Find the area $ A $ of a pyramid if side $a = 4$ and height $h = 3$.

$A = 44.8444$

STEP 1: find base area $ B $

To find base area $ B $ use formula:

$$ B = a^2 $$

After substituting $ a = 4 $ we have:

$$ B = 4^2 $$ $$ B = 16 $$

STEP 2: find slant height $ s $

To find slant height $ s $ use Pythagorean Theorem:

$$ h^2 + \frac{ a^2 }{ 4 } = s^2 $$

After substituting $ h = 3 $ and $ a = 4 $ we have:

$$ 3^2 + \frac{ 4^2 }{ 4 }= s^2 $$ $$ 9 + \frac{ 16 }{ 4 }= s^2 $$ $$ 9 + 4 = s^2 $$ $$ s^2 = 13 $$ $$ s = \sqrt{ 13 } $$

STEP 3: find lateral surface $ L $

To find lateral surface $ L $ use formula:

$$ L = 2 \cdot a \cdot s $$

After substituting $ a = 4 $ and $ s = \sqrt{ 13 } $ we have:

$$ L = 8 \cdot \sqrt{ 13 } $$$$ L = 8 \sqrt{ 13 } $$

STEP 4: find total surface $ A $

To find total surface $ A $ use formula:

$$ A = B + L $$

After substituting $ B = 16 $ and $ L = 8 \sqrt{ 13 } $ we have:

$$ A = 16 + 8 \sqrt{ 13 } $$ $$ A = 16 + 8 \sqrt{ 13 } $$ $$ A = 44.8444 $$

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