What is the area of an isosceles triangle with two equal sides of 10 cm and a base of 12 cm?

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What is the area of an isosceles triangle with two equal sides of 10 cm and a base of 12 cm?

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Answer 1 · 2015-11-21T01:04:33

Area $= 48$ $=48$ $c m^{2}$ $cm^2$

Explanation:

Since an isosceles triangle has two equal sides, if the triangle is split in half vertically, the length of the base on each side is:

$12$ $12$ $c m$ $cm$ $\div 2 =$ $-:2 =$ $6$ $6$ $c m$ $cm$

We can then use the Pythagorean theorem to find the height of the triangle.

The formula for the Pythagorean theorem is:

$a^{2} + b^{2} = c^{2}$ $a^2+b^2=c^2$

To solve for the height, substitute your known values into the equation and solve for $a$ $a$ :

where:
$a$ $a$ = height
$b$ $b$ = base
$c$ $c$ = hypotenuse

$a^{2} + b^{2} = c^{2}$ $a^2+b^2=c^2$
$a^{2} = c^{2} - b^{2}$ $a^2=c^2-b^2$
$a^{2} = {(10)}^{2} - {(6)}^{2}$ $a^2=(10)^2-(6)^2$
$a^{2} = (100) - (36)$ $a^2=(100)-(36)$
$a^{2} = 64$ $a^2=64$
$a = \sqrt{64}$ $a=sqrt(64)$
$a = 8$ $a=8$

Now that we have our known values, substitute the following into the formula for area of a triangle:

$b a s e = 12$ $base = 12$ $c m$ $cm$
$h e i g h t = 8$ $height = 8$ $c m$ $cm$

$A r e a = \frac{b a s e \cdot h e i g h t}{2}$ $Area=(base*height)/2$

$A r e a = \frac{(12) \cdot (8)}{2}$ $Area=((12)*(8))/2$

$A r e a = \frac{96}{2}$ $Area=(96)/(2)$

$A r e a = 48$ $Area=48$

$:.$ , the area is $48$ $cm^2$ .

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What is the area of an isosceles triangle with two equal sides of 10 cm and a base of 12 cm?

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