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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

In triangle ABC ; M is mid-point of AB, N mid-point of AC and D is any point in base BC. Then:

  1. MN bisects AD

  2. MN divides AD in the ratio 1:3

  3. MN divides AD in the ratio 1:2

  4. MN divides AD in the ratio 1:4
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In $\triangle ABC$, $M$ is mid point of $AB$ and $N$ is mid point of $AC$
$D$ is any point of BC
Now, Join AD and MN such that they met at O
In $\triangle ABC$
M is mid point of AB and N is mid point point of AC
Hence, $MN \parallel BC$ and $MN = \frac{1}{2} BC$

Now, In $\triangle ABD$
$MO \parallel BC$ and M is mid point of AB
Thus, $O$ is mid point of AD
Hence, $MN$ bisects $AD$

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 triangle $ABC$, angle $B$ is obtuse. $D$ and $E$ are mid-points of sides $AB$ and $BC$ respectively and $F$ is a point on side $AC$ such that $EF$ is parallel to $AB$. Then, $BEFD$ is a parallelogram. State True or False. 

  1. True

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

Given: $D$ is mid point of $AB$ and $E$ is mid point of $BC$, $F$ is any point on $AC$ and $EF \parallel AB$

Now, in $\triangle ABC$,
E is mid point of BC and $EF \parallel AB$
By Mid point Theorem, $F$ is mid point of $AC$

Also, D is mid point of AB and F is mid point of AC
Hence, by mid point theorem, $DF \parallel BE$
Since, $DF \parallel BE$ and $EF \parallel AB or BD$
Hence, BEFD is parallelogram.

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

If in two triangles $DEF$ and $PQR$, $\angle D=\angle Q$ and $\angle R=\angle E$, then which of the following is not true?

  1. $\cfrac{EF}{PR}=\cfrac{DF}{PQ}$
  2. $\cfrac{DE}{PQ}=\cfrac{EF}{RP}$
  3. $\cfrac{DE}{QR}=\cfrac{DF}{PQ}$
  4. $\cfrac{EF}{RP}=\cfrac{DE}{QR}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given triangles DEF and PQR, with angle D = angle Q and angle R = angle E. By AA similarity, triangle DEF is similar to triangle QRP. The corresponding sides are proportional: DE/QR = EF/RP = DF/QP. Option B claims DE/PQ = EF/RP, which is not necessarily true because PQ is not the corresponding side to DE.

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

If in the triangles $ABC$ and $DEF$, angle $A$ is equal to angle $E$, both are equal to ${40}^{o}$, $AB:ED=AC:EF$ and angle $F$ is ${65}^{o}$, then angle $B$ is:

  1. ${35}^{o}$
  2. ${65}^{o}$
  3. ${75}^{o}$
  4. ${85}^{o}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given triangle ABC and DEF. Angle A = Angle E = 40 degrees. AB/ED = AC/EF. This satisfies the SAS similarity criterion, so triangle ABC is similar to triangle EDF. Therefore, Angle B = Angle D and Angle C = Angle F. We are given Angle F = 65 degrees, so Angle C = 65 degrees. In triangle ABC, Angle A + Angle B + Angle C = 180. 40 + Angle B + 65 = 180 => Angle B = 180 - 105 = 75 degrees.

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

D is the mid point of the base BC of a triangle ABC. DM and DN are perpendiculars on AB and AC respectively. If $DM=DN$, the triangle  is

  1. Isosceles

  2. Equilateral

  3. Right angled

  4. Scalene
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given: D is mid point of BC. $DM \perp AB$ and $DN \perp AC$, $DM = DN$
Now, In $\triangle DMB$ and $\triangle DNC$,
$DM = DN$ (Given)
$\angle DMB = \angle DNC$ (Each $90^{\circ}$)
$BD = DC$ (D is mid point of BC)
Thus, $\triangle DMB \cong \triangle DNC$ (SAS rule)
Thus, $\angle B = \angle C$ (By cpct)
hence, $\triangle ABC$ is an Isosceles triangle.

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

If the medians of two equilateral triangles are in the ratio $3:2,$ then what is ratio of the sides$: ?$

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

Equilateral triangles are similar triangles.
In similar triangles, the ratio of their corresponding sides is the same as the ratio of their medians.
Hence, ratio of sides = $3: 2$

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

$\displaystyle \triangle ABC\sim \triangle PQR$ If ar(ABC)=2.25$\displaystyle m^{2}$ ar(PQR)=6.25$\displaystyle m^{2}$, PQ=0.5 m, then length of AB is

  1. $30 cm$
  2. $1.5 cm$
  3. $50 cm$
  4. $2 m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The area of two similar triangles are proportional to the square of their corresponding sides

$\therefore \dfrac{Ar(ABC)}{ar(PQR)}=\dfrac{(AB)^2}{(PQ)^2}$
$\Rightarrow \dfrac{2.25}{6.25}=\dfrac{(AB)^2}{(.5)^2}$
$\Rightarrow AB^2=\dfrac{2.25\times .5}{6.25}$
$\Rightarrow AB^2=.09$
$\Rightarrow AB=.3m=30cm$

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

If $\displaystyle \triangle ABC\sim \triangle DEF$ BC=4 cm, EF=5 cm and ar $\displaystyle \left ( \triangle ABC \right )=80cm2$,the ar$\displaystyle \left ( \triangle DEF \right )$ is

  1. $\displaystyle 120cm^{2}$
  2. $\displaystyle 125cm^{2}$
  3. $\displaystyle 150cm^{2}$
  4. $\displaystyle 200cm^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If two triangles are equals than the ratio of their square is equal to the ratio of their corresponding sides.

$\therefore \dfrac{arc(\triangle ABC)}{arc(\triangle DEF)}=\dfrac{BC^2}{EF^2}$
$\Rightarrow \dfrac{80}{arc(\triangle DEF)}=\dfrac{16}{25}$
$\Rightarrow arc(\triangle DEF)=\dfrac{80\times 25}{16}=125 cm^2$


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

If the ratio of the corresponding sides of two similar triangles is 2:3 then the ratio of their corresponding altitude is

  1. 3 : 5

  2. 16 : 81

  3. 4 : 9

  4. 2 : 3
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If two triangles are similar ,then the ratio of their corresponding sides and altitude are also same. Therefore the ratio of the altitude is 2:3.

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

If $\displaystyle \triangle ABC\cong \triangle RQP,\angle A=80^{\circ},\angle B=60^{\circ}$, then the value of $\displaystyle \angle P$ is

  1. $\displaystyle 60^{\circ}$
  2. $\displaystyle 50^{\circ}$
  3. $\displaystyle 40^{\circ}$
  4. $\displaystyle 80^{\circ}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If two triangles are equal then their corresponding angle are also equal

$\therefore \angle A=\angle R,$$\angle B=\angle Q$ and$ \angle C=\angle P$
In $\triangle ABC$
$\angle A+\angle B+\angle C=180^{\circ}$
$80^{\circ}+60^{\circ}+\angle C=180^{\circ}$
$\angle C=180^{\circ}-140^{\circ}=40^{\circ}$
$\therefore \angle P=40^{\circ}$