Find the area $ A $ of a pyramid if side $a = 6$ and height $h = 11$.

$A = 172.8211$

STEP 1: find base area $ B $

To find base area $ B $ use formula:

$$ B = a^2 $$

After substituting $ a = 6 $ we have:

$$ B = 6^2 $$ $$ B = 36 $$

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 = 11 $ and $ a = 6 $ we have:

$$ 11^2 + \frac{ 6^2 }{ 4 }= s^2 $$ $$ 121 + \frac{ 36 }{ 4 }= s^2 $$ $$ 121 + 9 = s^2 $$ $$ s^2 = 130 $$ $$ s = \sqrt{ 130 } $$

STEP 3: find lateral surface $ L $

To find lateral surface $ L $ use formula:

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

After substituting $ a = 6 $ and $ s = \sqrt{ 130 } $ we have:

$$ L = 12 \cdot \sqrt{ 130 } $$$$ L = 12 \sqrt{ 130 } $$

STEP 4: find total surface $ A $

To find total surface $ A $ use formula:

$$ A = B + L $$

After substituting $ B = 36 $ and $ L = 12 \sqrt{ 130 } $ we have:

$$ A = 36 + 12 \sqrt{ 130 } $$ $$ A = 36 + 12 \sqrt{ 130 } $$ $$ A = 172.8211 $$

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