Rs Aggarwal 2020 2021 Solutions for Class 9 Maths Chapter 14 Areas Of Triangles And Quadrilaterals are provided here with simple step-by-step explanations. These solutions for Areas Of Triangles And Quadrilaterals are extremely popular among Class 9 students for Maths Areas Of Triangles And Quadrilaterals 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 9 Maths Chapter 14 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 9 Maths are prepared by experts and are 100% accurate.

We have:
Base = 24 cm
Height = 14.5 cm

Now,

#### Page No 533:

Let the height of the triangle be h m.
∴ Base = 3h m
Now,
Area of the triangle =
We have:

Thus, we have:
Height = h = 300 m
Base = 3h = 900 m

#### Page No 533:

We know that the longest side is 42 cm.
Thus, we can find out the height of the triangle corresponding to 42 cm.
We have:

#### Page No 533:

We know that the smallest side is 18 cm.
Thus, we can find out the altitude of the triangle corresponding to 18 cm.
We have:

#### Page No 533:

We know that the longest side is 105 m.
Thus, we can find out the height of the triangle corresponding to 42 cm.

#### Page No 533:

Let the sides of the triangle be 5x m, 12x m and 13x m.
We know:
Perimeter = Sum of all sides
or, 150 = 5x + 12x + 13x
or, 30x = 150
or, x = 5
Thus, we obtain the sides of the triangle.
5$×$5 = 25 m
12$×$5 = 60 m
13$×$5 = 65 m

Now,

#### Page No 533:

Let the sides of the triangle be 25x m, 17x m and 12x m.
We know:
Perimeter = Sum of all sides
or, 540 = 25x + 17x + 12x
or, 54x = 540
or, x = 10
Thus, we obtain the sides of the triangle.
25$×$10 = 250 m
17$×$10 = 170 m
12$×$10 = 120 m

Now,

Cost of ploughing 1 m2 field = Rs 5
Cost of ploughing 9000 m2 field = .

#### Page No 533:

(ii) We can find out the height of the triangle corresponding to 154 m in the following manner:
We have:

#### Page No 533: Let $△$PQR be an isosceles triangle and PX$\perp$QR.
Now,

Also,

∴ Perimeter = 80 + 41 + 41  = 162 cm

#### Page No 533:

The ratio of the equal side to its base is 3 : 2.
$⇒$Ratio of sides = 3 : 3 : 2.
Let the three sides of triangle be 3x, 3x, 2x.
The perimeter of isosceles triangle = 32 cm.

Therefore, the three side of triangle are 3x, 3x, 2x = 12 cm, 12 cm, 8 cm.
Let S be the semi-perimeter of the triangle. Then, $\mathrm{S}=\frac{1}{2}\left(12+12+8\right)=\frac{32}{2}=16$
Area of the triangle will be

Disclaimer: The answer does not match with the answer given in the book.

#### Page No 534:

Let ABC be any triangle with perimeter 50 cm.
Let the smallest side of the triangle be x.
Then the other sides be x + 4 and 2x − 6.

Now,
x + x + 4 + 2x − 6 = 50        (∵ perimeter is 50 cm)
⇒ 4x − 2 = 50
⇒ 4x = 50 + 2
⇒ 4x = 52
x = 13

∴ The sides of the triangle are of length 13 cm, 17 cm and 20 cm.
∴ Semi-perimeter of the triangle is

∴ By Heron's formula,

Hence, the area of the triangle is .

#### Page No 534:

The sides of the triangle are of length 13 m, 14 m and 15 m.
∴ Semi-perimeter of the triangle is

∴ By Heron's formula,

Now,
The rent of the wall with area 84 m2 per year = Rs 2000 × 84
= Rs 168000
The rent of the wall with area 84 m2 for 6 months = Rs $\frac{168000}{2}$
= Rs 84000

Hence, the rent paid by the company is Rs 84000.

#### Page No 534:

Let the equal sides of the isosceles triangle be a cm each.
∴ Base of the triangle, b = $\frac{3}{2}$a cm
(i) Perimeter = 42 cm
or, a + a + $\frac{3}{2}$a = 42
or, 2a +$\frac{3}{2}$a= 42

So, equal sides of the triangle are 12 cm each.
Also,
Base = $\frac{3}{2}$a =
(ii)

(iii)

#### Page No 534:

Thus, we have:
Perimeter = 3 × Side = 3 × 12 = 36 cm

Now, we have:

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Side of the equilateral triangle = 8 cm

(i)

(ii)

#### Page No 534:

Height of the equilateral triangle = 9 cm
Thus, we have:

