Question

find area of triangle having sides measuring 5 cm, 6 cm and 9 cm using heron's formula?​

Answer 1

Answer: 10√2 cm.

Explanation:

a, b, c = 5 cm, 6 cm, 9 cm.

∴ s = 1/2(a + b + c) = 1/2(5 + 6 + 9) = 1/2(20) = 10 cm

Heron's formula : $\sqrt{s(s - a)(s - b)(s - c)}$

∴ Area = √10(10 - 5)(10 - 6)(10 - 9)

=> Area = √10·5·4

=> Area = 10√2 cm² = 14.14 cm² approx.

More:

• Area of an equilateral triangle with side s = $\dfrac{\sqrt{3} s^2}{4}$
• Area of an isoscleles triangle with equal sides a and unequal side b = $\dfrac{b}{2} \sqrt{a^2 - \dfrac{b^2}{4}}$
• Area of a ∆ with given its height & base = $\dfrac{1}{2} bh$
• Area of a ∆ given its one angle C (corresponding to the side which is not given) and two sides = $\dfrac{1}{2} ab \sin C$
• Area of a ∆ given its coordinates = $\dfrac{1}{2} [x_1(y_2 - y_3) + x_2(y_1 - y_3) + x_3(y_1 - y_2)]$

Derivation:-

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Answer 2

this photo is the correct answer OK