Question

an isosceles triangle has perimeter 30cm and each of the equal sides is 20 cm find the area of the triangle solution

Answer 1

Equal sides of Triangle = 20 cm
Perimeter of Triangle = 30 cm
Now, a = 20 cm, b = 20 cm, c = ?
As we know that Perimeter of Triangle = sum of its sides
so,
30 = a+b+c
30 = 20+20+c
30 = 40 + c
or, 40 + c = 30
c = 30 - 40
c = -10
so sides of Triangle are
a = 20 cm
b = 20 cm
c = -10 cm
Now Area of Triangle = √s(s-a)(s-b)(s-c)
s = Perimeter/2
s = 30/2
s = 15 cm
Area = √15(15-20)(15-20)(15-{-10})
√15×(-5)×(-5)×(25)
√9375
25√15 cm^2 = Area of Triangle
or 96.82458 cm^2 = Area of Triangle
Hope it will help you

Answer 2

$\underline\mathfrak{Given:-}$

• Perimeter of an Isosceles triangle 30 cm.
• Two equal side is 20 cm.

$\large\underline\mathfrak{To \: find:-}$

• Area of triangle ....?

$\large\underline\mathfrak{Solutions:-}$

$\: \: \: \: \: \therefore \: \: \: Perimeter \: \: of \: \: an \: \: Isosceles \: \: Triangle \: \: = \: \: {2} \: \times \: a \: + \: b$

$\: \: \: \: \: \leadsto \: \: {30} \: \: = \: \: {2} \: \times \: {20} \: + \: b$

$\: \: \: \: \: \leadsto \: \: {30} \: \: = \: \: {40} \: + \: b$

$\: \: \: \: \: \leadsto \: \: {b} \: \: = \: \: {30} \: - \: {40}$

$\: \: \: \: \: \leadsto \: \: {b} \: \: = \: \: {-10} \: cm$

$\: \: \: \: \: \therefore \: \: Area \: \: of \: \: triangle \: \: by \: \: heroes's \: \: formula:-$

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

• $\: \: \: \: \: \leadsto \: \: S \: \: = \: \: \frac{a \: + \: b \: + \: c}{2}$

• a ⇢ 20cm

• b ⇢ 20cm

• c ⇢ -10cm

$\: \: \: \: \: \leadsto \: \: S \: \: = \: \: \frac{{20} \: + \: {20} \: + \: {(-10)}}{2}$

$\: \: \: \: \: \leadsto \: \: S \: \: = \: \: \frac{30}{2}$

$\: \: \: \: \: \leadsto \: \: S \: \: = \: \: {15}$

$\: \: \: \: \: \therefore \: \: \: Area \: \: of \: \: triangular \: \: field:-$

$\: \: \: \: \: \leadsto \: \: \sqrt{{15} \: ({15} \: - \: {20}) \: ({15} \: - \: {20}) \: ({15} \: - \: {(-10)})}$

$\: \: \: \: \: \leadsto \: \: \sqrt{{15} \: \times \: {(-5)} \: \times \: {(-5)} \: \times \: {(25)}}$

$\: \: \: \: \: \leadsto \: \: \sqrt{9375}$

$\: \: \: \: \: \leadsto \: \: {96.82} \: {cm}^{2}$

$\: \: \: \: \: Hence, \\ \: \: \: Area \: \: of \: \: triangle \: \: \leadsto \: \: {96.82} \: {cm}^{2}$

$\large\underline\mathfrak{More \: Information:-}$

• $\: \: \: \: \: Perimeter \: \: of \: \: an \: \: Isosceles \: \: Triangle \: \: = \: \: {2} \: \times \: a \: + \: b.$

• $\: \: \: \: \: Area \: \: of \: \: an \: \: isosceles \: \: triangle \: \: = \: \: \frac{1}{2} \: b \: \times \: h.$

• $\: \: \: \: \: The \: \: altitude \: \: of \: \: an \: \: isosceles \: \: triangle \: \: = \: \: \sqrt{{a}^{2} \: - \: \frac{{b}^{2}}{4}}.$

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