Tag: maths

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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

When we construct a triangle similar to a given triangle as per given scale factor, we construct on the basis of ...........

  1. SSS Similarity

  2. AAA similarity

  3. Basic proportionality theorem
  4. $A$ and $C$ are correct
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As we consider only sides, therefore, SSS similarity is used.
Option A is correct.  

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

Goldfish are sold at Rs.15 each. The rectangular coordinate graph showing the cost of 1 to 12 goldfish is:

  1. a straight line segment

  2. a set of horizontal parallel line segments

  3. a set of vertical parallel line segments

  4. a finite set of distinct points

  5. a straight line
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\angle MAB=\angle PAO\longrightarrow (1),\hspace{1mm} O\hspace{1mm} be\hspace{1mm} center\ \angle AMB=90°=\angle AOP\ (1)\Longrightarrow 90°-\angle MAB=90°-\angle PAO\ \angle MBA=\angle APO$

By AAA property,
$\triangle APO\sim \triangle ABM\ \cfrac { \bar { AP }  }{ \bar { AB }  } =\cfrac { \bar { AO }  }{ \bar { AM }  } \ \therefore \bar { AP } \cdot \bar { AM } =\bar { AO } \cdot \bar { AB } $

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

For $\triangle ABC$ and $\triangle PQR$, if $m\angle A=m\angle R $ and $m\angle C=m\angle Q$, then $ABC \longleftrightarrow $_________ is a similarity.

  1. $RQP$
  2. $PQR$
  3. $RPQ$
  4. $QPR$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For $\triangle ABC $ and $\triangle PQR$,
$m\angle A = m\angle R$
$m\angle C= m\angle Q$
$\therefore $ by AA criteria for similarity 

$ABC \longleftrightarrow RPQ $ is a similarity.

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

Say true or false.

If in two triangles, two angles of one triangle are respectively equal to the two angles of the other triangle, then the two triangles are similar.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If two angles of a triangle is equal to two angles of another triangle, then the third angle of both triangles will be equal.
$\therefore$By AAA Theorem of Similarity, the two triangles are similar.

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

$ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid point of $BC$. Ratio of the areas of triangle $ABC$ and $BDE$ is

  1. $2:1$
  2. $1:2$
  3. $4:1$
  4. $1:4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\triangle ABC \sim \triangle BDE$                            (both are equilateral triangles)


$\Rightarrow \triangle ABC : \triangle BDE = AB^2 : BD^2$

                                          $= AB^2 :  (\dfrac{1}{2} BC)^{2} $
                                          
                                          $ = AB^2 : \dfrac{1}{4} BC^2 $

                                          $= 4 : 1$           $(\because AB = BC)$
Hence proof.

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 $AP$.