Question

What is the area of a trapezoid ABCD with bases AB and CD ,
1) if: m∠C=m∠D=60°, AB = BC = 8 cm and when
2) m∠C=m∠D=45°, AB=6 cm, BC=9 root 2 cm

Answer 1

check the picture below on the top side.

we know that x = 4 = b, therefore,  using the 30-60-90 rule, h = 4√3, and DC = 4+8+4 = 16.

$\bf \textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} a,b=\stackrel{bases}{parallel~sides}\\ h=height\\[-0.5em] \hrulefill\\ a=8\\ b=\stackrel{DC}{16}\\ h=4\sqrt{3} \end{cases}\implies A=\cfrac{4\sqrt{3}(8+16)}{2} \\\\\\ A=2\sqrt{3}(24)\implies \boxed{A=48\sqrt{3}}$

now, check the picture below on the bottom side.

since we know x = 9, then b = 9, therefore DC = 9+6+9 = 24, and h = b = 9.

$\bf \textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} a,b=\stackrel{bases}{parallel~sides}\\ h=height\\[-0.5em] \hrulefill\\ a=6\\ b=\stackrel{DC}{24}\\ h=9 \end{cases}\implies A=\cfrac{9(6+24)}{2} \\\\\\ A=\cfrac{9(30)}{2}\implies \boxed{A=135}$

Answer 2

Answer:

135 cm^2

Step-by-step explanation:

see the pic, sry if its bad writing.