Tag: similarity of triangles

Questions Related to similarity of triangles

Two polygons of different number of side ...... be similar.

  1. Cannot

  2. Can

  3. Both A and B

  4. None


Correct Option: A
Explanation:

Ex: A triangle and a square are not similar figures.
Therefore, A  is the correct answer.

If the corresponding angles are equal then the two figures having samenumber of sides are said to be

  1. Figures

  2. Not similar

  3. None of the above

  4. Similar


Correct Option: D
Explanation:

The angles are of the same measure means the figures are similar. 
Therefore, D is the correct answer.

If the same photograph is printed in different sizes , we say it is

  1. Not similar

  2. Similar

  3. Common

  4. None


Correct Option: B
Explanation:

Photograph being same it is printed in the same shape but in passport size, card size, and mini size. They are said to be similar.
Therefore, B is the correct answer.

Two quadrilaterals, a square and a rectangle are not similar as they ......... in shape as well as size.

  1. Differ

  2. Are same

  3. Do not siffer

  4. Angles also differ


Correct Option: A
Explanation:

When two quadrilaterals having corresponding angles equal but their corresponding sides are not equal, such figures are not similar.
Therefore, A is the correct answer.

Ratio of two corresponding sides of two similar triangles is $4:9$. Then ratio of their area is ___.

  1. $\dfrac{16} {81}$

  2. $\dfrac{34} {81}$

  3. $\dfrac{81} {16}$

  4. $None\ of\ these$


Correct Option: A
Explanation:

Ratio of areas of two similar triangles is equal to the squares of the ratio of their sides.

Ratio of sides $=\dfrac{4}{9}$
Ratio of areas $=\left( \dfrac { 4 }{ 9 }  \right) ^{ 2 }=\dfrac { 16 }{ 81 } $

$\triangle PQR \sim \triangle XYZ, \dfrac{XY}{PQ}=\dfrac{3}{2}$ then $\dfrac{Area\ of\ \triangle PQR}{Area\ of\ \triangle XYZ}=$____.

  1. $\dfrac{9}{4}$

  2. $\dfrac{4}{9}$

  3. $\dfrac{3}{2}$

  4. $\dfrac{2}{3}$


Correct Option: B
Explanation:

$\dfrac{{XY}}{{PQ}} = \dfrac{3}{2}$


$ \Rightarrow \dfrac{{PQ}}{{XY}} = \dfrac{2}{3}$


Now,  $\dfrac{{Area{\rm{ of  }}\Delta {\rm{PQR}}}}{{Area{\rm{ of  }}\Delta {\rm{XYZ}}}} = {\left( {\dfrac{2}{3}} \right)^2} = \dfrac{4}{9}$

ABCD is a tetrahedron and O is any point. If the lines joining O to the vertices meet the opposite at P, Q, R and S, then $\frac{OP}{AP}+\frac{OQ}{BQ}+\frac{OR}{CR}+\frac{OS}{DS}=2$.

  1. True

  2. False


Correct Option: B

It is given that $\Delta ABC \sim \Delta PQR$ with $\dfrac{BC}{QR} = \dfrac{1}{3}$. Then $\dfrac{ar (\Delta PQR)}{ar (\Delta ABC)}$ is equal to

  1. $9$

  2. $3$

  3. $\dfrac{1}{3}$

  4. $\dfrac{1}{9}$


Correct Option: A
Explanation:
 If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

Since, $\Delta ABC \sim \Delta PQR$

$\therefore \dfrac{ar (\Delta PQR)}{ar (\Delta ABC)} = \dfrac{PR^2}{AC^2} = \dfrac{QR^2}{BC^2} = \dfrac{9}{1} =9 \ \ \ ..........  \left [ \therefore \dfrac{QR}{BC} = \dfrac{3}{1} \right ]$

$CM$ and $RN$ are respectively the medians of $\triangle {ABC}$ and $\triangle{PQR}$. If $\triangle {ABC}\sim \triangle{PQR}$, then
  $\cfrac{CM}{RN}=\cfrac{AB}{PQ}$

  1. True

  2. False


Correct Option: A

In a square $ABCD$, the bisector of the angle $BAC$ cut $BD$ at $X$ and $BC$ at $Y$ then triangles $ACY, ABX$ are similar.

  1. True

  2. False


Correct Option: A