Tag: the mid-point theorem

Questions Related to the mid-point theorem

In a triangle $ABC,D$ and  $E$ are the mid-points of $BC,CA$ respectively. If $AD=5,BC=BE=4$, then $CA=$

maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

  1. $5$

  2. $\sqrt{7}$

  3. $2\sqrt{7}$

  4. $5\sqrt{5}$

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Correct Option: C

Tangents PA and PB drawn to $ x^2+y^2=9 $ from any arbitrary point 'P ' on the line $ x+y=25 $. Locus of midpoint of chord AB is

maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

  1. $ 25(x^2+y^2)=9(x+y) $

  2. $ 25(x^2+y^2)=3(x+y) $

  3. $ 5(x^2+y^2)=3(x+y) $

  4. None of these

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Correct Option: A
Explanation:

Let the point on the line $x+y=25$ be $P(a,b)$
Thus equation of chord of contact AB from point P to the circle is given by,
$T  =0 \Rightarrow ax+by = 9$  (i)
Let mid point of AB be $R(h,k)$.
Now equation of chord AB with mid point R is given by,
$T = S _1 \Rightarrow hx+ky = h^2+k^2$ (ii)
Both line (i) and (ii) represents the same line AB
$\therefore \displaystyle \frac{a}{h}=\frac{b}{k} = \frac{9}{h^2+k^2}$
$\Rightarrow  a=\cfrac{9h}{h^2+k^2}, b = \cfrac{9k}{h^2+k^2}$
Also point $(a,b)$ lie on the line $x+y = 25$
$\Rightarrow a+b = 25 \Rightarrow 25(h^2+k^2) = 9(h+k)$
Hence required locus of $R(h,k)$ is given by, $25(x^2+y^2) = 9(x+y)$

If $m {a},\ m _{b},\ m _{c}$ are lengths of medians through the vertices $A,B, C$ of $\triangle ABC$ respectively, then length of side $b=$___ 

maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

  1. $\sqrt { { 2m } _{ a }^{ 2 }+{ 2m } _{ c }^{ 2 }-{ 2m } _{ b }^{ 2 } } $

  2. $\dfrac { 1 }{ 3 } \sqrt { { 2m } _{ a }^{ 2 }+{ 2m } _{ c }^{ 2 }-{ 2m } _{ b }^{ 2 } }$

  3. $\dfrac { 2 }{ 3 } \sqrt { { 2m } _{ a }^{ 2 }+{ 2m } _{ c }^{ 2 }-{ 2m } _{ b }^{ 2 } }$

  4. $\dfrac { 3 }{ 4 } \sqrt { { 2m } _{ a }^{ 2 }+{ 2m } _{ c }^{ 2 }-{ 2m } _{ b }^{ 2 } }$

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Correct Option: D

If the diagonals $KT$ and $EI$ of a parallelogram $KITE$ intersect at $O$ and $P,Q,R$ and $S$ are the midpoints of $KO,EO,TO$ and $IO$ respectively then the ratio of $(PQ+QR+RS+SP)$ to $(KE+ET+TI+IK)$ is

maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

  1. $1:4$

  2. $1:3$

  3. $1:1$

  4. $1:2$

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Correct Option: A

In $\triangle ABC, D, E$ and $F$ are the mid points of $BC, CA$ and $AB$ respectively, then, $BDEF$=________$ABC$

maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

  1. $2$

  2. $\dfrac{1}{2}$

  3. $\dfrac{1}{4}$

  4. $\dfrac{31}{4}$

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Correct Option: A

Consider $\Delta$ABC and $\Delta A {1}B _{1}C _{1}$ in such a way that $\bar { AB } =\bar { { A } _{ 1 }{ B } _{ 1 } } $ and M,N,$M _{1}N _{1}$ be that mid points of AB,BC, $A _{1}B _{1}$ and $B _{1}C _{1}$ respectively, then ____________.

maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

  1. $\bar { M{ M } _{ 1 } } =\bar { NN _{ 1 } } $

  2. $\bar { { CC } _{ 1 } } =\bar { MM _{ 1 } } $

  3. $\bar { { CC } _{ 1 } } =\bar { NN _{ 1 } } $

  4. $\bar { { MM } _{ 1 } } =\bar { BB _{ 1 } } $

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Correct Option: A

A triangle ABC in which AB=AC, M is a point on AB and N is a point on AC such that if BM=CN then AM=AN

maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

  1. True

  2. False

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Correct Option: A
Explanation:

Since M is a midpoint on AB and N is a midpoint on AC, we have

$ AB = AM

+ MB $   and $ AB = AN + NC $

Since, $

AB = AC $

$ =>

AM + MB = AN + NC $

$

=>  AM = AN $

In triangle $ ABC $, $ M $ is mid-point of $ AB $ and a straight line through $ M $ and parallel to $ BC $ cuts $ AC $ in $ N $. Find the lenghts of $ AN $ and $ MN $ if $ BC= 7 $ cm and $ AC= 5 $ cm.

maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

  1. $ AN= 2.5 $ cm and $ MN= 3.5 $ cm

  2. $ AN= 1.5 $ cm and $ MN= 3.5 $ cm

  3. $ AN= 2.5 $ cm and $ MN= 4.5 $ cm

  4. none of the above

Show answer
Correct Option: A
Explanation:

M is the mid point of AB and MN II BC. Thus, N is the mid point of AC and


$MN = \dfrac{1}{2} BC$ (Mid point theorem)

$MN = \dfrac{1}{2} (7) $

$MN = 3.5 cm$

Also, $AN = \dfrac{1}{2} AC$

$AN = \dfrac{1}{2} (5)$

$AN = 2.5 cm$

State true or false:

In triangle  $ ABC  $,  $ P  $ is the mid-point of side  $ BC  $. A line through $ P  $ and Parallel to  $ CA  $ meets  $ AB  $ at point  $ Q  $; and a line through  $ Q  $ and parallel to  $ BC $ meets median  $ AP  $ at point  $ R  $. Can it be concluded that,
$ AP= 2AR $ ?

maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

  1. True

  2. False

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Correct Option: A
Explanation:

Given: $\triangle ABC$, P is mid point of BC, $QR \parallel BC$ and $PQ \parallel AC$

Since, $ PQ \parallel AC$ and P is mid point of BC, thus, by converse of mid point theorem
Q is mid point of AB.

Now, In $\triangle ABP$
Since, $QR \parallel BP$ and Q is mid point of AB. thus, by converse of Mid point theorem
R is mid point of AP.
Hence, $AP = 2 AR$

State true or false:


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

maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

  1. True

  2. False

Show answer
Correct Option: A
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.