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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

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

  1. $1:4$
  2. $1:3$
  3. $1:1$
  4. $1:2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the midpoint theorem, each segment connecting the midpoints of the triangle formed by the origin and the vertices of the parallelogram is half the length of the corresponding side of the parallelogram. Summing these segments gives a perimeter for PQRQ equal to half the perimeter of KITE. The ratio of PQ+QR+RS+SP to KE+ET+TI+IK is therefore 1:2.

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.