MCQ
$\triangle\text{ABC}$ is such that $AB = 3\ cm, BC = 2\ cm$ and $CA = 2.5\ cm.$ If $\triangle\text{DEF}\sim\triangle\text{ABC}$ and $EF = 4\ cm,$ then perimeter of $\triangle\text{DEF}$ is:
- A$7.5\ cm.$
- ✓$15\ cm.$
- C$22.5\ cm.$
- D$30\ cm.$
✓
Answer
Correct option: B.
$15\ cm.$
$\triangle\text{DEF}\sim\triangle\text{ABC}$
$AB = 3\ cm, BC = 2\ cm, CA = 2.5\ cm, EF = 4\ cm.$
$\triangle\text{s}$ are similar
$\frac{\text{DE}}{\text{AB}}=\frac{\text{EF}}{\text{BC}}=\frac{\text{FD}}{\text{CA}}$
$\Rightarrow\frac{\text{DE}}{3}=\frac{4}{2}=\frac{\text{FD}}{2.5}$
Now $\frac{\text{DE}}{3}=\frac{4}{2}$
$\Rightarrow\text{DE}=\frac{3\times4}{2}=6\text{cm}$
and $\text{FD}=\frac{4}{2}$
$\Rightarrow\text{FD}=\frac{4\times2.5}{2}=5\text{cm}$
$\therefore$ Perimeter of $\triangle\text{DEF}$
$=6+4+5$
$=15\text{cm}.$
$AB = 3\ cm, BC = 2\ cm, CA = 2.5\ cm, EF = 4\ cm.$
$\triangle\text{s}$ are similar
$\frac{\text{DE}}{\text{AB}}=\frac{\text{EF}}{\text{BC}}=\frac{\text{FD}}{\text{CA}}$
$\Rightarrow\frac{\text{DE}}{3}=\frac{4}{2}=\frac{\text{FD}}{2.5}$
Now $\frac{\text{DE}}{3}=\frac{4}{2}$
$\Rightarrow\text{DE}=\frac{3\times4}{2}=6\text{cm}$
and $\text{FD}=\frac{4}{2}$
$\Rightarrow\text{FD}=\frac{4\times2.5}{2}=5\text{cm}$
$\therefore$ Perimeter of $\triangle\text{DEF}$
$=6+4+5$
$=15\text{cm}.$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If the median of the data $4, 7, x - 1, x - 3, 16, 25,$ written in ascending order, is $13$ then $x$ is equal to
→The total surface area of a cube is $864cm^2$. Its volume is:
→$\text{ABCD}$ is a rectangle whose three vertices are $B(4, 0), C(4, 3)$ and $D(0, 3)$. The length of one of its diagonal is:
→If $7\tan\theta=4$ then $\frac{(7\sin\theta-3\cos\theta)}{(7\sin\theta+3\cos\theta)} =?$
→If $A(1, 3), B(-1, 2), C(2, 5)$ and $D(x, 4)$ are the vertices of $\ce{a || gm ABCD}$ then the value of $x$ is:
→If the circumference of a circle and the perimeter of a square are equal, then:
If one root of $5x^2+ 13x + k = 0$ be the reciprocal of the other root, then the value of k is:
→In the figure, if $TP$ and $TQ$ are tangents drawn from an external point $T$ to a circle with centre $O$ such that$\angle\text{TQP}=60^\circ,$ then:
→If $\sin\text{A}+\sin^2\text{A}=1$ then $\cos^2\text{A}+\cos^4\text{A}=?$
→A cylinder with base radius of $8 \ cm$ and height of $2\ cm$ is melted to form a cone of height $6\ cm$. The radius of the cone is:
→