Tag: how to check for similarity in triangles?

Questions Related to how to check for similarity in triangles?

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 $D E = \frac { 1 } { 8 } B C$

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

  1. True

  2. False

Show answer
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 $\displaystyle \Delta CPD.$

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

  1. True

  2. False

Show answer
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 $.

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

  1. True

  2. False

Show answer
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 $


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

  1. True

  2. False

Show answer
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 $\displaystyle \Delta COD$

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

  1. True

  2. False

Show answer
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 $PB$.

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

  1. $20$

  2. $24$

  3. $36$

  4. $18$

Show answer
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