In an isosceles triangle with base `BC=12 cm and AB=AC=10 cm`. There is a rectangle PQRS inside the triangle whose base PS lies on BC such that PQ=SR=y and QR=PS=2x. Find the value of `x+(3y)/(4)`.

Answer

To solve the problem, we need to find the value of \( x + \frac{3y}{4} \) in the given isosceles triangle with the specified dimensions.

### Step-by-Step Solution:

1. **Draw the Triangle and Rectangle**:
- Draw triangle \( ABC \) with \( AB = AC = 10 \, \text{cm} \) and base \( BC = 12 \, \text{cm} \).
- Place rectangle \( PQRS \) inside triangle \( ABC \) such that \( PS \) lies on \( BC \) and \( PQ = SR = y \) and \( QR = PS = 2x \).

2. **Determine the Height of the Triangle**:
- Since \( ABC \) is an isosceles triangle, we can drop a perpendicular \( AD \) from vertex \( A \) to base \( BC \).
- The midpoint \( D \) divides \( BC \) into two equal segments, so \( BD = CD = 6 \, \text{cm} \).
- Using the Pythagorean theorem in triangle \( ABD \):
\[ AB^2 = AD^2 + BD^2 \]
\[ 10^2 = AD^2 + 6^2 \]
\[ 100 = AD^2 + 36 \]
\[ AD^2 = 64 \implies AD = 8 \, \text{cm} \]

3. **Establish Relationships in the Rectangle**:
- The height of rectangle \( PQRS \) is \( y \) and its base \( PS = QR = 2x \).
- The segments \( PD \) and \( BD \) can be expressed as \( PD = 6 - x \) because \( BD = 6 \, \text{cm} \).

4. **Use Similar Triangles**:
- Triangles \( BPQ \) and \( BDA \) are similar because they share angle \( B \) and both have a right angle.
- Set up the ratio of the sides:
\[ \frac{BP}{BD} = \frac{PQ}{AD} \]
Substituting the values:
\[ \frac{6 - x}{6} = \frac{y}{8} \]
- Cross-multiplying gives:
\[ 8(6 - x) = 6y \]
\[ 48 - 8x = 6y \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ 8x + 6y = 48 \]
- Dividing the entire equation by 2:
\[ 4x + 3y = 24 \]

6. **Express \( y \) in Terms of \( x \)**:
- From the equation \( 4x + 3y = 24 \), we can express \( y \):
\[ 3y = 24 - 4x \implies y = \frac{24 - 4x}{3} \]

7. **Substituting \( y \) into \( x + \frac{3y}{4} \)**:
- Now substitute \( y \) into \( x + \frac{3y}{4} \):
\[ x + \frac{3}{4} \left(\frac{24 - 4x}{3}\right) \]
Simplifying:
\[ x + \frac{24 - 4x}{4} = x + 6 - x = 6 \]

### Final Answer:
Thus, the value of \( x + \frac{3y}{4} \) is \( \boxed{6} \).