respectively. The area of the parallelogram is 30 $cm^{2}$

[False]
We know that
$Area\; of\; parallelogram = base \times height$
$= 10 \times 3.5$
$= 10 \times \frac{35}{10} =35cm^{2}$

Hence, area of parallelogram is $35cm^{2}$
Therefore the given statement is false.

Question:7

The area of a regular hexagon of side ‘a’ is the sum of the areas of the five equilateral triangles with side a.

[False]
According to question
$Area\; of\; regular\; hexagon = sum\; of\; area \; of\; five\; equilateral \; triangles$

We know that a regular hexagon is divided into 6 equilateral triangles by its diagonals.
$Area\; of\; 1\; equilateral\; triangle = \frac{\sqrt{3}}{4} \times a^{2}$
$Area\; of\; 6\; equilateral\; triangle =6\times \frac{\sqrt{3}}{4} \times a^{2}=\frac{3\sqrt{3}}{2}a^{2}$
$Area\; of\; 5\; equilateral\; triangle =5\times \frac{\sqrt{3}}{4} \times a^{2}=\frac{5\sqrt{3}}{2}a^{2}$
Therefore the given statement is false.

Question:8

The cost of levelling the ground in the form of a triangle having the sides 51 m, 37 m and 20 m at the rate of Rs $3\; per\; m^{2}$ is Rs 918.

[True]
$Let\; a = 51 \; m, b = 37\; m \; and \; c = 20\; m$

$S=\frac{51+37+20}{2}=\frac{108}{2}=54m$
We know that using Heron’s formula
$Area\; of \; triangle AB=\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$
$=\sqrt{54\left ( 54-51 \right )\left ( 54-37 \right )\left ( 54-20 \right )}$
$=\sqrt{54\times 3\times 17\times 34}$
$=\sqrt{9\times 3\times 2\times 3\times 17\times 17\times 2}$
$=\sqrt{3\times3\times 3\times 3\times 2\times 2\times 17\times 17}$
=$3\times 3\times 2\times 17$
$= 306m^{2}$
To find cost :
Cost of levelling 1 m2 = Rs 3
$\therefore Cost\; of\; levelling\; 306\; m^{2} = \; 3 \times 306 = Rs.\; 918$
Therefore the given statement is true.

Question:9

In a triangle, the sides are given as 11 cm, 12 cm and 13 cm. The length of the altitude is 10.25 cm corresponding to the side having length 12 cm.

True
$Let \; a\; =\; 11\; cm ;\; b\; =\; 12\; cm\; ;\; c = 13 cm$
$Semi perimeter (S) = \frac{a+b+c}{2}$
$= \frac{11+12+13}{2}$
$= \frac{36}{2}=18cm$

$(Area\; of\; triangle \; ABC)\; by\; Heron's\; formula$
$=\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$
$=\sqrt{18\left ( 18-11 \right )\left ( 18-12 \right )\left ( 18-13\right )}$
$= \sqrt{18\times 7\times 6\times 5}$
$= \sqrt{6\times 3\times 7\times 6\times 5}$
$= \sqrt{6^{2}\times 3\times 7\times 5}$
$= 6\sqrt{3\times 7\times 5}$
$= 6\sqrt{105}$ $\because \sqrt{105}=10.2469\simeq 10.25$
$= 6 \times 10.25$
$= 61.5 cm^{2}$
$Given\; altitude\; =\; 10.25\; cm\; and$
$Its \; corresponding\; base \; = \; 12\; cm$
$Area\; of \times triangle ABC = \frac{1}{2} \times Base \times corresponding\; Height$
$= \frac{1}{2} \times 12 \times 10.25$
$=61.5cm^{2}$
Hence the area obtained is the same.
Therefore the given statement is true.

## Exercise:12.3

Question:1

Find the cost of laying grass in a triangular field of sides 50 m, 65 m and 65 m at the rate of Rs$7\; per\; m^{2}$

$Rs.10500$
To find the cost, first we have to find the area of this triangular field.
$Let \; sides\; be\; a = 50\; cm ; b = 65\; cm ; c = 65 \; m$
$Semi\; perimeter(s) = \frac{a+b+c}{2}$
$= \frac{50+65+65}{2}$
$= \frac{180}{2}$
= 90
Area of Triangular field:
$By\; herons\; formula =\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$
$=\sqrt{90\left ( 90-50\right )\left ( 90-65 \right )\left ( 90-65 \right )}$
$=\sqrt{90\times 40\times 25\times 25}$
$=\sqrt{9\times 10\times 4\times 10\times \left ( 5^{} \right )^{2}\times \left ( 5 \right )^{2}}$
=$3\times 10\times 2\times 5^{2}$
= $30\times 2\times 25$
$= 1500m^{2}$
$Rate\; of\; laying \; grass = 7\; Rs \; per\; m^{2}$
$Cost\; of \; laying \; grass\; for\; 1500 m^{2} = Rs (7 \times 1500)$
$= 10500\; Rs$
$Hence\; the \; answer\; is\; 10500 Rs.$

Question:2

The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 13 m, 14 m and 15 m. The advertisements yield an earning of Rs $2000\; per\; m^{2}$ a year. A company hired one of its walls for 6 months. How much rent did it pay?

