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Units And Dimensions
UNITS & DIMENSIONS - AREAS
Area Of Triangle
* If a, b, c are the sides of the triangle, then the perimeter of triangle = (a + b + c) units.
* Area of the triangle =\( \sqrt {(s - a)(s - b)(s - c)} \)
The semi-perimeter of the triangle, s = \( \frac{{(a + b + c)}} {2} \)
* In a triangle if ‘b’ is the base and h is the height of the triangle then
Area of triangle = \( \frac{1} {2} \)× base × height
Similarly,
\( \frac{1} {2} \)× AC × BD \( \frac{1} {2} \)× BC × AD
* Base of the triangle = (2 Area)/height
* Height of the triangle = (2 Area)/base
Area of right angled triangle
* If a represents the side of an equilateral triangle, then its area =\( \left( {\sqrt {3a^2 } } \right)/4 \)
* Area of right angled triangle
A = \( \frac{1} {2} \)× BC × AB
= \( \frac{1} {2} \)× b × h
SOLVED EXAMPLES
1. Find the area and height of an equilateral triangle of side 12 cm. (\( \sqrt 3 \)= 1.73).
Solution:
Area of the triangle =\( \frac{{\sqrt 3 }} {4}a^2 \) square units
=\( \frac{{\sqrt 3 }} {4} \times 12 \times 12 \)
= 36\( \sqrt 3 \) cm²
= 36 × 1.732 cm²
= 62.28 cm²
Height of the triangle = \( \frac{{\sqrt 3 }} {2} \) a units
= \( \frac{{\sqrt 3 }} {2} \) × 12 cm
= 1.73 × 6 cm
= 10.38 cm
2. Find area of right angled triangle whose hypotenuse is 15 cm & one of the sides is 12 cm.
Solution:
AB² = AC² - BC² = 15² - 12² = 225 – 144= 81
Therefore, AB = 9 cm
Therefore, area of the triangle = \( \frac{1} {2} \)× base × height= × 12 × 9 = 54 cm²
3. The base and height of the triangle are in the ratio 3 : 2. If the area of the triangle is 243 cm² find the base and height of the triangle.
Solution:
Let the common ratio be x
Then height of triangle = 2x
And the base of triangle = 3x
Area of triangle = 243 cm²
Area of triangle = 1/2 × b × h =243
243 = 1/2 × 3x × 2x
3x² = 243
x² = 243/3
x = \( \sqrt {81} \)=\( \sqrt {9 \times 9} \)= 9
Therefore, height of triangle = 2 × 9 = 18 cm
Base of triangle = 3x = 3 × 9 = 27 cm