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The diagonal of a rectangle is 3 times as long as the width. If the length is 4 cm, what is the width of the rectangle?

Answer 1

Savannah Aguilar

#"width "=sqrt2#

#"the diagonal 'splits' the rectangle into 2 right triangles"#

#"let width "=w" then diagonal "=3w#

#"solve the triangle with legs w and 4 and hypotenuse 3w"#

#"using "color(blue)"Pythagoras' theorem"#

#rArr(3w)^2=w^2+4^2#

#rArr9w^2=w^2+16#

#"subtract "w^2" from both sides"#

#rArr8w^2=16#

#"divide both sides by 8"#

#rArrw^2=2#

#color(blue)"take the square root of both sides"#

#sqrt(w^2)=+-sqrt2larrcolor(blue)"note plus or minus"#

#rArrw=+-sqrt2#

#w>0rArrw=sqrt2#

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Answer 2

Aurora Ayala

#w= sqrt2# cm

Let #d# represent the diagonal Let #l# represent the length Let #w# represent the width

#d= 3w# #l=4 #

Since the angles in a rectangle are #90^@# each: #l^2+w^2= d^2#

Substitute: #4^2+w^2= (3w)^2#

#16+w^2= 9w^2#

Set the expression equal to 0: #8w^2-16=0#

Factor with GCF: #8(w^2-2)=0#

#w^2-2=0# #w= +-sqrt2# Since distance can't be negative: #w= sqrt2# cm

Solving for #d#: #d= 3*sqrt2# #d=3sqrt2# cm

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.