Find the expression for the area of the figure. Give your answer as a simplified radical.
Help me plzzz :0
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Answer:
Option B
Step-by-step explanation:
Please see attached picture for full solution.
Answer:
B
Step-by-step explanation:
Using the rule of radicals
[tex]\sqrt{a}[/tex] × [tex]\sqrt{b}[/tex] ⇔ [tex]\sqrt{ab}[/tex]
The area (A) of a triangle is calculated as
A = [tex]\frac{1}{2}[/tex] bh ( b is the base and h the perpendicular height ), thus
A = [tex]\frac{1}{2}[/tex] × [tex]\sqrt{21y}[/tex] × [tex]\sqrt{3y^5}[/tex]
= [tex]\frac{1}{2}[/tex] × [tex]\sqrt{63y^6}[/tex]
= [tex]\frac{1}{2}[/tex] × [tex]\sqrt{63}[/tex] × [tex]\sqrt{y^6}[/tex]
= [tex]\frac{1}{2}[/tex] × [tex]\sqrt{9(7)}[/tex] × [tex]\sqrt{(y^3)^2}[/tex]
= [tex]\frac{1}{2}[/tex] × 3 × [tex]\sqrt{7}[/tex] × y³
= [tex]\frac{3}{2}[/tex] y³[tex]\sqrt{7}[/tex] → B