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Geometry and Trigonometry - Area and perimeter of 2D shapes

Grade 10IB

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 a rectangle, it is given by P=2(l+w)P = 2(l + w), where ll is length and ww is width.

Rectangle with length and width labels.
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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} b h. Note that the height must be perpendicular to the base.

Triangle with labeled base and perpendicular height.
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A sector is a portion of a circle enclosed by two radii and an arc. Its area is proportional to the central angle θ\theta: Area=θ360∘×πr2Area = \frac{\theta}{360^\circ} \times \pi r^2.

A circle sector showing radius and central angle theta.
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A parallelogram's area is calculated using A=b×hA = b \times h, where hh is the vertical (perpendicular) height, not the slant length.

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Composite shapes are figures made up of two or more simple shapes. To find their area, divide the figure into basic shapes like rectangles and triangles and sum their areas.

📐Formulae

AreaSquare=s2Area_{Square} = s^2

PerimeterSquare=4sPerimeter_{Square} = 4s

AreaRectangle=l×wArea_{Rectangle} = l \times w

PerimeterRectangle=2(l+w)Perimeter_{Rectangle} = 2(l + w)

AreaTriangle=12×b×hArea_{Triangle} = \frac{1}{2} \times b \times h

AreaParallelogram=b×hArea_{Parallelogram} = b \times h

AreaTrapezium=12(a+b)hArea_{Trapezium} = \frac{1}{2}(a + b)h

CircumferenceCircle=2πrCircumference_{Circle} = 2 \pi r

AreaCircle=πr2Area_{Circle} = \pi r^2

Arc Length=θ360∘×2πrArc\ Length = \frac{\theta}{360^\circ} \times 2 \pi r

AreaSector=θ360∘×πr2Area_{Sector} = \frac{\theta}{360^\circ} \times \pi r^2

💡Examples

Problem 1:

Calculate the area of a trapezium where the parallel sides are 12 cm12 \text{ cm} and 18 cm18 \text{ cm}, and the perpendicular height is 7 cm7 \text{ cm}.

Solution:

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

Explanation:

Identify the parallel sides a=12a = 12 and b=18b = 18 and the height h=7h = 7. Substitute these values into the trapezium area formula.

Problem 2:

Find the perimeter of a semicircle with a radius of 14 cm14 \text{ cm}. Take π≈227\pi \approx \frac{22}{7}.

Solution:

Perimeter=Arc Length+DiameterPerimeter = \text{Arc Length} + \text{Diameter} Perimeter=180∘360∘(2πr)+2rPerimeter = \frac{180^\circ}{360^\circ}(2 \pi r) + 2r Perimeter=πr+2rPerimeter = \pi r + 2r Perimeter=(227×14)+(2×14)Perimeter = (\frac{22}{7} \times 14) + (2 \times 14) Perimeter=44+28=72 cmPerimeter = 44 + 28 = 72 \text{ cm}

Explanation:

The perimeter of a semicircle consists of the curved arc (half the circumference) plus the straight diameter. We calculate the arc length πr\pi r and add 2r2r to get the total boundary length.

Problem 3:

A sector of a circle has a radius of 6 cm6 \text{ cm} and a central angle of 60∘60^\circ. Find its area in terms of π\pi.

Solution:

Area=θ360∘×πr2Area = \frac{\theta}{360^\circ} \times \pi r^2 Area=60360×π×62Area = \frac{60}{360} \times \pi \times 6^2 Area=16×36πArea = \frac{1}{6} \times 36\pi Area=6π cm2Area = 6\pi \text{ cm}^2

Explanation:

The area of a sector is a fraction of the total area of the circle. Using the ratio of the central angle to 360∘360^\circ, we multiply it by the area of the full circle πr2\pi r^2.

Problem 4:

Calculate the area of a composite shape consisting of a rectangle with dimensions 10 cm10 \text{ cm} by 6 cm6 \text{ cm} and a right-angled triangle attached to one of the 6 cm6 \text{ cm} sides with a base extension of 4 cm4 \text{ cm}.

Composite shape showing a rectangle and a triangle joined together.

Solution:

AreaRectangle=10×6=60 cm2Area_{Rectangle} = 10 \times 6 = 60 \text{ cm}^2 AreaTriangle=12×4×6=12 cm2Area_{Triangle} = \frac{1}{2} \times 4 \times 6 = 12 \text{ cm}^2 Total Area=60+12=72 cm2Total\ Area = 60 + 12 = 72 \text{ cm}^2

Explanation:

Divide the shape into a rectangle and a triangle. Calculate their areas separately using l×wl \times w and 12bh\frac{1}{2} b h, then add them together.

Problem 5:

A circular track has an inner radius of 7 m7 \text{ m} and an outer radius of 10 m10 \text{ m}. Find the area of the track path.

Two concentric circles representing a circular track with radius 7m and 10m.

Solution:

AreaOuter=π×102=100πArea_{Outer} = \pi \times 10^2 = 100\pi AreaInner=π×72=49πArea_{Inner} = \pi \times 7^2 = 49\pi AreaPath=100π−49π=51π≈160.22 m2Area_{Path} = 100\pi - 49\pi = 51\pi \approx 160.22 \text{ m}^2

Explanation:

To find the area of a ring (annulus), subtract the area of the smaller inner circle from the area of the larger outer circle.