Tag: mid-point and its converse

Questions Related to mid-point and its converse

 In any triangle are  the circumcentre, the centroid, the nine point centre and the orthocentro are all collinear ?

maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

  1. True

  2. False

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Correct Option: B

Mid-point theorem states that:

maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

  1. The line segment joining the mid-points of two sides of a triangle is not parallel to the third side and equal to half the length of the third side.

  2. The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to one-third the length of the third side.

  3. The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to the length of the third side.

  4. The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half the length of the third side.

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Correct Option: D
Explanation:

Mid-point theorem states that

The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is congruent to one half of the third side.
Hence, option D is correct.

In $\Delta ABC$, AB$ =5$cm, $BC=8$cm and $CA=7$cm. If D and E are respectively, the mid-points of AB and BC, then determine the length of DE.

maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

  1. $3.5$ cm

  2. $2.5$ cm

  3. $2.8$ cm

  4. $2.0$ cm

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Correct Option: A
Explanation:

Mid-Point Theorem : 
The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half the third side.

On applying the midpoint theorem, we get
 $DE = \cfrac {AC}{2} = \cfrac 72 = 3.5$cm

Suppose $ABCD$ is a rhombus. A straight line passing through $C$ meet $AD$ which is produced at $P$ and meet $AB$ produced at $Q$. Therefore if $DP=\dfrac {1}{2}AB$, then find the ratio between $BQ$ and $AB$?

maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

  1. $2:1$

  2. $1:1$

  3. $1:3$

  4. $3:1$

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Correct Option: A
Explanation:

$ ABCD $ is a rhombus. $ AB =BC = CD = DA $


$ \displaystyle \frac{DP}{AB} = \frac{1}{2} \Rightarrow DP =1 AB =2 $

In rhombus  $ \angle \theta _1 = \angle \theta _2 $

$ \angle Q $ is common for the $ \triangle BCQ \, and \, \triangle APQ $

$ \because \angle APQ = \angle BCQ $

$ \because \triangle BCQ $ is similar to $ \triangle APQ $ by $ AAA $ property. 

$ \displaystyle \frac{AP}{BC} = \frac{AQ}{BQ} = \frac{AD + DP}{BC} = \frac{3}{2} $

$ \displaystyle \frac{AQ}{BQ} = \frac{3}{2} \Rightarrow \frac{AB + BQ}{BQ} = \frac{3}{2} $

$ \displaystyle \frac{AB}{BQ} = \frac{3}{2} - 1 = \frac{1}{2} $

$ \displaystyle \frac{BQ}{AB} = \frac{2}{1} $

The median $AD$ of the triangle $ABC$ is bisected at $E$, $BE$ meets $AC$ in $F$, then $AF:AC$ is equal to ?

maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

  1. $3:4$

  2. $1:3$

  3. $1:2$

  4. $none\ of\ these$

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Correct Option: A

The incentre of the triangle formed by $(0, 0, 0), (3, 0, 0), (0, 3, 0)$.

maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

  1. $(\dfrac{3}{4}, \dfrac{3}{4}, 0)$

  2. $(1, \dfrac{3}{4}, 0)$

  3. $(0, 1, 1)$

  4. $(1, 1, 1)$

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Correct Option: B
Explanation:

using formula incenter $x= \dfrac{ax _{1}+bx _{2}+cx _{3}}{a+b+c}$ 


    $y=\dfrac{ay _{1}+by _{2}+cy _{3}}{a+b+c}$ 
    $z=\dfrac{az _{1}+bz _{2}+cx _{3}}{a+b+c}$
    $a=5$    $x=\dfrac{5(0)+4(3)+3(0)}{5+4+3}$       ;     $x=1$


    $b=4$      $y=\dfrac{5(0)+4(0)+3(3)}{5+4+3}$       ; $y=\dfrac{3}{4}$


    $c=3$      $z=\dfrac{5(0)+4(0)+3(0)}{5+4+3}$        ; $z=0$

 incentre =$(1,\dfrac{3}{4},0)$ 

The mid-points of the sides of a triangle are $D(6,1),E(3,5)$ and $F(-1,-2)$ then vertex opposite to D is 

maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

  1. $(-4,2)$

  2. $(-4,5)$

  3. $(2,2)$

  4. $(10,8)$

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Correct Option: A

ABC is an isosceles triangle with AB=AC. D,E, F are mid point of sides BC,AB and AC respectively then line segment $A D \perp E F$ and is bisected by it.

maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

  1. True

  2. False

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Correct Option: A

Each side of $\triangle ABC$ is 12 units. D is the foot of the perpendicular dropped from A on BC and E is the mid point of AD. The length of BE in the same units is: 

maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

  1. $\sqrt{18}$

  2. $\sqrt{28}$

  3. 6

  4. 7.93

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Correct Option: D

In $ABC,E$ and $F$ are mid points of sides $AB$ and $AC$ respectively then $EF // BC$

maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

  1. True

  2. False

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Correct Option: A