Tag: properties of parallel lines and their transversal

Questions Related to properties of parallel lines and their transversal

In $\triangle A B C$, D is a point on AB such that $A D = \frac { 1 } { 4 } A B$ and E is a point on AC such that $A E = \frac { 1 } { 4 } A C$ then $D E = \frac { 1 } { 8 } B C$

  1. True

  2. False


Correct Option: B

State true or false:

In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at point P, then
$\displaystyle \Delta APB$ is similar to $\displaystyle \Delta CPD.$

  1. True

  2. False


Correct Option: A
Explanation:

In $\triangle$ APB and $\triangle$ CPD,
$\angle APB = \angle CPD$ (Vertically opposite angles)
$\angle ABP = \angle CDP$ (Alternate angles of parallel sides AB and CD)
$\angle BAP = \angle DCP$ (Alternate angles of parallel sides AB and CD)
Hence, $\triangle APB \sim \triangle CPD$ (AAA rule)

State true or false:

In parallelogram $ ABCD $. $ E $ is the mid-point of $ AB $ and $ AP $ is parallel to $ EC $ which meets $ DC $ at point $ O $ and $ BC $ produced at $ P $. Hence 
$ O $ is mid-point of $ AP $.

  1. True

  2. False


Correct Option: A
Explanation:

In $\triangle$s, APB and ECB,

$\angle ABP = \angle EBC $ (Common angle)

$\angle PAB = \angle CEB$ (Corresponding angles of parallel lines)

$\angle APB = \angle ECB $ (Third angle of the triangle)

Thus $\triangle APB \sim \triangle ECB$ (AAA rule)

Hence, $\dfrac{AB}{EB} = \dfrac{BP}{BC}$ (Corresponding sides of similar triangles)

$2 = \dfrac{BP}{BC}$

$BP = 2 BC$

Now, in $\triangle$s $OPC$ and $APB,$

$\angle OPC = \angle APB$ (Common angle)

$\angle POC = \angle PAB$ (Corresponding angles of parallel lines)

$\angle PCO = \angle PBA$ (Third angle of a triangle)

$\triangle OPC \sim \triangle APB$ (AAA rule)

hence, $\dfrac{PC}{BP} = \dfrac{OP}{AP}$  (Corresponding sides)

$\dfrac{1}{2} = \dfrac{OP}{AP}$ 

$OP = \dfrac{1}{2} AP$

hence, $O$ is the midpoint of $AP$.

State true or false:

In parallelogram $ ABCD $. $ E $ is the mid-point of $ AB $ and $ AP $ is parallel to $ EC $ which meets $ DC $ at point $ O $ and $ BC $ produced at $ P $. Hence
$ BP= 2AD $


  1. True

  2. False


Correct Option: A
Explanation:

In $\triangle$s, APB and ECB,
$\angle ABP = \angle EBC $ (Common angle)
$\angle PAB = \angle CEB$ (Corresponding angles of parallel lines)
$\angle APB = \angle ECB $ (Third angle of the triangle)
Thus $\triangle APB \sim \triangle ECB$ 
Hence, $\frac{AB}{EB} = \frac{BP}{BC}$ (Corresponding sides of similar triangles)
$2 = \frac{BP}{BC}$
$BP = 2 BC$
$BP = 2 AD$  (BC = AD)

In quadrilateral ABCD, the diagonals AC and BD intersect each at point O. If $AO=2CO$ and $BO=2DO$; Then,

$\displaystyle \Delta AOB$ is similar to $\displaystyle \Delta COD$

  1. True

  2. False


Correct Option: B
Explanation:

Given: $AO = 2 CO$ or $\dfrac{AO}{CO} = 2$
Also given, $BO = 2 DO$ or $\dfrac{BO}{DO} = 2$
In $\triangle AOB$ and $\triangle COD$, we know 
$\angle AOB = \angle COD$
$\dfrac{AO}{CO} = \dfrac{BO}{DO}$
Thus, $\triangle AOB \sim \triangle COD$ (SAS rule)

$\angle BAC$ of triangle $ABC$ is obtuse and $AB=AC$. $P$ is a point in $BC$ such that $PC= 12$ cm. $ PQ $ and $PR$ are perpendiculars to sides $AB$ and $AC$ respectively. If $PQ= 15$ cm and $=9$ cm; find the length of $PB$.

  1. $20$

  2. $24$

  3. $36$

  4. $18$


Correct Option: A
Explanation:

Given: $AB = AC$, $PQ \perp AB$ and $PR \perp AC$
Since, $AB = AC$
$\angle ABC = \angle ACB$...(I) (Isosceles triangle property)

Now, In $\triangle PBQ$ and $\triangle PRC$
$\angle PBQ = \angle PCR$ (From I)
$\angle PQB = \angle PRC$ (Each $90^{\circ}$)
$\angle QPB = \angle RPC$ (Third angle)
Thus, $\triangle QPB \sim \triangle RPC$ (AAA rule)
Hence, $\dfrac{PQ}{PR} = \dfrac{PB}{PC}$
$\dfrac{15}{9} = \dfrac{PB}{12}$
$PB = \dfrac{15 \times 12}{9}$
$PB = 20$ cm

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.