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

A C = 20 , A B = 2 BC P y t ha g ore an t h eore m , w e ha v e : ∣ A C ∣ 2 = ( 2∣ BC ∣ ) 2 + ∣ BC ∣ 2 2 0 2 = 4∣ BC ∣ 2 + ∣ BC ∣ 2 400 = 5∣ BC ∣ 2 / : 5
∣ BC ∣ 2 = 80 ∣ BC ∣ = 80 ​ = 16 ⋅ 5 ​ = 4 5 ​ ∣ A B ∣ = 2 ⋅ ∣ BC ∣ = 2 ⋅ 4 5 ​ = 8 5 ​ A re a = ∣ A B ∣ ⋅ ∣ A ∣ A re a = 8 ⋅ 5 ​ ⋅ 4 ⋅ 5 ​ = 32 ⋅ 5 = 160 A n s w er : A re a A BC D a rec t an g l e = 160

Answered by Lilith | 2024-06-10

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

Answered by tadvisohil886 | 2024-06-10

The area of rectangle ABCD is calculated to be 160 square units by applying the Pythagorean theorem and substituting the given relationships between the sides. By finding the lengths of sides AB and BC, we then multiplied them to find the area. The final result confirms that the area is 160 square units.
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Answered by Lilith | 2024-12-23