Rs Aggarwal 2020 2021 Solutions for Class 10 Maths Chapter 15 Perimeter And Area Of Plane Figures are provided here with simple step-by-step explanations. These solutions for Perimeter And Area Of Plane Figures are extremely popular among Class 10 students for Maths Perimeter And Area Of Plane Figures Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 10 Maths Chapter 15 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

#### Page No 683:

Given:
Area of a triangle =

#### Page No 683:

Let the sides of the triangle be ​a = 20 cm, b = 34 cm and c = 42 cm.
Let s be the semi-perimeter of the triangle.

Length of the longest side is 42 cm.

Area of a triangle =$\frac{1}{2}×b×h$

The height corresponding to the longest side is 16 cm.

#### Page No 683:

Let the sides of triangle be ​a = 18 cm, b = 24 cm and c = 30 cm.
Let s be the semi-perimeter of the triangle.

The smallest side is 18 cm long. This is the base.

Now, area of a triangle =$\frac{1}{2}×b×h$

The height corresponding to the smallest side is 24 cm.

#### Page No 683:

Let the sides of a triangle be 5x m ,12x m and 13x m.

Since, perimeter is the sum of all the sides,

The  lengths of the sides are:

#### Page No 683:

The perimeter of a right-angled triangle = 40 cm
Therefore , a+b+c= 40 cm
Hypotenuse = 17 cm
Therefore, c = 17 cm
a+b+c= 40 cm
a+b+17 = 40
a+b = 23
b = 23 $-$ a...........(i)
Now, using Pythagoras' theorem, we have:

Substituting the value of a=15, in equation(i) we get:
b = 23-a
= 23 - 15
= 8 cm

If we had chosen , then,

In any case,

#### Page No 683:

Given:
Area of the triangle =
Let the sides of the triangle be a, b and c, where a is the height, b is the base and c is hypotenuse of the triangle.
$a-b=7\mathrm{cm}$
a = 7 + b.......(1)

Side of a triangle cannot be negative.
Therefore, b = 8 cm.

Substituting the value of b = 8 cm, in equation (1):
a = 7+8 = 15 cm

Now,  a = 15 cm, b = 8 cm

Now, in the given right triangle, we have to find third side.

So, the  third side is 17 cm.

Perimeter of a triangle = $a+b+c$.
Therefore, required perimeter of the triangle .

#### Page No 683:

Given:
Area of triangle =
Let the sides be a and b, where a is the height and b is the base of triangle.

a = 2 + b.......(1)

Side of a triangle cannot be negative.

Therefore, b = 6 cm.

Substituting the value of b = 6 cm in equation(1), we get:
a = 2+6 = 8 cm

Now,  a = 8 cm, b = 6 cm

In the given right triangle we have to find third side. Using the relation

So, the third side is 10 cm.

So, perimeter of the triangle = a b + c
= 8+6+10
​=24 cm

#### Page No 683:

Let the side of the equilateral triangle be x cm.

#### Page No 683:

Area of equilateral triangle =

Area of equilateral triangle = $\left(\frac{\sqrt{3}}{4}×{a}^{2}\right)$, where a is the length of the side.

Perimeter of a triangle = 3a

#### Page No 683:

Area of the equilateral triangle =
Area of an equilateral triangle =$\left(\frac{\sqrt{3}}{4}×{a}^{2}\right)$, where a is the length of the side.

Height of triangle = $\frac{\sqrt{3}}{2}×a$

#### Page No 683:

Base = 48 cm
Hypotenuse =50 cm
First we will find the height of the triangle; let the height be 'p'.

Area of the triangle=$\frac{1}{2}×\mathrm{base}×\mathrm{height}$

#### Page No 683:

Hypotenuse = 65 cm
​Base = 60 cm
In a right-angled triangle,

Area of triangle = $\frac{1}{2}×\mathrm{base}×\mathrm{perpendicular}$

#### Page No 684:

​Height = 6 cm
Area=?
In a right-angled triangle, the centre of the circumcircle is the mid-point of the hypotenuse.

