!10325|281x500 [Link Silindi]
The question involves finding the unknown lengths (x) in four different triangle problems.
Answer:
Below, I will solve each question one by one, explaining the reasoning and steps clearly.
1) First Triangle
Given:
- Triangle ( \triangle ABC ) with ( AB = 3 ), ( AC = 5 ), ( BC = 9 + x ).
- Point (D) on (BC), with (BD = 9) and (DC = x).
- (AD) is the bisector of angle (A).
Solution:
By the angle bisector theorem:
\frac{BD}{DC} = \frac{AB}{AC}
Substitute values:
\frac{9}{x} = \frac{3}{5}
Solve for (x):
9 \times 5 = 3 \times x \implies 45 = 3x \implies x = \frac{45}{3} = 15
2) Second Triangle
Given:
- Triangle ( \triangle ABC ) with (AB = x), (AC = 12), (BC = 8+3=11).
- (D) is on (BC) such that (BD = 3), (DC = 8).
- (AD) is the bisector of angle (A).
- (AD = 9).
Solution:
By the angle bisector theorem:
\frac{BD}{DC} = \frac{AB}{AC}
Put the known values:
\frac{3}{8} = \frac{x}{12}
Multiply both sides by 12:
x = \frac{3}{8} \times 12 = \frac{36}{8} = 4.5
3) Third Triangle
Given:
- Triangle ( \triangle ABC ), (AB = 6), (AC = 9), (BC=15).
- (D) lies on (AB) such that (BD = x), (AD = 6 - x).
- (DC) is the angle bisector of (\angle C).
- (DC = ? = x).
Solution:
Since (DC) is the bisector of (\angle C), by the angle bisector theorem we get:
\frac{BD}{DA} = \frac{BC}{AC}
Note: Above relation uses sides opposite to angles. But since (D) divides (AB), the theorem applies on segment division proportional to opposite sides.
So:
\frac{x}{6 - x} = \frac{15}{9} = \frac{5}{3}
Now solve for (x):
3x = 5(6 - x) \implies 3x = 30 - 5x \implies 3x + 5x = 30 \implies 8x = 30 \implies x = \frac{30}{8} = 3.75
4) Fourth Triangle
Given:
- Right triangle ( \triangle ABC ), right angle at (B).
- (BD = 6), (DC = 10), (AD) is altitude from (A) to (BC).
- Find the length of (AC).
Solution:
Since (AD) is altitude to hypotenuse (BC), by right triangle properties:
BD \times DC = AD^2
However, (AD) is the altitude, and is not given. The hypotenuse (BC = BD + DC = 6 + 10 = 16).
Also, because (B) is the right angle, side (AC) is the hypotenuse.
Using the Pythagorean theorem:
AB^2 + BC^2 = AC^2
But (AB) length is not given. Instead, we can use the property of the foot of the altitude dividing the hypotenuse.
The length (AC) can be found by using the relation:
AC^2 = BC \times DC
But it is the side opposite (B), so the lengths correspond.
More explicitly, in right triangle, the leg adjacent to (D) satisfies:
AB^2 = BD \times BC
Similarly:
AC^2 = DC \times BC
Given:
BC = 16, \quad DC = 10
Then,
AC^2 = 16 \times 10 = 160
So,
AC = \sqrt{160} = \sqrt{16 \times 10} = 4 \sqrt{10}
Summary Table
| Problem No | Known Values | Formula Used | Result |
|---|---|---|---|
| 1 | (BD=9), (AB=3), (AC=5) | (\frac{BD}{DC} = \frac{AB}{AC}) | (x=15) |
| 2 | (BD=3), (DC=8), (AC=12) | (\frac{BD}{DC} = \frac{AB}{AC}) | (x=4.5) |
| 3 | (AB=6), (BC=15), (AC=9) | (\frac{BD}{DA} = \frac{BC}{AC}) | (x=3.75) |
| 4 | Right triangle, (BD=6), (DC=10), (BC=16) | (AC^2 = BC \times DC) | (AC=4\sqrt{10}) |
Final Notes:
- Problems 1, 2, and 3 are solved using the angle bisector theorem, which states the ratio of the two divided sides equals the ratio of the other two adjacent sides.
- Problem 4 is a right triangle with altitude to hypotenuse, using geometric mean properties.
I hope this detailed explanation helps! If you want, I can provide more detailed steps or diagrams.