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Mensuration - Area and Perimeter of Plane Figures (Triangle, Quadrilaterals)

Grade 9ICSE

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

🔑Concepts

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The perimeter of a plane figure is the total length of its boundary. For a triangle with sides aa, bb, and cc, the perimeter is P=a+b+cP = a + b + c. For a rectangle with length ll and breadth bb, it is P=2(l+b)P = 2(l + b).

Rectangle showing length and breadth labels for perimeter calculation.
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Heron's Formula is used to find the area of a triangle when the lengths of all three sides are known. First, calculate the semi-perimeter s=a+b+c2s = \frac{a + b + c}{2}, then the Area A=s(s−a)(s−b)(s−c)A = \sqrt{s(s - a)(s - b)(s - c)}.

Triangle with sides labeled a, b, and c for Heron's formula.
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A Parallelogram's area is the product of its base and its corresponding height (altitude). Area=b×hArea = b \times h. A Rhombus, being a special parallelogram, also has its area defined by its diagonals: Area=12×d1×d2Area = \frac{1}{2} \times d_1 \times d_2.

Parallelogram showing base and height.
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A Trapezium is a quadrilateral with one pair of parallel sides. Its area is given by half the sum of the parallel sides multiplied by the perpendicular distance between them: Area=12(a+b)×hArea = \frac{1}{2} (a + b) \times h.

Trapezium with parallel sides a and b and height h.

📐Formulae

Perimeter of a Triangle: P=a+b+cP = a + b + c

Area of a Triangle (General): A=12×base×heightA = \frac{1}{2} \times \text{base} \times \text{height}

Semi-perimeter (ss): s=a+b+c2s = \frac{a + b + c}{2}

Heron's Formula: Area=s(s−a)(s−b)(s−c)\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}

Area of an Equilateral Triangle: A=34a2A = \frac{\sqrt{3}}{4} a^2

Area of a Rectangle: A=l×bA = l \times b

Perimeter of a Rectangle: P=2(l+b)P = 2(l + b)

Area of a Square: A=a2A = a^2 or A=12d2A = \frac{1}{2} d^2 (where dd is the diagonal)

Area of a Parallelogram: A=base×heightA = \text{base} \times \text{height}

Area of a Rhombus: A=12×d1×d2A = \frac{1}{2} \times d_1 \times d_2

Area of a Trapezium: A=12×(a+b)×hA = \frac{1}{2} \times (a + b) \times h

💡Examples

Problem 1:

Find the area of a triangle whose sides are 13 cm13 \text{ cm}, 14 cm14 \text{ cm}, and 15 cm15 \text{ cm}.

Solution:

  1. Find the semi-perimeter ss: s=13+14+152=422=21 cms = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21 \text{ cm}
  2. Apply Heron's Formula: Area=21(21−13)(21−14)(21−15)\text{Area} = \sqrt{21(21-13)(21-14)(21-15)} Area=21×8×7×6\text{Area} = \sqrt{21 \times 8 \times 7 \times 6} Area=7056=84 cm2\text{Area} = \sqrt{7056} = 84 \text{ cm}^2

Explanation:

Since all three sides of the triangle are given and it is a scalene triangle, Heron's formula is the most direct method to find the area.

Problem 2:

The area of a rhombus is 96 cm296 \text{ cm}^2 and one of its diagonals is 12 cm12 \text{ cm}. Find the length of the other diagonal and the length of its side.

Solution:

  1. Find the second diagonal (d2d_2): Area=12×d1×d2\text{Area} = \frac{1}{2} \times d_1 \times d_2 96=12×12×d296 = \frac{1}{2} \times 12 \times d_2 96=6×d2  ⟹  d2=16 cm96 = 6 \times d_2 \implies d_2 = 16 \text{ cm}
  2. Find the side (aa) using the property that diagonals bisect at right angles: side=(d12)2+(d22)2\text{side} = \sqrt{(\frac{d_1}{2})^2 + (\frac{d_2}{2})^2} a=62+82=36+64=100=10 cma = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ cm}

Explanation:

We use the area formula for a rhombus to find the missing diagonal. Then, we use the Pythagorean theorem on one of the four internal right-angled triangles formed by the diagonals to find the side length.

Problem 3:

Find the area of a trapezium where the parallel sides are 25 cm25 \text{ cm} and 13 cm13 \text{ cm}, and the non-parallel sides are 10 cm10 \text{ cm} each.

Isosceles trapezium with sides 25, 13, and 10.

Solution:

  1. Let the parallel sides be a=25 cma = 25 \text{ cm} and b=13 cmb = 13 \text{ cm}.
  2. The non-parallel sides are equal, so it is an isosceles trapezium. Draw perpendiculars from the ends of the shorter side to the longer side.
  3. The length of the base of the triangle formed on each side is 25−132=122=6 cm\frac{25 - 13}{2} = \frac{12}{2} = 6 \text{ cm}.
  4. Using Pythagoras theorem for the height hh: h2+62=102h^2 + 6^2 = 10^2 h2+36=100h^2 + 36 = 100 h2=64  ⟹  h=8 cmh^2 = 64 \implies h = 8 \text{ cm}
  5. Area of Trapezium: Area=12×(25+13)×8Area = \frac{1}{2} \times (25 + 13) \times 8 Area=12×38×8=152 cm2Area = \frac{1}{2} \times 38 \times 8 = 152 \text{ cm}^2

Explanation:

To find the area, we first find the height by utilizing the properties of an isosceles trapezium and applying the Pythagoras theorem on the right-angled triangle formed by the height and the non-parallel side.

Problem 4:

The perimeter of a rectangular field is 120 m120 \text{ m} and its length is 35 m35 \text{ m}. Find the area of the field and the length of its diagonal.

Rectangle with length 35m and diagonal d.

Solution:

  1. Given Perimeter P=120 mP = 120 \text{ m} and Length l=35 ml = 35 \text{ m}.
  2. Formula for Perimeter: 2(l+b)=1202(l + b) = 120 2(35+b)=1202(35 + b) = 120 35+b=60  ⟹  b=25 m35 + b = 60 \implies b = 25 \text{ m}
  3. Area of the rectangle: Area=l×b=35×25=875 m2Area = l \times b = 35 \times 25 = 875 \text{ m}^2
  4. Length of the diagonal dd: d=l2+b2d = \sqrt{l^2 + b^2} d=352+252=1225+625d = \sqrt{35^2 + 25^2} = \sqrt{1225 + 625} d=1850≈43.01 md = \sqrt{1850} \approx 43.01 \text{ m}

Explanation:

First, find the breadth using the perimeter formula. Then, use the breadth to calculate the area and the Pythagorean diagonal of the rectangle.