In an isosceles triangle with legs that are 1 unit long, the angles are 45 degrees, 67.5 degrees and 67.5 degrees What is its area?

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In an isosceles triangle with legs that are 1 unit long, the angles are 45 degrees, 67.5 degrees and 67.5 degrees What is its area?

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Answer 1 · 2015-11-28T20:15:52

approximately $0.35$ $0.35$ square units

Explanation:

To find the area, we first need to find the height of the triangle, since the formula for area of a triangle is :

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

First, we divide the isosceles triangle into $2$ $2$ right triangles.

![http://study.com/academy/lesson/http://what-is-an-isosceles-triangle-definition-properties-theorem.html](https://useruploads.socratic.org/pi3GFDHTpyD6Ace9h01Q_illustrationisoscelestriangletheorem5.png)

Since we know that all right triangles have one $90^{\circ}$ $90^@$ angle and that all triangles have a $180^{\circ}$ $180^@$ sum of interior angles, then $∠ C A D$ $/_CAD$ must be:

$∠ C A D = 180^{\circ} - 90^{\circ} - {67.5}^{\circ}$ $/_CAD=180^@-90^@-67.5^@$
$∠ C A D = {22.5}^{\circ}$ $/_CAD=22.5^@$

Using the Law of Sines, we can calculate the height of the right triangle:

$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$ $a/sinA=b/sinB=c/sinC$

$\frac{1}{{sin 90}^{\circ}} = \frac{b}{{sin 67.5}^{\circ}}$ $1/(sin90^@)=b/sin67.5^@$

$b \cdot {sin 90}^{\circ} = 1 \cdot sin 67.5$ $b*sin90^@=1*sin67.5$

$b \cdot 1 = 0.92$ $b*1=0.92$

$b = 0.92$ $b=0.92$

Since we do not yet know the base length of the right triangle, we can also use the Law of Sines to find the base:

$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$ $a/sinA=b/sinB=c/sinC$

$\frac{1}{{sin 90}^{\circ}} = \frac{c}{{sin 22.5}^{\circ}}$ $1/(sin90^@)=c/sin22.5^@$

$c \cdot {sin 90}^{\circ} = 1 \cdot sin 22.5$ $c*sin90^@=1*sin22.5$

$c \cdot 1 = 0.38$ $c*1=0.38$

$c = 0.38$ $c=0.38$

To find the base of the whole triangle, multiply the right triangle's base length by $2$ $2$ :

$c = 0.38 \cdot 2$ $c=0.38*2$
$c = 0.76$ $c=0.76$

Now that we have the base length and the height of the whole triangle, we can substitute these values into the formula for area of a triangle:

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

$A r e a_{triangle} = \frac{(0.76) \cdot (0.92)}{2}$ $Area_"triangle"=((0.76)*(0.92))/2$

$A r e a_{triangle} = \frac{0.7}{2}$ $Area_"triangle"=0.7/2$

$A r e a_{triangle} \approx 0.35$ $Area_"triangle"~~0.35$

$:.$ , the area of the triangle is approximately $0.35$ square units.

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In an isosceles triangle with legs that are 1 unit long, the angles are 45 degrees, 67.5 degrees and 67.5 degrees What is its area?

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