Perimeter and Area of Plane Figures Class 9 RS Aggarwal Exe-17A Goyal Brothers ICSE Maths Solutions

Perimeter and Area of Plane Figures Class 9 RS Aggarwal Exe-17A Goyal Brothers ICSE Maths Solutions Ch-17. In this article you will learn how to solve problems on Perimeter and Area of Triangle. Visit official Website CISCE for detail information about ICSE Board Class-9.

Perimeter and Area of Plane Figures Class 9 RS Aggarwal Exe-17A Goyal Brothers ICSE Maths Solutions

Perimeter and Area of Plane Figures Class 9 RS Aggarwal Exe-17A Goyal Brothers ICSE Maths Solutions Ch-17

Board ICSE
Subject Maths
Class 9th
Chapter-17 Perimeter and Area of Plane Figures
Writer RS Aggrawal
Topics Area and Perimeter of Triangle
Academic Session 2024-2025

How to Find Area of Any Triangle

For any triangle with base (b) and height (h):
Area = (b × h) / 2

Heron’s Formula
For any triangle with sides a, b, and c:
1. Calculate semi-perimeter (s) = (a + b + c) / 2
2. Area = √[s(s-a)(s-b)(s-c)]

Right Triangle
For a right triangle with legs a and b, and hypotenuse c:
Area = (a × b) / 2

Equilateral Triangle
For an equilateral triangle with side a:
Area = (√3 / 4) × a^2

Isosceles Triangle
For an isosceles triangle with base b and equal sides a:
1. Calculate height (h) = √(a^2 – (b/2)^2)
2. Area = (b × h) / 2

Direct Formula to Calculate Area of Any Triangle

  • Area of a Triangle = A = ½ (b × h) square units
  • Area of an Equilateral Triangle = A = (√3)/4 × side2
  • Area of Isosceles Triangle = b/4  √ (4a2– b2) where a is equal side

Exercise- 17A  (Area and Perimeter of Triangle)

Class 9 RS Aggarwal Goyal Brothers ICSE Maths Solutions Ch-17

Que-1: Find the area of triangle whose base is 15cm and the corresponding height is 9.6 cm.

Sol:  Base = 15 cm
height = 9.6 cm
Area of a triangle = (1/2) × base × height
= (1/2) × 15 × 9.6 = 72 cm²

Que-2: Find the area of the triangle whose sides are 13cm, 14cm and 15cm. Also, find the height of the triangle, corresponding to the longest side.

Sol:  Let, the sides of the triangle be a, b, c.
a = 13 cm
b = 14 cm
c = 15 cm
s = (a+ b + c)/2 = (13 + 14 + 15)/2 = 42/2 = 21
as we know sides of the triangle , we use heron’s formula.
area of triangle = √{s(s-a)(s-b)(s-c)}
= √{21(21-13)(21-14)(21-15)}
= √{21.8.7.6}
= √7056 = 84 cm²
area of triangle = 84 cm sq.
largest side = 15 cm ( given in the question )
Area = (1/2) × b × h
84 = (1/2) × 15 × h
h = (84×2)/15
h = 11.2 cm.

Que-3: Find the area of the triangle whose sides are 30cm, 24cm and 18cm. Also, find the length of the altitude corresponding to the smallest side of the triangle.

Sol:  Since the sides of the triangle are 18 cm, 24 cm and 30 cm respectively.
s = (18+24+30)/2
= 36
Hence the area of the triangle is
A  = √{s(s-a)(s-b)(s-c)}
= √{36(36-18)(36-24)(36-30)}
= √{36×18×12×6}
= √46656
= 216 sq.cm.
Again
A = (1/2) base × altitude
Hence
216 = (1/2)×18×h
h = 24 cm

Que-4: The lengths of the sides of a triangle are in the ratio 3:4:5 and its perimeter is 144cm. Find the area of a triangle.