$[RS. 84000]$
$Here \; sides\; of \; triangular\; walls\; are \; a = 14 m, b = 15 m \; and \; c = 13 m$

We have to find the area of this triangle
Using Heron’s formula, semi-perimeter:
$S=\frac{a+b+c}{2}=\frac{14+15+13}{2}=\frac{42}{2}=21m$
$The\; area\; of\; triangular\; wall =\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$
$=\sqrt{21\left ( 21-13\right )\left ( 21-14 \right )\left ( 21-15 \right )}$
$={\sqrt{21\times 8\times 7\times 6}}$
$= \sqrt{3\times 7\times 2\times 2\times 2\times 7\times 2\times 3}$
$= \sqrt{\left ( 2\times 2\times 2\times 2 \right )\times \left ( 3\times 3\times 7\times 7 \right )}$
$= 2\times 2\times 3\times 7=84m^{2}$
Given that the advertisements yield earning per m2 for 1 year=RS 2000
$Earnings\; per\; m^{2} \; per\; month =Rs.\left ( \frac{2000}{12} \right )$
$Earnings\; per\; m^{2}\; for\; 6\; months =Rs.\; \left ( \frac{2000}{12}\times 6 \right )$
$Earnings\; for\; 84 m^{2}\; for\; 6 \; months =Rs\left ( \frac{2000}{2}\times 84 \right ) Rent \; the\; company\; has\; to\; pay$
$Rent \; the\; company \; has\; to\; pay = Rs. 2000 \times 42 = Rs. \; 84000$
$Hence\; company\; has\; to\; pay\; Rs.\; 84000.$

Question:3

From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The lengths of the perpendiculars are 14 cm, 10 cm and 6 cm. Find the area of the triangle.

$[300\sqrt{3}cm^{2} ]$
Here $\Delta$ABC is equilateral triangle i.e.
AB = BC = CA = x (Let)

Now we can see here three perpendicular
$OX \perp AB, OZ \perp BC, OY \perp AC$
$OX = 10 cm, OY = 14 cm\; and\; OZ = 6 cm$
$So\; that\; area\; of\; \Delta ABC = Ar(\Delta AOC) + Ar(\Delta BOC) + Ar(\Delta AOB)$
$\frac{\sqrt{3}}{4} side^{2}=\frac{1}{2} \times base \times height + \frac{1}{2} \times base \times height + \frac{1}{2} \times base \times height$
$[\Theta \Delta ABC\; is\; equilateral\; \Delta ,so\; area\; of\; equilateral\; \Delta = \frac{\sqrt{3}}{4}side^{2} ]$
$\Rightarrow \frac{^{\sqrt{3}}}{4}\times x^{2}=\frac{1}{2}\times AC\times OY+\frac{1}{2}\times BC\times OZ+\frac{1}{2}\times AB\times OX$
$Taking\; \frac{1}{2} \; common$
$\frac{\sqrt{3}}{4}\times x^{2}=\frac{1}{2}\times \left [ AC\times OY+BC\times OZ+AB\times OX \right ]$
$\frac{\sqrt{3}}{4}\times x^{2}=\frac{1}{2}\times \left [ AC\times OY+AC\times OZ+AC\times OX \right ]$
[$\Theta$ AB = BC = AC, $\Delta$ABC equilateral triangle]
$\frac{\sqrt{3}}{4}\times x^{2}=\frac{1}{2}AC\times \left [ OY+OZ+OX \right ]$
$\frac{\sqrt{3}}{4}\times x^{2}=\frac{1}{2}AC\times \left [ 14+6+10 \right ]$
$\frac{\sqrt{3}}{4}\times x^{2}=\frac{1}{2}AC\times \left [ 30 \right ]$
$\frac{\sqrt{3}}{4}\times x^{2}=15\times AC$
$\sqrt{3}\times x^{2}=15\times x\times 4\; \; \; \; \; \; [\because AC = BC = AC = x]$
$\frac{x^{2}}{x}=\frac{15\times 4}{\sqrt{3}}$
$x=\frac{60}{\sqrt{3}}cm$
$Hence \; the\; length\; of\; the\; sides\; of\; \Delta ABC\; is\; \frac{60}{\sqrt{3}}cm$
$Area\; of\; \Delta ABC =\frac{\sqrt{3}}{4}\times \left ( \frac{60}{\sqrt{3}} \right )^{2}=\frac{\sqrt{3}}{4}\times \frac{60}{\sqrt{3}}\times \frac{60}{\sqrt{3}}$
$=\frac{15\times 60}{\sqrt{3}}=\frac{900}{\sqrt{3}}$
On rationalisation
$\frac{900}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{900\times \sqrt{3}}{3}=300\sqrt{3}cm^{2}$

Question:4

The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. Find the area of the triangle.