Also,

#### Page No 534: Let $△$PQR be a right-angled triangle and PQ$\perp$QR.
Now,

#### Page No 534:

In right angled ∆ABD,
AB2 = AD2 + DB2          (Pythagoras Theorem)
AB2 = 122 + 162
AB2 = 144 + 256
AB2 = 400
AB = 20 cm

Area of ∆ADB = $\frac{1}{2}×DB×AD$
= $\frac{1}{2}×16×12$
= 96 cm2               ....(1)

In ∆ACB,
The sides of the triangle are of length 20 cm, 52 cm and 48 cm.
∴ Semi-perimeter of the triangle is

∴ By Heron's formula,

Now,
Area of the shaded region = Area of ∆ACB − Area of ∆ADB
=
480 − 96
= 384 cm2

Hence, the area of the shaded region in the given figure is 384 cm2.

#### Page No 534:

In the given figure, ABCD is a quadrilateral with sides of length 6 cm, 8 cm, 12 cm and 14 cm respectively and the angle between the first two sides is a right angle. Join AC.

In right angled ∆ABC,
AC2 = AB2 + BC2          (Pythagoras Theorem)
AC2 = 62 + 82
AC2 = 36 + 64
AC2 = 100
AC = 10 cm

Area of ∆ABC = $\frac{1}{2}×AB×BC$
= $\frac{1}{2}×6×8$
= 24 cm2               ....(1)

In ∆ACD,
The sides of the triangle are of length 10 cm, 12 cm and 14 cm.
∴ Semi-perimeter of the triangle is

∴ By Heron's formula,

Thus,
Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD
= (24 + 58.8) cm2
= 82.8 cm2

Hence, the area of quadrilateral ABCD is 82.8 cm2.

#### Page No 534:

We know that $△$ABD is a right-angled triangle.
∴

Now,
Area of quadrilateral ABCD  = Area of $△$ABD + Area of $△$BCD
= (60 + 54) cm2 =114 cm2
And,
Perimeter of quadrilateral ABCD = AB + BC + CD + AD = 17 + 8 + 12 + 9 = 46 cm

#### Page No 535:

In right angled ∆ABC,
BC2 = AB2 + AC2          (Pythagoras Theorem)
BC2 = 212 + 202
BC2 = 441 + 400
BC2 = 841
BC = 29 cm

Area of ∆ABC = $\frac{1}{2}×AB×AC$
= $\frac{1}{2}×21×20$
= 210 cm2               ....(1)

In ∆ACD,
The sides of the triangle are of length 20 cm, 34 cm and 42 cm.
∴ Semi-perimeter of the triangle is

∴ By Heron's formula,

Thus,
Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD
= (210 + 336) cm2
= 546 cm2

Also,
Perimeter of quadrilateral ABCD = (34 + 42 + 29 + 21) cm
= 126 cm

Hence, the perimeter and area of quadrilateral ABCD is 126 cm and 546 cm2, respectively.

#### Page No 535:

We know that $△$BAD is a right-angled triangle.

∴

Also, we know that $△$BDC is an equilateral triangle.

Now,
Area of quadrilateral ABCD = Area of $△$ABD + Area of $△$BDC
= (120 + 292.37) cm2 = 412.37 cm2
Perimeter of ABCD = AB + BC + CD + DA = 10 + 26+ 26 + 24 = 86 cm

#### Page No 535:

We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) =

#### Page No 535:

We know that a diagonal divides a parallelogram into two triangles of equal areas.
∴ Area of parallelogram ABCD = 2(Area of triangle ABC) =

#### Page No 535:

In the given figure, ABCD is a trapezium with parallel sides AB and CD. Let the length of CD be x.
Then, the length of AB be x + 4.

Area of trapezium =
$⇒475=\frac{1}{2}×\left(x+x+4\right)×19\phantom{\rule{0ex}{0ex}}⇒475×2=19\left(2x+4\right)\phantom{\rule{0ex}{0ex}}⇒950=38x+76\phantom{\rule{0ex}{0ex}}⇒38x=950-76\phantom{\rule{0ex}{0ex}}⇒38x=874\phantom{\rule{0ex}{0ex}}⇒x=\frac{874}{38}\phantom{\rule{0ex}{0ex}}⇒x=23\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

∴ The length of CD is 23 cm and the length of AB is 27 cm.

Hence, the lengths of two parallel sides is 23 cm and 27 cm.

#### Page No 535:

In ∆ABC,
The sides of the triangle are of length 7.5 cm, 6.5 cm and 7 cm.
∴ Semi-perimeter of the triangle is

∴ By Heron's formula,

Now,
Area of parallelogram DBCE = Area of ∆ABC
= 21 cm2

Also,
Area of parallelogram DBCE = base × height

Hence, the height DL of the parallelogram is 3 cm.

