Geometry Worksheets | Similarity Worksheets - Free Printable

To solve the problems involving similar right triangles, we will use the properties of similar triangles and the Pythagorean theorem where necessary. Let's go through each problem step by step.

Problem 1:

The triangle is an isosceles right triangle with legs of length \( x \) and hypotenuse of length 49.

Using the Pythagorean theorem:
\[
x^2 + x^2 = 49^2
\]
\[
2x^2 = 2401
\]
\[
x^2 = 1200.5
\]
\[
x = \sqrt{1200.5}
\]

However, this approach seems incorrect because the problem likely involves simpler radical forms. Let's re-evaluate using the properties of isosceles right triangles:
\[
x = \frac{49}{\sqrt{2}} = \frac{49\sqrt{2}}{2}
\]

So, the answer is:
\[
x = \frac{49\sqrt{2}}{2}
\]

Problem 2:

The triangle has one leg of length 81 and the other leg of length 60. We need to find the hypotenuse \( x \).

Using the Pythagorean theorem:
\[
x^2 = 81^2 + 60^2
\]
\[
x^2 = 6561 + 3600
\]
\[
x^2 = 10161
\]
\[
x = \sqrt{10161}
\]

This can be simplified further if possible, but it appears to be in simplest radical form:
\[
x = \sqrt{10161}
\]

Problem 3:

The triangle has one leg of length 16 and the hypotenuse of length 25. We need to find the other leg \( x \).

Using the Pythagorean theorem:
\[
25^2 = 16^2 + x^2
\]
\[
625 = 256 + x^2
\]
\[
x^2 = 369
\]
\[
x = \sqrt{369}
\]

This can be simplified further:
\[
x = \sqrt{9 \times 41} = 3\sqrt{41}
\]

So, the answer is:
\[
x = 3\sqrt{41}
\]

Problem 4:

The triangle has one leg of length 9 and the hypotenuse of length 64. We need to find the other leg \( x \).

Using the Pythagorean theorem:
\[
64^2 = 9^2 + x^2
\]
\[
4096 = 81 + x^2
\]
\[
x^2 = 4015
\]
\[
x = \sqrt{4015}
\]

This is already in simplest radical form:
\[
x = \sqrt{4015}
\]

Problem 5:

The triangle has one leg of length 36 and the hypotenuse of length 40. We need to find the other leg \( x \).

Using the Pythagorean theorem:
\[
40^2 = 36^2 + x^2
\]
\[
1600 = 1296 + x^2
\]
\[
x^2 = 304
\]
\[
x = \sqrt{304}
\]

This can be simplified further:
\[
x = \sqrt{16 \times 19} = 4\sqrt{19}
\]

So, the answer is:
\[
x = 4\sqrt{19}
\]

Problem 6:

The triangle has one leg of length 4 and the other leg of length 12. We need to find the hypotenuse \( x \).

Using the Pythagorean theorem:
\[
x^2 = 4^2 + 12^2
\]
\[
x^2 = 16 + 144
\]
\[
x^2 = 160
\]
\[
x = \sqrt{160}
\]

This can be simplified further:
\[
x = \sqrt{16 \times 10} = 4\sqrt{10}
\]

So, the answer is:
\[
x = 4\sqrt{10}
\]

Problem 7:

The triangle has one leg of length 64 and the hypotenuse of length 73. We need to find the other leg \( x \).

Using the Pythagorean theorem:
\[
73^2 = 64^2 + x^2
\]
\[
5329 = 4096 + x^2
\]
\[
x^2 = 1233
\]
\[
x = \sqrt{1233}
\]

This is already in simplest radical form:
\[
x = \sqrt{1233}
\]

Problem 8:

The triangle has one leg of length 25 and the hypotenuse of length 41. We need to find the other leg \( x \).

Using the Pythagorean theorem:
\[
41^2 = 25^2 + x^2
\]
\[
1681 = 625 + x^2
\]
\[
x^2 = 1056
\]
\[
x = \sqrt{1056}
\]

This can be simplified further:
\[
x = \sqrt{16 \times 66} = 4\sqrt{66}
\]

So, the answer is:
\[
x = 4\sqrt{66}
\]

Final Answers:

\[
\boxed{\frac{49\sqrt{2}}{2}, \sqrt{10161}, 3\sqrt{41}, \sqrt{4015}, 4\sqrt{19}, 4\sqrt{10}, \sqrt{1233}, 4\sqrt{66}}
\]