$[32\sqrt{2}cm^{2} ]$
$Given\; perimeter\; of\; isosceles\; triangle = 32 cm$
$We\; have\; ratio\; of\; equal\; side\; and\; base = 3 : 2$
$Thus,\; let\; the \; sides\; of\; triangle\; be\; AB = AC = 3x\; and\; BC = 2x$
$\therefore Perimeter \; of\; a\; triangle = 32 cm$
$3x + 3x + 2x = 32$
$8x = 32$
$x=\frac{32}{8}$
$x = 4 cm$
$So, AC = AB = 3 \times 4 cm\; and\; BC = 2 \times 4 cm$
$AC = AB = 12 cm\; and \; BC = 8 cm$

$Now, a = 8\; cm, b = 12\; cm, c = 12 \; cm$
Using Heron’s formula
$S= \frac{a+b+c}{2}=\frac{8+12+12}{2}=\frac{32}{2}=16cm$
$\therefore\; Area \; of\; isosceles\; triangle\; \Delta ABC =\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$
$=\sqrt{16\left ( 16-8 \right )\left ( 16-12 \right )\left ( 16-12 \right )}$
$= \sqrt{16\times 8\times 4\times 4}$
$= \sqrt{4\times 4\times 4\times 2\times 4\times 4}$
$=4\times 4 \sqrt{ 2\times 2\times 2}=4\times 4\times 2\sqrt{2}=32\sqrt{2}\; cm^{2}$
$Hence,\; the\; area\; of\; isosceles \; triangle\; is\; 32\sqrt{2}\; cm^{2}$

Question:5

$Area = 180 cm^{2}$
$Length\; of\; the\; altitude \; from\; vertex\; A\; on\; the\; side\; DC = 15\; cm$

$Area \; of\; parallelogram = 2 \times area \; of\; \Delta DBC$
$So \; here\; \Delta DBC\; sides\ having\; DB = 25 cm, BC = 17cm\;and\;CD = 12 cm$
Using Heron’s formula
$S=\frac{25+17+12}{2}=\frac{54}{2}=27cm$
$Area \; of\; \; triangle\; \Delta DBC =\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$
$=\sqrt{27\left ( 27-25 \right )\left ( 27-17 \right )\left ( 27-12 \right )}$
$= \sqrt{27\times 2\times 10\times 15}$
$= \sqrt{9\times 3\times 2\times 5\times 2\times 5\times 3}$
$= \sqrt{3\times 3\times 3\times 3\times 5\times 5\times 2\times 2}$
$= 2\times 3\times 3\times 5=90\; cm^{2}$
$\\ Area\; of\; parallelogram \; ABCD = 2 \times Ar(\Delta DBC) = 2 \times 90 cm^{2} = 180 cm^{2}.$
$We \; know \; that\; area\; of\; parallelogram = base \times height$
$= base\; CD \times length\; of\; the \; altitude\; from \; vertex\; A \; on\; the\; side\; DC$
$180\; cm^{2} = 12\; cm \; (length\; of\; the \; altitude\; from\; vertex\; A\; on\; the\; side\; DC)$
$\\Length\; of\; the \; altitude\; from \; vertex\; A\; on\; the\; side\; DC=180/12 = 15 cm.$
Hence, length of the altitude from vertex A on the side DC = 15 cm

Question:6

A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. Find the area of the parallelogram.

$[600\sqrt{15}m^{2} ]$
We know that ABCD is parallelogram
So the opposite sides are equal.
AB = CD = 60 m, AD = BC = 40 m

$Thus, Area\; of\; parallelogram = 2 \times Ar(\Delta ABC)$
In $\Delta$ABC, using heron’s formula
$Semi perimeter, S =\frac{60+40+80}{2}=\frac{180}{2}=90m$
$Area \; of\; \; triangle\; \Delta ABC =\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$
$=\sqrt{90\left ( 90-60 \right )\left ( 90-40 \right )\left ( 90-80 \right )}$
$=\sqrt{90\times 30\times 50\times 10}$
$=\sqrt{3\times 30\times 30\times 10\times 10\times 5}$
$= 30\times 10\sqrt{3\times 5}=300\sqrt{15}\; m^{2}$
$Hence,\; area\; of\; parallelogram =2\times 300\sqrt{15}\; m^{2}=600\sqrt{15}\; m^{2}$

Question:7

The perimeter of a triangular field is 420 m and its sides are in the ratio 6 : 7 : 8. Find the area of the triangular field.

