Two corners of an isosceles triangle are at #(5 ,4 )# and #(9 ,2 )#. If the triangle's area is #36 #, what are the lengths of the triangle's sides?

Question

Two corners of an isosceles triangle are at #(5  ,4   )# and #(9   ,2   )#.  If the triangle's area is #36   #, what are the lengths of the triangle's sides?

Sadie Allen · Accepted Answer

Tony B s=sqrt( ( (2sqrt(5))/2)^2 +(36/sqrt(5))^2# #=>s=sqrt( ( 5 +(36^2)/5)# #s=sqrt( (25+36^2)/5) = sqrt(1321/5)# #s~~16.254#

Ella Armstrong · Answer

Support of the decision that the given points are for the base of the triangle.

Suppose the coordinates give were not for the base of an Isosceles triangle but for one of the other two sides. Then we would have:

Where

#x=2sqrt(5)xxsin(theta) #
#h=2sqrt(5)xxcos(theta)#

Given that Area# = 36 = x xx h#

Thus we have:

#" "color(blue)(36=(2sqrt(5)color(white)(.))^2( sin(theta)cos(theta)))#

#color(brown)("Using the Trig Identity of "sin(2theta)=2sin(theta)cos(theta))#

#" "color(brown)(36=20( sin(theta)cos(theta)) ->)color(blue)(36=20/2sin(2theta))#

#=> sin(2theta)=72/20#

But #" "-1<=sin(2theta)<=+1#

and #72/20>+1# so there is a#" "color(red)(underline("contradiction"))#

Implying that this scenario of #2sqrt(5)# not being the base is false.

#color(magenta)("The length of "2sqrt(5)" applies to the base of the triangle")#

Claire Akin · Answer

undefined Given the coordinates of two corners of the isosceles triangle, we can find the length of its base and height. Then, using the formula for the area of a triangle ($ A = \frac{1}{2} \times \text{base} \times \text{height} $), we can find the lengths of the triangle's sides.

Let's denote the third vertex of the triangle as $ (x, y) $. Since the triangle is isosceles, the distance from this vertex to each of the other vertices should be equal.

The length of the base ($ b $) can be calculated using the distance formula:

$$ b = \sqrt{(9 - 5)^2 + (2 - 4)^2} $$

$$ b = \sqrt{(4)^2 + (-2)^2} $$

$$ b = \sqrt{16 + 4} $$

$$ b = \sqrt{20} $$

$$ b = 2\sqrt{5} $$

The height ($ h $) of the triangle can be calculated by finding the perpendicular distance from the third vertex to the line containing the base. We use the formula for the distance between a point and a line:

$$ h = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} $$

Where $ (x_1, y_1) $ is the coordinates of the third vertex and $ Ax + By + C = 0 $ is the equation of the line containing the base.

Using the given points (5, 4) and (9, 2), we can find the equation of the line containing the base. The slope of this line is:

$$ m = \frac{2 - 4}{9 - 5} = \frac{-2}{4} = -\frac{1}{2} $$

The equation of the line can be expressed as:

$$ y - 4 = -\frac{1}{2}(x - 5) $$

$$ y - 4 = -\frac{1}{2}x + \frac{5}{2} $$

$$ 2y - 8 = -x + 5 $$

$$ x + 2y - 13 = 0 $$

Now, we can use the formula for the distance between a point and a line:

$$ h = \frac{|x_1 + 2y_1 - 13|}{\sqrt{1^2 + 2^2}} $$

$$ h = \frac{|x_1 + 2y_1 - 13|}{\sqrt{5}} $$

Given that the area ($ A $) of the triangle is 36, and $ A = \frac{1}{2}bh $, we have:

$$ 36 = \frac{1}{2} \times 2\sqrt{5} \times \frac{|x_1 + 2y_1 - 13|}{\sqrt{5}} $$

$$ 36 = |x_1 + 2y_1 - 13| $$

$$ |x_1 + 2y_1 - 13| = 36 $$

Since the triangle is isosceles, the distance from the third vertex to the line containing the base should be 36 units. Therefore:

$$ x_1 + 2y_1 - 13 = 36 $$ or $$ x_1 + 2y_1 - 13 = -36 $$

$$ x_1 + 2y_1 = 49 $$ or $$ x_1 + 2y_1 = -23 $$

Using these equations, we can find the coordinates of the third vertex. Let's take the first equation:

$$ x_1 + 2y_1 = 49 $$

Given that $ (x_1, y_1) $ lies on the line, we have:

$$ x_1 + 2y_1 = 49 $$

$$ x_1 + 2(4) = 49 $$

$$ x_1 + 8 = 49 $$

$$ x_1 = 41 $$

Now, substitute $ x_1 = 41 $ into one of the given equations:

$$ 41 + 2y_1 = 49 $$

$$ 2y_1 = 8 $$

$$ y_1 = 4 $$

So, the third vertex is $ (41, 4) $. Now, we can find the lengths of the sides of the triangle using the distance formula:

$$ \text{Side 1} = \sqrt{(41 - 5)^2 + (4 - 4)^2} = \sqrt{36^2} = 36 $$

$$ \text{Side 2} = \sqrt{(41 - 9)^2 + (4 - 2)^2} = \sqrt{32^2 + 2^2} = \sqrt{1040} $$

$$ \text{Side 3} = \sqrt{(9 - 5)^2 + (2 - 4)^2} = \sqrt{16 + 4} = \sqrt{20} $$

So, the lengths of the triangle's sides are 36, $ \sqrt{1040} $, and $ \sqrt{20} $.