Question

Draw five triangles and measure their sides. Check-in each case, if the sum of the lengths of any two sides is always less than the third side.

Answer 1

Step 1. Draw a triangle $\mathrm{ABC}$ with sides $\mathrm{AB}=3\mathrm{cm},\mathrm{BC}=3.5\mathrm{cm},\mathrm{and}\mathrm{AC}=3.9\mathrm{cm}$.

Now, add any two sides $\left(\mathrm{say}\mathrm{AB}\mathrm{and}\mathrm{BC}\right)$ of the triangle $\mathrm{ABC}$:

$$\begin{gathered} AB+BC=3+3.5 \\ =6.5\mathrm{cm}>AC \end{gathered}$$

Step 2. Draw a triangle $\mathrm{ABC}$ with sides $\mathrm{AB}=3\mathrm{cm},\mathrm{BC}=4\mathrm{cm},\mathrm{and}\mathrm{AC}=5\mathrm{cm}$.

Now, add the other two sides$\left(\mathrm{say}\mathrm{AB}\mathrm{and}\mathrm{BC}\right)$ of the $△\mathrm{ABC}$:

$$\begin{gathered} AB+BC=3+4 \\ =7\mathrm{cm}>AC \end{gathered}$$

Step 3. Draw a triangle $\mathrm{ABC}$ with sides $\mathrm{AB}=2.5\mathrm{cm},\mathrm{BC}=2.5\mathrm{cm},\mathrm{and}\mathrm{AC}=2.5\mathrm{cm}$.

Now, add any two sides$\left(\mathrm{say}\mathrm{AB}\mathrm{and}\mathrm{BC}\right)$ say of the $△\mathrm{ABC}$:

$$\begin{gathered} AB+BC=2.5+2.5 \\ =5\mathrm{cm}>AC \end{gathered}$$

Step 4. Draw a triangle $\mathrm{ABC}$ with sides $\mathrm{AB}=3\mathrm{cm},\mathrm{BC}=3.5\mathrm{cm},\mathrm{and}\mathrm{AC}=4\mathrm{cm}$.

Now, add any two sides$\left(\mathrm{say}\mathrm{AB}\mathrm{and}\mathrm{BC}\right)$ of the $△\mathrm{ABC}$.

$$\begin{gathered} AB+BC=3+3.5 \\ =6.5\mathrm{cm}>AC \end{gathered}$$

Step 5. Draw a triangle $\mathrm{ABC}$ with sides $\mathrm{AB}=2\mathrm{cm},\mathrm{BC}=4\mathrm{cm},\mathrm{and}\mathrm{AC}=4.5\mathrm{cm}$.

Add any two sides$\left(\mathrm{say}\mathrm{AB}\mathrm{and}\mathrm{BC}\right)$ of the triangle $△\mathrm{ABC}$:

$$\begin{gathered} AB+BC=2+4 \\ =6\mathrm{cm}>AC \end{gathered}$$

It is observed that in each case, the sum of the lengths of any two sides is greater than the third side.

Hence, the sum of the lengths of any two sides can never be less than the third side.