$[2100\sqrt{15}\; m^{2} ]$
Given perimeter of a triangular field = 420 m and ratio of sides = 6 : 7 : 8
Let the sides of triangular field be = 6x, 7x and 8x

$\therefore Perimeter\; of\; triangular\; field = 6x + 7x +\; 8x$
$420 = 6x + 7x + 8x$
$420 = 21x$
$x=\frac{420}{21}$
$x = 20$
$Then\; sides\; are\; 6 \times 20 = 120m, 7 \times 20 = 140m\; and \; 8 \times 20 = 160m$
Using Heron’s formula for finding the area of $\Delta$ABC
a = 120m, b = 140m, c = 160m
$S=\frac{a+b+c}{2}=\frac{120+140+160}{2}=\frac{420}{2}=210m$
$Area \; of\; \; triangle\; \Delta ABC =\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$
$=\sqrt{210\left ( 210-120 \right )\left ( 210-140 \right )\left ( 210-160\right )}$
$= \sqrt{210\times 90\times 70\times 50}$
$= \sqrt{7\times 30\times 3\times 30\times 7\times 10\times 5\times 10}$
$= \sqrt{7\times 7\times 10\times 10\times 30\times 30\times 3\times 5}$
= $2100\sqrt{15}\; m^{2}$

Question:8

The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order) respectively, and the angle between the first two sides is a right angle. Find its area.

$\left [ 24\; cm^{2}+24\sqrt{6}cm^{2} \right ]$
Here we have, AB = 6 cm, BC = 8 cm, CD = 12 cm and AD = 14 cm
According to question, the angle between AB and BC is $90^{\circ}$

Join AC
In Right $\Delta$ABC, [By Pythagoras theorem]
$AC^{2} = (6)^{2} + (8)^{2}$
$AC^{2} = 36 + 64$
$AC = \sqrt{100}$
$AC = 10 cm$
$Area\; of \; quadrilateral \: ABCD = Ar(\Delta ABC) + Ar(\Delta ACD)$
$Now, we\: find \; area\; of \; \Delta ABC = \frac{1}{2} \times base \times height$
$= \frac{1}{2}\times BC\times AB= \frac{1}{2}\times 8\times 6= (4 \times 6) cm^{2}$
$Area of \Delta ABC = 24 cm^{2}$
$In \Delta ACD, AC = a = 10 cm, CD = b = 12 cm, AD = c = 14 cm$
$S= \frac{a+b+c}{2}= \frac{10+12+14}{2}= \frac{36}{2}= 18cm$
$Using\; Heron's \; formula \; area \; of \Delta ACD=\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$
$= \sqrt{18\left ( 18-10 \right )\left ( 18-12 \right )\left ( 18-14 \right )}$
$= \sqrt{18\times 8\times 6\times 4}$
$= \sqrt{9\times 2\times 4\times 2\times 3\times 2\times 4}$
$= \sqrt{3\times 3\times 2\times 2\times 3\times 2\times 4\times 4}$
$= 3\times 2\times 4\times \sqrt{3\times 2}= 24\sqrt{6}\; cm^{2}$
$Hence, area \; of\; quadrilateral \; ABCD = 24cm^{2}+24\sqrt{6}\; cm^{2}$

Question:9

A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs 5 per $cm^{2}$. Find the cost of painting.

[Rs. 960]
Let ABCD be a rhombus thus AB = BC = CD = DA = x (Let)

$We \; have\; perimeter \; of \; rhombus = 40 cm$
$\Rightarrow AB + BC + CD + DA = 40 cm$
$\Rightarrow x + x + x + x = 40 cm$
$\Rightarrow 4x = 40 cm$
$\Rightarrow x = \frac{40}{4} cm$
$\Rightarrow x = 10 cm$
$\therefore sides \; of \; rhombus \; AB = BC = CD = DA = 10 cm$
Area of rhombus = 2 $\times$ Ar($\Delta$ABC) [diagonal of rhombus divides it into two triangles of equal area]
Now, we find area of triangle using Heron’s formula
$In\; \Delta ABC, S=\frac{a+b+c}{2}= \frac{10+10+12}{2}= \frac{32}{2}= 16cm$
$Area \; of\; \Delta ABC =\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$
$=\sqrt{16\left ( 16-10 \right )\left ( 16-10 \right )\left ( 16-12 \right )}$
$= \sqrt{16\times 6\times 6\times 4}$
$= \sqrt{8\times 8\times 6\times 6\times 4}$
$= \sqrt{4\times 4\times 6\times 6\times 4}$
$= 4\times 6\sqrt{2\times 2}$
$= 4\times 6\times 2= 48cm^{2}$
Now, Area of rhombus = 2 $\times$ Ar($\Delta$ABC)
$= 2\times 40cm^{2}= 96cm^{2}$
We find the cost of painting
Thus,
$\because cost\; of \; painting\; the\; sheet \; of\; 1 cm^{2} = Rs.\; 5$
$\therefore cost \; of\; painting\; the \; sheet\; of \; 96\; cm^{2}= Rs.\; 96 \times 5 = Rs.\; 480$
Hence, the cost of the painting both sides of the sheet = 2 $\times$ 480 = Rs. 960.

