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Mensuration - Area and Perimeter of 2D Shapes

Grade 12A Level

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

🔑Concepts

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The perimeter is the total distance around the boundary of a 2D shape. For rectilinear shapes, it is the sum of all side lengths. For a circle, this distance is called the circumference.

Rectangle showing length and width for perimeter calculation
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The area of a triangle is half the product of its base and its perpendicular height (A=12bhA = \frac{1}{2}bh). Note that the height must be measured at right angles to the base.

Triangle showing base and perpendicular height
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Compound shapes are made up of two or more basic shapes. To find the total area, split the shape into rectangles, triangles, or circles, calculate their individual areas, and add them together.

L-shaped compound figure split into two rectangles
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A sector is a portion of a circle defined by two radii and an arc. Its area and arc length are proportional to the central angle θ\theta as a fraction of 360∘360^{\circ}.

📐Formulae

Area of Rectangle=l×w\text{Area of Rectangle} = l \times w

Area of Triangle=12×b×h\text{Area of Triangle} = \frac{1}{2} \times b \times h

Area of Parallelogram=b×h\text{Area of Parallelogram} = b \times h

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

Circumference of Circle=2πr or πd\text{Circumference of Circle} = 2\pi r \text{ or } \pi d

Area of Circle=πr2\text{Area of Circle} = \pi r^2

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

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

💡Examples

Problem 1:

Calculate the area of a trapezium where the parallel sides are 12 cm and 18 cm, and the perpendicular height is 7 cm.

Solution:

A=12(12+18)×7=12(30)×7=15×7=105 cm2A = \frac{1}{2}(12 + 18) \times 7 = \frac{1}{2}(30) \times 7 = 15 \times 7 = 105\text{ cm}^2

Explanation:

Identify the parallel sides 'a' and 'b' and the height 'h'. Substitute them into the trapezium formula: Area = 1/2(sum of parallel sides) × height.

Problem 2:

Find the perimeter of a sector with a radius of 10 cm and a central angle of 72∘72^{\circ}. (Take π=3.142\pi = 3.142)

Solution:

Arc Length=72360×2×3.142×10=12.568 cm\text{Arc Length} = \frac{72}{360} \times 2 \times 3.142 \times 10 = 12.568\text{ cm}. Perimeter=Arc Length+2r=12.568+20=32.568 cm\text{Perimeter} = \text{Arc Length} + 2r = 12.568 + 20 = 32.568\text{ cm}.

Explanation:

To find the perimeter of a sector, you must calculate the arc length first and then add the two radii that form the 'v' shape of the sector.

Problem 3:

A circular hole of radius 3 cm is cut out of a square piece of metal with side length 10 cm. Find the area of the remaining metal.

Solution:

Area of Square=10×10=100 cm2\text{Area of Square} = 10 \times 10 = 100\text{ cm}^2. Area of Circle=π×32=9π≈28.27 cm2\text{Area of Circle} = \pi \times 3^2 = 9\pi \approx 28.27\text{ cm}^2. Remaining Area=100−28.27=71.73 cm2\text{Remaining Area} = 100 - 28.27 = 71.73\text{ cm}^2.

Explanation:

This is a compound area problem involving subtraction. Calculate the total area of the outer shape (square) and subtract the area of the shape removed (circle).

Problem 4:

A running track consists of a rectangle with semi-circular ends. The rectangle has a length of 100100 m and a width of 6464 m. Calculate the total distance around the inside of the track. (Use π=3.14\pi = 3.14)

Running track with rectangular center and semi-circular ends

Solution:

Perimeter=(2×Length)+(2×Semi-circumference)\text{Perimeter} = (2 \times \text{Length}) + (2 \times \text{Semi-circumference}) Perimeter=(2×100)+(2×12×π×d)\text{Perimeter} = (2 \times 100) + (2 \times \frac{1}{2} \times \pi \times d) Perimeter=200+(3.14×64)\text{Perimeter} = 200 + (3.14 \times 64) Perimeter=200+200.96\text{Perimeter} = 200 + 200.96 Perimeter=400.96 m\text{Perimeter} = 400.96 \text{ m}

Explanation:

The boundary consists of two straight lengths of 100100 m and two semi-circular arcs. The two semi-circles combine to form one full circle with a diameter equal to the width of the rectangle (6464 m).

Problem 5:

Calculate the area of a parallelogram with a base of 1515 cm and a slanted side of 1010 cm, where the angle between the base and the slanted side is 30∘30^{\circ}.

Parallelogram with base 15cm and slanted side 10cm at 30 degrees

Solution:

First, find the perpendicular height hh using trigonometry: h=10×sin⁡(30∘)=10×0.5=5 cmh = 10 \times \sin(30^{\circ}) = 10 \times 0.5 = 5 \text{ cm} Now, calculate the area: Area=base×height\text{Area} = \text{base} \times \text{height} Area=15×5=75 cm2\text{Area} = 15 \times 5 = 75 \text{ cm}^2

Explanation:

The area of a parallelogram is base×heightbase \times height. If the perpendicular height is not given, it can be calculated using the slanted side and the interior angle using sin⁡(θ)=OppositeHypotenuse\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}.