Math Ncert Exemplar 2019 Solutions for Class 9 Maths Chapter 12 Heron S Formula are provided here with simple step-by-step explanations. These solutions for Heron S Formula are extremely popular among Class 9 students for Maths Heron S Formula Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Math Ncert Exemplar 2019 Book of Class 9 Maths Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Math Ncert Exemplar 2019 Solutions. All Math Ncert Exemplar 2019 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Question 1:

An isosceles right triangle has area 8 cm2. The length of its hypotenuse is

(A) $\sqrt{32}$ cm

(B) $\sqrt{16}$ cm

(C) $\sqrt{48}$ cm

(D) $\sqrt{24}$ cm

Let base and height be $x$.
Area = 8 cm2

Length of hypotenuse = $\sqrt{{4}^{2}+{4}^{2}}$
=

Hence, the correct answer is option A.

#### Question 2:

The perimeter of an equilateral triangle is 60 m. The area is

(A)

(B)

(C)

(D)

Let the length of side of an equilateral triangle be $x$.
Perimeter of equilateral triangle = $3×x$

Hence, the correct answer is option D.

#### 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

By heron's formula,
Area of triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$
Where s$\frac{a+b+c}{2}$
We have, a = 56 cm, b = 60 cm and c = 52 cm.
Now,
$s=\frac{56+60+52}{2}=84$

Hence, the correct answer is option C.

#### 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

Hence, the correct answer is option A.

#### Question 5:

The length of each side of an equilateral triangle having an area of $9\sqrt{3}$ cm2 is
(A) 8 cm
(B) 36 cm
(C) 4 cm
(D) 6 cm

Area of equilateral triangle = $\frac{\sqrt{3}}{4}×{\left(\mathrm{side}\right)}^{2}$

Hence, the correct answer is option D.

#### Question 6:

If the area of an equilateral triangle is $16\sqrt{3}$ cm2, then the perimeter of the triangle is
(A) 48 cm
(B) 24 cm
(C) 12 cm
(D) 36 cm

Area of equilateral triangle = $\frac{\sqrt{3}}{4}×{\left(\mathrm{side}\right)}^{2}$
$⇒16\sqrt{3}=\frac{\sqrt{3}}{4}×{\left(\mathrm{side}\right)}^{2}$
$⇒{\left(\mathrm{side}\right)}^{2}=64$

Hence, the correct answer is option B.

#### 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

Let ABC be a triangle with AB = 35 cm, BC = 54 cm, CA = 61 cm. Now, semi perimeter (s)

Also, Area of triangle = $\frac{1}{2}×\mathrm{base}×\mathrm{altitude}$
$⇒420\sqrt{5}=\frac{1}{2}×35×$Altitude
⇒ Altitude = $24\sqrt{5}\mathrm{cm}$

Hence, the correct answer is option C.

#### 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)

(B)

(C)

(D)

Area of the triangle by heron's formula = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$
Where, s$\frac{a+b+c}{2}$
We have, a = 4 cm, b = 4 cm and c = 2 cm.
Now,
s

Hence, the correct answer is option A.

#### 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 cm2 is
(A) Rs 2.00
(B) Rs 2.16
(C) Rs 2.48
(D) Rs 3.00

Area of triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$
Where, s$\frac{a+b+c}{2}$
Now,
s
Here,

Hence, the correct answer is option B.

#### Question 1:

Write True or False and justify your answer:
The area of a triangle with base 4 cm and height 6 cm is 24 cm2.

False,

Hence, the area of triangle will be 12 cm2.

#### Question 2:

Write True or False and justify your answer:
The area of ∆ABC is 8 cm2 in which AB = AC = 4 cm and ∠A = 90º.

True

We have, ABC is a right-angled triangle at A, where AB = AC = 4 cm. Here, area of $∆\mathrm{ABC}$ is 8 cm2.

#### Question 3:

Write True or False and justify your answer:
The area of the isosceles triangle is $\frac{5}{4}\sqrt{11}$ cm2, if the perimeter is 11 cm and the base is 5 cm.

True
Base = 5 cm, Perimeter = 11 cm.
Let other equal sides be $x$.

By Heron's formula
Area of triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$
Here, s$\frac{a+b+c}{2}$
Now,
s

Hence, area of triangle is $\frac{5\sqrt{11}}{4}$ cm2.

#### Question 4:

Write True or False and justify your answer:
The area of the equilateral triangle is $20\sqrt{3}$ cm2 whose each side is 8 cm.

False

Hence, the area of an equilateral triangle is $16\sqrt{3}{\mathrm{cm}}^{2}$.