Question:10

$[114\; cm^{2}]$
$Firstly\; we\; have\; side \; PS = 12 m, SR = 13, QR = 7 m$

Join RT
So here PT = PS – ST
PT = 12 m – 5 m
PT = 7 m
and ST = PS – PT
$ST = \left ( 12-7 \right )m$
ST = 5 m
Now, In $\Delta$STR, Using Pythagoras theorem
$We\; get, (SR)^{2} = (ST)^{2} + (TR)^{2}$
$(13)^{2} = (5)^{2} + (TR)^{2}$
$169 = 25 + (TR)^{2}\; \; \; \; \; \; \; \; \; \; \; [(13)^{2} = 169 \; and\; (5)^{2} = 25]$
$169 - 25 = (TR)^{2}$
$144 = (TR)^{2}\; \; \; \; \; \; \; \; \; \; \; \; [ \sqrt{144} = 12]$
$TR = 12cm\; \; \; \; \; \; \; \; \; \; \; \; [ \sqrt{12\times 12} = 12]$
Now, we can find the area of trapezium
$Area \; of\; trapezium = \frac{1}{2} \times [sum\; of \; parallel\; sides] \times height$
$= \frac{1}{2}\times \left [ 12+7 \right ]\times 12$
$= 19\times 6\; cm^{2}$
$= 114\; cm^{2}$
$Hence, the\; area\; of\; trapezium\; is\; 114\; cm^{2}.$

## Exercise 12.4

Question:1

$Red = 242 \; cm^{2}$
$Yellow = 484\; cm^{2}$
$Green = 373.14 \; cm^{2}$
$Given,\; ABCD\; is\; a\; square.$
We know that all sides of a square are equal
AB = BC = CD = DA and
$\angle A = \angle B = \angle C = \angle D = 90^{\circ} [\because \; All\; angles\; of\; a\; square\; are\; 90^{\circ}]$
$In \Delta ABC,\; using \; Pythagoras \; theorem$
$(AC)^{2} = (AB)^{2}+ (BC)^{2}$
$(44)^{2} = (AB)^{2}+ (BC)^{2}\; \; \; \; \; \; \; \; [\Theta \; AB = BC\; equal\; sides]$
$44\times 44=2\left ( AB \right )^{2}$
$\frac{44\times 44}{2}=\left ( AB \right )^{2}$
$(AB)^{2} = 22 \times 44$
Taking square root on both sides
$\sqrt{\left ( AB \right )^{2}}=\sqrt{22\times 44}$
$AB=\sqrt{22\times 2\times 22}$
$AB=22\sqrt{2}$
$AB = BC = CD = DA =22\sqrt{2}\; cm$
$Now,\; Area\; of \; square\; ABCD = (side)^{2}$
$=\left ( 22\sqrt{2}\right )^{2}=22\times 22\times \sqrt{2}\times \sqrt{2}$
$= 484\times \sqrt{2\times 2}=484\times 2$
$Area\; of\; square\; ABCD = 968\; cm^{2}$
But square ABCD is divided into four coloured squares.
$So,\; area\; of\; Yellow \; I =\frac{968}{4}= 242 cm^{2}$
$Area\; of\; Yellow \; II =\frac{968}{4}= 242 \; cm^{2}$
$Area\; of\; Green \; III =\frac{968}{4}= 242 \; cm^{2}$
$Area\; of\; Red \; IV =\frac{968}{4}= 242 \; cm^{2}$
$Total\; yellow\; area\; = 242\; cm^{2} + 242\; cm^{2} = 484\; cm^{2}$
We have to find the lower triangle of green colour as well.
$Let\; a = 20\; cm, b = 20\; cm, c = 14\; cm$
$Semi\; perimeter(s) = \frac{a+b+c}{2}$
$= \frac{20+20+14}{2}$
$= \frac{54}{2}=27$
Area of Triangular field:
$By\; heron's\; formula =\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$
$=\sqrt{27\left ( 27-20\right )\left ( 27-20 \right )\left ( 27-14 \right )}$
$= \sqrt{3\times 3\times 3\times 7\times 7\times 13}$
$= 21\sqrt{3\times 13}$
$= 131.14\; cm^{2}$
$So\; total\; green\; area\; = 242 + 131.14 = 373.14\; cm^{2}$
$Hence,\; paper\; required$
$Red = 242\; cm^{2}$
$Yellow = 484\; cm^{2}$
$green = 373.14\; cm^{2}$

