Savvas Learning Co Geometry Student Edition Chapter 8 Right Triangles And Trigonometry Exercise 8.1 The Pythagorean Theorem And Its Converse

Savvas Learning Co Geometry Student Edition Chapter 8 Right Triangles And Trigonometry Exercise 8.1 The Pythagorean Theorem And Its Converse

 

Page 495 Exercise 1 Problem 1

Given:

To find – The value of the variable.

Using the Pythagorean theorem find the value of the given variable.

We have

x² = 35² + 12²

x² = 1225 + 144

x²  = 1369

x = \(\sqrt{1369}\)

x = 37

The value is x = 37.

 

Page 495  Exercise 2  Problem 2

Given:

To find – The value of the variable.

Using the Pythagorean theorem find the value of the given variable.

We have

x² = 9² + 7²

x² = 81 + 49

x = \(\sqrt{130}\)

x ≈ 11.40

The value is x ≈ 11.40

 

Page 495  Exercise 3  Problem 3

Given:

To find – The value of the variable.

Using the Pythagorean theorem find the value of the given variable.

We have

x² =  5² − 3²

x²  = 25 − 9

x = \(\sqrt{16}\)

x = 4

The value is x = 4.

 

Page 495 Exercise 4 Problem 4

Given:

To find – The value of the variable.

Using the Pythagorean theorem find the value of the given variable.

We have

x² = 13² − 11²

x² = 169 − 121

x = \(\sqrt{48}\)

x = 4\(\sqrt{3}\).

The value is  x = 4\(\sqrt{3}\).

 

Page 495  Exercise 5  Problem 5

Given: Describe the conditions that a set of three numbers must meet in order to form a Pythagorean triple.

To find –  The condition for a Pythagorean triple.

The condition that must be met for three numbers to form a Pythagorean triple is that the sum of the squares of the two numbers should be equal to the square of the third number.

The condition that must be met for three numbers to form a Pythagorean triple is that the sum of the squares of the two numbers should be equal to the square of the third number.

 

Page 495  Exercise 6  Problem 6

Given:

To find – Describe the error.

Friend made a mistake in choosing the hypotenuse. Instead of 30,34 should have been chosen for the hypotenuse.

Choose  34  as hypotenuse, and 16,30 for legs.

We have

16² + 30² = 34²

256 + 900 = 1156

1156 = 1156

The friend made a mistake in choosing the hypotenuse. Instead of 30,34 should have been chosen for the hypotenuse. We see that the sum of the squares of the lengths of the legs is equal to the square of the hypotenuse. So the triangle is a right-angled triangle.

 

Page 495  Exercise 7  Problem 7

Given:

To find – The value of the variable.

Using the Pythagorean theorem find the value of the given variable.

We have

x² = 16² + 30² 

x²  = 256 + 900

x² = 1156

x = \(\sqrt{1156}\)

x = 34

The value is x = 34.

 

Page 495  Exercise 8  Problem 8

Given: A right-angle triangle

To Find – The value of x

We have, a or leg₁ = 16 and b or leg₂ = 12

So by applying the Pythagorean theorem

​⇒   a² + b²  = x²

⇒   16² + 12² =  x²

⇒   256 + 144 = x²

⇒   400 = x²

​⇒  \(\sqrt{400}\)

=  \(\sqrt{x^2}\)  (Take positive square root)

⇒  x =  20

So the length of the hypotenuse or the value of x is 20.

The length of the hypotenuse (value of x) of a right-angle triangle with the other two sides being 16 and 12 is 20.

 

Page 495  Exercise 9  Problem 9

Given: A right-angle triangle

To Find – The value of x

We have, a or leg₁  = 65 and b or leg₂  = 72

So by applying the Pythagorean theorem

​⇒  a²  + b² =  x²

⇒  65² + 72² =  x²

⇒   4225 + 5184  =  x²

⇒  9409 = x²

⇒  x = 97    (Take positive square root)

So the value of x is 97.

The length of the hypotenuse (value of x) of a right-angle triangle with the other two sides being 65 and 72 is 97.

 

Page 495 Exercise 10 Problem 10

Given: A right-angle triangle

To Find – The value of x

We have, a or leg₁ = 15 and b or leg₂ =  8

So by applying the Pythagorean theorem

⇒  a²  + b² = x²

⇒  15² + 8² = x²

⇒  225 + 64 = x²

⇒  289 = x²

​⇒  x = 17 (Take positive square root)

So the value of x is 17.

