Question

Using Heron's Formula, find the area of an isosceles
triangle, the measure of one of its equal sides being
a units and the third side 2b
units.​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

ANSWER

Area of the isosceles triangle=

Area of the isosceles triangle= s(s−a)(s−b)(s−c)

Area of the isosceles triangle= s(s−a)(s−b)(s−c)

Area of the isosceles triangle= s(s−a)(s−b)(s−c)

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s=

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c =

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c = 2

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c = 2a+a+2b

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c = 2a+a+2b

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c = 2a+a+2b =a+b

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c = 2a+a+2b =a+b arΔ=

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c = 2a+a+2b =a+b arΔ= (a+b)(a+b−a)(a+b−a)(a+b−2b)

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c = 2a+a+2b =a+b arΔ= (a+b)(a+b−a)(a+b−a)(a+b−2b)

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c = 2a+a+2b =a+b arΔ= (a+b)(a+b−a)(a+b−a)(a+b−2b) =

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c = 2a+a+2b =a+b arΔ= (a+b)(a+b−a)(a+b−a)(a+b−2b) = b

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c = 2a+a+2b =a+b arΔ= (a+b)(a+b−a)(a+b−a)(a+b−2b) = b 2

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c = 2a+a+2b =a+b arΔ= (a+b)(a+b−a)(a+b−a)(a+b−2b) = b 2 (a+b)(a−b)

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c = 2a+a+2b =a+b arΔ= (a+b)(a+b−a)(a+b−a)(a+b−2b) = b 2 (a+b)(a−b)

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c = 2a+a+2b =a+b arΔ= (a+b)(a+b−a)(a+b−a)(a+b−2b) = b 2 (a+b)(a−b)

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c = 2a+a+2b =a+b arΔ= (a+b)(a+b−a)(a+b−a)(a+b−2b) = b 2 (a+b)(a−b) =b

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c = 2a+a+2b =a+b arΔ= (a+b)(a+b−a)(a+b−a)(a+b−2b) = b 2 (a+b)(a−b) =b a

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c = 2a+a+2b =a+b arΔ= (a+b)(a+b−a)(a+b−a)(a+b−2b) = b 2 (a+b)(a−b) =b a 2

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c = 2a+a+2b =a+b arΔ= (a+b)(a+b−a)(a+b−a)(a+b−2b) = b 2 (a+b)(a−b) =b a 2 −b

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c = 2a+a+2b =a+b arΔ= (a+b)(a+b−a)(a+b−a)(a+b−2b) = b 2 (a+b)(a−b) =b a 2 −b 2

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c = 2a+a+2b =a+b arΔ= (a+b)(a+b−a)(a+b−a)(a+b−2b) = b 2 (a+b)(a−b) =b a 2 −b 2

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c = 2a+a+2b =a+b arΔ= (a+b)(a+b−a)(a+b−a)(a+b−2b) = b 2 (a+b)(a−b) =b a 2 −b 2

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c = 2a+a+2b =a+b arΔ= (a+b)(a+b−a)(a+b−a)(a+b−2b) = b 2 (a+b)(a−b) =b a 2 −b 2 sq.units

Area of the isosceles triangle= s(s−a)(s−b)(s−c) s= 2a+b+c = 2a+a+2b =a+b arΔ= (a+b)(a+b−a)(a+b−a)(a+b−2b) = b 2 (a+b)(a−b) =b a 2 −b 2 sq.unitsAnswered By

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