#### Page No 535:

In the given figure, ABCD is a trapezium having parallel sides 90 m and 30 m. Draw DE perpendicular to AB, such that DE = BC.

AD2 = AE2 + ED2          (Pythagoras Theorem)
⇒ 1002 = (90 − 30)2 + ED2
⇒ 10000 = 3600 + ED2
ED2 = 10000 − 3600
ED2 = 6400
ED = 80 m

Thus, the height of the trapezium = 80 m        ...(1)

Now,
Area of trapezium =
= $\frac{1}{2}×\left(90+30\right)×80$
= 4800 m2

The cost to plough per m2 = Rs 5
The cost to plough 4800 m2 = Rs 5 × 4800
= Rs 24000

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

#### Page No 536:

Let ABCD be 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, the length of the inner rectangle = 40 − 3 − 3 = 34 m and the breath of the inner rectangle = 15 − 2 − 2 = 11 m.

∴ Area of the inner rectangle PQRS = Length × Breath
= 34 × 11
= 374 m2

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

#### Page No 536:

Let the sides of rhombus be of length x cm. Perimeter of rhombus = 4x
⇒ 40 = 4x
x = 10 cm

Now,
In ∆ABC,
The sides of the triangle are of length 10 cm, 10 cm and 12 cm.
∴ Semi-perimeter of the triangle is

∴ By Heron's formula,

The sides of the triangle are of length 10 cm, 10 cm and 12 cm.
∴ Semi-perimeter of the triangle is

∴ By Heron's formula,

∴ Area of the rhombus = Area of ∆ABC + Area of ∆ADC
= 48 + 48
= 96 cm2

The cost to paint per cm2 = Rs 5
The cost to paint 96 cm2 = Rs 5 × 96
= Rs 480
The cost to paint both sides of the sheet = Rs 2 × 480
= Rs 960

Hence, the total cost of painting is Rs 960.

#### Page No 536:

Let the semi-perimeter of the triangle be s.
Let the sides of the triangle be a, b and c.
Given: sa = 8, sb = 7 and sc = 5      ....(1)

Adding all three equations, we get
3s − (a + b + c) = 8 + 7 + 5
⇒ 3s − (a + b + c) = 20
⇒ 3s − 2s = 20
s = 20 cm                  ...(2)

∴ By Heron's formula,

Hence, the area of the triangle is .

#### Page No 536:

Area of one triangular-shaped tile can be found in the following manner:

Now,
Area of 16 triangular-shaped tiles =
Cost of polishing tiles of area 1 cm2 = Rs 1
Cost of polishing tiles of area 1536 cm2 =

#### Page No 536:

We know that the triangle is an isosceles triangle.
Thus, we can find out the area of one triangular piece of cloth.

Now,
Area of 1 triangular piece of cloth = 490 cm2
Area of 12 triangular pieces of cloth =

#### Page No 536:

In the given figure, ABCD is a square with diagonal 44 cm.
AB = BC = CD = DA.        ....(1)

In right angled ∆ABC,
AC2 = AB2 + BC2          (Pythagoras Theorem)
⇒ 442 = 2AB2
⇒ 1936 = 2AB2
AB2 = $\frac{1936}{2}$
AB2 = 968
AB = $22\sqrt{2}$ cm      ...(2)

∴ Sides of square = AB = BC = CD = DA = $22\sqrt{2}$ cm

Area of square ABCD = (side)2
= ($22\sqrt{2}$)2
= 968 cm2          ...(3)

Area of red portion =
Area of yellow portion =
Area of green portion =

Now, in ∆AEF,
The sides of the triangle are of length 20 cm, 20 cm and 14 cm.
∴ Semi-perimeter of the triangle is

∴ By Heron's formula,

Total area of the green portion = 242 + 131.04 = 373.04 cm2

Hence, the paper required of each shade to make a kite is red paper 242 cm2, yellow paper 484 cm2 and green paper 373.04 cm2.

#### Page No 537:

Area of rectangle ABCD = Length × Breath
= 75 × 4
= 300 m2

Area of rectangle PQRS = Length × Breath
= 60 × 4
= 240 m2

Area of square EFGH = (side)2
= (4)2
= 16 m2

∴ Area of the footpath = Area of rectangle ABCD + Area of rectangle PQRS − Area of square EFGH
= 300 + 240 − 16
= 524 m2

The cost of gravelling the road per m2 = Rs 50
The cost of gravelling the roads 524 m2 = Rs 50 × 524
= Rs 26200

Hence, the total cost of gravelling the roads at Rs 50 per m2 is Rs 26200.

#### Page No 537:

The top and the bottom of the canal are parallel to each other.
Let the height of the trapezium be h.