Sol:   Let the sides of triangles are 3x,4x and 5x.
And perimeter of the triangle is 144cm.
So , 3x+4x+5x = 144
12x = 144
x = 12
so, the sides are triangle are 3(12) = 36cm,  4(12) = 48cm ,  5(12) = 60cm
so , the longest side is 60 cm.
Area of triangle = √{s(s−a)(s−b)(s−c)}
here , s = 72 and a = 36, b = 48 c = 60
= √{72(72−36)(72−48)(72−60)}
= √{72(36)(24)(12)}
= √746496
= 864 cm²

Que-5: The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the triangle. Also, find the cost of cultivating the field at Rs24.60 per 100 m².

Sol:  The sides of a triangle are in the ratio 25 : 17 : 12
Let the sides of a triangle are a = 25x, b = 17 x and c = 12x say.
Perimeter = 25 = a + b + c = 540 cm
⇒ 25x + 17x + 12x = 540 cm
⇒ 54x = 540cm
⇒ x = 540/54
⇒ x = 10 𝑐𝑚
∴ The sides of a triangle are a = 250 cm, b = 170 cm and c = 120 cm
Now, Semi perimeter s = (a+b+c)/2
= 540/2 = 270cm
∴The aera of the triangle = √{s(s-a)(s-b)(s-c)}
= √{270(270-250)(270-170)(270-120)}
= √{27(20)(100)(150)}
= √{(9000)(9000)}
= 9000 cm²
The area of triangle = 9000 cm²
100 cm² cost = 24.60
1 cm² = 24.60/100
9000 cm² = (24.60/100) × 9000
= RS 2214.

Que-6: The base of a triangular field is twice its altitude. If the cost of cultivating the field at Rs14.50 per 100 m² is Rs52200, find its base and altitude.

Sol:  Altitude = x
base = 2x
Area = (1/2) b × h
= (1/2) × 2x × x = x²
we know total cost is 52200
the rate of cultivating is 14.50 per 100 m
Area = (52200×100)/(14.50)
Area = 360000
x² = 360000
x = 600 = base
2x = 1200 = altitude

 Que-7: The perimeter of a right triangle is 60 cm and its hypotenuse is 25 cm. Find the area of the triangle.

Sol:  Let ABC be a right-angled triangle.
Since the perimeter of the right triangle is 60 cm,
AB + BC +CA = 60 cm
⇒ AB + BC + 25 = 60
⇒ AB + BC = 35 cm          …..(1)
In ∆ABC,
AB2 + BC2 = CA2
⇒ (AB + BC)2 − 2(AB)(BC) = (25)2
⇒ (35)2 − 2(AB)(BC) = (25)2                                     [From (1)]
⇒ (35 − 25)(35 + 25) = 2(AB)(BC)’
⇒ (AB)(BC) = 300
Now,
Area of ∆ABC= (1/2)×AB×BC
= (1/2)×300 = 150 cm²
Hence, the area of the triangle is 150 cm².

Que-8: Find the length of hypotenuse of an isosceles right angled triangle, having an area of 200 cm².

Sol: Let the equal sides of isosceles triangle be x cm
AB = BC = x
Area of triangle = 1/2 × BC × AB
⇒ 200 = 1/2 × x × x
⇒ x2 = 400
⇒ x = 20 cm
AB = BC = 20 cm
By using Pythagoras theorem
AB2 + BC2 = AC2
⇒ 202 + 20= AC2
⇒ AC = 20√2 = 28.28 cm

Que-9: Calculate the area and the height of an equilateral triangle whose perimeter is 60 cm.

Sol:  Each side of the triangle is 60/3 =  20 cm
Hence the area of the equilateral triangle is given by
A = 34 x 202
   = 100√3
= 173.2 sq.cm
The height h of the triangle is given by
12 x 20 x h = 173.2
h = 17.32 cm.

Que-10: Find the perimeter and the area of an equilateral triangle whose height is 12 cm. Write your answers, correct to two decimal places.