Now, base = 16 cm and height = 6 cm

Area of the triangle =$\frac{1}{2}×base×height\phantom{\rule{0ex}{0ex}}$

#### Page No 684:

In a right isosceles triangle, $\mathrm{base}=\mathrm{height}=a$

Therefore,

Further, given that area of isosceles right triangle = 200 cm2

In an isosceles right triangle, two sides are equal ('a') and the third side is the hypotenuse, i.e. 'c

Therefore, c = $\sqrt{{a}^{2}+{a}^{2}}$

Perimeter of the triangle = $a+a+c$
​                                       $=20+20+28.2$

= 68.2 cm

The length of the hypotenuse is 28.2 cm and the perimeter of the triangle is 68.2 cm.

#### Page No 684:

Given  :
Base = 80 cm
Area = 360 ${\mathrm{cm}}^{2}$
Area of an isosceles triangle = $\left(\frac{1}{4}b\sqrt{4{a}^{2}-{b}^{2}}\right)$

Squaring both the sides, we get:

Perimeter $=\left(2a+b\right)$

So, the perimeter of the triangle is 162 cm.

#### Page No 684:

Let the height of the triangle be h cm.
Each of the equal sides measures and b = 12 cm (base).

Now,
Area of the triangle = Area of the isosceles triangle
$\frac{1}{2}×\mathrm{base}×\mathrm{height}=\frac{1}{4}×b\sqrt{4{a}^{2}-{b}^{2}}$

Area of the triangle = $\frac{1}{2}×b×h$

#### Page No 684:

Let:
Length of each of the equal sides of the isosceles right-angled triangle = a = 10 cm
And,
Base = Height = a

The hypotenuse of an isosceles right-angled triangle can be obtained using Pythagoras' theorem.

If h denotes the hypotenuse, we have:

∴ Perimeter of ​the isosceles right-angled triangle = $2a+\sqrt{2}a$

#### Page No 684:

Given:
Side of equilateral triangle ABC = 10 cm
BD = 8 cm
Area of $\mathrm{equilateral}∆ABC=\frac{\sqrt{3}}{4}{\mathrm{a}}^{2}$ (where a = 10 cm)

Area of triangle $△$BCD = $\frac{1}{2}×b×h$

Area of the shaded region =
= 43.30 $-$ 24
= 19.3 cm2

#### Page No 691:

So, the length of the plot is 24 m and its area is 384 m2.

#### Page No 691:

Let the breadth of the rectangular park be.
∴ Length of the rectangular park$=l=2b$
Perimeter = 840 m

#### Page No 692:

One side of the rectangle = 12 cm
Diagonal of the rectangle = 37 cm

The diagonal of a rectangle forms the hypotenuse of a right-angled triangle. The other two sides of the triangle are the length and the breadth of the rectangle.

Now, using Pythagoras' theorem, we have:

Thus, we have:
Length  = 35 cm

#### Page No 692:

Area of the rectangular plot = 462 m2
Length (l) = 28 m

Perimeter of the plot = $2\left(l+b\right)$

#### Page No 692:

Let the length and breadth of the rectangular lawn be 5x m and 3x m, respectively.

Given:
Area of the rectangular lawn =

Thus, we have:

Cost of fencing 1 m lawn = Rs 65
∴ Cost of fencing 240 m lawn =

#### Page No 692:

Hence, the cost of covering the floor with carpet is ₹172.80.

Disclaimer: The answer given in the textbook is incorrect. The same has been rectified above.

#### Page No 692:

The length and breadth of floor of a rectangular hall is 24 m and 18 m respectively.
The area of rectangular floor is
The length and breadth of carpet is 2.5 m and 80 cm or 0.8 m respectively.
The area of carpet is

So, the number of such carpets required to cover the floor of rectangular hall is =

#### Page No 692:

Length of the stone = 6 dm = 0.6 m
Breadth of the stone = 5 dm = 0.5 m

Thus, 1800 stones will be required to pave the verandah.

