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Measuring Space: Perimeter and Area - Area of a Triangle

Grade 9CBSE

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

🔑Concepts

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For a right-angled triangle, the area can be calculated directly using the legs of the triangle as base and height. The area is given by 12×base×height\frac{1}{2} \times \text{base} \times \text{height}.

Right-angled triangle showing base and height
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In an isosceles triangle, the altitude (height) drawn from the vertex between the equal sides bisects the base. This allows the use of the Pythagorean theorem to find the height before calculating the area.

Isosceles triangle with altitude bisecting the base
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Heron's Formula is used when the lengths of all three sides are known but the height is not given. First, calculate the semi-perimeter s=a+b+c2s = \frac{a+b+c}{2}, then use Area=s(s−a)(s−b)(s−c)\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}.

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The area of a triangle remains the same regardless of which side is chosen as the base, provided the corresponding altitude (perpendicular distance from the opposite vertex) is used.

📐Formulae

Area of a triangle=12×base×height\text{Area of a triangle} = \frac{1}{2} \times \text{base} \times \text{height}

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

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

Area of an equilateral triangle=34a2\text{Area of an equilateral triangle} = \frac{\sqrt{3}}{4}a^2

Perimeter of a triangle=a+b+c\text{Perimeter of a triangle} = a + b + c

💡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. Calculate 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=(3×7)×(2×2×2)×7×(2×3)\text{Area} = \sqrt{(3 \times 7) \times (2 \times 2 \times 2) \times 7 \times (2 \times 3)} Area=24×32×72\text{Area} = \sqrt{2^4 \times 3^2 \times 7^2} Area=22×3×7=4×21=84 cm2\text{Area} = 2^2 \times 3 \times 7 = 4 \times 21 = 84\text{ cm}^2

Explanation:

Since all three sides are known, Heron's formula is the most efficient method. We first find the semi-perimeter ss and then substitute s,a,b,s, a, b, and cc into the radical expression.

Problem 2:

The sides of a triangular plot are in the ratio of 3:5:73 : 5 : 7 and its perimeter is 300 m300\text{ m}. Find its area.

Solution:

  1. Let the sides be 3x3x, 5x5x, and 7x7x.
  2. Given perimeter =300 m= 300\text{ m}, so: 3x+5x+7x=3003x + 5x + 7x = 300 15x=300  ⟹  x=2015x = 300 \implies x = 20
  3. The sides are: a=3×20=60 ma = 3 \times 20 = 60\text{ m}, b=5×20=100 mb = 5 \times 20 = 100\text{ m}, c=7×20=140 mc = 7 \times 20 = 140\text{ m}.
  4. Calculate ss: s=3002=150 ms = \frac{300}{2} = 150\text{ m}
  5. Apply Heron's Formula: Area=150(150−60)(150−100)(150−140)\text{Area} = \sqrt{150(150 - 60)(150 - 100)(150 - 140)} Area=150×90×50×10\text{Area} = \sqrt{150 \times 90 \times 50 \times 10} Area=6750000=15003 m2\text{Area} = \sqrt{6750000} = 1500\sqrt{3}\text{ m}^2

Explanation:

First, find the actual side lengths using the given ratio and perimeter. Once the sides are determined, use Heron's formula to calculate the area.

Problem 3:

Find the area of an equilateral triangle with side 10 cm10\text{ cm}. (Use 3=1.732\sqrt{3} = 1.732)

Solution:

  1. Side a=10 cma = 10\text{ cm}.
  2. Use the area formula for an equilateral triangle: Area=34a2\text{Area} = \frac{\sqrt{3}}{4}a^2 Area=34×(10)2\text{Area} = \frac{\sqrt{3}}{4} \times (10)^2 Area=34×100=253\text{Area} = \frac{\sqrt{3}}{4} \times 100 = 25\sqrt{3}
  3. Substitute the value of 3\sqrt{3}: Area=25×1.732=43.3 cm2\text{Area} = 25 \times 1.732 = 43.3\text{ cm}^2

Explanation:

For equilateral triangles, using the specific formula 34a2\frac{\sqrt{3}}{4}a^2 is faster than the general Heron's formula.

Problem 4:

An isosceles triangle has a perimeter of 32 cm32\text{ cm} and its base is 12 cm12\text{ cm}. Find the area of the triangle.

Isosceles triangle with sides 10, 10, 12

Solution:

  1. Let the equal sides be aa and base be b=12 cmb = 12\text{ cm}.
  2. Perimeter =2a+b=32  ⟹  2a+12=32  ⟹  2a=20  ⟹  a=10 cm= 2a + b = 32 \implies 2a + 12 = 32 \implies 2a = 20 \implies a = 10\text{ cm}.
  3. The semi-perimeter s=10+10+122=322=16 cms = \frac{10 + 10 + 12}{2} = \frac{32}{2} = 16\text{ cm}.
  4. Using Heron's Formula: Area=s(s−a)(s−a)(s−b)\text{Area} = \sqrt{s(s-a)(s-a)(s-b)} Area=16(16−10)(16−10)(16−12)\text{Area} = \sqrt{16(16-10)(16-10)(16-12)} Area=16×6×6×4\text{Area} = \sqrt{16 \times 6 \times 6 \times 4} Area=4×6×2=48 cm2\text{Area} = 4 \times 6 \times 2 = 48\text{ cm}^2.

Explanation:

First find the length of the equal sides using the perimeter. Then, apply Heron's formula using the three known sides (10,10,1210, 10, 12).

Problem 5:

The sides of a triangle are 5 cm5\text{ cm}, 12 cm12\text{ cm}, and 13 cm13\text{ cm}. Calculate its area using the semi-perimeter method.

Triangle with sides 5, 12, 13

Solution:

  1. Sides are a=5,b=12,c=13a = 5, b = 12, c = 13.
  2. s=5+12+132=302=15 cms = \frac{5 + 12 + 13}{2} = \frac{30}{2} = 15\text{ cm}.
  3. Area=s(s−a)(s−b)(s−c)\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} Area=15(15−5)(15−12)(15−13)\text{Area} = \sqrt{15(15-5)(15-12)(15-13)} Area=15×10×3×2\text{Area} = \sqrt{15 \times 10 \times 3 \times 2} Area=900=30 cm2\text{Area} = \sqrt{900} = 30\text{ cm}^2.

Explanation:

This is a right-angled triangle (52+122=1325^2 + 12^2 = 13^2), but Heron's formula works for any triangle. We calculate the semi-perimeter and substitute the values into the formula.