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ACT Math : How to find the perimeter of a 45/45/90 right isosceles triangle

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Example Question #2 : 45/45/90 Right Isosceles Triangles

What is the perimeter of an isosceles right triangle with an hypotenuse of length $20\:cm$ ?

Possible Answers:

$10\sqrt{2} + 20\:cm$

$40\sqrt{2}\:cm$

$60\:cm$

$80\sqrt{2}\:cm$

$20\sqrt{2}+20\:cm$

Correct answer:

$20\sqrt{2}+20\:cm$

Explanation:

Your right triangle is a 45-45-90 triangle. It thus looks like this:

_tri41

Now, you know that you also have a reference triangle for 45-45-90 triangles. This is:

Triangle454590

This means that you can set up a ratio to find . It would be:

$\frac{x}{1}=\frac{20}{\sqrt{2}}$

Your triangle thus could be drawn like this:

_tri42

Now, notice that you can rationalize the denominator of $\frac{20}{\sqrt{2}}$ :

$\frac{20}{\sqrt{2}} = \frac{20}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}}=\frac{20\sqrt{2}}{2}=10\sqrt{2}$

Thus, the perimeter of your figure is:

$10\sqrt{2} + 10\sqrt{2} + 20 = 20\sqrt{2}+20\:cm$

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Example Question #2 : How To Find The Perimeter Of A 45/45/90 Right Isosceles Triangle

What is the perimeter of an isosceles right triangle with an area of $72x^2\:in^2$ ?

Possible Answers:

$36x^2\sqrt{2}\:in$

$36x\sqrt{2}\:in$

$144x^2\:in$

$144x^2 + 12\sqrt{2}\:in$

$24x+12x\sqrt{2}\:in$

Correct answer:

$24x+12x\sqrt{2}\:in$

Explanation:

Recall that an isosceles right triangle is also a 45-45-90 triangle. Your reference figure for such a shape is:

Triangle454590 or _tri51

Now, you know that the area of a triangle is:

$A=\frac{1}{2}bh$

For this triangle, though, the base and height are the same. So it is:

$A=\frac{1}{2}x^2$

Now, we have to be careful, given that our area contains . Let's use , for "side length":

$72x^2=\frac{1}{2}s^2$

s^2=144x^2

Thus, $s=12x\:in$ . Now based on the reference figure above, you can easily see that your triangle is:

_tri71

Therefore, your perimeter is:

$12x+12x+12x\sqrt{2} = 24x+12x\sqrt{2}\:in$

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Example Question #202 : Plane Geometry

A tree is feet tall and is planted in the center of a circular bed with a radius of feet. If you want to stabalize the tree with ropes going from its midpoint to the border of the bed, how long will each rope measure?

Possible Answers:

24.5 ft

15 ft

13 ft

19 ft

12 ft

Correct answer:

13 ft

Explanation:

This is a right triangle where the rope is the hypotenuse. One leg is the radius of the circle, 5 feet. The other leg is half of the tree's height, 12 feet. We can now use the Pythagorean Theorem a^2+b^2=c^2 giving us 5^2+12^2=169 . If c^2=169 then $c=\sqrt{169}=13$ .

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Example Question #141 : Triangles

An isosceles right triangle has a hypotenuse of length . What is the perimeter of this triangle, in terms of ?

Possible Answers:

$x+ x\sqrt 2$

$\frac{4}{3}x$

$\frac{2}{3}x$

$x+ x\sqrt3$

Correct answer:

$x+ x\sqrt 2$

Explanation:

The ratio of sides to hypotenuse of an isosceles right triangle is always $x: x: x\sqrt2$ . With this in mind, setting as our hypotenuse means we must have leg lengths equal to:

$\frac{x}{\sqrt2} = \frac{x\sqrt2}{2}$

Since the perimeter has two of these legs, we just need to multiply this by and add the result to our hypothesis:

$P\Delta = x + 2\left(\frac{x\sqrt2}{2}\right) = x+ x\sqrt2$

So, our perimeter in terms of is:

$x+ x\sqrt 2$

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ACT Math : How to find the perimeter of a 45/45/90 right isosceles triangle

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #2 : 45/45/90 Right Isosceles Triangles

Example Question #2 : How To Find The Perimeter Of A 45/45/90 Right Isosceles Triangle

Example Question #202 : Plane Geometry

Example Question #141 : Triangles

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