Two corners of an isosceles triangle are at #(4 ,2 )# and #(1 ,5 )#. If the triangle's area is #64 #, what are the lengths of the triangle's sides?

Answer 1

#color(blue)(a=b=sqrt(32930)/6 and c=3sqrt(2)#

Let #A=(4,2)# and #B=(1,5)#
If #AB# is the base of an isosceles triangle then #C=(x,y)# is the vertex at the altitude.
Let The sides be # a,b,c#, #a=b#

Let h be the height, bisecting AB and passing through point C:

Length #AB = sqrt((4-1)^2+(2-5)^2)=sqrt(18)=3sqrt(2)#
To find #h#. We are given area equals 64:
#1/2AB*h=64#
#1/2(3sqrt(2))h=64=>h=(64sqrt(2))/3#

By Pythagoras' theorem:

#a=b=sqrt(((3sqrt(2))/2)^2+((64sqrt(2))/3)^2)=sqrt(32930)/6#

So the lengths of the sides are:

#color(blue)(a=b=sqrt(32930)/6 and c=3sqrt(2)#
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Answer 2

To find the lengths of the sides of the isosceles triangle, we first need to determine the coordinates of the third vertex. Since the triangle is isosceles, the third vertex will be directly below the midpoint of the base.

Let's find the midpoint of the base:

Midpoint = (\left(\frac{{x_1 + x_2}}{2}, \frac{{y_1 + y_2}}{2}\right))

Midpoint = (\left(\frac{{4 + 1}}{2}, \frac{{2 + 5}}{2}\right))

Midpoint = (\left(\frac{5}{2}, \frac{7}{2}\right))

Now, we have the midpoint of the base. To find the length of the altitude (height) of the triangle, we can calculate the distance between the midpoint and the top vertex.

Distance formula:

(d = \sqrt{{(x_2 - x_1)^2 + (y_2 - y_1)^2}})

(d = \sqrt{{\left(\frac{5}{2} - 4\right)^2 + \left(\frac{7}{2} - 2\right)^2}})

(d = \sqrt{{\left(\frac{-3}{2}\right)^2 + \left(\frac{3}{2}\right)^2}})

(d = \sqrt{{\frac{9}{4} + \frac{9}{4}}})

(d = \sqrt{{\frac{18}{4}}})

(d = \sqrt{{\frac{9}{2}}})

(d = \frac{3\sqrt{2}}{2})

Now, we can use the formula for the area of a triangle:

Area = (\frac{1}{2} \times \text{base} \times \text{height})

64 = (\frac{1}{2} \times \text{base} \times \frac{3\sqrt{2}}{2})

Base = (\frac{{2 \times 64}}{{3\sqrt{2}}})

Base = (\frac{{128}}{{3\sqrt{2}}})

Now, we have the base of the triangle. Since the triangle is isosceles, the lengths of the other two sides are equal to the base. Therefore, the lengths of the sides of the triangle are (\frac{{128}}{{3\sqrt{2}}}) and (\frac{{128}}{{3\sqrt{2}}}).

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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.

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.

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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.

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