Respuesta :

Answer:

P = 43.3 units

Step-by-step explanation:

the area (A) of the triangle is calculated as

A = [tex]\frac{1}{2}[/tex] bh ( b is the base and h the height )

given A = 56 units²

with b = [tex]\frac{1}{3}[/tex] h

then

[tex]\frac{1}{2}[/tex] bh = 56 ( multiply both sides by 2 to clear the fraction )

bh = 112 , so

[tex]\frac{1}{3}[/tex] h × h = 112 ( multiply both sides by 3 to clear the fraction )

h² = 336 ( take square root of both sides )

h = [tex]\sqrt{336}[/tex] ≈ 18.3

then b = [tex]\frac{1}{3}[/tex] h = [tex]\frac{1}{3}[/tex] × 18.3 = 6.1

the height h from the vertex to the base bisects the base ( altitude ), that is

[tex]\frac{1}{2}[/tex] b = [tex]\frac{1}{2}[/tex] × 6.1 = 3.05

this divide the isosceles triangle into 2 right triangles

let the leg of the triangle be x

then using Pythagoras' identity in the right triangle formed

x² = 18.3² + 3.05² = 334.89 + 9.3 = 334.19 ( take square root of both sides )

x = [tex]\sqrt{334.19}[/tex] ≈ 18.6

Then perimeter (P) is

P = 2x + b = 2(18.6) + 6.1 = 37.2 + 6.1 = 43.3 units

semsee45

Answer:

43.3 units

Step-by-step explanation:

Base and height of the triangle

[tex]\text{Area of a triangle}=\rm \dfrac{1}{2}bh[/tex]

where:

  • b is the base
  • h is the height

Given:

  • Area = 56 units²
  • [tex]\rm b=\dfrac{1}{3}h[/tex]

Substitute the given values into the formula and solve for h:

[tex]\begin{aligned}\text{Area of a triangle} & = \rm \dfrac{1}{2}bh\\\\\implies \rm 56 & = \rm \dfrac{1}{2} \cdot \dfrac{1}{3}h \cdot h\\\\\rm 336 & = \rm h^2\\\\\rm h & = \rm 4\sqrt{21}\:units \end{aligned}[/tex]

Substitute the found value of h into [tex]\rm b=\dfrac{1}{3}h[/tex] and solve for b:

[tex]\implies \rm b=\dfrac{1}{3}\cdot 4\sqrt{21}=\dfrac{4\sqrt{21}}{3}\:units[/tex]

Length of one side of the isosceles triangle

[tex]\begin{aligned}\rm side\:length & =\rm \sqrt{\left(\dfrac{b}{2}\right)^2+h^2}\\\\\ \implies \text{side length}& =\rm \sqrt{\left(\dfrac{4\sqrt{21}}{6}\right)^2+(4\sqrt21})^2}\\\\& =\rm \sqrt{\dfrac{28}{3}+336}\\\\& =\rm \sqrt{\dfrac{1036}{3}}\\\\& = \rm \dfrac{2\sqrt{777}}{3}\:units\end{aligned}[/tex]

Total Perimeter

[tex]\begin{aligned}\rm Total\:perimeter & = \rm base + 2\:sides\\\\\implies \text{Total perimeter}& = \rm \dfrac{4\sqrt{21}}{3}+2 \cdot \dfrac{2\sqrt{777}}{3}\\\\& = \rm 43.2763939...\\\\& = \rm 43.3\:units\:(nearest\:tenth)\end{aligned}[/tex]

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