The length of the hypotenuse (value of x) of a right-angle triangle with the other two sides being 15 and 8 is 17.

 

Page 495  Exercise 11  Problem 11

Given: A set of numbers- 4, 5, 6

To Find – Does the set of numbers form a Pythagorean triple

We have, a set of whole numbers- 4, 5, 6

Let, a = 4, b = 5, c = 6

Substituting the value of a, b, and c in a² + b² = c²

​⇒   4² + 5² = 6² 

⇒  16 + 25 = 36

⇒  41 ≠ 36
​
As L.H.S. is not equal to R.H.S., therefore, the sets 4, 5, 6 do not satisfy the equation a² + b²  = c²

So the set 4, 5, 6 is not a Pythagorean triple.

The set of numbers 4, 5, 6 is not a Pythagorean triple as it does not satisfy the equation a² + b²  = c².

 

Page 495  Exercise 12  Problem 12

Given: A set of numbers- 10, 24, 26.

To Find – Does the given set of numbers form a Pythagorean triple

We have, a set of numbers- 10, 24, 26

Let, a = 10,b = 24, c = 26

Substituting the values of a, b, c in the equation a² + b²  = c²

​⇒  a² + b²  = c² 

⇒  10²  + 24²  = 26²

⇒  100 + 576 = 676

⇒  676 = 676
​
As L.H.S. is equal to R.H.S. therefore, the set of numbers 10, 24, 26 satisfy the equation a² + b² = c²

Hence, the set of numbers 10, 24, 26 is a Pythagorean triple.

The set of numbers 10, 24, 26 is a Pythagorean triple because it satisfies the equation a² + b² = c²

 

Page 495  Exercise 13  Problem 13

Given: A set of numbers- 10, 24, 26.

To Find – Does the given set of numbers form a Pythagorean triple

We have, a set of numbers- 10, 24, 26

Let, a = 10, b = 24, c = 26

Substituting the values of a, b, and c in the equation a² + b² = c²

​⇒  a²  + b²  = c²

⇒  10² + 24² = 26²

⇒  100 + 576  =  676

⇒  676 = 676
​
As L.H.S. is equal to R.H.S., therefore, the set of numbers 10, 24, 26 satisfies the equation a² + b² = c²

Hence, the set of numbers 10, 24, 26 is a Pythagorean Triple.

The set of numbers 10, 24, 26 is a Pythagorean triple because it satisfies the equation a²+ b² = c².

 

Page 496  Exercise 14  Problem 14

Given: A set of numbers- 15, 20, 25

To Find – Does the given set of numbers form a Pythagorean triple

We have, a set of numbers- 15, 20, 25

Let, a = 15,b = 20,c = 25

Substituting the values of a, b, c in the equation a² + b² = c²

​⇒   a² + b²  = c²

⇒  15² + 20² =  25²

⇒  225 + 400 = 625

⇒  625 = 625
​
As L.H.S. is equal to R.H.S., therefore, the set of numbers 15, 20, 25 satisfies the equation a² + b² = c²

Hence, the set of numbers 15, 20, 25 is a Pythagorean Triple.

The set of numbers 15, 20, 25 is a Pythagorean Triple because it satisfies the equation  a² + b² = c²

 

Page 496  Exercise 15  Problem 15

Given: A right-angle triangle with two sides

To Find – The value of x

In the given right-angle triangle

Base (a) = x units

Height (b) = 4 units

Hypotenuse (c) = 6 units

(leg₁ )²+(leg₂)² = (Hypotenuse)²               (Pythagorean theorem)

Substitute the values of a, b, c in the Pythagorean Theorem

​⇒   a² + b² = c²

⇒   x ²+ 4²  = 6²

⇒    x² + 16 = 36

⇒   x² + 16 − 16 = 36 − 16

⇒     x² = 20

⇒    x = \(\sqrt{20}\)

⇒    x = \(\sqrt{4(5)}\)

⇒   x = 2\(\sqrt{5}\)

The value of x in the simplest radical form is 2\(\sqrt{5}\).

 

Page 496  Exercise 16  Problem 16

Given: A right-angle triangle with two sides

To Find – The value of x in simplest radical form

In the given right-angle triangle

Base (a) = x units

Height (b) = 4 units

Hypotenuse (c) = 7 units

Substituting the values in the Pythagorean theorem

(leg1)² + (leg2)² =(Hypotenuse)²

​⇒   x² + 4² = 7²

⇒   x² + 16 = 49

⇒   x²  + 16 − 16 = 49 − 16

⇒   x² = 33

⇒   x = \(\sqrt{33}\)

​The value of x in the simplest radical form is \(\sqrt{33}\).

