Chapter 12, Problem 10RP

a.

To determine

To calculate: The lateral area and the total area of the pyramid ABCD

Answer to Problem 10RP

The lateral area and the total area of the solid ABCD are $180$ unit ${}^2$ and $180+25\sqrt{3}$ unit ${}^2$

Explanation of Solution

Given:

The measures of the given lengths are as follows:

AB = 13

AD = 10

Concept used:

Here, we are using the formula:

Lateral surface area of the pyramid = $\frac{3}{2}\times b\times h$

(Where, “ b ” is base and “ h ” is height)

Total surface area of pyramid = Lateral surface area + $\frac{\sqrt{3}}{4}\times a^2$

(Where, “ a ” is side)

Calculation:

  

Here, In $\Delta ABE$

Using Pythagoras theorem

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

  ${\left(AB\right)}^2={\left(BE\right)}^2+{\left(AE\right)}^2$

  ${\left(13\right)}^2={\left(BE\right)}^2+{\left(5\right)}^2$

  $BE=\sqrt{169-25}$

  $BE=\sqrt{144}$

  $BE=12$

Lateral surface area of pyramid = $\frac{3}{2}\times b\times h$

(Where, “ b ” is base and “ h ” is height)

  → $\frac{3}{2}\times 10\times 12$

  → $180$

Now, Total surface area of pyramid = Lateral surface area + $\frac{\sqrt{3}}{4}\times a^2$

(Where, “ a ” is side)

  → $180+\sqrt{3}\times \frac{100}{4}$

  → $180+25\sqrt{3}$

Conclusion: Hence, the lateral area and the total area of the solid ABCD are $180$ unit ${}^2$ and $180+25\sqrt{3}$ unit ${}^2$

b.

To determine

To calculate: The lateral area and the total area of the cone PQRS

Answer to Problem 10RP

The lateral area and the total area of the cone PQRS are $60\pi$ and $96\pi$

Explanation of Solution

Given:

The measures of the given lengths are as follows:

AD = $8$

AC = $10$

Concept used:

Here, we are using the formula:

Lateral surface area of the pyramid = $\frac{1}{2}\times circumference\times slantheight$

Total surface area of pyramid = $\pi r^2+\pi rl$

(Where, “ r ” is radius and “ l ” is slant height)

Calculation:

  

Here, Lateral surface area of the pyramid = $\frac{1}{2}\times circumference\times slantheight$

  → $\frac{1}{2}\times 2\pi r\times l$

  → $\pi rl$

→

  → $6\pi \left(20\right)$

  → $120\pi$

And, Volume of cylinder = $\frac{\pi }{2}\times r^2\times h$

(Where, “ r ” is $6$ and “ h ” is $20$ )

  → $\frac{\pi }{2}\times {\left(6\right)}^2\times 20$

  → $\frac{\pi }{2}\times 36\times 20$

  → $360\pi$

Conclusion: Hence, the volume and the total area of the right cylinder are $360\pi$ unit ${}^3$ and $156\pi$ unit ${}^2$