## suppose you have an isosceles triangle

# 2 Answers

^{o}

Find: A

_{ΔABC }

**= (1/2)AB*BC*sin45**

*A*_{ΔABC}^{o}= (1/2)*[(√2)/2] =

**≈ 0.3535534 ft**

*(√2)/4 ft*^{2}^{2}

in general:

**A = (1/2) r**^{2}sin(t^{o}) unit^{2}Dalia,

For the first triangle using the fact that it is an isosceles with two equal sides of 1ft forming an angle of 45°, you know first off that this cannot be a right triangle. The equal sides must form a 90° angle for an isosceles right triangle.

Step 1: For me I now know that I draw the triangle (for visual clarity) with the known angle on top with both sides going down and away from the original point, then connect at the bottom.

Step 2: You also know that the remaining angles added together = 135° and must be equal ... so they both equal 67.5° each. (the picture now shows you have all 3 angles and 2 sides label the triangle using capitals A, B, & C for the angles starting at the top
and going counter clockwise and label opposite sides lower case a, b, & c).

Step 3: Bisect the triangle from the top point to the base creating 2 right triangles, call that side d. Each of these triangles are exactly the same so when we find the area of 1 we will know the total area is 2 times.

step 4: Get ready to use some trigonometry angle B = 67.5° and side c = 1, using this we can calculate the other 2 sides which will give us the area.

Step 5: Sin B = O/H ... Sin 67.5° = d/1 ... Sin 67.5° = d = .92388 ft

Step 6: Cos B = A/H ... Cos 67.5° = .5d/1 ... Cos 67.5° = .5d (divide both sides by .5) ... (Cos 67.5°)/.5 = d = .382683/.5 = .765367 ft

Step 7: A = (b x h) / 2 = (.92388 ft x .765367 ft) / 2 = .707107 sqft / 2 = .353553 sqft

Step 8: Multiply the above by 2 to get your total area as mentioned above (or just use the number before it is divided by 2 = .707107 sqft

I know that was a lot and there are likely other ways to do this but this was the easiest way I know. For the other triangle if you place the letters as needed and use this you should be able to follow through to get that the area of the triangle is r2 x {(Sin(180°
- t) / 2) x (Cos(180° - t) / 2)}

I hope this all helps you Dalia.