Area of trapezium =
⇒ 640 = $\frac{1}{2}×\left(10+6\right)×h$
⇒ 640 = $8×h$
h = $\frac{640}{8}$
h = 80 m

Hence, the depth of the canal is 80 m.

#### Page No 537:

In the given figure, ABCD is the trapezium. Draw a line BE parallel to AD.

In ∆BCE,
The sides of the triangle are of length 15 m, 13 m and 14 m.
∴ Semi-perimeter of the triangle is

∴ By Heron's formula,

Also,
Area of ∆BCE = $\frac{1}{2}×\mathrm{Base}×\mathrm{Height}$

∴ Height of ∆BCE = Height of the parallelogram ABED = 12 m

Now,
Area of the parallelogram ABED = Base × Height
= 11 × 12
= 132 m2                     ...(2)

∴ Area of the trapezium = Area of the parallelogram ABED + Area of the triangle BCE
= 132 + 84
= 216 m2

Hence, the area of a trapezium is 216 m2.

#### Page No 537:

Let the length of the parallel sides be x and x − 8.
The height of the trapezium = 24 cm

Area of trapezium =
⇒ 312 = $\frac{1}{2}×\left(x+x-8\right)×24$
⇒ 312 = 12(2x − 8)
⇒ 2x − 8 = $\frac{312}{12}$
⇒ 2x − 8 = 26
⇒ 2x = 26 + 8
⇒ 2x = 34
x = 17 cm

Hence, the lengths of the parallel sides are 17 cm and 9 cm.

#### Page No 537:

Diagonals d1 and d2 of the rhombus measure 120 m and 44 m, respectively.

Base of the parallelogram = 66 m

Now,
Area of the rhombus = Area of the parallelogram

Hence, the measure of the altitude of the parallelogram is 40 m.

#### Page No 537:

It is given that,
Sides of the square = 40 m
Altitude of the parallelogram = 25 m

Now,
Area of the parallelogram = Area of the square

Hence, the length of the corresponding base of the parallelogram is 64 m.

#### Page No 537:

It is given that,
The sides of rhombus = 20 cm.
One of the diagonal = 24 cm. In ∆ABC,
The sides of the triangle are of length 20 cm, 20 cm and 24 cm.
∴ Semi-perimeter of the triangle is

∴ By Heron's formula,

In ∆ACD,
The sides of the triangle are of length 20 cm, 20 cm and 24 cm.
∴ Semi-perimeter of the triangle is

∴ By Heron's formula,

∴ Area of the rhombus = Area of ∆ABC + Area of ∆ACD
= 192 + 192
= 384 cm2

Hence, the area of a rhombus is 384 cm2.

#### Page No 537:

It is given that,
Area of rhombus = 480 cm2.
One of the diagonal = 48 cm.

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

Hence, the length of the other diagonal is 20 cm.

(ii) We know that the diagonals of the rhombus bisect each other at right angles. In right angled ∆ABO,
AB2 = AO2 + OB2          (Pythagoras Theorem)
AB2 = 242 + 102
AB2 = 576 + 100
AB2 = 676
AB = 26 cm

Hence, the length of each of the sides of the rhombus is 26 cm.

(iii) Perimeter of the rhombus = 4 × side
= 4 × 26
= 104 cm

Hence, the perimeter of the rhombus is 104 cm.

(b) 30 cm2

(a) 96 cm2

(b)

(b)

(c) 4 cm

#### Page No 540:

(b) 50 cm2
Here, the base and height of the triangle are 10 cm and 10 cm, respectively.
Thus, we have:

(b)

(a)

(c) 384 m2

#### Page No 541:

(b) 750 cm2

Let the sides of the triangle be 5x cm, 12x cm and 13x cm.
Perimeter = Sum of all sides
or, 150 = 5x + 12x + 13x
or, 30x = 150
or, x = 5
Thus, the sides of the triangle are 5$×$5 cm, 12$×$5 cm and 13$×$5 cm, i.e., 25 cm, 60 cm and 65 cm.

Now,

#### Page No 541:

(a) 24 cm

The smallest side is 18 cm.
Hence, the altitude of the triangle corresponding to 18 cm is given by:

#### Page No 541:

(b) 36 cm Let $△$PQR be an isosceles triangle and PX$\perp$QR.
Now,

∴ Perimeter = (10 + 10 + 16) cm = 36 cm

#### Page No 541:

(a) 36 cm

Now,
Perimeter
= 3 × Side = 3 × 12 = 36 cm

(c) 60 cm2

#### Page No 541:

(c) 336 cm2 Let $△$PQR be a right-angled triangle and PQ$\perp$QR.
Now,