Question

Find the expression for the area of the figure. Give your answer as a simplified radical.

Answer 1

Answer:

B

Step-by-step explanation:

Using the rule of radicals

$\sqrt{a}$ × $\sqrt{b}$ ⇔ $\sqrt{ab}$

The area (A) of a triangle is calculated as

A = $\frac{1}{2}$ bh ( b is the base and h the perpendicular height ), thus

A = $\frac{1}{2}$ × $\sqrt{21y}$ × $\sqrt{3y^5}$

   = $\frac{1}{2}$ × $\sqrt{63y^6}$

   = $\frac{1}{2}$ × $\sqrt{63}$ × $\sqrt{y^6}$

   = $\frac{1}{2}$ × $\sqrt{9(7)}$ × $\sqrt{(y^3)^2}$

   = $\frac{1}{2}$ × 3 × $\sqrt{7}$ × y³

   = $\frac{3}{2}$ y³$\sqrt{7}$ → B

Answer 2

Answer:

Option B

Step-by-step explanation:

Please see attached picture for full solution.