#### Page No 692:

Area of the rectangle = 192 cm2
Perimeter of the rectangle = 56 cm

$\mathrm{Perimeter}=2\left(\mathrm{length}+\mathrm{breadth}\right)\phantom{\rule{0ex}{0ex}}⇒56=2\left(l+b\right)\phantom{\rule{0ex}{0ex}}⇒l+b=28\phantom{\rule{0ex}{0ex}}⇒l=28-b$

Thus, we have;

We will take length as 16 cm and breath as 12 cm because length is greater than breadth by convention.

#### Page No 692:

The field is planted with grass, with 2.5 m uncovered on its sides.

The field is shown in the given figure.

Thus, we have;
Length of the area planted with grass =

Width of the area planted with grass =

Area of the rectangular region planted with grass =

#### Page No 692:

The plot with the gravel path is shown in the figure.

Area of the rectangular plot = $l×b$
Area of the rectangular plot =
Length of the park including the path = 125 + 6 = 131 m
Breadth of the park including the path = 78 + 6 = 84 m
Area of the plot including the path

Area of the path = $11004-9750$
= 1254 m2
Cost of gravelling 1 m2 of the path = Rs 75
∴ Cost of gravelling 1254 m2 of the path = $1254×75$
= Rs 94050

#### Page No 692:

(i) Area of the rectangular field =

Let the width of the path be x m. The path is shown in the following diagram:

Length of the park excluding the path = (54 $-$ 2x) m
Breadth of the park excluding the path = (35 $-$ 2x ) m

Thus, we have:
Area of the path = 420 m2

The width of the path cannot be more than the breadth of the rectangular field.
∴ x = 2.5 m

​Thus, the path is 2.5 m wide.

(ii) Let the width of the border be x m.
The length and breadth of the carpet are 8 m and 5 m, respectively.
Area of the carpet =
Length of the carpet without border = $\left(8-2x\right)$
Breadth of carpet without border = $\left(5-2x\right)$
Area of the border = 12 m2
Area of the carpet without border = $\left(8-2x\right)\left(5-2x\right)$

Because the border cannot be wider than the entire carpet, the width of the carpet is $\frac{1}{2}$ m, i.e., 50 cm.

#### Page No 692:

Let the length and breadth of the garden be 9x m and 5x m, respectively,
Now,
Area of the garden = $\left(9x×5x\right)=45{x}^{2}$
Length of the garden excluding the path = (
Breadth of the garden excluding the path = $\left(5x-7\right)$
Area of the path = $45{x}^{2}-\left[\left(9x-7\right)\left(5x-7\right)\right]$

Thus, we have:
Length =

#### Page No 692:

Width of the room left uncovered = 0.25 m
Now,
Length of the room to be carpeted =
Breadth of the room be carpeted =

Area to be carpeted =

Breadth of the carpet = 80 cm = 0.8 m
We know:
Area of the room = Area of the carpet

Cost of 1 m carpet = Rs 80
Cost of 16.5 m carpet =

#### Page No 692:

It is given that the dimensions of rectangular park is 50 m × 40 m.
∴ Area of the rectangular park = 50 × 40 = 2000 m2
Area of the grass surrounding the pond = 1184 m2
Now,
Area of the rectangular pond
= Area of the rectangular park − Area of the grass surrounding the rectangular pond
= 2000 − 1184
= 816 m2
Let the uniform width of the surrounding grass be x.
∴ Length of the rectangular pond = (50 − 2x) m
Breadth of the rectangular pond = (40 − 2x) m
Now,
Area of rectangular pond = 816 m2
∴ (50 − 2x) × (40 − 2x) = 816
⇒ 2000 − 80− 100x + 4x2 = 816
⇒ 4x2 − 180x + 2000 − 816 = 0
⇒ 4x2 − 180+ 1184 = 0
⇒ x2 − 45+ 296 = 0
x2 − 37x − 8x + 296 = 0
⇒ x(− 37) − 8(− 37) = 0
⇒ (x − 8)(x − 37) =  0
⇒ x − 8 = 0 or x − 37 = 0
x = 8 or x = 37
For x = 37,
Length of rectangular pond = 50 − 2 × 37 = −24 m, which is not possible
So, x ≠ 37
Therefore, x = 8.
When x = 8,
Length of the rectangular pond = 50 − 2 × 8 = 50 − 16 = 34 m
Breadth of the rectangular pond = 40 − 2 × 8 = 40 − 16 = 24 m.

