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Mensuration: Area and Perimeter - Calculate and compare areas of rectangles, parallelograms, and triangles

Grade 9CBSE

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

🔑Concepts

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The area of a rectangle is the product of its length and breadth, while the perimeter is the total boundary length. For a rectangle with length ll and breadth bb, Area = l×bl \times b and Perimeter = 2(l+b)2(l + b).

A rectangle showing length and breadth labels.
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A parallelogram's area is determined by its base (bb) and its vertical height (hh). Any side can be chosen as the base, but the height must be the perpendicular distance to that base: Area = b×hb \times h.

A parallelogram with a perpendicular height line drawn from the top vertex to the base.
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The area of a triangle is exactly half the area of a parallelogram with the same base and height. Formula: Area = 12×b×h\frac{1}{2} \times b \times h. This applies to all triangles regardless of their shape.

A triangle with base b and altitude h.
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When a triangle and a parallelogram share the same base and lie between the same parallel lines, the area of the triangle is half the area of the parallelogram.

A triangle and parallelogram sharing a base between two parallel lines.

📐Formulae

AreaRectangle=l×bArea_{Rectangle} = l \times b

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

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

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

s=a+b+c2s = \frac{a + b + c}{2}

AreaΔ=s(s−a)(s−b)(s−c)Area_{\Delta} = \sqrt{s(s-a)(s-b)(s-c)}

💡Examples

Problem 1:

A rectangular field has a length of 120120 mm and a breadth of 8080 mm. Calculate the area of the field and the cost of fencing it at the rate of ₹25₹25 per meter.

Solution:

Length l=120l = 120 mm, Breadth b=80b = 80 mm. Area=l×b=120×80=9600Area = l \times b = 120 \times 80 = 9600 m2m^2. Perimeter=2(l+b)=2(120+80)=2(200)=400Perimeter = 2(l + b) = 2(120 + 80) = 2(200) = 400 mm. Cost of fencing = Perimeter×Rate=400×25=₹10,000Perimeter \times Rate = 400 \times 25 = ₹10,000.

Explanation:

First, the area is found by multiplying dimensions. Then, the perimeter is calculated to find the total length of the fence, which is multiplied by the unit rate.

Problem 2:

A parallelogram and a triangle are on the same base of 1414 cmcm and between the same parallel lines. If the height of the parallelogram is 1010 cmcm, find the area of both shapes and compare them.

Solution:

Base b=14b = 14 cmcm, Height h=10h = 10 cmcm. AreaParallelogram=b×h=14×10=140Area_{Parallelogram} = b \times h = 14 \times 10 = 140 cm2cm^2. AreaTriangle=12×b×h=12×14×10=70Area_{Triangle} = \frac{1}{2} \times b \times h = \frac{1}{2} \times 14 \times 10 = 70 cm2cm^2. Comparison: AreaTriangleAreaParallelogram=70140=12\frac{Area_{Triangle}}{Area_{Parallelogram}} = \frac{70}{140} = \frac{1}{2}.

Explanation:

This demonstrates that for the same base and height, a triangle's area is exactly half that of a parallelogram.

Problem 3:

Find the area of a triangle whose sides are 1313 cmcm, 1414 cmcm, and 1515 cmcm using Heron's formula.

Solution:

Let a=13,b=14,c=15a = 13, b = 14, c = 15. s=13+14+152=422=21 cms = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21 \text{ cm} Area=21(21−13)(21−14)(21−15)Area = \sqrt{21(21-13)(21-14)(21-15)} Area=21×8×7×6Area = \sqrt{21 \times 8 \times 7 \times 6} Area=7056=84 cm2Area = \sqrt{7056} = 84 \text{ cm}^2

Explanation:

When the height is unknown but all three sides are given, Heron's formula is used by first calculating the semi-perimeter ss and then the square root of the product of ss and its differences from each side.

Problem 4:

A square and a rectangle have the same perimeter. If the side of the square is 6060 mm and the length of the rectangle is 8080 mm, find the breadth of the rectangle and determine which shape has a larger area.

Comparison of a square and a rectangle with equal perimeters.

Solution:

  1. Perimeter of square = 4×side=4×60=2404 \times side = 4 \times 60 = 240 mm.
  2. Since perimeters are equal, Perimeter of rectangle = 2(l+b)=2402(l + b) = 240 mm.
  3. 2(80+b)=240⇒80+b=120⇒b=402(80 + b) = 240 \Rightarrow 80 + b = 120 \Rightarrow b = 40 mm.
  4. Area of square = side2=60×60=3600side^2 = 60 \times 60 = 3600 m2m^2.
  5. Area of rectangle = l×b=80×40=3200l \times b = 80 \times 40 = 3200 m2m^2.
  6. Comparison: 3600>32003600 > 3200, so the square has a larger area.

Explanation:

Calculated the perimeter of the square first to find the missing breadth of the rectangle using the equality condition, then computed both areas to compare.

Problem 5:

A triangular park has sides 99 mm, 1212 mm, and 1515 mm. Find the cost of leveling the park at RsRs 1010 per m2m^2.

A right-angled triangle representing the park with sides 9, 12, and 15.

Solution:

  1. Calculate semi-perimeter s=9+12+152=362=18s = \frac{9 + 12 + 15}{2} = \frac{36}{2} = 18 mm.
  2. Use Heron's formula: Area=s(s−a)(s−b)(s−c)Area = \sqrt{s(s-a)(s-b)(s-c)}
  3. Area=18(18−9)(18−12)(18−15)Area = \sqrt{18(18-9)(18-12)(18-15)}
  4. Area=18×9×6×3=2916=54Area = \sqrt{18 \times 9 \times 6 \times 3} = \sqrt{2916} = 54 m2m^2.
  5. Cost = Area×Rate=54×10=RsArea \times Rate = 54 \times 10 = Rs 540540.

Explanation:

Since the height was not given, Heron's formula was used to find the area from the three sides, which was then multiplied by the unit rate to find total cost.