Tag: proving the mid-point theorem

Questions Related to proving the mid-point theorem

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

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

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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.

The straight line joining the mid-points of the opposite sides of a parallelogram divides it into two parallelogram of equal area  

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

In a $\triangle DEF$; $A,B$ and $C$ are the mid-points of $EF,FD$ and $DE$ respectively. If the area of $\triangle DEF$ is $14.4{ cm }^{ 2 }$, then find the area of $\triangle {ABC}$.

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

  1. $1.75$cm

  2. $2.54$cm

  3. $3.2$cm

  4. $3.6$cm

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Correct Option: D
Explanation:

Fact: Using mid-point theorem $\dfrac{\Delta{DEF}}{\Delta{ABC}}=4$


Here $\Delta{DEF}=14.4$ cm$^2$ is given 
Hence are of triangle $ABC $ is given by $ \dfrac{14.4}{4}=3.6$ cm$^2$ 

In a $\triangle ABC$, if $D, E, F$ are the midpoints of the sides $BC, CA, AB$ respectively then $\overline {AD} + \overline {BE} + \overline {CF} =$

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

  1. $\overline {0}$

  2. $\overline {AE}$

  3. $\overline {BD}$

  4. $\overline {CE}$

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Correct Option: A

A cross section at the midpoint of the middle piece of a human sperm will show

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

  1. Centriole, mitochondria and 9 +2 arrangement of microtubules.

  2. Centriole and mitochondria

  3. Mitochondria and 9+2 arrangement of microtubules.

  4. 9+2 arrangement of microtubules only.

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Correct Option: C
Explanation:

The middle piece is the tubular structure in which mitochondria are spirally arranged and it also has the beginning part of the flagellum. The sperm tail or the flagellum is based upon unique 9+2 arrangement. This arrangement refers to the nine peripheral, symmetrically arranged microtubule doublets.
Thus, the cross-section of the middle piece of sperm will show mitochondria and 9+2 arrangement of microtubules.
Hence, the correct answer is option (C), 'Mitochondria and 9+2 arrangement of microtubules'.