Question:2

The perimeter of a triangle is 50 cm. One side of a triangle is 4 cm longer than the smaller side and the third side is 6 cm less than twice the smaller side. Find the area of the triangle.

$\left [ 20\sqrt{30}\; cm^{2} \right ]$
Let the smaller side of the triangle be x cm
Let BC = x cm

According to question,
One side of a triangle is 4 cm longer than the smaller side
Let this side be AC = x + 4
Also, third side is 6 cm less than twice the smaller side
Let this side be AB = (2x - 6) cm
Given perimeter of $\Delta$ABC = 50 cm
x + x + 4 + 2x - 6 = 50
$4x - 2 = 50$
$4x = 50+2$
$4x = 52$
$x = \frac{52}{4}$
$x = 13 \; cm$
$So\; the\; side \; AC = (x + 4) = (13 + 4) = 17\; cm$
$Side\; AB = \left ( 2x-6 \right )\; cm= \left ( 26-6 \right )\; cm= 20\; cm$
Now in $\Delta$ABC, a = 13 cm, b = 17 cm, and c = 20 cm
Using Heron’s formula
$S= \frac{a+b+c}{2}= \frac{13+17+20}{2}= \frac{50}{2}= 25\; cm$
$Area \; of\; \Delta ABC =\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$
$=\sqrt{25\left ( 25-13 \right )\left ( 25-17 \right )\left ( 25-20 \right )}$
$=\sqrt{25\times 12\times 8\times 5}$
$=\sqrt{5\times 5\times 2\times 2\times 3\times 2\times 2\times 2\times 5}$
$=\sqrt{5\times 5\times 5\times 2\times 2\times 2\times 2\times 2\times 3}$
$= 5\times 2\times 2\sqrt{30}$
$Area \; of \; \Delta ABC = 20\sqrt{30}\; cm^{2}$
$Hence\; the \; area\; of\; triangle\; is \; 20\sqrt{30}\; cm^{2}$

Question:3

The area of a trapezium is 475 $cm^{2}$ and the height is 19 cm. Find the lengths of its two parallel sides if one side is 4 cm greater than the other.

$\left [ 23 \; cm, 27\; cm \right ]$
Let the smaller parallel side be CD = x cm

Then other parallel side AB = (x + 4) cm
Given, area of trapezium = 475 $cm^{2}$
Height DE = 19 cm
$We \; know \; that, \; Area \; of \: trapezium = \frac{1}{2} \times height \times (sum\; of \; parallel \; sides)$
$475= \frac{1}{2}\times DE\times \left ( DC+AB \right )$
$475\times 2= 19\times \left ( x+x+4 \right )$
$\frac{475\times 2}{19}= 2x+4$
25 $\times$ 2 = 2x + 4
50 = 2x + 4
50 - 4 = 2x
46 = 2x
$x= \frac{46}{2}$
x = 23 cm
So the smaller side CD is 23 cm and other parallel side AB is (23 + 4) cm = 27 cm

Question:4

A rectangular plot is given for constructing a house, having a measurement of 40 m long and 15 m in the front. According to the laws, a minimum of 3 m, wide space should be left in the front and back each and 2 m wide space on each of other sides. Find the largest area where house can be constructed.

$[374\; m^{2}]$
Let ABCD be the rectangular plot,
AB = 40 cm, AD = 15 cm

Given that minimum of 3 m wide space should be left in the front and back
$length \; of \; PQ = [AB - (3 + 3)] \; m$
$PQ = [40 - 6] \; m$
$PQ = 34 \; m$
Similarly, RS = 34 m
Given that 2 m wide space on each of other sides is to be left
$Length\; of \; PS = [AD - (2 + 2)] \; m$
$PS = [15 - 4] \; m$
$PS = 11 \; m \; and$
$QR = 11 \; m$
So here PQRS is another rectangle formed in the rectangle ABCD
So, Area of rectangle PQRS = length $\times$ breadth
$= PQ\times PS= \left ( 34\times 11 \right )\; m^{2}= 374\; m^{2}$
Hence the area of house can be constructed in 374 $m^{2}$

Question:5

A field is in the shape of a trapezium having parallel sides 90 m and 30 m. These sides meet the third side at right angles. The length of the fourth side is 100 m. If it costs Rs 4 to plough 1$m^{2}$ of the field, find the total cost of ploughing the field.

