ANGLE BISECTOR! Find X Value by using the Angle Bisector Theorem & Law of Cosines | Simple Tutorial - Free Printable

Yes, we can find the value of $ x $, which represents the height (altitude) from the apex of the triangle to the base of length 12. The triangle has:

- Two sides: 10 and 8
- Base: 12
- Height: $ x $ (unknown)

We are to find $ x $ using two methods.

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🔷 Method 1: Using Area with Heron’s Formula

We'll compute the area of the triangle using Heron’s formula, then use that area to solve for $ x $, since area = $ \frac{1}{2} \times \text{base} \times \text{height} $.

#### Step 1: Label the sides
Let:
- $ a = 10 $
- $ b = 8 $
- $ c = 12 $

#### Step 2: Compute the semi-perimeter $ s $
$$
s = \frac{a + b + c}{2} = \frac{10 + 8 + 12}{2} = \frac{30}{2} = 15
$$

#### Step 3: Use Heron’s formula for area
$$
\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}
$$
$$
= \sqrt{15(15 - 10)(15 - 8)(15 - 12)} = \sqrt{15 \times 5 \times 7 \times 3}
$$
$$
= \sqrt{15 \times 5 \times 7 \times 3} = \sqrt{(15 \times 3) \times (5 \times 7)} = \sqrt{45 \times 35}
$$
$$
= \sqrt{1575}
$$

Now simplify $ \sqrt{1575} $:
$$
1575 = 25 \times 63 = 25 \times 9 \times 7 = 5^2 \times 3^2 \times 7
$$
$$
\Rightarrow \sqrt{1575} = 5 \times 3 \times \sqrt{7} = 15\sqrt{7}
$$

So, Area = $ 15\sqrt{7} $

#### Step 4: Use area formula with base and height
$$
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times x = 6x
$$
Set equal:
$$
6x = 15\sqrt{7}
\Rightarrow x = \frac{15\sqrt{7}}{6} = \frac{5\sqrt{7}}{2}
$$

✔ So, $ x = \frac{5\sqrt{7}}{2} $ (exact value)

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🔷 Method 2: Using the Pythagorean Theorem (Splitting the Triangle)

Let’s drop the altitude $ x $ from the top vertex to the base of 12. This splits the base into two segments, say $ a $ and $ b $, such that:
$$
a + b = 12
$$

Now, this creates two right triangles:
- Left triangle: hypotenuse = 10, one leg = $ x $, other leg = $ a $
- Right triangle: hypotenuse = 8, one leg = $ x $, other leg = $ b $

Apply the Pythagorean theorem to both:

1. $ a^2 + x^2 = 10^2 = 100 $
2. $ b^2 + x^2 = 8^2 = 64 $

Also, $ a + b = 12 $

Let’s solve this system.

From equation (1):
$$
a^2 = 100 - x^2
\quad \text{(i)}
$$

From equation (2):
$$
b^2 = 64 - x^2
\quad \text{(ii)}
$$

But $ b = 12 - a $, so substitute into (ii):

$$
(12 - a)^2 = 64 - x^2
\Rightarrow 144 - 24a + a^2 = 64 - x^2
$$

Now substitute $ a^2 = 100 - x^2 $ from (i):

$$
144 - 24a + (100 - x^2) = 64 - x^2
$$

Simplify:
$$
144 - 24a + 100 - x^2 = 64 - x^2
\Rightarrow 244 - 24a - x^2 = 64 - x^2
$$

Cancel $ -x^2 $ from both sides:
$$
244 - 24a = 64
\Rightarrow 24a = 244 - 64 = 180
\Rightarrow a = \frac{180}{24} = 7.5
$$

Then $ b = 12 - 7.5 = 4.5 $

Now plug back into equation (i):
$$
a^2 + x^2 = 100
\Rightarrow (7.5)^2 + x^2 = 100
\Rightarrow 56.25 + x^2 = 100
\Rightarrow x^2 = 43.75
$$

Now convert 43.75 to fraction:
$$
43.75 = \frac{175}{4}
\Rightarrow x = \sqrt{\frac{175}{4}} = \frac{\sqrt{175}}{2}
$$

Simplify $ \sqrt{175} $:
$$
175 = 25 \times 7 \Rightarrow \sqrt{175} = 5\sqrt{7}
\Rightarrow x = \frac{5\sqrt{7}}{2}
$$

✔ Same result!

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✔ Final Answer:

$$
\boxed{x = \frac{5\sqrt{7}}{2}}
$$

This is approximately:
$$
\frac{5 \times 2.6458}{2} \approx \frac{13.229}{2} \approx 6.614
$$

So, $ x \approx 6.61 $ units.

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🎯 Summary of Two Methods:

1. Heron’s Formula: Used to find area first, then solved for height.
2. Pythagorean Theorem: Split the triangle, used algebra to find the segments, then found height.

Both lead to the same exact answer: $ \boxed{\frac{5\sqrt{7}}{2}} $.