In a triangle `ABC,` medians `AD and CE` are drawn. If `AD=5, angle DAC=pi/8 and angle ACE=pi/4`then the area of the triangle `ABC` is equal to `(5a)/b,` then `a+b` is equal to

Answer

To solve the problem, we need to find the area of triangle ABC given the lengths of the medians and the angles. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We have a triangle ABC with medians AD and CE. We know:
- Length of median AD = 5
- Angle DAC = π/8
- Angle ACE = π/4

We need to find the area of triangle ABC in the form \( \frac{5a}{b} \) and then determine \( a + b \).

### Step 2: Identify the Centroid
Let O be the centroid of triangle ABC. The centroid divides each median in the ratio 2:1. Therefore, AO = \( \frac{2}{3} \) of AD.

\[ AO = \frac{2}{3} \times 5 = \frac{10}{3} \]

### Step 3: Use the Sine Rule in Triangle AOC
In triangle AOC, we can apply the sine rule. The sides opposite angles DAC and ACE are AO and OC respectively.

Using the sine rule:
\[ \frac{OC}{\sin(\angle DAC)} = \frac{AO}{\sin(\angle ACE)} \]

Substituting the known values:
\[ \frac{OC}{\sin(\frac{\pi}{8})} = \frac{\frac{10}{3}}{\sin(\frac{\pi}{4})} \]

Since \( \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \):
\[ \frac{OC}{\sin(\frac{\pi}{8})} = \frac{\frac{10}{3}}{\frac{\sqrt{2}}{2}} = \frac{20}{3\sqrt{2}} \]

Thus,
\[ OC = \frac{20}{3\sqrt{2}} \sin(\frac{\pi}{8}) \]

### Step 4: Calculate the Area of Triangle AOC
The area of triangle AOC can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times AO \times OC \times \sin(\angle AOC) \]

Since \( \angle AOC = \pi - (\angle DAC + \angle ACE) = \pi - \left(\frac{\pi}{8} + \frac{\pi}{4}\right) = \pi - \frac{3\pi}{8} = \frac{5\pi}{8} \):
\[ \sin(\angle AOC) = \sin\left(\frac{5\pi}{8}\right) = \cos\left(\frac{\pi}{8}\right) \]

Now substituting the values:
\[ \text{Area}_{AOC} = \frac{1}{2} \times \frac{10}{3} \times \left(\frac{20}{3\sqrt{2}} \sin\left(\frac{\pi}{8}\right)\right) \times \cos\left(\frac{\pi}{8}\right) \]

### Step 5: Simplify the Area of Triangle AOC
\[ \text{Area}_{AOC} = \frac{100}{6\sqrt{2}} \sin\left(\frac{\pi}{8}\right) \cos\left(\frac{\pi}{8}\right) \]
Using the identity \( 2 \sin A \cos A = \sin(2A) \):
\[ \text{Area}_{AOC} = \frac{50}{6\sqrt{2}} \sin\left(\frac{\pi}{4}\right) = \frac{50}{6\sqrt{2}} \cdot \frac{\sqrt{2}}{2} = \frac{25}{6} \]

### Step 6: Find the Area of Triangle ABC
The area of triangle ABC is three times the area of triangle AOC:
\[ \text{Area}_{ABC} = 3 \times \frac{25}{6} = \frac{75}{6} = \frac{25}{2} \]

### Step 7: Express the Area in the Required Form
We need to express the area in the form \( \frac{5a}{b} \):
\[ \frac{25}{2} = \frac{5 \times 5}{2} \]
Here, \( a = 5 \) and \( b = 2 \).

### Step 8: Calculate \( a + b \)
Now, we calculate \( a + b \):
\[ a + b = 5 + 2 = 7 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{7} \]