In a triangle ABC, `AB = 7 cm, AC = 5 sqrt2 cm " also " angle A = 135^@` and point E and F are the mid points of side AB and AC respectively then what will be the length of EF?

Options

A. A) 12 cm
B. B) 14 cm
C. C) 13/2 cm
D. D) 16 cm

Answer

To find the length of EF in triangle ABC, where AB = 7 cm, AC = 5√2 cm, and angle A = 135°, we can follow these steps:

### Step 1: Use the Cosine Rule to find BC
We start by applying the cosine rule, which states:
\[ BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(A) \]

### Step 2: Substitute the known values
Given:
- \( AB = 7 \, \text{cm} \)
- \( AC = 5\sqrt{2} \, \text{cm} \)
- \( \angle A = 135^\circ \)

First, we need to calculate \( \cos(135^\circ) \):
\[ \cos(135^\circ) = \cos(90^\circ + 45^\circ) = -\sin(45^\circ) = -\frac{1}{\sqrt{2}} \]

Now substitute the values into the cosine rule:
\[ BC^2 = 7^2 + (5\sqrt{2})^2 - 2 \cdot 7 \cdot 5\sqrt{2} \cdot \left(-\frac{1}{\sqrt{2}}\right) \]

### Step 3: Calculate each term
Calculating each term:
- \( 7^2 = 49 \)
- \( (5\sqrt{2})^2 = 50 \)
- \( -2 \cdot 7 \cdot 5\sqrt{2} \cdot \left(-\frac{1}{\sqrt{2}}\right) = 2 \cdot 7 \cdot 5 = 70 \)

Putting it all together:
\[ BC^2 = 49 + 50 + 70 = 169 \]

### Step 4: Solve for BC
Taking the square root:
\[ BC = \sqrt{169} = 13 \, \text{cm} \]

### Step 5: Use the Midpoint Theorem to find EF
According to the midpoint theorem, the length of EF (the line segment joining the midpoints of sides AB and AC) is half the length of BC:
\[ EF = \frac{1}{2} \cdot BC = \frac{1}{2} \cdot 13 = 6.5 \, \text{cm} \]

### Final Answer
The length of EF is \( 6.5 \, \text{cm} \).

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