$\left [ Rs. \; 19200 \right ]$
Given, ABCD is trapezium having parallel side AB = 90 m, CD = 30 m

Draw DE parallel to CB
So, BE = 30 m
Now, AE = (AB - EB)
AE = (90 - 30) m
AE = 60 m
So, in right triangle $\Delta$AED
$\left ( AD \right )^{2}= (AE)^{2} + (DE)^{2} \: \; \; \; \; \; [Using Pythagoras theorem]$
$\left ( 100 \right )^{2}= (60)^{2} + (DE)^{2}$
$10000= 3600+(DE)^{2}$
$10000- 3600= (DE)^{2}$
$6400= (DE)^{2}$
Taking square root on both sides
$\sqrt{6400}= \sqrt{(DE)^{2}}$
DE = 80 m
$We\; know \; that\; the \; area \; of \; trapezium \; ABCD = \frac{1}{2} \times (sum \; of \; parallel \; sides) \times height$
$= \frac{1}{2} \times (AB+CD) \times DE$
$= \frac{1}{2} \times (90+30) \times 80$
$= 120\times 40= 4800\; m^{2}$
$\because \; cost \; of\; ploughing \; 1 m^{2} \; field = Rs \; 4$
$\therefore \; cost \; of \; ploughing \; 4800 \; m^{2}\; field = 4800 \times 4 = Rs. \; 19200$
Hence the total cost of ploughing the field is Rs. 19200.

Question:6

$[3 cm]$

AB = 7.5 cm, AC = 6.5 cm, BC = 7 cm

Let a = 7.5 cm, b = 6.5 cm, c = 7 cm

$Now, S=\frac{a+b+c}{2}=\frac{7.5+6.5+7}{2}=\frac{21}{2}=10.5\; cm$

$Area\; of\; \Delta ABC,\; By\; heron's\; formula =\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$

$=\sqrt{10.5\left ( 10.5-7.5 \right )\left ( 10.5-6.5 \right )\left ( 10.5-7 \right )}$

$=\sqrt{10.5\times 3\times 4\times 3.5}$

$=\sqrt{\frac{105}{10}\times 3\times 4\times \frac{35}{10}}$

$=\sqrt{21\times 3\times 7}=\sqrt{3\times 7\times 3\times 7}=3\times 7=21\; cm^{2}$

$Now,\; we\; find\; the \; length\; DF \; of\; parallelogram\: DBCE$

$Area\; of\; parallelogram = base \times height = BC \times DF$

$Area\; of\; parallelogram = 7DF$

According to question,

$Area\; of\; \Delta ABC = Area \; of\; parallelogram DBCE$

$21 = 7DF$

$\frac{21}{7}= DF$

$DF = 3\; cm$

$Hence\; height\; of\; parallelogram\; is\; 3\; cm.$

Question:7

$[QC = 45\; cm, PD = 40\; cm]$
$Given, BC = 51\; cm \; and \; CD = 25\; cm$
$Area \; of\; rectangle\; ABCD = (51 \times 25) cm^{2}$
$In\; trapezium\; PQCD\; , parallel\; sides\; QC\; and\; PD\; are\; in\; the\; ratio\, 9:8$
Let length of QC = 9x and PD = 8x
$We\; have\: , Area\: of\: trapezium\: PQCD = \frac{5}{6}th\: part\; of\; Area(ABCD)$
$\frac{1}{2} \times (Sum\; of\; ||\; sides) \times height =\frac{5}{6} \times length \times breadth$
$\frac{1}{2}\times (PD + QC) \times CD = \frac{5}{6}\times BC \times CD$
$\frac{1}{2} \times (8x + 9x) \times 25 = \frac{5}{6}\times 51 \times 25$
$\frac{1}{2}\times 17x \times 25 = \frac{5}{6}\times 51 \times 25$
$\frac{25}{2}\times 17x = \frac{5}{6}\times 51 \times 25$
$x = \frac{5}{6}\times 51 \times 25\times \frac{1}{17}\times \frac{2}{25}$
$x = 5$
$PD = 8x$
$= 8\times 5$
$= 40\; cm$
$QC = 9x$
= 9 x 5 = 45

