The sides of an equilateral triangle are(2p-q+5) and(2q-p+2)  If area of the triangle is root3/4.k  then k= 
class 9 correct ans

To solve for \( k \) in the area expression \( \frac{\sqrt{3}}{4}k \) for the equilateral triangle, we first note the property that in an equilateral triangle, all sides are equal.

Given side lengths:
1. \( 2p - q + 5 \)
2. \( 2q - p + 2 \)

Since the triangle is equilateral, we set the sides equal to each other:

\[
2p - q + 5 = 2q - p + 2
\]

Now, let's solve this equation.

1. **Rearrange the equation:**

\[
2p + p = 2q + q + 2 - 5
\]

\[
3p = 3q - 3
\]

\[
p = q - 1
\]

2. Substitute \( p = q - 1 \) back into one of the side equations to find the side length in terms of \( q \):

Substitute into \( 2p - q + 5 \):

\[
2(q - 1) - q + 5 = 2q - 2 - q + 5 = q + 3
\]

So, the side length of the equilateral triangle is \( q + 3 \).

3. **Calculate the area using the formula for the area of an equilateral triangle:**

\[
\text{Area} = \frac{\sqrt{3}}{4} \times \text{side}^2
\]

Given that the area is \( \frac{\sqrt{3}}{4} \times k \), equating this with the area from the side:

\[
\frac{\sqrt{3}}{4} \times (q + 3)^2 = \frac{\sqrt{3}}{4} \times k
\]

Therefore,

\[
k = (q + 3)^2
\]

Thus, the value of \( k \) is \( (q + 3)^2 \).

### Would you like more details, or do you have any further questions?

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**Related Questions:**

1. What is the value of \( q \) if \( p = 2 \)?
2. Can you find \( p \) and \( q \) if \( k = 25 \)?
3. What is the perimeter of the triangle if \( q = 2 \)?
4. How does the area change if \( q \) increases by 1?
5. Can you solve the equation if given \( p \) and \( q \) explicitly?

**Tip:** When solving for the sides of an equilateral triangle, remember that setting the sides equal often simplifies the problem quickly.

Learn how to calculate the area of an equilateral triangle when side lengths are given as algebraic expressions. This problem involves applying the formula for the area of an equilateral triangle and solving algebraic equations. Suitable for Class 9 students.

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