Problem 1 :
Calculate the area of the triangle.
Solution :
area of triangle = 1/2 × b × h
Base (b) = 10 cm
Height (h) = 5 cm
= 1/2 × 10 × 5
= 5 × 5
= 25 cm^{2}
So, the area of a triangle is 25 cm^{2}.
Problem 2 :
A triangle has base length of 14 cm. The perpendicular height is 9 cm. Find the area of the triangle.
Solution :
area of triangle = 1/2 × b × h
Base (b) = 14 cm
Height (h) = 9 cm
= 1/2 × 14 × 9
= 7 × 9
= 63 cm^{2}
So, the area of a triangle is 63 cm^{2}.
Problem 3 :
Work out the area of the right - angled triangle.
Solution :
area of triangle = 1/2 × b × h
Base (b) = 7 cm
Height (h) = 4 cm
= 1/2 × 7 × 4
= 7 × 2
= 14 cm^{2}
So, the area of a triangle is 14 cm^{2}.
Problem 4 :
The area of the triangle is 20 cm^{2}, find x.
Solution :
area of triangle = 1/2 × b × h
area of triangle = 20 cm^{2}
Base (b) = 5 cm
Height (h) = x
20 = 1/2 × 5 × x
x = (20 × 2)/5
x = 40/5
x = 8 cm
Problem 5 :
The area of the triangle is 30 cm^{2}, find y.
Solution :
area of triangle = 1/2 × b × h
area of triangle = 30 cm^{2}
Base (b) = y
Height (h) = 6 cm
30 = 1/2 × y × 6
y = (30 × 2)/6
y = 60/6
y = 10 cm
Problem 6 :
The area of the triangle is 12 cm^{2}, find z.
Solution :
area of triangle = 1/2 × b × h
area of triangle = 12 cm^{2}
Base (b) = 6 cm
Height (h) = z
12 = 1/2 × 6 × z
z = (12 × 2)/6
z = 24/6
z = 4 cm
Problem 7 :
The area of the triangle is 56 cm^{2}, find a.
Solution :
area of triangle = 1/2 × b × h
area of triangle = 56 cm^{2}
Base (b) = a
Height (h) = 8 cm
56 = 1/2 × a × 8
a = (56 × 2)/8
a = 112/8
a = 14 cm^{2}
Problem 8 :
The diagram below shows a farmer's field.
The farmer wants to plant a new crop. Each sack of seed covers 30m^{2}. The cost of each sack is $6.
Work out the cost to buy enough seed to cover the field.
Solution :
Area of rectangle = l × w
length = 80 m
width = 30 m
Area of rectangle = 80 × 30
= 2400 m^{2}
Area of triangle = 1/2 × b × h
= 1/2 × 20 × 20
= 400/2
= 200 m^{2}
Total area = area of rectangle + area of triangle
= 2400 + 200
= 2600 m^{2}
Each sack of seed covers 30 m^{2}
= 2600/30
= 86.6
= 87
Cost of each sack = $6
So, 87 sack cost = 87 × 6
= $522
Problem 9 :
ABCD and WXYZ are squares.
Calculate the area of the shaded square WXYZ.
Solution :
By observing the figure,
Using Pythagorean theorem
WZ^{2} = AZ^{2} + AW^{2}
WZ^{2}= 8^{2} + 2^{2}
= 64 + 4
WZ^{2 }= 68
WZ = √68
Squaring on each sides.
(WZ)^{2} = (√68)^{2}
(WZ)^{2} = 68 cm
So, the area of the shaded square WXYZ is 68 cm.
Problem 10 :
The diagram below shows a garden.
The garden has a triangular vegetable patch and the rest of the garden is grass calculate the area of the garden that is grass.
Solution :
Area of rectangle = l × w
length = 8 m
width = 12 m
Area of rectangle = 8 × 12
= 96 m^{2}
Area of triangle = 1/2 × b × h
base = 5 m
height = 4 m
= 1/2 × 5 × 4
= 20/2
= 10 m^{2}
area of garden = area of rectangle - area of triangle
= 96 - 10
area of garden = 86 m^{2}
Problem 11 :
A logo consists of a rectangle and an isosceles triangle.
Calculate the area of the logo.
Solution :
Area of rectangle = l × w
length = 6 cm
width = 20 cm
Area of rectangle = 6 × 20
= 120 cm^{2}
Area of isosceles triangle = 1/2 × b × h
base = 6 cm
height = 5 cm
= 1/2 × 6 × 5
= 30/2
= 15 cm^{2}
area of the logo = area of rectangle + area of isosceles triangle
= 120 + 15
area of the logo = 135 cm^{2}
So, area of the logo is 135 cm^{2}.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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