Question

find the area of a triangle whose hypotenuse is 41cm and its perpendicular is 9cm.
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Answer 1

Answer:

$180 \: {cm}^{2} .$

Step-by-step explanation:

$base \: = \: \sqrt{ {41}^{2} - {9}^{2} } \\ = \sqrt{(41 - 9)(41 + 9)} \\ = \sqrt{32 \times 50} = \sqrt{8 \times 4 \times 50} \\ = \sqrt{8 \times 200} \\ = \sqrt{1600} \\ = 40 \: cm \\ \\ base = \: 40 \: cm \\ \\ \\ area \: = \: \frac{1}{2} \times (40) \times (9) \\ \\ = 20 \times 9 \\ = 180 \: {cm}^{2} .$