sebuah persegi panjang berukuran panjang = (5 akar 3 - 2 akar 5) cm dan lebar = (20 + akar 15) cm. Hitunglah : a. luas persegi panjang tsb b. panjang diagonalny

Panjang = (5√3 - 2√5) cm
Lebar = (20 + √15)

a. Luas Persegi Panjang

[tex]L=(5 \sqrt{3}-2 \sqrt{5})\times(20+ \sqrt{15}) \\ \\ L=(5 \sqrt{3}\times20)+(5 \sqrt{3}\times \sqrt{15} )-(2 \sqrt{5}\times20)-(2 \sqrt{5}\times \sqrt{15} ) \\ \\ L=(100 \sqrt{3})+(5 \sqrt{45})-(40 \sqrt{5})-(2 \sqrt{75}) \\ \\ L= (100 \sqrt{3})+(5 \sqrt{9\times5})-(40 \sqrt{5})-(2 \sqrt{25\times3})\\ \\ L= (100 \sqrt{3})+(15 \sqrt{5})-(40 \sqrt{5})-(10 \sqrt{3}) \\ \\ L=(90 \sqrt{3}-25 \sqrt{5})\ \text{cm}^2 [/tex]

b. Panjang Diagonal

[tex]D= \sqrt{(5 \sqrt{3}-2 \sqrt{5})^2+(20+ \sqrt{15})^2} \\ \\ D= \sqrt{(5 \sqrt{3})^2-(20 \sqrt{15})+( 2 \sqrt{5})^2+20^2+ (40\sqrt{15})+(\sqrt{15})^2} \\ \\ D=\sqrt{75-(20 \sqrt{15})+20+400+ (40\sqrt{15})+15} \\ \\ D=\sqrt{(75+20+400+15)+(40-20) \sqrt{15}}\\ \\ D=\sqrt{510+20 \sqrt{15}}\ \text{cm}[/tex]

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