Sol:  Height of the equilateral triangle = 12 cm
Height of triangle = (√3/2) × side of an equilateral triangle
Area of an equilateral triangle = (√3/4) × side2
Side of the triangle = (2/√3) × 12 = 8√3
Perimeter = 8√3 + 8√3 + 8√3
= 24√3 = 41.56 cm
Area of the triangle = (√3/4) × (8√3)2
⇒ (√3/4) × 64 × 3
= √3 × 16 × 3
= 48√3 = 83.14 cm2

Que-11: The length of two sides of a right triangle containing the right angle differ by 2cm. If the area of the triangle is 24 cm², find the perimeter of the triangle.

Sol:  Given: Area of triangle 24 cm²
Let the sides be a and b, where a is the height and b is the base of triangle
a-b = 2cm
a = 2+b               ………….(1)
Area of triangle = (1/2)×b×h
⇒ 24 = (1/2)×b×(2+b)
⇒ 48 = b+(1/2)b²
⇒ 48 = 2b+b²
⇒ b²+2b-48 = 0
⇒ (b+8)(b-6) = 0
⇒ b = -8 or 6
Side of a triangle cannot e negative.
Therefore, b = cm
Substituting the value of b=6 cm in equation (1), we get :
a = 2+6 = 8cm
Now, a = 8cm,  b = 6cm
In the given right triangle we have to find third side. Using the relation
(Hyp)² = (Oneside)² + (otherside)²
⇒ Hyp² = 8² + 6²
⇒ Hyp² = 64 + 36
⇒ Hyp² = 100
⇒ Hyp = 10cm
So, the third side is 10 cm So,
perimeter of the triangle a+b+c
= 8+6+10
= 24 cm.

Que-12: The sides of a right angled triangle containing the right angle are (5x) cm and (3x-1) cm. If its area is 60 cm², find its perimeter.

Sol:  Area of the right triangle ABC
= {5x(3x-1)}/2
{5x(3x-1)}/2 = 60
⇒ 15x2 – 5x = 120
⇒ 3x2 – x = 24
⇒ 3x2 – x – 24 = 0
⇒ 3x2 – 9x + 8x – 24 = 0
⇒ 3x(x – 3) + 8(x – 3) = 0
⇒ (x – 3) (3x + 8) = 0
⇒ x – 3 = 0 or 3x + 8 = 0
⇒ x = 3 or x = -8/3
Hence, sides are AB = 3x – 1 = 8 cm
BC = 5x = 15 cm
Also from AC
= √{(AB)²+(BC)²}
= √(64+225)
= √289
= 17 cm.
Perimeter = sum of all sides
= 8 + 17 + 15
= 40 cm.

Que-13: Each of the equal sides of an isosceles triangle is 2cm more than its height and the base of the triangle is 12cm. Find the area of the triangle.

Sol:   Let the height of the triangle be h cm.
Each of the equal sides measures  a = (h+2)cm  and  b = 12cm(base)
Now, Area of the triangle = Area of the isosceles triangle
= (1/2) × base × height = (1/4)×b√(4a²-b²)
⇒ (1/2)×12×h = (1/4)×12×√{4(h+2)²-144}
⇒ 6h = 3√{4h²+16h-144}
⇒ 2h = √{4h²+16h+16-144}
On squaring both the sides, we get:
⇒ 4h² = 4h²+16h+16-144
⇒16h-128=0
⇒ h = 8
Area of the triangle = (1/2)×b×h
= (1/2)×12×8
= 48 cm²

Que-14: Find the area of an isosceles triangle, each of whose equal side is 13 cm and base 24 cm.

Sol: Let the equal sides of the isosceles triangles be a cm and the base be of b cm.
Then according to the problem, a = 13  and  b = 24.
Now the area of the isosceles triangle will be
= (1/2)×b×√{a²−(b²/2)}
= (1/2)×24×√{13²−12²}
= 12×5 = 60.
So the area of the isosceles triangles be 60 cm².

