Tag: properties of parallel lines and their transversal

Quizzes (2) Questions (42)

Questions Related to properties of parallel lines and their transversal

Multiple choice maths properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity

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
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By the Basic Proportionality Theorem converse, if AD/AB = AE/AC = 1/4, then DE is parallel to BC and triangle ADE is similar to triangle ABC with a scale factor of 1/4. Therefore, DE = (1/4)BC, not 1/8.

Multiple choice maths properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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)

Multiple choice maths properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity
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
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $AP$.

Multiple choice maths properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity
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
Reveal answer Fill a bubble to check yourself
A Correct answer
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)

Multiple choice maths properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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)

Multiple choice maths properties of parallel lines and their transversal how to check for similarity in triangles? criteria for similarity of triangles criteria for triangle similarity

$\angle BAC$ 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. $18$
Reveal answer Fill a bubble to check yourself
A Correct answer
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

Multiple choice maths properties of parallel lines and their transversal introduction to shapes similarity of triangles introduction to similar triangles

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.