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How do you find legs in a $45 ° - 45 ° - 90 °$ triangle when its hypotenuse is $11$ ?

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7.66

7.77

7.67

6.77

answer is B.

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Detailed Solution

Let the figure be -
Question Image

Hypotenuse

= 11

(Given)

∠ A = 45 °

(Given)

∠ B = 90 °

(Given)

∠ C = 45 °

(Given)
Let the legs or side

AB

and

BC

are

= x

ΔABC

-
By pythagoras theorem,

{(hypotenuse)}^{2} = {(perpendicular)}^{2} + {(base)}^{2}

{(AC)}^{2} = {(AB)}^{2} + {(BC)}^{2}

{(11)}^{2} = x^{2} + x^{2}

121 = 2 x^{2}

\frac{121}{2} = x^{2}

x^{2} = \frac{121}{2}

x = \sqrt{\frac{121}{2}}

x = \frac{11}{\sqrt{2}}

x = \frac{11 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} \Rightarrow x = \frac{11 \sqrt{2}}{2}

(∵ multiplying numerator and denominator by

\sqrt{2}

)

x = \frac{11 \times 1.414}{2}

(∵ \sqrt{2} = 1.414)

x = 7.77

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How do you find legs in a 45°-45°-90° triangle when its hypotenuse is 11?

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How do you find legs in a $45 ° - 45 ° - 90 °$ triangle when its hypotenuse is $11$ ?