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The area of rectangle ABCD with vertices A(0, 0), B(0, 4), C(x, 4) and D(x, 0) is \(16\) square units. If x > 0, what is the value of x?

 Apr 13, 2022
 #1
avatar+578 
+3

From the information given above, we can say that the width of the rectangle is \(4\)

 

The area of a rectangle is \(length \cdot width\)

 

That is, \(4 \cdot length = 16\)

 

That gives us length = \(16/4 = 4\)

 

From that, we can tell that the value of \(x\) must be \(4\).

 Apr 13, 2022
 #2
avatar+578 
+3

I'm pretty sure that 4 is the answer, but....hold up

 

Wouldn't that make it a square though??

 

Can anyone please confirm?

Vinculum  Apr 13, 2022
 #4
avatar+2455 
+1

But, a square is technically a rectangle... So your answer would still be right. 

BuilderBoi  Apr 13, 2022
 #3
avatar+115 
-3

Yes i think so

 Apr 13, 2022
 #5
avatar+124598 
+1

Note that  a square is a special case of a  rectangle....but a rectangle isn't necessarily a square  smiley smiley smiley

 

Since AB  = 4 ....then  it must also be that BC = 4

 

So....the area  =  (AB) (BC) =  (4)(4)  =  16

 

cool cool cool  

 Apr 13, 2022
 #6
avatar+115 
-2

Good Job vin

 Apr 13, 2022

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