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tatuchka [14]
3 years ago
7

Can I please have help with question 2) a and b

Mathematics
1 answer:
Ulleksa [173]3 years ago
7 0

Answer:

\frac{10\sqrt{6} }{3} and \frac{11\sqrt{3} }{3}

Step-by-step explanation:

(a)

A = \frac{1}{2} × 4\sqrt{2} × \frac{5}{\sqrt{3} }

   = 2\sqrt{2} × \frac{5}{\sqrt{3} }

   = \frac{10\sqrt{2} }{\sqrt{3} } × \frac{\sqrt{3} }{\sqrt{3} } ← rationalise the denominator

= \frac{10\sqrt{6} }{3} units²

(b)

Using Pythagoras' identity in the right triangle

let hypotenuse be h , then

h² = (4\sqrt{2} )² + (\frac{5}{\sqrt{3} } )²

    = 32 + \frac{25}{3}

    = \frac{96}{3} + \frac{25}{3}

    = \frac{121}{3} ( take the square root of both sides )

h = \sqrt{\frac{121}{3} } = \frac{11}{\sqrt{3} } × \frac{\sqrt{3} }{\sqrt{3} } ← rationalise the denominator

h = \frac{11\sqrt{3} }{3}

 

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