Que-15: The base of an isosceles triangle is 18cm and its area is 108 cm². Find its perimeter.

Sol: Base (b) = 18 cm
Area (A) = 108 cm2
A = bh/2
P = 2a+b
h = √{a²-(b²/4)}
Solving for P
P = b+√{b²+16(A/b)²}
P = 18+√{18²+16(108/18)²}
P = 48 cm.

Que-16: In the given figure, ΔABC is an equilateral triangle having each side equal to 10cm and ΔPBC is right angled at P in which PB = 8cm. Find the area of the shaded region.

Sol:  Given:
Side if equilateral triangle  ABC = 10 cm
8 = BD cm
Area of equilateral ΔABC = (√3/4)a²  (where a=10 cm  )
Area of equilateral ΔABC = (√3/4)×10²
= 25√3
= 25×1.732
= 43.30 cm²
In the right , ΔBDC we have:
BC² = BD²+CD²
⇒ 10² = 8²+CD²
⇒ CD² = 10²-8²
⇒ CD² = 36
⇒ CD = 6
Area of triangle ΔBCD = (1/2)×b×h
= (1/2)×8×6
= 24 cm²
Area of the shaded region = Area of ΔABC – Area of ΔBDCArea of ΔABC –  Area of ΔBDC
= 43.30 – 24
= 19.3 cm²

Que-17: If the area of an equilateral triangle is 81√3 cm², find its perimeter.

Sol:  We know that area is √3/4 a2
area of the equilateral triangle = 81√3
√3/4 a2 = 81√3
1/4 a2 = 81
a2 = 81(4)
a2 = 324
a = 18cm
Then the perimeter of the equilateral triangle
= 3×a
= 3×18
= 54cm.

Que-18: The base of a right-angled triangle is 24cm and its hypotenuse is 25cm. Find the area of a triangle.

Sol:  Base = 24 cm and Hypotenuse = 25 cm
Using Pythagoras theorem
(Hypotenuse)2 = (Base)2 + (Perpendicular)2
252 = 242 + (Perpendicular)2
Perpendicular = √(625 – 576) = √49 = 7 cm
Area of the triangle = (1/2) × Base × height = (1/2) × 24 × 7 = 84 cm2

Que-19: The altitude drawn to the base of an isosceles triangle is 8cm and the perimeter is 32 cm. Find the area of the triangle.

Sol:  Let the equal sides of the isosceles triangle be x
So, base = 32 – 2x
As we in an isosceles triangle the altitude drawn to the base bisects the base
So, according to that,
BD = DC = BC/2 = 16 – x
Now,
82 + (16 – x)2 = x2
⇒ 64 + 256 + x2 – 32x = x2
⇒ 32x = 320
⇒ x = 10
So, base is 32 – 20
⇒ 12 cm
Now,
Area = 1/2 × 8 × 12
⇒ 48 cm2

Que-20: The area of a triangle is 216 cm² and its side are in the ratio 3:4:5. Find the perimeter of the triangle.

Sol:  Sides ratio of a triangle = 3 : 4 : 5
Area of the triangle = 216 cm2
Sides ratio of triangle = 3x : 4x : 5x
⇒ s = (3x + 4x + 5x)/2
⇒ s = 12x/2 = 6x
Area of triangle = √[6x×(6x−3x)×(6x−4x)×(6x−5x)]
⇒ 216 = √[6x×3x×2x×x]
⇒ 216 = 6x²
⇒ 6x² = 216
⇒ x² = 216/6
⇒ x = √36
⇒ x = 6
∴ Perimeter of the triangle = 3x + 4x + 5x = 12x = 12 × 6 = 72 cm

– : End of Perimeter and Area of Plane Figures Class 9 RS Aggarwal Exe-17A Goyal Brothers ICSE Maths Solutions :–

Return to : – RS Aggarwal Solutions for ICSE Class-9 Mathematics

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