Problem 1 :
Find the area of:
Solution:
Given, ∠A = 40˚, b = 9 cm, c = 10 cm
Area of triangle = 1/2 × b × c × sin A
= 1/2 × 9 × 10 × sin 40˚
= 1/2 × 9 × 10 × 0.64
A = 28.9 cm^{2}
Problem 2 :
Find the area of the triangle given below.
Solution:
Given, ∠A = 82˚, b = 25 km, c = 31 km
Area of triangle = 1/2 × b × c × sin A
= 1/2 × 25 × 31 × sin 82˚
= 1/2 × 25 × 31` × 0.99
= 383.625 km^{2}
A = 384 km^{2}
Problem 3 :
Find the area of the triangle given below.
Solution:
Given, ∠A = 2π/3, b = 10.2 cm, c = 6.4 cm
Area of triangle = 1/2 × b × c × sin A
= 1/2 × 10.2 × 6.4 × sin 2π/3
= 1/2 × 10.2 × 6.4` × √3/2
= 28.26 cm^{2}
A = 28.3 cm^{2}
Problem 2 :
If triangle ABC has area 150 cm^{2}, find the value of x:
Solution:
Area of triangle = 1/2 × a × c × sin B
150 = 1/2 × 17 × x × sin 68˚
300 = 17x × 0.927
17x = 300/0.927
17x = 323.6
x = 323.6/17
x = 19.0
So, the value of x is 19.0.
Problem 3 :
A parallelogram has two adjacent sides of length 4 cm and 6 cm respectively. If the included angle measures 52˚, find the area of the parallelogram.
Solution:
Given, x = 4 cm and y = 6 cm
Area of Rhombus = xy sin θ
= 4 × 6 × sin 52˚
= 24 × 0.78
= 18.91 cm^{2}
So, area of parallelogram is 18.9 cm².
Problem 4 :
A rhombus has sides of length 12 cm and an angle 72˚. Find its area.
Solution:
Given, side length = 12 cm, θ = 72˚
Area of Rhombus = a² sin θ
= (12)² sin (72˚)
= 144 × 0.95
= 136.95 cm²
So, area of rhombus is 137 cm².
Problem 5 :
Find the area of a regular hexagon with sides of length 12 cm.
Solution:
Given, side length = 12 cm
Area of regular hexagon = 3/2 (√3) a²
= 3/2 (√3) (12)²
= 3/2 (√3) (144)
= 216 × 1.732
= 374.112
So, area of regular hexagon is 374 cm².
Problem 6 :
A rhombus has an area of 50 cm² and an internal angle of size 63˚. Find the length of its sides.
Solution:
Given, Area of rhombus = 50cm²
Internal angle θ = 63˚
Area of Rhombus = a² sin θ
50 = a² sin 63˚
a² = 50/sin 63˚
a² = 50/0.89
a² = 56.17
a = 7.49 cm
So, side length of the rhombus is 7.49 cm.
Problem 7 :
A regular pentagonal garden plot has centre of symmetry O and an area of 338 m^{2}. Find the distance OA.
Solution:
Area of each triangular sector = 338/5
Angle subtended at centre = 360/5 = 72˚
Let circum radius of pentagon = r
Side of pentagon = 2rsin 36˚
Apothem = rcos 36˚
Area of each sector = 1/2 (2r sin 36˚) (r cos 36˚)
= r^{2}/2sin 72˚ = 338/5
r^{2} = 338/5 × 2/sin 72˚
r^{2} = 142.16
r = 11.9 m
Find the possible values of the included angle of a triangle with:
Problem 8 :
Sides 5 cm and 8 cm and area 15 cm^{2}
Solution:
Area of triangle = 1/2 × a × b × sin θ
15 = 1/2 × 5 × 8 × sin θ
30 = 40 × sin θ
sin θ = 30/40
sin θ = 0.75
θ = sin^{-1} (0.75)
θ = 48.6˚
Problem 9 :
Sides 45 km and 53 km and area 800 km^{2}
Solution :
Area of triangle = 1/2 × a × b × sin θ
800 = 1/2 × 45 × 53 × sin θ
1600 = 2385 × sin θ
sin θ = 1600/2385
sin θ = 0.67
θ = sin^{-1} (0.67)
θ = 42.1
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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