Question

ABCD is a rectangle with AC=20 and AB=2BC. What is the area of rectangle ABCD?

Answer 1

Let AB = 2x and BC be x.

By pythagoras theorum,

20² = x² + (2x)²

400 = x² + 4x²
400 = 5x²
80 = x²

x= √80 = 4√5 , thus, 2x = 8√5

Now, area = AB x AC

= 4√5 * 8√5

= 32 * 5

= 160

Thus, the area of rectangle ABCD is 160 units

Answer 2

$AC= 20 , \ AB=2BC \\\\ Pythagorean\ theorem,\ we \ have : \\ \\ |AC|^2 = (2|BC|)^2+ |BC|^2 \\ \\20^2 = 4|BC| ^2+ |BC|^2 \\ \\400 =5|BC|^2\ \ /:5$

$|BC|^2=80 \\ \\|BC|=\sqrt{80}=\sqrt{16\cdot 5}=4\sqrt{5}\\ \\|AB|=2\cdot |BC| =2\cdot 4\sqrt{5}=8\sqrt{5}\\ \\Area = |AB|\cdot |A|\\ \\Area=8\cdot \sqrt{5}\cdot 4\cdot \sqrt{5}=32 \cdot 5 = 160 \\ \\ Answer : Area \ ABCD \ a \ rectangle = 160$