#### Page No 692:

The length and breadth of the lawn are 80 m and 64 m, respectively.
The layout of the roads is shown in the figure below:

Area of the road ABCD =
Area of the road PQRS =
Clearly, the area EFGH is common in both the roads.
Area EFGH =
Area of the roads = $400+320-25$
=
Given:
Cost of gravelling 1 m2 area = Rs 40
∴ Cost of gravelling 695 m2 area = $695×40$
= Rs 27,800

#### Page No 693:

The room has four walls to be painted.

Now,
Area of the two doors =

Area of the four windows =

The walls have to be painted; the doors and windows are not to be painted.

∴ Total area to be painted
Cost for painting 1 m2 = Rs 35
Cost for painting 300 m2 =

#### Page No 693:

So, the dimensions of the room are .

#### Page No 693:

Area of the square = $\frac{1}{2}×{\mathrm{Diagonal}}^{2}$

=$\frac{1}{2}×24×24$
=
Now, let the side of the square be x m.
Thus, we have:
$\mathrm{Area}={\mathrm{Side}}^{2}\phantom{\rule{0ex}{0ex}}⇒288={x}^{2}\phantom{\rule{0ex}{0ex}}⇒x=12\sqrt{2}\phantom{\rule{0ex}{0ex}}⇒x=16.92$

Perimeter = $4×\mathrm{Side}$

Thus, the perimeter of the square plot is 67.68 m.

#### Page No 693:

Area of the square = 128 cm2

Now,
Area = ${\mathrm{Side}}^{2}$

Perimeter = 4(Side)

Thus, the perimeter of the square is 45.24 cm.

#### Page No 693:

We know that 1 hectare = 10000 m2.
So, Area of square field = 8 hectares = 80000 m2.

Now, distance to be travelled = 400 m
speed is given to be 4 km/h =
Therefore,

#### Page No 693:

Cost of fencing the lawn = Rs 28000
Let l be the length of each side of the lawn. Then, the perimeter is 4l.
We know:
$\mathrm{Cost}=\mathrm{Rate}×\mathrm{Perimeter}\phantom{\rule{0ex}{0ex}}⇒28000=14×4\mathrm{l}$

Area of the square lawn

Cost of mowing 100 m2 of the lawn = Rs 54
Cost of mowing 1 m2 of the lawn

∴ Cost of mowing 250000 m2 of the lawn

#### Page No 693:

So, the area of the quadrilateral ABCD is 252 cm2.

#### Page No 693:

$∆$BDC is an equilateral triangle with side a= 26 cm.
Area of

By using Pythagoras' theorem in the right-angled triangle $∆$DAB, we get:

Area of $∆ABD=\frac{1}{2}×b×h$

Area of the quadrilateral = Area of $∆$BCD + Area of $∆$ABD

= 412.37 cm2

Perimeter of the quadrilateral = AB + BC + CD + AD
= 24 + 10 + 26 + 26
= 86 cm

#### Page No 693:

In the right-angled $∆$ACB:

Perimeter = $AB+BC+CD+AD$
= $17+8+12+9$
= 46 cm

Area of $∆ABC=\frac{1}{2}\left(b×h\right)\phantom{\rule{0ex}{0ex}}$

=$\frac{1}{2}\left(8×15\right)$
=

Area of $∆ADC=\frac{1}{2}×b×h$

∴ Area of the quadrilateral = Area of $∆$ABC + Area of $∆$ADC
= 60 + 54
= 114 cm2

