What is the height of a rectangle whose base is 40 inches and whose diagonal is 41 inches?

Question

I need these informations below:

Draw a picture of the problem and include labels:

Write and solve the equation:

Answer the question:

 

Answer 1

Draw a rectangle.  Label the height as 40 inches and the diagonal as 41 inches. 

 

Visually you are trying to find the third side of a right triangle which means we can use the Pythagorean Thereom:

 

a² + b² =  c²

 

 

c always represents the hypotenuse of a right triangle, which in this case is 41 inches--the diagonal.

 

 

a and b represent the "legs" or the two shorter sides of the right triangle.

 

 

Plug in a= 40  and c = 41 into the formula:     40²  + b²  = 41²   and solve.

 

 

 

1600  +  b² = 1681;      b² = 1681 - 1600;    b² =  81...........take the square root of both sides and your

 

answer is b = 9

Answer 2

You must draw a right triangle and label the hypotenuse as 41 and the longest leg (horizontal in this case, since we are to determine the height) as 40. We do not know the height so call it h.

 

Now the Pythagorean (sp /) Theorem states that the square of the hyp. is equal to the sum of the squares of the two legs.

 

so, 41 ^ 2 = 40 ^ 2 + h ^ 2

 

41 ^ 2 = 1681   40 ^ 2 = 1600     Therefore h ^2 must be equal to 81.   This means that h = 9.

 

The height is 9 inches.