Question:8

$[The\; area\; of\; the\; design\; is\; 1632\; cm^{2}\; and\; the\; remaining \; area\; of\; the \; tile \; is\; 1868\; cm^{2}.]$
We have dimensions of rectangle tile as 50 cm × 70 cm
We know that area of rectangle = length × breadth
Given sides of triangular design: 26 cm, 17 cm, 25 cm
To find the area using Heron’s formula
Let, a = 26 cm, b = 17 cm, c = 25 cm
$S=\frac{a+b+c}{2}=\frac{26+17+25}{2}=\frac{68}{2}=34\; cm$
$Area\; of\; triangle =\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$
$=\sqrt{34\left ( 34-26 \right )\left ( 34-17 \right )\left ( 34-25 \right )}$
$= \sqrt{34\times 8\times 17\times 9}$
$= \sqrt{17\times 2\times 2\times 2\times 2\times 17\times 3\times 3}$
$= 2\times 2\times 3\times 17$
Area of $\Delta$ABC = 204 $cm^{2}$
But we have 8 triangles of equal area
So area of design = 8 × area of one $\Delta$
= 8 × 204 = 1632 $cm^{2}$
Remaining area of tile = Area of tile - Area of design
= (3500 – 1632) $cm^{2}$ = 1868 $cm^{2}$
Hence the area of the design is 1632 $cm^{2}$ and the remaining area of the tile is 1868 $cm^{2}$.

## Important Topics for NCERT Exemplar Solutions Class 9 Maths Chapter 12:

Key topics covered in NCERT exemplar Class 9 Maths solutions chapter 12 are:

• How to find out are of a triangle using base length and height.
• How to find out are of a triangle using Heron’s Formula.
• NCERT exemplar Class 9 Maths solutions chapter 10 explain how Heron’s formula can be used to find out the area of a quadrilateral.

## NCERT Class 9 Maths Exemplar Solutions for Other Chapters:

 Chapter 1 Number System Chapter 2 Polynomials Chapter 3 Coordinate geometry Chapter 4 Linear equations in Two Variable Chapter 5 Introduction to Euclid’s Geometry Chapter 6 Lines and Angles Chapter 7 Triangles Chapter 8 Quadrilaterals Chapter 9 Area of Parallelograms and Triangles Chapter 10 Circles Chapter 11 Constructions Chapter 13 Surface Areas and Volumes Chapter 14 Statistics and Probability

## Features of NCERT Exemplar Class 9 Maths Solutions Chapter 12:

These Class 9 Maths NCERT exemplar chapter 12 solutions provide a basic knowledge of how Heron’s formula can be used to find out the area of a triangle without calculating any angle or any other distance except the length of sides. The proof of this formula can be obtained in a book named Metrica. This formula can be proved by the use of trigonometry in modern mathematics. The practice problems provided in the exemplar are explained in a highly elaborate way through Class 9 Maths NCERT exemplar solutions chapter 12 Heron’s formula and prove to be adequate for solving other books such as NCERT Class 9 Maths, RD Sharma Class 9 Maths, Mathematics Pearson Class 9, RS Aggarwal Class 9 Maths etcetera.

NCERT exemplar Class 9 Maths solutions chapter 12 pdf download allows the users to use the pdf version of these solutions and can be used while attempting NCERT exemplar Class 9 Maths chapter 12 in an offline environment. The web page can be downloaded using any existing options to download the web page.

### Check the Solutions of Questions Given in the Book

 Chapter No. Chapter Name Chapter 1 Number Systems Chapter 2 Polynomials Chapter 3 Coordinate Geometry Chapter 4 Linear Equations In Two Variables Chapter 5 Introduction to Euclid's Geometry Chapter 6 Lines And Angles Chapter 7 Triangles Chapter 8 Quadrilaterals Chapter 9 Areas of Parallelograms 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

### Also Check NCERT Books and NCERT Syllabus here

1. Is Heron’s formula applicable in finding the area of a scalene triangle?

Yes, we can find out the area of any triangle if three sides of a triangle are known by using the heron’s formula.

2. Is Heron Formula have any limitation to find out area of triangle?

No, we can use heron’s formula to find out the area of any triangle if sides are given.

3. If we know the perimeter of an equilateral triangle, can we find out the area of this triangle?

In an equilateral triangle, all sides have equal length, therefore if we know the perimeter of triangle, we know each side length. Now by using hero formula we can find out area of this equilateral triangle.

4. How we can find out area of triangle?

We can find out area of triangle by two methods. If we know the base length and height of triangle then half of their product will give the area of triangle. If we know three sides of the triangle, we can use hero formula to find out area of triangle

5. What is weightage of the chapter on Heron’s Formula in the final examination?

This chapter concludes to around 5-7% marks of the final paper. Generally, the type of questions that can be expected from this chapter is MCQs and short answer-type questions.  NCERT exemplar Class 9 Maths solutions chapter 12 is adequate to practice, understand and score well in the examinations.

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