Find the Surface Area pyramid (6cm)(7cm)(8cm)

 $\begin{array}{r} h = & 6 cm \\ l = & 8 cm \\ w = & 7 cm \end{array}$

Step 1

The surface area of a pyramid is equal to the sum of the areas of each side of the pyramid. The base of the pyramid has area $l w$ , and $s_{l}$ and $s_{w}$ represent the slant height on the length and slant height on the width.

$(l e n g t h) \cdot (w i d t h) + (w i d t h) \cdot s_{l} + (l e n g t h) \cdot s_{w}$

Step 2

Substitute the values of the length $l = 8$ , the width $w = 7$ , and the height $h = 6$ into the formula for surface area of a pyramid.

$8 \cdot 7 + 7 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(6)}^{2}} + 8 \cdot \sqrt{{(\frac{7}{2})}^{2} + {(6)}^{2}}$

Step 3

Simplify each term.

Multiply $8$ by $7$ .

$56 + 7 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(6)}^{2}} + 8 \cdot \sqrt{{(\frac{7}{2})}^{2} + {(6)}^{2}}$

Divide $8$ by $2$ .

$56 + 7 \cdot \sqrt{4^{2} + {(6)}^{2}} + 8 \cdot \sqrt{{(\frac{7}{2})}^{2} + {(6)}^{2}}$

Raise $4$ to the power of $2$ .

$56 + 7 \cdot \sqrt{16 + {(6)}^{2}} + 8 \cdot \sqrt{{(\frac{7}{2})}^{2} + {(6)}^{2}}$

Raise $6$ to the power of $2$ .

$56 + 7 \cdot \sqrt{16 + 36} + 8 \cdot \sqrt{{(\frac{7}{2})}^{2} + {(6)}^{2}}$

Add $16$ and $36$ .

$56 + 7 \cdot \sqrt{52} + 8 \cdot \sqrt{{(\frac{7}{2})}^{2} + {(6)}^{2}}$

Rewrite $52$ as $2^{2} \cdot 13$ .

Factor $4$ out of $52$ .

$56 + 7 \cdot \sqrt{4 (13)} + 8 \cdot \sqrt{{(\frac{7}{2})}^{2} + {(6)}^{2}}$

Rewrite $4$ as $2^{2}$ .

$56 + 7 \cdot \sqrt{2^{2} \cdot 13} + 8 \cdot \sqrt{{(\frac{7}{2})}^{2} + {(6)}^{2}}$

Pull terms out from under the radical.

$56 + 7 \cdot (2 \sqrt{13}) + 8 \cdot \sqrt{{(\frac{7}{2})}^{2} + {(6)}^{2}}$

Multiply $2$ by $7$ .

$56 + 14 \sqrt{13} + 8 \cdot \sqrt{{(\frac{7}{2})}^{2} + {(6)}^{2}}$

Apply the product rule to $\frac{7}{2}$ .

$56 + 14 \sqrt{13} + 8 \cdot \sqrt{\frac{7^{2}}{2^{2}} + {(6)}^{2}}$

Raise $7$ to the power of $2$ .

$56 + 14 \sqrt{13} + 8 \cdot \sqrt{\frac{49}{2^{2}} + {(6)}^{2}}$

Raise $2$ to the power of $2$ .

$56 + 14 \sqrt{13} + 8 \cdot \sqrt{\frac{49}{4} + {(6)}^{2}}$

Raise $6$ to the power of $2$ .

$56 + 14 \sqrt{13} + 8 \cdot \sqrt{\frac{49}{4} + 36}$

To write $36$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$56 + 14 \sqrt{13} + 8 \cdot \sqrt{\frac{49}{4} + 36 \cdot \frac{4}{4}}$

Combine $36$ and $\frac{4}{4}$ .

$56 + 14 \sqrt{13} + 8 \cdot \sqrt{\frac{49}{4} + \frac{36 \cdot 4}{4}}$

Combine the numerators over the common denominator.

$56 + 14 \sqrt{13} + 8 \cdot \sqrt{\frac{49 + 36 \cdot 4}{4}}$

Simplify the numerator.

Multiply $36$ by $4$ .

$56 + 14 \sqrt{13} + 8 \cdot \sqrt{\frac{49 + 144}{4}}$

Add $49$ and $144$ .

$56 + 14 \sqrt{13} + 8 \cdot \sqrt{\frac{193}{4}}$

Rewrite $\sqrt{\frac{193}{4}}$ as $\frac{\sqrt{193}}{\sqrt{4}}$ .

$56 + 14 \sqrt{13} + 8 \cdot \frac{\sqrt{193}}{\sqrt{4}}$

Simplify the denominator.

Rewrite $4$ as $2^{2}$ .

$56 + 14 \sqrt{13} + 8 \cdot \frac{\sqrt{193}}{\sqrt{2^{2}}}$

Pull terms out from under the radical, assuming positive real numbers.

$56 + 14 \sqrt{13} + 8 \cdot \frac{\sqrt{193}}{2}$

Cancel the common factor of $2$ .

Factor $2$ out of $8$ .

$56 + 14 \sqrt{13} + 2 (4) \cdot \frac{\sqrt{193}}{2}$

Cancel the common factor.

$56 + 14 \sqrt{13} + 2 \cdot 4 \cdot \frac{\sqrt{193}}{2}$

Rewrite the expression.

$56 + 14 \sqrt{13} + 4 \cdot \sqrt{193}$

$56 + 14 \sqrt{13} + 4 \sqrt{193}$

Step 4

Calculate the approximate solution to $4$ decimal places.

$162.0475 {(cm)}^{2}$

Find the Surface Area pyramid (6cm)(7cm)(8cm)

Related Questions