Question

Two similar triangles ABC and DEF such that the area of △ ABC = 64 sq. unit and area of △DEF =121 sq. unit.

Statement (1): \[\frac{\text{Perimeter of } \triangle DEF}{\text{Perimeter of } \triangle ABC} = \frac{8}{11}.\]

Statement (2): The ratio of perimeters of two similar triangles is equal to the ratio of their areas.

पर्याय

• Both the statements are true.

• Both the statements are false.

• Statement 1 is true, and statement 2 is false.

• Statement 1 is false, and statement 2 is true.

Answer 1

Both the statements are false.

Explanation:

We know that,

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

∴ \[\frac{\text{Area of } \triangle DEF}{\text{Area of } \triangle ABC} = \frac{DE^2}{AB^2}\]

⇒ \[\frac{121}{64} = \frac{DE^2}{AB^2}\]

⇒ \[\frac{AB^2}{DE^2} = \frac{64}{121}\]

⇒ \[\frac{AB}{DE} = \sqrt{\frac{64}{121}}\]

⇒ \[\frac{AB}{DE} = \frac{8}{11}\]

Since, corresponding sides of similar triangle are proportional.

⇒ \[\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}\]

We know that,

For any two or more equal ratios, each ratio is equal to the ratio between sum of their antecedents and sum of their consequents.

⇒ \[\frac{AB}{DE} = \frac{AB + BC + AC}{DE + EF + DF}\]

⇒ \[\frac{\text{Perimeter of } \triangle ABC}{\text{Perimeter of } \triangle DEF} = \frac{AB}{DE}\]

⇒ \[\frac{\text{Perimeter of } \triangle DEF}{\text{Perimeter of } \triangle ABC} = \frac{DE}{AB} = \frac{11}{8}\]

So, statement 1 is false.

The ratio of perimeters of two similar triangles is equal to the ratio of their corresponding sides.

So, statement 2 is false.