Tag: mid-point and its converse

Questions Related to mid-point and its converse

If the lengths of the medians $AD, BE$ and $CF$ of the triangle $ABC$, are $6,8,10$ respectively, then

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

  1. $AD$ and $BE$ are perpendicular

  2. $BE$ and $CF$ are perpendicular

  3. area of $\Delta ABC=32$

  4. area of $\Delta DEF=8$

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Correct Option: D
Explanation:

$\begin{array}{l}AD = \sqrt {2A{B^2} + 2A{C^2} - B{C^2}}  = 6\2A{B^2} + 2A{C^2} - B{C^2} = 36\BE = \sqrt {2A{B^2} + 2B{C^2} - A{C^2}}  = 8\2A{B^2} + 2B{C^2} - A{C^2} = 64\CF = \sqrt {2A{C^2} + 2B{C^2} - A{B^2}}  = 10\2A{C^2} + 2B{C^2} - A{B^2} = 100\A{B^2} = x\A{C^2} = y\B{C^2} = z\2x + 2y - z = 36\2x + 2z - y = 64\2y + 2z - x = 100\x = A{B^2} = \frac{{100}}{9}\y = A{C^2} = \frac{{208}}{9}\z = B{C^2} = \frac{{292}}{9}\AD = 6,BE = 8,CF = 10\in,\Delta ABE\AD \bot BE\area,\Delta BEC = 16\area,\Delta ABE = 16 + 16 = 32\\frac{{area,\Delta ABE}}{{area,\Delta DEF}} = 4\\frac{{32}}{{area,\Delta DEF}} = 4\area,\Delta DEF = 8\end{array}$

In $\triangle ABC , \angle B=90^0$ and D is the mid-point of BC then
$BC^2=4(AD^2-AB^2)$

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
The area of rectangle ABCD with vertices A (0,0), B (5,0), C (5, 6) and D (0,6) is 
maths mid-point and its converse proving the mid-point theorem the mid-point theorem mid point theorem

  1. $25cm^2$

  2. $20cm^2$

  3. $30cm^2$

  4. None of these.

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Correct Option: C
Explanation:

The vertices of rectangle are $A(0,0),B(5,0),C(5,6),D(0,6)$

From the points 

$\implies $ Length $=6-0=6$

$\implies $ Breadth $=5-0=5$

Area of rectangle is $5\times 6=30$

If a line cuts sides $BC, CA$ and $AB$ of $\triangle ABC$ at $P, Q, R$ respectively then " $\dfrac {BP}{PC}\cdot \dfrac {CQ}{QA}\cdot \dfrac {AR}{RB} = 1$. "  that statement is ?

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 $ABC,D$ 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

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

Show answer
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