How do you find the area of a rectangle whose diagonal measure 12 inches and whose width measures 6 inches?

1 Answer
Apr 28, 2016

Answer:

Area = width x height

#=> area = 6sqrt3xx6 = 36sqrt(3)" inches"^2#

Explanation:

Tony B

Let the width be #w#
Let the height be #h#
Let the diagonal be #d#

Using Pythagoras

#d^2=w^2+h^2#

We need to determine the value of #h#

Subtract #w^2# from both sides

#d^2-w^2=w^2-w^2+h^2#

But #w^2-w^2=0#

#d^2-w^2=h^2#

Square root both sides

#sqrt(h^2)=sqrt(d^2-w^2)#

But #sqrt(h^2)=h#

#h=sqrt(d^2-w^2)#

Substituting known values

#h=sqrt(12^2-6^2)#

#h=sqrt(108)~~10.39# This is only an approximate value.
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Finding the exact value

Tony B

#h=sqrt(s^2xx3^2xx3)#

#h=2xx3xxsqrt(3)#

#h=6sqrt(3)" "# as an exact value
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Area = width x height

#=> area = 6sqrt3xx6 = 36sqrt(3)" inches"^2#