Problem 1 :
Find the area of an equilateral triangle with side 10 cm.
Solution:
Given, side length = 10 cm
Area of an equilateral triangle = (√3/4) a^{2}
= (√3/4) × 10^{2}
= (√3/4) × 100
A = 25√3 cm^{2}
Problem 2 :
If the angles of a triangle are in the ratio 2: 3: 4 find the angles of the triangle.
Solution:
Let us consider the angles as 2x, 3x and 4x.
2x + 3x + 4x = 9x
Angle sum of triangle = 180˚
9x = 180˚
x = 180/9
x = 20˚
So, 2x = 2 × 20 = 40˚
3x = 3 × 20 = 60˚
4x = 4 × 20 = 80˚
Therefore, the angles are 40˚, 60˚, 80˚.
Problem 3 :
The area of a triangle is 48 cm^{2}. If its base is 12 cm, find its altitude.
Solution:
Given, area of a triangle = 48 cm^{2}
base = 12 cm
Area of triangle = 1/2 × base × altitude
48 = 1/2 × 12 × altitude
altitude = 48/6
altitude = 8 cm
Problem 4 :
The hypotenuse of an isosceles right triangle is 10 cm. Find its area.
Solution:
Let sides are x.
Base^{2} + Height^{2} = Hypotenuse^{2}
x^{2} + x^{2} = 10^{2}
2x^{2} = 100
x^{2} = 50
Area of triangle = 1/2 × base × height
= 1/2 × x × x
= x^{2}/2
= 50/2
A = 25 cm^{2}
Hence, the area of isosceles triangle is 25 cm^{2}.
Problem 5 :
A rhombus has perimeter 120 m and one of its diagonal is 50 m. Find the area of the rhombus.
Solution:
Perimeter of rhombus = 120 m
d_{1} + d_{2} = 120 m
50 + d_{2} = 120 m
d_{2} = 120 - 50
d_{2} = 70 m
Area of rhombus = 1/2 × d_{1} × d_{2}
= 1/2 × 50 × 70
= 1750 m^{2}
So, the area of rhombus is 1750 m² .
Problem 6 :
The area of an equilateral triangle is 2√3 cm^{2}. Find its perimeter.
Solution:
Given, Area of an equilateral triangle = 2√3 cm^{2}
Area of an equilateral triangle = √3/4 a²
√3/4 a² = 2√3
a² = 2 × 4
a² = 8
a = 2√2
Perimeter of an equilateral triangle = 3a
= 3 × 2√2
= 6√2 cm
Problem 7 :
The area of an equilateral triangle is 81√3 cm², find its height.
Solution:
Given, Area of an equilateral triangle = 2√3 cm2
Area of an equilateral triangle = √3/4 a²
√3/4 a² = 81√3
a² = 81 × 4
a² = 324
a = 18 cm
Height of an equilateral triangle = √3/2 (a)
= √3/2 (18)
= 9√3 cm
So, height of an equilateral triangle is 9√3 cm.
Problem 8 :
Find the area of a triangle whose sides are 13 cm, 14 cm and 15 cm respectively.
Solution:
Let sides are 13 cm, 14 cm and 15 cm.
s = (a + b + c) / 2
= (13 + 14 + 15) / 2
= 42/2
s = 21
Area of triangle = √ (s(s - a) (s - b) (s - c))
= √ 21(21 - 13) (21 - 14) (21 - 15)
= √ 21 × 8 × 7 × 6
= √7056
A = 84 cm²
So, the area of triangle is 84 cm².
Problem 9 :
Find the area of an isosceles triangle whose equal sides measure 13 cm each and the base measures 24 cm.
Solution:
Area of an isosceles triangle = (1/4) b√(4a² - b²)
= 1/4 × 24√(4(13)² - (24)²)
= 6√ (4 × 169 - 576)
= 6√676 - 576
= 6√100
= 6 × 10
= 60 cm²
So, Area of an isosceles triangle is 60 cm².
Problem 10 :
Prove that the area of a quadrilateral ABCD is 4(√3 + 2√2)m². AB = 6 cm, BC = 6 cm, CD = 4 cm and AD = 4 cm and diagonal AC = 4 cm.
Solution:
∆ ABC, AB = 6 cm, BC = 6 cm, AC = 4 cm
s = (a + b + c) / 2
= (6 + 6 + 4) / 2
= 16/2
s = 8
Area of triangle = √ (s(s - a) (s - b) (s - c))
= √ 8(8 - 6) (8 - 6) (8 - 4)
= √ 8 × 2 × 2 × 4
= √128
A = 8√2 cm²
∆ ADC, AC = 4 cm, AD = 4 cm, CD = 4 cm
s = (a + b + c) / 2
= (4 + 4 + 4) / 2
= 12/2
s = 6
Area of triangle = √ (s(s - a) (s - b) (s - c))
= √ 6(6 - 4) (6 - 4) (6 - 4)
= √ 6 × 2 × 2 × 2
= √48
A = 4√3 cm²
Total area of quadrilateral ABCD = 8√2 + 4√3 cm²
= 4(√3 + 2√2) cm²
Hence, it is proved.
Problem 11 :
The perimeter of a right angles is 144 cm and its hypotenuse measures 65 cm. Find the length of other sides and calculate its area. Verify the result using Heron's formula.
Solution:
x + y + 65 = 144
x + y = 144 - 65
x + y = 79 ---> (1)
x² + y² = (65)²
x² + y² = 4225
Now,
(x + y)² = x² + y² + 2xy
2xy = (x + y)² - (x² + y²)
= (79)² - 4225
= 6241 - 4225
= 2016
Then, x - y = ± √ (x² + y² - 2xy)
x - y = ± √ (4225 - 2016)
= ± √2209
x - y = ± 47 ---> (2)
Solving (1) and (2), we get
x = 63 cm, y = 16 cm (or)
x = 16 cm, y = 63 cm
Area of triangle = 1/2 × b × h
= 1/2 × 63 × 16
A = 504 cm²
Now, verify using Heron's formula,
s = (a + b + c) / 2
= (65 + 63 + 16) / 2
= 144/2
s = 72
Area of triangle = √ (s(s - a) (s - b) (s - c))
= √ 72(72 - 65) (72 - 63) (72 - 16)
= √ 72 × 7 × 9 × 56
= √254016
A = 504 cm²
So, it is verified.
Problem 12 :
Find the perimeter of an isosceles right triangle having an area equal to 200 cm².
Solution:
Let ABC be an isosceles right angled triangle with right angle at B such that AB = BC = a cm
Area of triangle = 1/2 × b × h
200 = 1/2 × a × a
200 = a²/2
a² = 400
a = 20 cm
By using Pythagorean Theorem,
AC² = AB² + BC²
AC² = a² + a²
AC² = 2a²
AC = a√2 cm
= 20√2 cm
Perimeter = AB + BC + AC
= 20 + 20 + 20√2
= 68.2 cm
Problem 13 :
The area of an isosceles right triangle is 128 cm². Find the length of its hypotenuse.
Solution:
Let the base and height of the isosceles triangle be x.
Area = 1/2 × base × height
128 = 1/2 × x × x
x² = 256
x = 16
Using Pythagorean Theorem,
AC² = AB² + BC²
AC² = 16² + 16²
AC² = 256 + 256
AC² = 512
AC = 22.62 cm
So, the length of hypotenuse is 22.62 cm.
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