Question

$\Delta \mathrm{ABC}$ is such that $\mathrm{AB}=3 \mathrm{~cm}, \mathrm{BC}=2 \mathrm{~cm}, \mathrm{CA}=2.5 \mathrm{~cm}$. If $\Delta \mathrm{ABC} \sim \Delta \mathrm{DE F}$ and $4 \mathrm{~cm}$, then perimeter of $\triangle \mathrm{DE F}$ is
(a) $7.5 \mathrm{~cm}$
(b) $15 \mathrm{~cm}$
(c) $22.5 \mathrm{~cm}$
(d) $30 \mathrm{~cm}$

Answer 1

Similar Triangles

Let me write the complete question for a better understanding.

Complete question:

ΔABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. If ΔDEF ~ ΔABC and EF = 4 cm, then perimeter of ΔDEF is:

(a) 7.5 cm

(b) 15 cm

(c) 22.5 cm

(d) 30 cm

Genral concept:

Perimeter - The perimeter of a triangle means the sum of all three sides.

BPT - If two triangles are similar, then their corresponding angles are congruent and corresponding sides are in equal proportion.

The given values are,

$\longrightarrow \triangle ABC \sim \triangle D EF$

$\longrightarrow AB = 3 \; cm$

$\longrightarrow BC = 2 \; cm$

$\longrightarrow CA = 2.5 \; cm$

$\longrightarrow EF = 4 \; cm$

We need to find the perimeter of triangle DEF.

Solution:

Since, ΔDEF ~ ΔABC.

We know that if two triangles are similar then their corresponding sides are in equal proportion.

$\implies \dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF}$

Substituting the known values in it, we obtain the following results:

$\implies \dfrac{3}{DE} = \dfrac{2}{4} = \dfrac{2.5}{DF}$

$\implies \dfrac{3}{DE} = \dfrac{1}{2} = \dfrac{2.5}{DF}$

Now from here we should solve for each values so that we can get the values of DE and DF respectively. Solving for side DE:

$\implies \dfrac{3}{DE} = \dfrac{1}{2}$

$\implies DE \times 1 = 3 \times 2$

$\implies DE = 6$

Now similarly solving for the side DF:

$\implies \dfrac{1}{2} = \dfrac{2.5}{DF}$

$\implies DF \times 1 = 2 \times 2.5$

$\implies DF = 5$

Now, we know that the sum of all sides of the triangle is equal to the perimeter of triangle. Here we have to find the perimeter of ΔDEF, the sides of the triangle is DE, EF, and DF respectively. And we have also got the values of all the sides.

$\implies Perimeter_{(\triangle D EF)} = DE + EF + DF$

So by substituting the known values in the perimeter of triangle formula. we get the following results:

$\implies Perimeter_{(\triangle D EF)} = 6 + 4 + 5$

$\implies \boxed{\mathrm{Perimeter_{(\triangle D EF)} = 15 \; cm}}$

Hence the perimeter of triangle DEF is 15cm.

$\rule{300}{2}$

Conclusion:

We were given the value of one side in triangle DEF and needed to fund the values of two sides. So first we funded the values of DE and DF side with the help of BPT [Basic Proportionality Theorem] which was needed for perimeter of triangle.

We then used the perimeter of triangle formula and after substituting the sides values we got the final result.