Question

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Answer 1

$find \: lengh \: bc \: because \: find \: the \: base \: area \: \\ bc = ad \\ from \: ac = 15mm \\ ab = 12mm \\ from Pythagoras \: {c}^{2} = {a}^{2} + {b}^{2} \\ is \: {(ac)}^{2} = {(bc)}^{2} + {(ab)}^{2} \\ 15 ^{2} = {(bc)}^{2} + {12}^{2} \\ {(bc)}^{2} = 15^{2} - {12}^{2} \\ bc = \sqrt{{15}^{2} - {12}^{2} } \\ bc = \sqrt{225 - 144} \\ bc = \sqrt{81} \\ bc = 9 \: or \: - 9 \\ bc = 9 \\ base \: area \: = bc \: \times \: ab \\ = 9 \times 12 = 108 {mm}^{2} \\ volume \: pyramid \: = \: \frac{1}{3} \times base \: area \times hight \\ = \frac{1}{3} \times 108 {mm}^{2} \times 13mm \\ = 468 {mm}^{3}$