bythehill6 bythehill6
  • 03-03-2020
  • Mathematics
contestada

Find the area of the isosceles triangle with legs 12 and base 17. Simplify all radicals

a2+b2=c2

Respuesta :

kudzordzifrancis
kudzordzifrancis kudzordzifrancis
  • 11-03-2020

Answer:

[tex]Area= \frac{17 \sqrt{865} }{4} \: {units}^{2} [/tex]

Step-by-step explanation:

From Pythagoras theorem,

[tex] {h}^{2} = {12}^{2} + { (\frac{17}{2} )}^{2} [/tex]

[tex]{h}^{2} = \frac{865}{4} [/tex]

[tex]h = \frac{ \sqrt{865} }{2} [/tex]

The area of the isosceles triangle is

[tex]Area= \frac{1}{2} bh[/tex]

[tex]Area= \frac{1}{2} \times 17 \times \frac{ \sqrt{865} }{2} = \frac{17 \sqrt{865} }{4} [/tex]

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