Chapter 7.5, Problem 59PPS

(a)

To determine

To Find:

The length of the red segment using the Pythagorean Theorem.

Answer to Problem 59PPS

  $\text{Red side = }\sqrt{2}\text{ units}$

Explanation of Solution

Given Information:

Use the Pythagorean Theorem to prove that the length of the red segment is $\sqrt{2}\text{ units}$ .

  

Concept Used :

Pythagorean Theorem:

The Pythagorean theorem, is a fundamental relation among the three sides of a right triangle. It states that the area of the square whose side is the hypotenuse is equal to the sum of the areas of the squares on the other two sides. 

  

Calculation:

• Given

  

• $\begin{array}{l}\text{Using Pythagoream Theorem :}\\ \text{a=1}\\ \text{b=1}\\ \text{c=}\sqrt{a^2+b^2}=\sqrt{1^2+1^2}=\sqrt{2}\end{array}$
• $\text{Red side = }\sqrt{2}\text{ units}$

(b)

To determine

To Draw:

Extend the segment to represent $\sqrt{2}+\sqrt{2}$ .

Explanation of Solution

Given Information:

  

Concept Used :

Pythagorean Theorem:

The Pythagorean theorem, is a fundamental relation among the three sides of a right triangle. It states that the area of the square whose side is the hypotenuse is equal to the sum of the areas of the squares on the other two sides. 

  

Calculation:

• Given



• Extend the red segment to the same length ahead.

(c)

To determine

To Prove:

  $\sqrt{2}+\sqrt{2}\ne \sqrt{2+2}$

Explanation of Solution

Concept Used :

In an isosceles right angled triangle , the length of the hypotenuse is greater than any one the side .

Calculation:

• Given

  

• The new triangle is an isosceles right angled triangle with hypotenuse $\sqrt{2}+\sqrt{2}$ .
• So, as we know that the length of the hypotenuse is greater than the length of each side.
• Length of the equal sides of the new triangle is 2 units each.
• So,
$\begin{array}{l}\sqrt{2}+\sqrt{2}>2=\sqrt{2^2}\\ \Rightarrow \sqrt{2}+\sqrt{2}>\sqrt{2+2}\\ \Rightarrow \sqrt{2}+\sqrt{2}\ne \sqrt{2+2}\end{array}$

Hence , proved.

(d)

To determine

To Graph:

A square in the dot paper whose sides are $\sqrt{2}\text{ units}$

Explanation of Solution

Concept Used :

Pythagorean Theorem:

The Pythagorean theorem, is a fundamental relation among the three sides of a right triangle. It states that the area of the square whose side is the hypotenuse is equal to the sum of the areas of the squares on the other two sides. 

  

Calculation:

• The sides of the required square are $\sqrt{2}$ units.
• Using the Pythagorean theorem , we can draw a right angled triangle with sides 1 unit and hence ,the hypotenuse will be $\sqrt{1^2+1^2}=\sqrt{2}\text{ units as seen in part (a)}$
• So, we draw 4 such triangles with sides 1 units to form the square with side $\sqrt{2}$ units.
• The diagonals of the square will be 2 units.
• 

  

(e)

To determine

To Prove:

The area of the square is $\sqrt{2}\cdot \sqrt{2}=2\text{ square units}$

Explanation of Solution

Concept Used :

  $\begin{array}{l}\text{Area of a triangle with base b and height h is }\\ \frac{1}{2}\cdot b\cdot h\end{array}$

Calculation:

• The square is made up of 4 triangles each of whose base is 1 unit and height is 1 unit.
• 
• $\begin{array}{l}\text{The area of one such triangle is :}\\ A=\frac{1}{2}\cdot b\cdot h\\ \text{Substituting the values b=1 and h=1 in the above equation:}\\ A=\frac{1}{2}(1)(1)=\frac{1}{2}\\ \text{There are 4 such triangles , so adding the areas of all triangles:}\\ \text{Area of square = 4}\cdot \frac{1}{2}=2=\sqrt{2}\cdot \sqrt{2}={(}^s\end{array}$
• Hence proved