How do you find the area of an isosceles triangle if the two equal sides are 10cm and the base is 12cm?

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How do you find the area of an isosceles triangle if the two equal sides are 10cm and the base is 12cm?

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Answer 1 · 2015-07-22T05:22:41

I found: $48"cm"^2$

Explanation:

Considering:
enter image source here
applying Pythagoras on half triangle you get:
$h^2+6^2=10^2$
$h=8cm$
So $Area=(basexxheight)/2=(12xx8)/2=48"cm"^2$

Answer 2 · 2015-07-22T10:26:20

Area $= 48$ sq. cm.
$color(white)("XXXX")$ $color(white)("XXXX")$ (Using Heron's formula)

Explanation:

As an alternate solution method:

Heron's Formula for the area of a triangle with sides $a, b, c$
$color(white)("XXXX")$ $A = sqrt(s(s-a)(s-b)(s-c))$
$color(white)("XXXX")$ $color(white)("XXXX")$ where $s$ is the semi-perimeter (i.e. $s= (a+b+c)/2$

In this case:
$color(white)("XXXX")$ $s = 16$

$color(white)("XXXX")$ $A = sqrt(16(6)(6)(4))$
$color(white)("XXXX")$ $color(white)("XXXX")$ $= sqrt(4^2*6^2*2^2)$
$color(white)("XXXX")$ $color(white)("XXXX")$ $=48$

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How do you find the area of an isosceles triangle if the two equal sides are 10cm and the base is 12cm?

2 Answers

Explanation:

Explanation:

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