#### Page No 693:

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

Area of $∆$ABD = $\sqrt{48\left(48-42\right)\left(48-20\right)\left(48-34\right)}$

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

Area of $∆$BDC = $\sqrt{35\left(35-29\right)\left(35-20\right)\left(35-21\right)}$

∴ Area of quadrilateral ABCD = Area of $∆$ABD + Area of $∆$BDC
​                                                  = 336 + 210
= 546 cm2

#### Page No 694:

Given:
Base = 25 cm
Height = 16.8 cm
∴ Area of the parallelogram

#### Page No 694:

Longer side = 32 cm
Shorter side = 24 cm
Let the distance between the shorter sides be x cm.
Area of a parallelogram =
=
or, $32×17.4=24×x$

or,

∴ Distance between the shorter sides = 23.2 cm

#### Page No 694:

Area of the parallelogram = 392 m2
Let the base of the parallelogram be b m.
Given:
Height of the parallelogram is twice the base.
∴ Height = 2b m
​Area of a parallelogram =

∴ Base = 14 m
Altitude =

#### Page No 694:

Parallelogram ABCD is made up of congruent $∆$ABC and $∆$ADC.

Area of triangle ABC$\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$   (Here, s is the semiperimeter.)

Thus, we have:

Now,
Area of the parallelogram =

#### Page No 694:

Area of the rhombus = $\frac{1}{2}×{d}_{1×}{d}_{2}$, where d1 and d2 are the lengths of the diagonals.

Side of the rhombus = $\frac{1}{2}\sqrt{{{d}_{1}}^{2}+{{d}_{2}}^{2}}$

Perimeter of the rhombus = $4a$
= $4×17$
= 68 cm

#### Page No 694:

Perimeter of a rhombus = 4a    (Here, a is the side of the rhombus.)

(i) Given:
One of the diagonals is 18 cm long.

Thus, we have:

∴ Length of the other diagonal = 24 cm

(ii) Area of the rhombus$=\frac{1}{2}{d}_{1}×{d}_{2}$

#### Page No 694:

i) Area of a rhombus$=\frac{1}{2}×{d}_{1}×{d}_{2}$, where d1 and d2 are the lengths of the diagonals.

∴ Length of the other diagonal = 20 cm

(ii) Side = $\frac{1}{2}\sqrt{{{d}_{1}}^{2}+{{d}_{2}}^{2}}$

∴ Length of the side of the rhombus = 26 cm

(iii) Perimeter of the rhombus = $4×\mathrm{Side}$

#### Page No 694:

So, the area of the trapezium is 84 cm2.

#### Page No 694:

Area of the canal =
Area of trapezium =

Therefore, the depth of the canal is 80 m.

#### Page No 694:

Draw $DE\parallel BC$ and DL perpendicular to AB.
The opposite sides of quadrilateral DEBC are parallel. Hence, DEBC is a parallelogram.
DE = BC = 13 m
Also,

For $∆DAE$:
Let:
AE = a =14 m
DE = b = 13 m
DA = c =15 m

Thus, we have:

Area of

Area of $∆DAE=\frac{1}{2}×AE×DL$

Area of trapezium =

#### Page No 697:

(d) 30 m

Let the length of the rectangle be x m.

Length cannot be negative.
∴ Length = = 30 m

#### Page No 697:

(b) 2520 m2

Let the breadth of the field be x m.
∴ Length = (x + 23) m

Now,
Perimeter
Thus, we have:
$4x+46=206\phantom{\rule{0ex}{0ex}}⇒4x=206-46=160\phantom{\rule{0ex}{0ex}}⇒x=\frac{160}{4}=40$
∴ Breadth = x = 40 m
Length = + 23
Area

#### Page No 697:

(a) 108 m2

Length of the rectangular field = 12 m
Diagonal = 15 m

${\mathrm{Diagonal}}^{2}={\mathrm{Length}}^{2}+{\mathrm{Breadth}}^{2}$

∴ Area of the field

#### Page No 697:

Hence, the correct answer is option (c).