 

Page 496  Exercise 17  Problem 17

Given: A right-angle triangle with two sides

To Find  –  The value of x in simplest radical form

In the given right-angle triangle

leg₁ = 16, leg₂ = x,  Hypotenuse = 19

Substituting the values in the Pythagorean Theorem
(leg₁)² +(leg₂)² = (Hypotenuse)²

​⇒   16²  + x² = 19²

⇒   256 + x² = 361

⇒   x² = 361 − 256

⇒   x²  = 105

⇒  x = \(\sqrt{105}\)

​The value of x in the simplest radical form is \(\sqrt{105}\).

 

Page 496  Exercise 18  Problem 18

Given: A right-angle triangle

To Find –  The value of x in simplest radical form

In the given right-angle triangle

leg₁ = x, leg₂ = x, Hypotenuse = 6

Substituting the values in the Pythagorean Theorem

​⇒   (leg₁)²  + (leg₂ )² = (Hypotenuse)

⇒   x²  + x² = 6²

⇒  2x² = 36

⇒   \(\frac{2 x^2}{2}=\frac{36}{2}\)

⇒   x² = 18

⇒   x = \(\sqrt{18}\)

⇒   x  =  \(\sqrt{9(2)}\)

⇒   x =  3\(\sqrt{2}\)

The value of x in the simplest radical form is 3\(\sqrt{2}\)

 

Page 496  Exercise 19  Problem 19

Given: A right-angle triangle

To find – The value of x in the simplest radical form

In the given right-angle triangle

leg₁ = 5, leg₂ = x , Hypotenuse = 10

Substituting the values in the Pythagorean Theorem

​⇒  (leg1)² + (leg2)² = (Hypotenuse)²

⇒   5² + x²  = 10²

⇒   25 + x²  = 100

⇒   25 − 25 + x² = 100 − 25

⇒    x² = 75

⇒    x = \(\sqrt{75}\)

⇒    x = \(\sqrt{25(3)}\)

⇒   x = 5\(\sqrt{3}\)

The value of x in the simplest radical form is 5\(\sqrt{3}\).

 

Page 496  Exercise 20  Problem 20

​Given: Length of the ladder is 15 -ft, Base of the ladder from the house is 5 -ft

To Find –  Height on the house that the ladder reach

As the ladder is leaning against the house so the height of the ladder will be the hypotenuse, i.e., 15 − ft.

The base of the ladder is 5 ft from the house, so the base will be  5 ft.  b = 5ft

Let the height on the house that the ladder reach be x ft.

So according to the Pythagorean Theorem

​⇒   Base²+ Height² =  Hypotenuse²

⇒   5² + x² = 152

⇒   25 + x² = 225

⇒  x² = 225 − 25

⇒  x = \(\sqrt{200}\)

⇒  x = 14.142
​
⇒  x = 14.1ft (Roud off to nearest tenth)

Hence the height on the house that the ladder reach is 14.1 ft.

When a 15 − ft ladder leans against the house and the base of the ladder is 5 − ft from the house then the height on the house that the ladder reach is 14.1 ft.

 

Page 496 Exercise 21 Problem 21

Given: A 24 m long walkway forms a diagonal of a square playground.

To Find –  Length of a side of the playground

As the walkway divides the square into two right-angle triangles, and the walkway (Diagonal) is 24 m long so the Pythagorean Theorem can be applied.

As each side of a square is equal so let leg₁  = leg₂ = x meters and the walkway will be the hypotenuse.

Substituting the values in the Pythagorean Theorem

⇒  (leg₁)² + (leg₂)² = (hypotenuse)²

⇒    x² + x² = 24²

⇒   2x² = 576

⇒   \(\frac{2 x^2}{2}=\frac{576}{2}\)

⇒   x²  = 288

⇒   x = \(\sqrt{288}\)

⇒   x≈16.97
​
⇒  x≈17.0 meters (Round off)

Hence, the length of each side of the playground is approximately 17 meters.

If a 24 m walkway forms one diagonal of a square playground then, the length of a side of the playground is approximately 17 meters.

