Given \sqrt{27}+\sqrt{108} 27

+
108


=27^{\frac{1}{2}}+108^{\frac{1}{2}} =27
2
1


+108
2
1


3^{\frac{1}{2}}\cdot9^{\frac{1}{2}} +3^{\frac{1}{2}}\cdot36^{\frac{1}{2 }}3
2
1


⋅9
2
1


+3
2
1


⋅36
2
1


3\cdot3^{\frac{1}{2}}+6\cdot3^{\fra c{1}{2}}3⋅3
2
1


+6⋅3
2
1


18\cdot3^{\frac{1}{2}}18⋅3
2
1

2y= 3 2 x−2 y = − 1 3 x + 1 y=− 3 1 x+1

What is the measure of one interior angle of a pentagon

Given: ae ≅ ce ; de ≅ be prove: abcd is a parallelogram. we have that ab || dc. by a similar argument used to prove that △aeb ≅ △ced, we can show that △ ≅ △ceb by. so, ∠cad ≅ ∠ by cpctc. therefore, ad || bc by the converse of the theorem. since both pair of opposite sides are parallel, quadrilateral abcd is a parallelogram.