#### 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 cm2.

Ture,
To find area of rhombus we divide it into two triangles. By Heron's formula,
Area of triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$, where, s$\frac{a+b+c}{2}$
Now,
s

Hence, the area of the rhombus is 96 cm2.

#### 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 cm2.

False,

Hence, the area of the parallelogram is 35 cm2.

#### 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.

False,

We know that a regular hexagon is divided into six equilateral triangles.
Area of a regular hexagon is the sum of the areas of six equilateral triangles.

#### 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 m2 is Rs 918.

True,

Let a = 51 m, b = 37 m, c = 20 m.

By Heron's formula
Area of triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$
Here, s$\frac{a+b+c}{2}$
Now, s$\frac{51+37+20}{2}=54$ m

Hence, cost of the Painting is ₹918.

#### 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.

True,

Let, a = 11 cm, b = 12 cm and c = 13 cm.
By Heron's formula
Area of triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$
Here, s$\frac{a+b+c}{2}$
Now, s

$\therefore$ Area of triangle = $\frac{1}{2}×$ base $×$ height
⇒ 61.5 = $\frac{1}{2}×12×$ height
⇒ Height = $\frac{61.5×2}{12}$
⇒ Height = 10.25 cm

Hence, length of altitude is 10.25 cm.

#### 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 m2.

Given: sides of triangle, a = 50 m, b = 65 m, c = 65 m
By Heron's formula
Area of triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$
Here, s$\frac{a+b+c}{2}$
Now,
s

Hence, cost of laying grass is Rs. 10500.

#### 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 m2 a year. A company hired one of its walls for 6 months. How much rent did it pay?

Given: sides of triangle a = 13 m, b = 14 cm, c = 15 m
From Heron's formula, we have
Area = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$
Where s$\frac{a+b+c}{2}$
Now,
s

Advertisements yield is ₹2000 for 1 year
For 6 months =$\frac{2000}{2}=₹1000$

Hence, the company has to pay ₹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.

Let the side of an equilateral triangle be a cm. Here OP, OQ and OR are perpendicular or side AB, BC and CA of equilateral triangle.

Area of equilateral triangle $△\mathrm{ABC}$ = Area of

$⇒\frac{\sqrt{3}}{4}{a}^{2}=15a$

Hence, the area of triangle is .

#### Question 4:

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

Given: Perimeter of an isosceles triangle = 32 cm
Ratio of equal side to base = 3 : 2
Let equal side be $3x$ and base be $2x$.

$⇒3x+3x+2x=32\phantom{\rule{0ex}{0ex}}⇒x=4$

$\therefore$ Equal sides = 12 cm, Base = 8 cm.
Let a = 12 cm, b = 12 cm, c = 8 cm.
From Heron's formula.
Area of triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$
Here s$\frac{a+b+c}{2}$
Now,
s

Hence, the area of the triangle is .

#### 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. In $△\mathrm{BCD}$, let a = 12 cm, b = 17 cm, c = 25 cm
From Heron's formula,
Area of triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$
Where s$\frac{a+b+c}{2}$
Now,
s

Let the altitude of parallelogram be h.
Also, area of parallelogram ABCD = Base $×$ Altitude
⇒ 180 = 12 $×$ h
h = 15 cm

Hence, area of parallelogram is 180 cm2, and altitude is 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. In $△\mathrm{ACB}$, let a = 60 m, b = 40 m, c = 80 m.
From Heron's formula,
Area of triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$
Here, s
Now,

Hence, the area of parallelogram is .

#### 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.

Given: Perimeter of triangle = 420 m
Ratio of sides length = 6 : 7 : 8.
Let the side length be

Hence, sides of triangle are 120 cm, 140 cm, 160 cm.
By Heron's formula
Area of triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$
Here, s$\frac{a+b+c}{2}$
Now, s

Hence, the area of the triangular field is .

#### 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.

Given: ABCD is a quadrilateral, AB = 6 cm, BC = 8 cm, CD = 12 cm, DA = 14 cm. Here, ABC is a right triangle at B.
By Pythagoras theorem,

Now,
Area of quadrilateral ABCD = Area of ($△\mathrm{ABC}+△\mathrm{ACD}$)

By Heron's formula
Area of $△\mathrm{ACD}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$
Here, a = 14 cm, b = 12 cm, c = 10 cm
s

Hence, the area of the quadrilateral is .

