Tag: congruence

Questions Related to congruence

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

Assume that, $\Delta RST \sim \Delta XYZ$. Complete the following statement.


$\displaystyle \frac{RT}{XY} = \frac{- -}{YZ}, \frac{RS}{XY} = \frac{ST}{- -}, \frac{XY}{ - -} = \frac{YZ}{ST}$

  1. ST, YZ, RT

  2. ST, YZ, RS

  3. YT, YS, RZ

  4. ST, YZ, RZ


Correct Option: B
It is given that $\triangle FED\sim \triangle STU$. Is it true to say that $\cfrac{DE}{UT}=\cfrac{EF}{TS}$? 
  1. Yes

  2. No

  3. Cannot say

  4. None


Correct Option: B
Explanation:

$\triangle FED \sim \triangle STU$
The corresponding sides of both the triangles are $F\leftrightarrow S$, $E\leftrightarrow T$, $D\leftrightarrow  U$. With this correspondence,
$\cfrac{EF}{ST}=\cfrac{DE}{TU}$

Consider the following statements:
(1) If three sides of triangle are equal to three sides of another triangle, then the triangles are congruent.
(2) If three angles of a triangle are respectively equal to three angles of another triangle, then the two triangles are congruent.

Of these statements,

  1. $(1)$ is correct and $(2)$ is false

  2. Both $(1)$ and $(2)$ are false

  3. Both $(1)$ and $(2)$ are correct

  4. $(1)$ is false and $(2)$ is correct


Correct Option: A
Explanation:

If three sides of triangle are equal to other three sides of the triangle, then the two triangles are congruent by SSS rule.

If three angles of the triangle are equal to other three sides of the triangle, then the two triangles are similar by AAA rule but not congruent.

If in trianges $ABC$ and $DEF$, $\cfrac{AB}{DE}=\cfrac{BC}{FD}$, then they will be similar, when:

  1. $\angle B=\angle E$

  2. $\angle A=\angle D$

  3. $\angle B=\angle D$

  4. $\angle A=\angle F$


Correct Option: C
Explanation:

In $\triangle ABC$ and $\triangle DEF$,
$\dfrac{AB}{DE} = \dfrac{BC}{FD}$ (Given)

The angle between these sides are $\angle B$ and $\angle D$. Thus, If the containing angles are equal. The triangles will be similar..

In $\triangle PQR,$ $PQ=4$ cm, $QR=3$ cm, and $RP=3.5$ cm. $\triangle DEF$ is similar to $\triangle PQR.$ If $EF=9$ cm, then what is the perimeter of $\triangle DEF: ?$

  1. $10.5$ cm

  2. $21$ cm

  3. $31.5$ cm

  4. Cannot be determined as data is insufficient


Correct Option: C
Explanation:

$PQ = 4, QR = 3$ and $RP = 3.5$
Also, $EF = 9$
Now, perimeter of $\triangle PQR = PQ + QR + RP = 4 + 3 + 3.5 = 10.5$
Given, $\triangle DEF \sim \triangle PQR$
$\dfrac{EF}{QR} = \dfrac{Perimeter(\triangle DEF)}{Perimeter(\triangle PQR)}$
$\dfrac{9}{3} = \dfrac{Perimeter(\triangle DEF)}{10.5}$
Perimeter $(\triangle DEF) = 31.5$ cm

The perimeter of two similar triangles are $24$ cm and $16$ cm, respectively. If one side of the first triangle is $10$ cm, then the corresponding side of the second triangle is

  1. $9$ cm

  2. $\dfrac{20}3$ cm

  3. $\dfrac{16}3$ cm

  4. $5$ cm


Correct Option: B
Explanation:
In similar triangles, ratio of the sides is equal to the ratio of the perimeters.
Thus, $\dfrac{p _1}{p _2} = \dfrac{s _1}{s _2}$
$\dfrac{24}{16} = \dfrac{10}{s _2}$
$s _2 = \dfrac{20}{3}$
Thus, side of the other triangle is $\dfrac{20}{3}$ cm.