NCERT Exemplar Class 9 Maths Solutions Chapter 12 Herons Formula

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) $\sqrt{24}\; cm$

Answer:

An isosceles right triangle is given.
According to the definition of a right triangle, one angle should be 90^o
According to the definition of isosceles trian angle any two sides equal.
i.e., are AB = BC
Suppose equal sides of the 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 [$\because$ 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) $100\sqrt{3}\; m^{2}$

Answer:

Given the perimeter of the equilateral triangle = 60 m
Suppose the sides of an equilateral triangle, AB = BC = CA = x m
We know that the perimeter of an 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

Answer:

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

Answer:

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, the area of equity lateral 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

Answer:

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


$9\sqrt{3}= \frac{\sqrt{3}}{4}\times \left ( side \right )^{2}$
$9\sqrt{3}= \frac{\sqrt{3}}{4}\times \left ( a \right )^{2}$ [$\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, the side of an 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

Answer:

Given area of equilateral triangle = $16\sqrt{3}cm^{2}$
Suppose the side of the equilateral triangle is = 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 $cm$

Answer:

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

$S= \frac{a+b+c}{2}$
$S= \frac{35+54+61}{2}$
$\Rightarrow S= 75cm$
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 a base 2of 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) $4\sqrt{15}cm^{2}$

Answer:

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, the area of the 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

Answer:

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, the cost of painting is Rs. 2.16.
Hence, option (B) is correct.

Question 1 Write True or False and justify your answer:
The area of a triangle with a base of 4 cm and height of 6 cm is 24 $cm^{2}$

Answer:

Given a base of 4 cm and a height of 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 Write True or False and justify your answer:
The area of $\Delta$ABC is 8 $cm^{^{2}}$ in which AB = AC = 4 cm and $\angle$A = $90^{\circ}$

Answer:

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 Write True or False and justify your answer:
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.

Answer:

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

We know that
$Perimeter= x+x+5 \; \; \; \; \; \; \; \; \; \; \left [ \therefore Let\; equal\; sides\; of\; triangle= x \right ]$
$11cm-5cm= 2x$
$\Rightarrow 6cm= 2x\Rightarrow x= \frac{6}{2}cm$
$\Rightarrow x=3cm$
Now, in the right triangle ADB
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 Write True or False and justify your answer:
The area of the equilateral triangle is$20\sqrt{3}cm^{2}$ whose each side is 8 cm.

Answer:

[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 Write True or False and justify your answer:
If the side of a rhombus is 10 cm and one diagonal is 16 cm, the area of the rhombus is 96 $cm^{2}$.

Answer:

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

We know that the diagonals of a rhombus intersect each other at a 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, the area of a rhombus is 96 $cm^{2}$
Therefore, the given statement is true.

Question 6 Write True or False and justify your answer:
The base and the corresponding altitude of a parallelogram are 10 cm and 3.5 cm, respectively. The area of the parallelogram is 30 $cm^{2}$

Answer:

We know that
$Area\; of\; parallelogram = base \times height$
$= 10 \times 3.5$
$= 10 \times \frac{35}{10} =35cm^{2}$
Hence, the area of a parallelogram is $35cm^{2}$
Therefore, the given statement is false.

Question 7 Write True or False and justify your answer:
The area of a regular hexagon of side ‘a’ is the sum of the areas of the five equilateral triangles with side a.

Answer:

According to the 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 Write True or False and justify your answer:
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.

Answer:

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

$S=\frac{51+37+20}{2}=\frac{108}{2}=54m$
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 Write True or False and justify your answer:
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.

Answer:

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

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

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$7\; per\; m^{2}$

Answer:

$Rs.10500$
To find the cost, firs,t 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?

Answer:

$Here \; sides\; of \; triangular\; walls\; are \; a = 14 m, b = 15 m \; and \; c = 13 m$

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 earnings 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.

Answer:

Here $\Delta$ABC is an equilateral triangle i.e.
AB = BC = CA = x (Let)

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

Answer:

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

$Now, a = 8\; cm, b = 12\; cm, c = 12 \; cm$
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 Find the area of a parallelogram given in the figure. Also, find the length of the altitude from vertex A on the side DC.

