Question

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

Answer 1

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