 

Page 496  Exercise 22  Problem 22

Given: A triangle

To Find – If the triangle is a right-triangle

Hypotenuse = 25, leg₁  = 24, leg₂ = 8

For checking, if the triangle is a right triangle apply the Pythagorean theorem

​⇒  (leg₁)² + (leg₂ )² = (Hypotenuse)²

⇒   24² + 8² = 25²

⇒  576 + 64 = 625

⇒   640 ≠ 625
​
As L.H.S. is not equal to R.H.S., therefore it does not satisfy the Pythagorean Theorem and hence it is not a right triangle.

The triangle is not a right triangle because it does not satisfy the Pythagorean theorem, 24² + 8² ≠25².

 

Page 496  Exercise 23  Problem 23

Given: The triangle.

To Find – If the triangle is a right-angled triangle or not.

Using the Pythagoras theorem, we will find if the triangle is a right-angled triangle.

Consider the triangle

Pythagoras theorem states that the square of the hypotenuse is equal to the sum of the squares of other sides.

Let,​c = 65, a = 56 , b = 33

​Now, using Pythagoras theorem:

​c² = 65²

c² = 4225
​
And,​a² + b² = 56² + 33²

a² + b² = 4225.
​
Since,c² = a2 + b²  the triangle is a right-angled triangle.

The triangle is a right triangle.

 

Page 496  Exercise 24  Problem 24

Given: The lengths of the sides of a triangle are 4,5,6.

To Find – The type of triangle.

In order to determine the type of triangle, we need to observe the inequality of the Pythagoras theorem

Consider the lengths of the three sides of a triangle , 4,5,6

Now, if  a² + b² = c² the triangle is a right angle, if a² + b² > c² triangle is acute, if a² + b² < c² triangle is obtuse.

Now, the highest length is the hypotenuse.

So, ​a = 4 , b = 5 , c = 6

​So, a² + b² = 4² + 5²

a² + b² =16 + 25

a² + b² = 41

​c² = 6²

c² = 36.

As a² + b² >c² the triangle is an acute triangle.

 

Page 496  Exercise 25  Problem 25

Given: The length of the sides of a triangle is  0.3,0.4,0.6 .

To Find – The type of triangle.

In order to determine the type of triangle, we need to observe the inequality of the Pythagoras theorem.

Consider the lengths of the three sides of a triangle  , 0.3,0.4,0.6.

Now, if a2 + b2 = c2 the triangle is a right triangle, if a² + b² > c² triangle is acute, if a² + b² < c² triangle is obtuse.

The highest length is the hypotenuse. So,​a = 0.3 b = 0.4 c = 0.6
​
Now,​a² + b² = 0.3² + 0.4²

a² + b² = 0.25 and c² = 0.62

c² = 0.36
​
As a² + b² <c²  the triangle is an obtuse triangle.

The triangle with lengths with sides 0.3,0.4,0.6 is an obtuse triangle.

 

Page 496  Exercise 26  Problem 26

Given: The length of the sides of a triangle is11,12,15.

To Find – The type of triangle.

In order to determine the type of triangle, we need to observe the inequality of the Pythagoras theorem.

Consider the sides of the triangle , 11,12,15.

Now, if a²  + b²  = c² the triangle is a right-angle triangle, if a²  +b² > c²  triangle is acute, if a²+ b²  <c²  triangle is obtuse.

The highest length is the hypotenuse. So,​a = 11 b = 12 c = 15

So,​a² + b² = 11² + 12²

a² + b² = 121 + 144

a² + b² = 265.
​
And,​c²= 15²

c² = 225

As a²  + b² > c² the triangle is an acute triangle.

The triangle with lengths of sides 11,12,15 is an acute triangle.

 

Page 496  Exercise 27  Problem 27

Given: The length of the sides of a triangle is \(\sqrt{3}\),2,3.

 To find – The type of triangle.

In order to determine the type of triangle, we need to observe the inequality of the Pythagoras theorem.

Consider the sides of the triangle ,  \(\sqrt{3}\),2,3.

Now, if a² + b² = c² triangle is a right angle triangle, if a² + b² > c² the triangle is an acute triangle and if a² + b² < c² the triangle is an obtuse triangle.

The highest length in a triangle is the hypotenuse.

So, a =   \(\sqrt{3}\) , b= 2, c= 3

Now, a² + b² = 7 and

​c² = 3²

c² = 9

As a² + b² < c²  the triangle is an obtuse triangle.

The triangle with the sides  \(\sqrt{3}\) ,2,3 is an obtuse triangle.