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MODEL PAPER 2 (BASIC) — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsMODEL PAPER 2 (BASIC)1 Mark
MCQ
$\triangle ABC$ is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. If $\triangle D E F \sim \triangle A B C$ and EF = 4 cm, then perimeter of $\triangle DEF$ is
  • A
    30 cm
  • B
    15 cm
  • C
    22.5 cm
  • D
    7.5 cm
✓

Answer

(B) 15 cm
Explanation:
Image
$\triangle DEF \sim \triangle ABC$
AB = 3CM, BC = 2 CM, CA = 2.5 CM, EF = 4CM
Since $\triangle^{\prime}$ s are similar, we have
$\frac{ DE }{ AB }=\frac{ EF }{ BC }=\frac{ FD }{ CA }$
$\Rightarrow \frac{ DE }{3}=\frac{4}{2}=\frac{ FD }{2.5}$
Now $\frac{ DE }{3}=\frac{4}{2}$
$\Rightarrow DE =\frac{3 \times 4}{2}=6 cm$
and $FD =\frac{4}{2} \Rightarrow FD =\frac{4 \times 2.5}{2}=5 cm$
perimeter of $\triangle D E F$
= 6 + 4 + 5 = 15cm

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