Two corners of an isosceles triangle are at #(4 ,9 )# and #(9 ,3 )#. If the triangle's area is #64 #, what are the lengths of the triangle's sides?

Question

Two corners of an isosceles triangle are at #(4   ,9   )# and #(9  ,3    )#.  If the triangle's area is #64   #, what are the lengths of the triangle's sides?

Wyatt Anderson · Accepted Answer

<p><img alt="enter image source here" src="https://tutor.hix.ai The sides are:<br>
Base, <mathjax>#b = bar(AB) = 7.8#</mathjax><br>
Equal sides, <mathjax>#bar(AC) = bar(BC) = 16.8 # #A_Delta = 1/2 bh = 64 #
Using the distance formula find b...
#b = sqrt((x_2-x_1)^2 + (y_2 - y_1)^2)#
#x_1 = 4; x_2 = 9; y_1 = 9; y_2 = 3 #
substitute and find h:
#b = sqrt(25 + 36) = sqrt(61) ~~ 7.81#
#h = 2(64)/sqrt(61) = 16.4#
Now using Pythagoras theorem find the sides, #barAC#:
#barAC = sqrt(61/4 + 128^2/61) = sqrt((3,721 + 65,536)/2) = 16.8#

Sophia Alfred · Answer

undefined To find the lengths of the sides of the isosceles triangle, follow these steps:

1. Calculate the distance between the two given points to find the base of the triangle.
2. Use the formula for the area of a triangle to find the height.
3. Calculate the lengths of the equal sides using the Pythagorean theorem.

1. Distance between the two points:
$$ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

Using the given points $(4, 9)$ and $(9, 3)$:
$$ \text{Distance} = \sqrt{(9 - 4)^2 + (3 - 9)^2} $$
$$ \text{Distance} = \sqrt{5^2 + (-6)^2} $$
$$ \text{Distance} = \sqrt{25 + 36} $$
$$ \text{Distance} = \sqrt{61} $$

2. Area of the triangle:
$$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $$

Given that the area is 64 and the base is $\sqrt{61}$:
$$ 64 = \frac{1}{2} \times \sqrt{61} \times \text{height} $$
$$ 128 = \sqrt{61} \times \text{height} $$
$$ \text{height} = \frac{128}{\sqrt{61}} $$

3. Using the Pythagorean theorem to find the lengths of the equal sides:
Let $ s $ be the length of each equal side.
$$ s^2 = \left( \frac{\sqrt{61}}{2} \right)^2 + \left( \frac{128}{\sqrt{61}} \right)^2 $$

Solving for $ s $:
$$ s^2 = \frac{61}{4} + \frac{16384}{61} $$
$$ s^2 = \frac{15284}{61} $$
$$ s = \sqrt{\frac{15284}{61}} $$
$$ s = \sqrt{250.56} $$
$$ s = 15.82 $$ (rounded to two decimal places)

Therefore, the lengths of the triangle's sides are approximately 15.82 units each.

Aurora Ayala · Answer

undefined To find the lengths of the sides of the isosceles triangle, we first need to determine the length of the base and the height of the triangle. We can find the length of the base by calculating the distance between the given points (4, 9) and (9, 3) using the distance formula:

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

Plugging in the coordinates, we get:

$$ d = \sqrt{(9 - 4)^2 + (3 - 9)^2} $$

$$ d = \sqrt{5^2 + (-6)^2} $$

$$ d = \sqrt{25 + 36} $$

$$ d = \sqrt{61} $$

Now, we know the length of the base of the triangle is $ \sqrt{61} $.

To find the height of the triangle, we can use the formula for the area of a triangle:

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

Given that the area of the triangle is 64, we can plug in the values:

$$ 64 = \frac{1}{2} \times \sqrt{61} \times \text{height} $$

$$ 128 = \sqrt{61} \times \text{height} $$

$$ \text{height} = \frac{128}{\sqrt{61}} $$

Now, we have the length of the base and the height of the triangle. With this information, we can use the Pythagorean theorem to find the lengths of the other two sides of the triangle. Since the triangle is isosceles, these two sides will be equal.

Let's denote the length of one of these equal sides as $ s $. Then, using the Pythagorean theorem:

$$ s^2 = \left(\frac{\sqrt{61}}{2}\right)^2 + \left(\frac{128}{\sqrt{61}}\right)^2 $$

$$ s^2 = \frac{61}{4} + \frac{16384}{61} $$

$$ s^2 = \frac{61^2}{4 \times 61} + \frac{16384}{61} $$

$$ s^2 = \frac{61^2 + 16384}{61} $$

$$ s^2 = \frac{3721 + 16384}{61} $$

$$ s^2 = \frac{20105}{61} $$

$$ s = \sqrt{\frac{20105}{61}} $$

So, the lengths of the sides of the triangle are approximately $ \sqrt{\frac{20105}{61}} $.