#### Page No 697:

(d) 64 cm

Let the breadth of the rectangle be x cm.
∴ Length of the rectangle = 3x cm
We know:
$\mathrm{Diagonal}=\sqrt{\left(\mathrm{Length}{\right)}^{2}+\left(\mathrm{Breadth}{\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒8\sqrt{10}=\sqrt{{x}^{2}+\left(3x{\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒8\sqrt{10}=\sqrt{{x}^{2}+9{x}^{2}}\phantom{\rule{0ex}{0ex}}⇒8\sqrt{10}=x\sqrt{10}\phantom{\rule{0ex}{0ex}}⇒x=8$

Now,
Breadth of the rectangle = x = 8 cm
Length of the rectangle = 3x = 24 cm
Perimeter of the rectangle

#### Page No 697:

(d) 4% decrease

Let:
Length$=x$
breadth$=y$
Area$=xy$

Now,
New length$=x+20%x=x+\frac{1}{5}x=\frac{6}{5}x$

New breadth$=y-20%y=y-\frac{1}{5}y=\frac{4}{5}y$

New area$=\frac{6}{5}x×\frac{4}{5}y=\frac{24}{25}xy$

Difference in the areas$=xy-\frac{24}{25}xy=\frac{1}{25}xy$

Difference in percentage$=\left[\left(\frac{\frac{1}{25}xy}{xy}\right)×100\right]%=4%$

#### Page No 697:

(a) 264 m2

Length of the ground including the path
Breadth of the ground including the path

Total area (including the path)

Area of the field

Area of the path

#### Page No 697:

(b) 100 cm2

A diagonal of a square forms the hypotenuse of a right-angled triangle with base and height equal to side a.

∴ Area of the square

#### Page No 697:

(d) 110 m

Let the diagonal of the square field be d m.

In case of a square field, ${d}^{2}=2{a}^{2}$, where a is the side of the square field.
Now,

∴ $d=\sqrt{2×6050}=\sqrt{12100}=110$

#### Page No 697:

(c) 100 m

Disclaimer :- The length cannot be in hectare So we used is as area of the square.
Area of the square field

The diagonal divides the square into two isosceles right-angled triangles.

Using Pythagoras' theorem, we have:

${\mathrm{Diagonal}}^{2}={a}^{2}+{a}^{2}=2{a}^{2}$
Area of a square = ${a}^{2}$

∴ Diagonal​

#### Page No 697:

(b) 12 cm

Area of an equilateral triangle $=\frac{\sqrt{3}}{4}{a}^{2}$ (where a is the length of the side)
Thus, we have:

Perimeter of the equilateral triangle = 3a

#### Page No 697:

(c) $16\sqrt{3}{\mathrm{cm}}^{2}$

Let the side of the equilateral triangle be a.
Given:
a = 8 cm
Now,
Area of the equilateral triangle

#### Page No 698:

Hence, the correct option is (b).

#### Page No 698:

Hence, the correct answer is option (c).

#### Page No 698:

Hence, the correct answer is option (b).

#### Page No 698:

Hence, the correct answer is option (c).

#### Page No 698:

(c) $4:\sqrt{3}$

Let:
Length of the side of the square = Length of the side of the equilateral triangle = a unit

Now,

Area of the square

Area of the equilateral triangle

#### Page No 698:

(b)

Area of a circle$=\mathrm{\pi }{r}^{2}$

The radius of the circle is equal to the side of the equilateral triangle.

r = a (Here, a is the side of the equilateral triangle.)

∴ Area of the equilateral triangle

#### Page No 698:

Hence, the correct answer is option (c).

#### Page No 698:

Hence, the correct answer is option (d).

#### Page No 701:

(b) 114 sq cm

Using Pythagoras' theorem in $∆$ABC, we get:

#### Page No 701:

(a) 306 m2

In the given figure, AECD is a rectangle.

Length AE = Length CD = 28 m

Now,

Also,
AD = CE = 9 m

In the given figure, if DA is perpendicular to AE, then it can be solved, otherwise it cannot be solved.

