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Mensuration - Perimeter and area of 2D shapes

Grade 11A Level

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The perimeter of a 2D shape is the total distance around its boundary. For polygons, it is the sum of all side lengths. For a circle, it is called the circumference (C=2πrC = 2\pi r).

A rectangle showing length l and width w for perimeter calculation.
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Area measures the surface inside a 2D shape. The area of a parallelogram is calculated using the base and the perpendicular height (A=b×hA = b \times h). Note that the height must be at a right angle to the base.

Parallelogram showing base and perpendicular height.
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A sector is a portion of a circle defined by two radii and an arc. Its area is proportional to the central angle θ\theta: Area=θ360×πr2\text{Area} = \frac{\theta}{360} \times \pi r^2.

A circle with a 45 degree sector highlighted.
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The area of a trapezium is found by taking the average of the parallel sides (aa and bb) and multiplying by the perpendicular height (hh): A=12(a+b)hA = \frac{1}{2}(a+b)h.

Trapezium showing parallel sides a, b and height h.

📐Formulae

Square: Area=s2,Perimeter=4s\text{Square: Area} = s^2, \text{Perimeter} = 4s

Rectangle: Area=l×w,Perimeter=2(l+w)\text{Rectangle: Area} = l \times w, \text{Perimeter} = 2(l + w)

Triangle: Area=12×base×height\text{Triangle: Area} = \frac{1}{2} \times \text{base} \times \text{height}

Triangle (Trigonometric): Area=12absin⁡(C)\text{Triangle (Trigonometric): Area} = \frac{1}{2}ab \sin(C)

Parallelogram: Area=base×height\text{Parallelogram: Area} = \text{base} \times \text{height}

Trapezium: Area=12(a+b)h\text{Trapezium: Area} = \frac{1}{2}(a + b)h

Circle: Area=πr2,Circumference=2πr\text{Circle: Area} = \pi r^2, \text{Circumference} = 2\pi r

Arc Length=θ360×2πr\text{Arc Length} = \frac{\theta}{360} \times 2\pi r

Sector Area=θ360×πr2\text{Sector Area} = \frac{\theta}{360} \times \pi r^2

💡Examples

Problem 1:

A sector of a circle has a radius of 6 cm and a central angle of 60°. Calculate the area of the sector and the length of the arc. (Use π = 3.142)

Solution:

Arc Length = (60/360) * 2 * 3.142 * 6 = 6.284 cm. Sector Area = (60/360) * 3.142 * 6^2 = 18.852 cm².

Explanation:

To find the arc length and sector area, we multiply the total circumference and total area of the circle by the fraction of the circle represented by the angle (60/360).

Problem 2:

A trapezium has parallel sides of length 8 cm and 12 cm. If the area of the trapezium is 50 cm², find its perpendicular height.

Solution:

50 = 1/2 * (8 + 12) * h => 50 = 1/2 * 20 * h => 50 = 10h => h = 5 cm.

Explanation:

Substitute the known values into the area of a trapezium formula: Area = 1/2(a+b)h. Solve the resulting linear equation for the unknown height (h).

Problem 3:

Calculate the area of a triangle where two sides are 7 cm and 10 cm, and the included angle between them is 30°.

Solution:

Area = 1/2 * 7 * 10 * sin(30°) = 1/2 * 70 * 0.5 = 17.5 cm².

Explanation:

When the perpendicular height is not given but an angle is, use the trigonometric area formula Area = 1/2 ab sin(C).

Problem 4:

A compound shape is formed by a rectangle of length 12 cm12\text{ cm} and width 8 cm8\text{ cm}, with a semi-circle attached to one of the shorter sides. Find the total area of the shape. (Take π=3.142\pi = 3.142)

A rectangle with a semi-circle on its right side.

Solution:

Area of rectangle=l×w=12×8=96 cm2\text{Area of rectangle} = l \times w = 12 \times 8 = 96\text{ cm}^2 Radius of semi-circle=w2=82=4 cm\text{Radius of semi-circle} = \frac{w}{2} = \frac{8}{2} = 4\text{ cm} Area of semi-circle=12πr2=12×3.142×42=25.136 cm2\text{Area of semi-circle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \times 3.142 \times 4^2 = 25.136\text{ cm}^2 Total Area=96+25.136=121.136 cm2\text{Total Area} = 96 + 25.136 = 121.136\text{ cm}^2

Explanation:

To find the area of a compound shape, divide it into basic shapes (a rectangle and a semi-circle), calculate their individual areas, and add them together. The diameter of the semi-circle matches the width of the rectangle.

Problem 5:

A running track consists of a rectangle with semi-circular ends. If the rectangle has a length of 100 m100\text{ m} and the semi-circles have a diameter of 60 m60\text{ m}, calculate the total perimeter of the track. (Take π=3.142\pi = 3.142)

A stadium track shape showing straight sides and circular ends.

Solution:

Length of two straight sides=2×100=200 m\text{Length of two straight sides} = 2 \times 100 = 200\text{ m} Circumference of two semi-circles (one full circle)=πd=3.142×60=188.52 m\text{Circumference of two semi-circles (one full circle)} = \pi d = 3.142 \times 60 = 188.52\text{ m} Total Perimeter=200+188.52=388.52 m\text{Total Perimeter} = 200 + 188.52 = 388.52\text{ m}

Explanation:

The perimeter of the track consists of the two straight lengths of the rectangle and the two curved arcs. Since there are two semi-circles of the same diameter, they combine to form the circumference of one full circle.

Perimeter and area of 2D shapes Grade 11 Notes & Examples