Answer:
$Length\; of\; the\; altitude \; from\; vertex\; A\; on\; the\; side\; DC = 15\; cm$

We know $A B=C D$ and $A D=B C \quad[\because A B C D$ is a parallelogram $]$

$Area \; of\; parallelogram = 2 \times area \; of\; \Delta DBC$
$So \; here\; \Delta DBC\; sides\ having\; DB = 25 cm, BC = 17cm\;and\;CD = 12 cm$
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, the 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.

Answer:

We know that ABCD is a 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: 8. Find the area of the triangular field.

Answer:

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.

Answer:

Here we have, AB = 6 cm, BC = 8 cm, CD = 12 cm and AD = 14 cm
According to the 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 a perimeter of 40 cm and one diagonal of 12 cm, is painted on both sides at the rate of Rs 5 per $cm^{2}$. Find the cost of painting.

Answer:

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 the area of the 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 on both sides of the sheet = 2 $\times$ 480 = Rs. 960.

Question 10 Find the area of the trapezium PQRS with height PQ given in Figure.

Answer:

$Firstly\; we\; have\; side \; PS = 12 m, SR = 13, QR = 7 m$

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 the trapezium
$Area \; of\; trapezium = \frac{1}{2} \times [sum\; of \; parallel\; sides] \times height$
$= \frac{1}{2}\times \left [ 12+7 \right ]\times 12$
$= 19\times 6\; cm^{2}$
$= 114\; cm^{2}$
$Hence, the\; area\; of\; trapezium\; is\; 114\; cm^{2}.$

Question 1 How much paper of each shade is needed to make a kite given in Figure, in which ABCD is a square with diagonal 44 cm.

Answer:

$Red = 242 \; cm^{2}$
$Yellow = 484\; cm^{2}$
$Green = 373.14 \; cm^{2}$
$Given,\; ABCD\; is\; a\; square.$
We know that all sides of a square are equal
AB = BC = CD = DA and
$\angle A = \angle B = \angle C = \angle D = 90^{\circ} [\because \; All\; angles\; of\; a\; square\; are\; 90^{\circ}]$
$In \Delta ABC,\; using \; Pythagoras \; theorem$
$(AC)^{2} = (AB)^{2}+ (BC)^{2}$
$(44)^{2} = (AB)^{2}+ (BC)^{2}\; \; \; \; \; \; \; \; [\because \; AB = BC\; equal\; sides]$
$44\times 44=2\left ( AB \right )^{2}$
$\frac{44\times 44}{2}=\left ( AB \right )^{2}$
$(AB)^{2} = 22 \times 44$
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.

Answer:

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

According to the question,
One side of a triangle is 4 cm longer than the smaller side
Let this side be AC = x + 4
Also, the 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.

Answer:

Let the smaller parallel side be CD = x cm

The 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 the 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 the other sides. Find the largest area where the house can be constructed.

Answer:

Let ABCD be the rectangular plot,
AB = 40 cm, AD = 15 cm

Given that a 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 the 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.

Answer:

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 In Figure, $\Delta$ABC has sides AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC, a parallelogram, DBCE of the same area as that of $\Delta$ABC is constructed. Find the height DF of the parallelogram.

Answer:

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 the 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 The dimensions of a rectangle ABCD are 51 cm × 25 cm. A trapezium PQCD with its parallel sides QC and PD in the ratio 9 8, is cut off from the rectangle as shown in the Figure. If the area of the trapezium PQCD is $\frac{5}{6}th$ part of the area of the rectangle, find the lengths QC and PD.

Answer:

$Given, BC = 51\; cm \; and \; CD = 25\; cm$
$Area \; of\; rectangle\; ABCD = (51 \times 25) cm^{2}$
$In\; trapezium\; PQCD\; , parallel\; sides\; QC\; and\; PD\; are\; in\; the\; ratio\, 9:8$
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