#### Page No 701:

(c) 12.5 cm

Let the sides of the triangle be 12x cm, 14x cm and 25x cm.
Thus, we have:

∴ Greatest side of the triangle

(c) 52 cm2

#### Page No 701:

Given:
Side of the equilateral triangle = 10 cm
Thus, we have:

#### Page No 701:

Area of an isosceles triangle:

#### Page No 701:

Let the rectangle ABCD represent the hall.

Using the Pythagorean theorem in the right-angled triangle ABC, we have:
${\mathrm{Diagonal}}^{2}={\mathrm{Length}}^{2}+{\mathrm{Breadth}}^{2}$

∴ Area of the hall

#### Page No 701:

The diagonal of a square forms the hypotenuse of an isosceles right triangle. The other two sides are the sides of the square of length a cm.

Using Pythagoras' theorem, we have:

$⇒$a$=\frac{24}{\sqrt{2}}$
Area of the square

#### Page No 701:

To find the area of the triangle, we will first find the semiperimeter of the triangle.

Thus, we have:

Now,

#### Page No 701:

Let the length and breadth of the lawn be , respectively.

Now,

Area of the lawn$=5x×3x=5{x}^{\mathit{2}}$

∴ Perimeter of the lawn
Total cost of fencing the lawn at Rs 20 per metre

#### Page No 701:

Given:
Sides are 20 cm each and one diagonal is of 24 cm.
The diagonal divides the rhombus into two congruent triangles, as shown in the figure below.

We will now use Hero's formula to find the area of triangle ABC.
First, we will find the semiperimeter.

Now,
Area of the rhombus

#### Page No 702:

We will divide the trapezium into a triangle and a parallelogram.

Difference in the lengths of parallel sides
We can represent this in the following figure:

Trapezium ABCD is divided into parallelogram AECD and triangle CEB.

1. Consider triangle CEB.
In triangle CEB, we have:

Using Hero's theorem, we will first evaluate the semiperimeter of triangle CEB and then evaluate its area.

Semiperimeter,

Also,

Area of triangle CEB$=\frac{1}{2}\left(\mathrm{Base}×\mathrm{Height}\right)$

Height of triangle CEB
1. Consider parallelogram AECD.
​​Area of parallelogram AECD

Area of trapezium ABCD

#### Page No 702:

The diagonal of a parallelogram divides it into two congruent triangles. Also, the area of the parallelogram is the sum of the areas of the triangles.

We will now use Hero's formula to calculate the area of triangle ABC.

Area of the parallelogram

#### Page No 702:

Given:
Cost of fencing = Rs 2800
Rate of fencing = Rs 14

Now,
Perimeter

Because the lawn is square, its perimeter .

Area of the lawn =

Cost for mowing the lawn per 100 m2= Rs 54

Cost for mowing the lawn per 1 m2

Total cost for mowing the lawn  per 2500 m2

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Quadrilateral ABCD is divided into triangles $∆$ABD and $∆$BCD.
We will now use Hero's formula.
For $∆$ABD:

For $∆$BCD:

Thus, we have:

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Area of the rhombus

Area of the parallelogram$=\mathrm{Base}×\mathrm{Height}=66×\mathrm{Height}$

Given:
The area of the rhombus is equal to the area of the parallelogram.

∴ Corresponding height of the parallelogram = 40 m

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Diagonals of a rhombus perpendicularly bisect each other. The statement can help us find a side of the rhombus. Consider the following figure.

ABCD is the rhombus and AC and BD are the diagonals. The diagonals intersect at point O.

We know:

Similarly,

Using Pythagoras' theorem in the right-angled triangle $∆$DOC, we get:

DC is a side of the rhombus.
We know that in a rhombus, all sides are equal.
∴ Perimeter of ABCD

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∴ Distance between the longer sides = 9 cm

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The field, which is represented as ABCD, is given below.

The area of the field is the sum of the areas of triangles ABC and ADC.

Area of the triangle ABC