#### 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 m2. Find the cost of painting.

Let the side length of rhombus be $x$.
Perimeter of rhombus = 4 $×$ side
⇒ 40 = 4 By Heron's formula,
Area of triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$
Here, s$\frac{a+b+c}{2}$
Now,
s

Hence, the cost of painting is ₹960.

#### Question 10:

Find the area of the trapezium PQRS with height PQ given in the figure Drawing a line parallel to PQ from R. By Pythagoras theorem,

Now,
TR = PQ = 12 cm

Hence, the area of the trapezium PQRS is 114 cm2.

#### Question 1:

How much paper of each shade is needed to make a kite given in the figure in which ABCD is a square with diagonal 44 cm. Area of yellow sheet = (Area of region 1) + (Area of region 2)

By Heron's formula,
Area of $△\mathrm{CEF}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$
Here, s$\frac{a+b+c}{2}$
Now,
s

Therefore, the area of yellow sheet is 484 cm2, area of red sheet is 242 cm2 and area of green sheet is 373.15 cm2.

#### 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.

Let the smaller side be $x$ cm.
Then, larger side =  and third side =
Perimeter = 50 cm (given)

Therefore, sides of the triangle are 13 cm, 17 cm, 20 cm.
Let a = 13, b = 17 cm, c = 20 cm

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

Hence, the area of the triangle is .

#### Question 3:

The area of a trapezium is 475 cm2 and the height is 19 cm. Find the lengths of its two parallel sides if one side is 4 cm greater than the other.

Let one side of trapezium be .
Then, other side =
Area of trapezium = 475 cm2

Hence, length of two parallel sides are 23 cm and 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. Here, length of inner rectangle, EF =
Breadth of inner rectangle, FG =
$\therefore$ Area of inner rectangle EFGH = EF × FG = 34 $×$11 = 374 m2

Hence, largest area where house can be construct is 374 m2.

#### 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 1m2 of the field, find the total cost of ploughing the field. In $△\mathrm{BCE}$,
Using Pythagoras theorem,
$\begin{array}{l}{\left(\mathrm{BC}\right)}^{2}={\left(\mathrm{EC}\right)}^{2}+{\left(\mathrm{BE}\right)}^{2}\\ ⇒{\left(100\right)}^{2}={\left(60\right)}^{2}+{\left(\mathrm{BE}\right)}^{2}\\ ⇒10000=3600+{\left(\mathrm{BE}\right)}^{2}\\ {⇒\left(\mathrm{BE}\right)}^{2}=6400={80}^{2}\\ ⇒\mathrm{BE}=80 \mathrm{cm}\end{array}$

Now, cost of ploughing 1 m2 = ₹5

Hence, the total cost of ploughing the field is ₹24000.

#### Question 6:

In the figure ∆ABC has sides AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC a parallelogram DBCE of same area as that of ∆ABC is constructed. Find the height DF of the parallelogram. Let us first calculate are of $△\mathrm{ABC}$.
Here, AB = a =7.5 cm, BC = b =7 cm, CA = c = 6.5 cm

Now, Area of  parallelogram BCED = Base $×$ height = 7 $×$DF  .....(2)
Area of triangle = Area of parallelogram (given)

Hence, the height of the 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. Given: Parallel sides of trapezium are in ratio 9 : 8
i.e., QC : PD = 9 : 8
Let QC be 9x and PD be 8x.

From question,
Area of trapezium =
$⇒\frac{1}{2}\left(9x+8x\right)×25=\frac{5}{6}×\left(51×25\right)\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}×17x×25=\frac{5}{6}×51×25\phantom{\rule{0ex}{0ex}}⇒x=\frac{5×51×25×2}{6×17×25}\phantom{\rule{0ex}{0ex}}⇒x=5$

#### Question 8:

A design is made on a rectangular tile of dimensions 50 cm × 70 cm as shown in the figure. The design shows 8 triangles, each of sides 26 cm, 17 cm and 25 cm. Find the total area of the design and the remaining area of the tile. Given: Dimension of rectangular tile is 50 cm $×$70 cm
$\therefore$ Area of rectangular tile = 50 $×$70 = 3500 cm2
Let sides of triangle in a design be a = 25 cm, b = 17 cm, c = 26 cm.

By Heron's formula, we have
Area of triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$
Here, s$\frac{a+b+c}{2},$
Now, s$\frac{25+17+26}{2}=34$

$\therefore$ Total area of 8 triangles = 8 $×$204 = 1632 cm2
Now,
Area of design = Area of 8 triangles = 1632 cm2

Hence, the total area of the design is 1632 cm2 and the remaining area of the tile is 1868 cm2.

View NCERT Solutions for all chapters of Class 9