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Geometry and Measure - Area, perimeter and volume

Grade 7Cambridge (IGCSE)

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

🔑Concepts

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Perimeter is the total distance around the outside of a 2D shape. For composite shapes, sum all external side lengths.

L-shaped composite polygon for perimeter calculation
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The Area of a circle is calculated using the radius rr with the formula A=πr2A = \pi r^2. The radius is half of the diameter dd.

Circle showing radius r
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A Trapezium has one pair of parallel sides (aa and bb). The area is the average of these sides multiplied by the perpendicular height hh: A=12(a+b)hA = \frac{1}{2}(a + b)h.

Trapezium with parallel sides a and b and height h
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Volume of a prism (like a cuboid) is the area of the cross-section multiplied by its length. For a cuboid, V=l×w×hV = l \times w \times h.

3D cuboid showing length, width, and height

📐Formulae

Perimeter of Rectangle=2(l+w)\text{Perimeter of Rectangle} = 2(l + w)

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

Volume of Cuboid=l×w×h\text{Volume of Cuboid} = l \times w \times h

Total Surface Area of Cuboid=2(lw+lh+wh)\text{Total Surface Area of Cuboid} = 2(lw + lh + wh)

💡Examples

Problem 1:

A triangle has a base of 14 cm and a perpendicular height of 9 cm. Calculate its area.

Solution:

63 cm263 \text{ cm}^2

Explanation:

Apply the area of a triangle formula: A=12×base×heightA = \frac{1}{2} \times \text{base} \times \text{height}. Calculation: 12×14×9=7×9=63\frac{1}{2} \times 14 \times 9 = 7 \times 9 = 63.

Problem 2:

Find the circumference of a circle with a diameter of 10 cm. (Use π≈3.14\pi \approx 3.14)

Solution:

31.4 cm31.4 \text{ cm}

Explanation:

Using the formula C=πdC = \pi d, multiply the diameter by π\pi: 3.14×10=31.43.14 \times 10 = 31.4.

Problem 3:

A cereal box (cuboid) has dimensions 20 cm by 5 cm by 30 cm. Find its volume.

Solution:

3000 cm33000 \text{ cm}^3

Explanation:

Volume of a cuboid is found by multiplying length, width, and height: V=20×5×30=100×30=3000V = 20 \times 5 \times 30 = 100 \times 30 = 3000.

Problem 4:

Calculate the area of a trapezium where the parallel sides are 8 cm and 12 cm, and the height is 5 cm.

Solution:

50 cm250 \text{ cm}^2

Explanation:

Use the formula A=12(a+b)hA = \frac{1}{2}(a+b)h. First, add the parallel sides (8+12=208 + 12 = 20). Then, A=12×20×5=10×5=50A = \frac{1}{2} \times 20 \times 5 = 10 \times 5 = 50.

Problem 5:

Calculate the area of a parallelogram with a base of 12 cm12 \text{ cm} and a perpendicular height of 7 cm7 \text{ cm}.

Parallelogram with base 12 and height 7

Solution:

Area=base×height\text{Area} = \text{base} \times \text{height} Area=12×7\text{Area} = 12 \times 7 Area=84 cm2\text{Area} = 84 \text{ cm}^2

Explanation:

To find the area of a parallelogram, multiply the length of the base by the perpendicular height. Do not use the slanted side length.

Problem 6:

Find the area of the shaded region: a large circle with radius 10 cm10 \text{ cm} has a smaller circle with radius 4 cm4 \text{ cm} removed from its center.

Concentric circles representing a shaded ring

Solution:

Area of large circle=π×102=100π\text{Area of large circle} = \pi \times 10^2 = 100\pi Area of small circle=π×42=16π\text{Area of small circle} = \pi \times 4^2 = 16\pi Shaded Area=100π−16π=84π\text{Shaded Area} = 100\pi - 16\pi = 84\pi Shaded Area≈84×3.14159≈263.89 cm2\text{Shaded Area} \approx 84 \times 3.14159 \approx 263.89 \text{ cm}^2

Explanation:

To find the area of a ring (annulus), calculate the area of the outer circle and subtract the area of the inner circle.