Question

An equilateral triangle with length 4a cm , 3b cm and a +b+c) cm
Find a and b hence find the length of the triangles

Answer 1

In an equilateral triangle, all three sides are equal. Given that the sides of the triangle are \(4a\), \(3b\), and \( (a + b + c) \), we can set up the following equations based on the property of equilateral triangles:

1. \( 4a = 3b \)
2. \( 4a = a + b + c \)
3. \( 3b = a + b + c \)

Now, let's solve these equations step by step.

### From Equation 1:

From \( 4a = 3b \), we can express \( b \) in terms of \( a \):

\[
b = \frac{4a}{3}
\]

### Substitute \( b \) into Equations 2 and 3:

**Substituting into Equation 2:**

Now, substitute \( b = \frac{4a}{3} \) into \( 4a = a + b + c \):

\[
4a = a + \frac{4a}{3} + c
\]

To combine terms, multiply through by 3 to eliminate the fraction:

\[
12a = 3a + 4a + 3c
\]

Simplifying gives:

\[
12a = 7a + 3c
\]

Rearranging gives:

\[
5a = 3c \quad \Rightarrow \quad c = \frac{5a}{3}
\]

**Substituting into Equation 3:**

Now substitute \( b = \frac{4a}{3} \) and \( c = \frac{5a}{3} \) into \( 3b = a + b + c \):

\[
3\left(\frac{4a}{3}\right) = a + \frac{4a}{3} + \frac{5a}{3}
\]

This simplifies to:

\[
4a = a + \frac{9a}{3}
\]

Converting \( \frac{9a}{3} \) to a simpler form gives:

\[
4a = a + 3a
\]

Indeed, this confirms:

\[
4a = 4a
\]

### Now, we can find the triangle side lengths

We know:
- \( b = \frac{4a}{3} \)
- \( c = \frac{5a}{3} \)

### Let's find the lengths:

To find the lengths, let's test the value of \( a \) and find \( b \):

Let \( a = 3 \), then:

\[
b = \frac{4 \times 3}{3} = 4
\]

So, the lengths of the triangle sides would be:

- \( 4a = 4 \times 3 = 12 \) cm
- \( 3b = 3 \times 4 = 12 \) cm
- \( a + b + c = 3 + 4 + \frac{5 \times 3}{3} = 3 + 4 + 5 = 12 \) cm

Thus, all side lengths of the equilateral triangle will confirm out to be 12 cm each:

### Conclusion

The lengths of the triangle sides are:

\[
\text{Each side length} = 12 \text{ cm}
\]