Tag: mid-point and its converse

Quizzes (2) Questions (54)

Questions Related to mid-point and its converse

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

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

  1. $AD$ and $BE$ are perpendicular
  2. $BE$ and $CF$ are perpendicular
  3. area of $\Delta ABC=32$
  4. area of $\Delta DEF=8$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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 ?

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is Menelaus' Theorem, which states that for a line intersecting the sides of a triangle, the product of the ratios of the segments is 1.

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.