Question

calcule a area.......​

Answer 1

c) $A=b \cdot h =\frac{\sqrt{7}+1}{\sqrt{8} -\sqrt{2}} \cdot \frac{\sqrt{7}-1}{\sqrt{8} +\sqrt{2}} =\frac{(\sqrt{7})^2-1^2}{(\sqrt{8})^2-(\sqrt{2})^2} =\frac{7-1}{8-2}=\frac{6}{6} =1$

d)

$b= \frac{1}{\sqrt{3}+1} \\\\h=\sqrt{3}-1\\\\A_1=b\cdot h=\frac{1}{\sqrt{3}+1} \cdot (\sqrt{3}-1)=\frac{\sqrt{3}-1}{\sqrt{3}+1} =\frac{\sqrt{3}-1}{\sqrt{3}+1} \cdot \frac{\sqrt{3}-1}{\sqrt{3}-1} =\frac{(\sqrt{3})^2 -2\cdot \cdot 1 \cdot \sqrt{3} +1^2}{(\sqrt{3})^2-1^2}\\=\frac{3 -2\sqrt{3} +1}{3-1}=\frac{4 -2\sqrt{3}}{2}=\frac{2 \cdot(2-\sqrt{3})}{2}=2 -\sqrt{3}$

$h= \frac{1}{\sqrt{3}+1} \\\\b=\sqrt{3}+1\\\\A_2=b\cdot h=\frac{1}{\sqrt{3}+1} \cdot (\sqrt{3}+1)=\frac{\sqrt{3}+1}{\sqrt{3}+1} =1$

$A_1+A_2=2-\sqrt{3} +1=3-\sqrt{3}$