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Two corners of an isosceles triangle are at #(5 ,2 )# and #(2 ,1 )#. If the triangle's area is #3 #, what are the lengths of the triangle's sides?

Answer 1

Three sides of triangle are #3.16(2dp), 2.47(2dp) ,2.47(2dp)# unit.

The base of the isosceles triangle, #b=sqrt( (x_1-x_2)^2+(y_1-y_2)^2) = sqrt( (5-2)^2+(2-1)^2) =sqrt10=3.16(2dp)unit #

The area of the isosceles triangle is #A_t = 1/2 * b *h =1/2*3.16 *h ; A_t=3 :. h = (2* A_t )/b = (2*3)/3.16=6/3.16= 1.90(2dp) unit#. Where #h# is the altitude of triangle.

The legs of the isosceles triangle are #l_1=l_2= sqrt(h^2+(b/2)^2)=sqrt(1.9^2+(3.16/2)^2) =2.47(2dp) unit#

Hence the length of three sides of triangle are #3.16(2dp), 2.47(2dp) ,2.47(2dp)# unit [Ans]

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Answer 2

Violet Aldrich

The lengths of the triangle's sides are approximately ( \sqrt{13} ), ( \sqrt{13} ), and ( 2\sqrt{13} ) units.

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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.