Question

each figure shows a triangle with its three angle bisectors intersecting at point p

1. ( pl = 6 ). find ( ph ).
2. find ( pv ) if ( pw = 3 ).
3. find ( pr ) if ( pa = 5 ).
4. find ( pd ) if ( pc = 8 ).
5. ( pv = 2 ) and ( hp = 3 ). find ( hy ).
6. find ( ap ) if ( pq = 1 ) and ( ar = 2 ).

Explanation:

Let's take problem 7: \( PY = 2 \) and \( HP = 3 \), find \( HY \). We can use the Pythagorean theorem here.

Step 1: Identify the right triangle

We have a right triangle \( \triangle HPY \) where \( PY = 2 \) (one leg), \( HP = 3 \) (another leg), and \( HY \) is the hypotenuse. The Pythagorean theorem states that for a right triangle with legs \( a \) and \( b \), and hypotenuse \( c \), \( a^2 + b^2 = c^2 \).

Step 2: Apply the Pythagorean theorem

Let \( a = PY = 2 \), \( b = HP = 3 \), and \( c = HY \). Then:

$$ HY^2 = PY^2 + HP^2 $$

Substitute the values:

$$ HY^2 = 2^2 + 3^2 $$

$$ HY^2 = 4 + 9 $$

$$ HY^2 = 13 $$

Take the square root of both sides:

$$ HY = \sqrt{13} $$

Answer:

\( \sqrt{13} \)