Tag: maths

Questions Related to maths

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

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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

Show answer
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.

In $\bigtriangleup : ABC$ , $E$ and $F$ are mid-points of sides $AB$ and $AC$ respectively. If $BF$ and $CE$ intersect each other at point $O$, then the $\bigtriangleup :OBC$ and quadrilateral $AEOF$ are equal in area.

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 that $E$ and $F$ are mid points. We have to prove that ${a} _{1}(BOC)={a} _{1}(AEOF)$

As $EF\parallel BC$ (By midpoint theorem.)
${ a } _{ 1 }(\triangle BEF)={ a } _{ 1 }(\triangle ACFE)$  ($\because $ Between the parallel lines.)
${ a } _{ 1 }(\triangle BEF)-{ a } _{ 1 }(\triangle EOF)={ a } _{ 1 }(\triangle CFE)-{ a } _{ 1 }(\triangle EOF)\ \therefore { a } _{ 1 }(\triangle BOE)={ a } _{ 1 }(\triangle FOC)\longrightarrow (i)$
${a} _{1}(AEOF)={a} _{1}(BOC)$
$\therefore$ The statement is true.

If $D, E, F$ are respectively the midpoints of the sides $AB, BC, CA$ of $\Delta ABC$ and the area of $\Delta ABC$ is $24\ sq.\ cm$, then the area of $\Delta DEF$ is:

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

  1. $24\ {cm}^{2}$

  2. $12\ {cm}^{2}$

  3. $8\ {cm}^{2}$

  4. $6\ {cm}^{2}$

Show answer
Correct Option: D
Explanation:

Given: 

$\Delta ABC,\ \ D, E$ and $F$ are mid points of $AB, BC, CA$ respectively.

In $\Delta ABC$
$F$ is mid point of $AC$ and $D$ is mid point of $AB$. 
Thus, by Mid point theorem, we get
$FD = \dfrac{1}{2} CB$, 
$FD = CE$ and $FD \parallel CE$     ...(1)
Similarly,
$DE =  FC$ and $DE \parallel FC$   ...(2)
$FE = DB$ and $FE \parallel DB$    ...(3) 

From (1), (2) and (3)
$\Box ADEF$, $\Box DBEF$, $\Box DECF$ are parallelograms.

The diagonal of a parallelogram divides the parallelogram into two congruent triangles.
Hence, $\Delta DEF \cong \Delta ADF$
$\Delta DEF \cong \Delta DBE$
$\Delta DEF \cong \Delta FEC$
Or, $\Delta DEF \cong \Delta ADF \cong \Delta ECF \cong \Delta ADF$
Thus, mid points divide the triangle into $4$ equal parts.

Now, 
$A (\Delta DEF) = \dfrac{1}{4} A (\Delta ABC)$

$A (\Delta DEF) = \dfrac{1}{4} (24)$

$A (\Delta DEF) = 6\ {cm}^2$

Hence, option D.

Suppose the triangle ABC has an obtuse angle at C and let D be the midpoint of side AC Suppose E is on BC such that the segment DE is parallel to AB. Consider the following three statements
i) E is the midpoint of BC
ii) The length of DE is half the length of AB
iii) DE bisects the altitude from C to AB

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

  1. only (i) is true

  2. only (i) and (ii) are true

  3. only (i) and (iii) are true

  4. all three are true

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Correct Option: D
Explanation:

The triangle ABC has height h and base l,
E is the midpoint of BC, line parallel to the base will be interect the triangle at the midpoint of the opposite side.
Triangle ADE is similar to triangle ABC, as all 3 angles are equal.
Therefore, the altitude of ADE is half of ABC, and DE will be half of BC.

Let $ABC$ be a triangle and let $P$ be an interior point such that $\angle BPC = 90$, $\angle BAP = \angle BCP$. Let $M, N$ be the mid-points of $AC, BC$ respectively. Suppose $BP = 2PM$. Then $A, P, N$ are collinear ?

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

If $\displaystyle \Delta ABC$ is an isosceles triangle and midpoints $D, E,$ and $F$ of $AB, BC,$ and $CA$ respectively are joined, then $\displaystyle \Delta DEF$ is:

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

  1. Equilateral

  2. Isosceles

  3. Scalene

  4. Right-angled

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Correct Option: B
Explanation:

Given: In $\triangle ABC, D, E$ and $F$ are midpoints of sides $AB, BC$ and $CA$.

$BE=EC$
$\therefore DF=\dfrac { 1 }{ 2 } BC$
$\therefore \dfrac { DF }{ BC } =\dfrac { 1 }{ 2 }$ ....... $\left( 1 \right) $

Similarly, $ \dfrac { DE }{ AC } =\dfrac { 1 }{ 2 } $ and $ \dfrac { EF }{ AB } =\dfrac {1 }{ 2 } $

$\Rightarrow \dfrac { DF }{ BC } =\dfrac { DE }{ AC } =\dfrac { EF }{ AB } =\dfrac { 1 }{ 2 } $
$\triangle ABC$ is propotional to $\triangle DEF$ as the sides of the triangles are proportional. 
So corresponding angles are equal.
Hence, $\triangle ABC\sim \triangle EDF$ [by SSS similarly theorm]
$\therefore \triangle DEF$ is isosceles.

M is the midpoint of $\displaystyle\overline{AB}$. The coordinates of A are $(-2,3)$ and the coordinates of M are $(1,0)$. Find the coordinates of B.

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

  1. $(-1/2, 3/2)$

  2. $(4,-3)$

  3. $(-4,3)$

  4. $(-5,6)$

  5. none of these

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Correct Option: B
Explanation:

Let $(x,y)$ be the coordinates of B.
According to question, M is mid point of A and B.
$\Rightarrow \dfrac{-2+x}2=1\Rightarrow x=4$
and $ \dfrac{3+y}2=0\Rightarrow y=-3$
Therefore B is $(4,-3)$
Option B is correct.