Tag: similarity of triangles

Questions Related to similarity of triangles

The sides of a triangle are $5$ cm, $6$ cm and $7$ cm. One more triangle is formed by joining the midpoints of the sides. The perimeter of the second triangle is:

  1. $18$ cm

  2. $12$ cm

  3. $9$ cm

  4. $6$ cm


Correct Option: C
Explanation:

Let the $\triangle ABC $ have sides $AB = 5$cm,

$BC = 6$ cm and $AC = 7$cm.
Let the midpoints of the sides AB and AC be points D and E respectively.
$\therefore \dfrac {AD}{DB} = \dfrac {AE}{EC}$         ...By B.P.T

$\therefore \dfrac {AD+DB}{DB} = \dfrac {AE+EC}{EC}$   ....By Componendo

$\therefore \dfrac {AB}{DB} = \dfrac {AC}{EC}$      ......(1)

Also, $\angle BAC \cong \angle DAE$    ....(2)

$\therefore \triangle ABC \sim \triangle ADE$      ....SAS test of similarity

$\therefore \dfrac {AB}{AD} = \dfrac {BC}{DE} = \dfrac {AC}{AE}$       ....C.S.S.T

But $\dfrac {AB}{AD} = \dfrac {AB}{\frac 12 AB} = \dfrac 12$

$\therefore \dfrac {BC}{DE} = \dfrac 12$


Perimeter $(\triangle ADE) = AD + DE + AE$ 
$ = \dfrac 12 AB + \dfrac 12 BC + \dfrac 12 AC$

$= \dfrac 12 \left(AB + BC + AC \right)$

$ = \dfrac 12 \times 18 = 9$ cm.

So, option C is correct.

Point L, M and N lie on the sides AB, BC and CA of the triangle ABC such that $\ell (AL) : \ell (LB) = \ell (BM) : \ell (MC) = \ell (CN) : \ell (NA) = m : n$, then the areas of the triangles LMN and ABC are in the ratio

  1. $\dfrac{m^2}{n^2}$

  2. $\dfrac{m^2 - mn + n^2}{(m + n)^2}$

  3. $\dfrac{m^2 - n^2}{m^2 + n^2}$

  4. $\dfrac{m^2 + n^2}{(m + n)^2}$


Correct Option: A

A man of height 1.8 metre is moving away from a lamp post at the  rate of 1.2 m/sec . If the height of the lamp post be 4.5 metre , then the rate at which the shadow of the  man is lengthening is 

  1. $0.4 m/sec$

  2. $0.8m/sec$

  3. $1.2m/sec$

  4. None of these


Correct Option: B