Chapter 9.6, Problem 16PSB

a.

To determine

To find:The value of x

Answer to Problem 16PSB

The value of x is $7$

Explanation of Solution

Given:

Measures of other sides are $15,17,6,25$ and $26$

Concept used:

Here, we are using Pythagoras theorem.

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

Calculation:

  

Here, the formula is

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

Let the missing side be x

In $\Delta ABF$

  ${\left(x\right)}^2={\left(17\right)}^2-{\left(15\right)}^2$

  $x=\sqrt{289-225}$

  $x=\sqrt{64}$

  $x=8$

Now, In $\Delta BCF$

  $x^2={\left(8\right)}^2+{\left(6\right)}^2$

  $x=\sqrt{64+36}$

  $x=\sqrt{100}$

  $x=10$

Now, In $\Delta FCD$

  $x^2={\left(26\right)}^2-{\left(10\right)}^2$

  $x=\sqrt{676-100}$

  $x=\sqrt{576}$

  $x=24$

Now, In $\Delta FDE$

  $x^2={\left(25\right)}^2-{\left(24\right)}^2$

  $x=\sqrt{625-576}$

  $x=\sqrt{49}$

  $x=7$

Conclusion: Hence, the value of x is $7$

b.

To determine

To find:The value of x and y

Answer to Problem 16PSB

The values of x and y are $20$ and $20$

Explanation of Solution

Given:

The measures of other sides are $12,3$ and $5$

Concept used:

Here, we are using Angle Angle similarity and Pythagoras theorem.

Calculation:

  

Here, In $\Delta ABE$ and $\Delta ACD$

  $\angle A=\angle A$ (common)

  $\angle ABE=\angle ACD$ (right angles)

So, $\Delta ABE\sim \Delta ACD$ (by using Angle Angle similarity)

Hence, $\frac{12}{15}=\frac{y}{y+5}$

  $12y+60=15y$

  $y=20$

Now, using Pythagoras theorem

  ${\left(x\right)}^2={\left(25\right)}^2-{\left(15\right)}^2$

  $x=\sqrt{625-225}$

  $x=\sqrt{400}$

  $x=20$

Conclusion: Hence, the values of x and y are $20$ and $20$