Tag: the mid-point theorem

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Questions Related to the mid-point theorem

Multiple choice maths mid-point and its converse proving the mid-point theorem the mid-point theorem 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 $CA=$

  1. $5$
  2. $\sqrt{7}$
  3. $2\sqrt{7}$
  4. $5\sqrt{5}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using Apollonius theorem on triangle ABC with median AD, 2(AD^2 + BD^2) = AB^2 + AC^2. With BE as a median to AC, 2(BE^2 + AE^2) = AB^2 + BC^2. Solving these equations with given values leads to AC = 2*sqrt(7).

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

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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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, $25(x^2+y^2) = 9(x+y)$

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

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 $b=$___ 

  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. $\dfrac { 3 }{ 4 } \sqrt { { 2m } _{ a }^{ 2 }+{ 2m } _{ c }^{ 2 }-{ 2m } _{ b }^{ 2 } }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The length of a median m_b is given by m_b = 1/2 * sqrt(2a^2 + 2c^2 - b^2). Rearranging this formula for side b gives b = 2/3 * sqrt(2m_a^2 + 2m_c^2 - m_b^2).

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

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 ____________.

  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 } } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given AB = A1B1 and M, N, M1, N1 are midpoints, the vectors MM1 and NN1 represent the displacement between the midpoints of the sides of two congruent triangles, which must be equal.

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

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.

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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, $AN = \dfrac{1}{2} AC$

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

$AN = 2.5 cm$

Multiple choice maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem
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 $ ?

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
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, $AP = 2 AR$

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

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. 

  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.