Chapter 11, Problem 23RP

Solution

a.

To calculate: The area of square

The area of the square is $338$ unit ${}^2$

Given:

The measure of the diagonal of the square is $26$

Concept used:

Here, we are using Pythagoras theorem:

  ${\left(hypotenuse\right)}^2={\left(perpendicular\right)}^2+{\left(base\right)}^2$

And, the formula to find square’s area which is

Area of square = $a^2$

(Where, “ a ” is side)

Calculation:

  

Here, In $\Delta BCD$

Using Pythagoras theorem,

  ${\left(hypotenuse\right)}^2={\left(perpendicular\right)}^2+{\left(base\right)}^2$

  ${\left(BC\right)}^2={\left(BD\right)}^2+{\left(CD\right)}^2$

  ${\left(26\right)}^2=a^2+a^2$

  $26\times 26=2s^2$

  $338=a^2$

  $a=18.38$

Now, Area of square = $a^2$

(Where, “ a ” is side)

➔${\left(18.38\right)}^2=338$

Conclusion: Hence, the area of the square is $338$ unit ${}^2$

b.

To calculate: The diagonal of the square

The diagonal of the square is $6$ unit.

Given:

Area of the square is 18 unit ${}^2$

Concept used:

Here, we are using Pythagoras theorem

  ${\left(hypotenuse\right)}^2={\left(perpendicular\right)}^2+{\left(base\right)}^2$

Calculation:

  

Here, In $\Delta BCD$

Using Pythagoras theorem,

  ${\left(hypotenuse\right)}^2={\left(perpendicular\right)}^2+{\left(base\right)}^2$

  ${\left(BC\right)}^2={\left(BD\right)}^2+{\left(CD\right)}^2$

  ${\left(d\right)}^2=a^2+a^2$

(Where, “ a ” is side and “ d ” is diagonal)

  $d^2=2a^2$

  $d^2=2\times 18$

($\because$ Area of square = $a^2$ and area of square is given in the question which is $18$ . So, $a^2$ will also be equal to $18$ )

  $d=\sqrt{36}=6$

Conclusion: Hence, the diagonal of the square is $6$