Tag: the mid-point theorem

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

State true or false:

$ D, E $ and $ F $ are the mid-points of the sides $ AB, BC $ and  $ CA $ of an isosceles $ \bigtriangleup ABC $  in which $ AB= BC $. then
 $ \bigtriangleup DEF $  is also isosceles.

  1. True

  2. False

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

AB = AC
Hence, $\angle ABC = \angle ACB$ (Isosceles triangle property)
Now, since, D and F are mid point of AB and AC respectively, thus DF II BC (Mid point theorem)
Hence, 
$\angle ADF = \angle ABC$ and $\angle AFD = \angle ACB$ (Corresponding angles)
Thus, 
$\angle ADF = \angle ABC = \angle AFD = \angle ACB$ 
Now, In $\triangle$ ADF and FEC
$\angle ADF = \angle FEC$ (Corresponding angles of parallel lines EF and AB)
$\angle AFD = \angle ACB $(Corresponding angles of parallel lines DF and BC)
AF = FC (F is the mid point of AC)
Thus $\triangle ADF \cong \triangle FEC$ (AAS rule)
Hence, AD = FE (corresponding sides of congruent triangles)
Similarly, we can prove, AF = DE
Since, AD = AF (half lengths of equal sides, AB and AC)
Thus, EF = DE or $\triangle$ DEF is an isosceles triangle.

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

In $\Delta ABC$, D and E are mid points of AB and BC respectively and $\angle ABC=90^o$, then

  1. $AE^2+CD^2=AC^2$
  2. $AE^2+CD^2=\frac {5}{4}AC^2$
  3. $AE^2+CD^2=\frac {3}{4}AC^2$
  4. $AE^2+CD^2=\frac {4}{5}AC^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In a right triangle ABC, using the Pythagorean theorem for triangles ABE and BCD, AE^2 = AB^2 + BE^2 and CD^2 = BC^2 + BD^2. Substituting BE = BC/2 and BD = AB/2, we get AE^2 + CD^2 = 5/4(AB^2 + BC^2), which equals 5/4(AC^2).

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

Fill in the blanks:
(i) The ling segment joining a vertex of a triangle to the midpoint of its opposite side is called a $\underline { P } $ of the triangle.
(ii) The perpendicular line segment from a vertex of a triangle to its opposite is called an $\underline { Q } $ of the triangle
(iii) A triangle has $\underline { R } $ altitudes and $\underline { S } $ medians

  1. $P-$ Altitude; $Q$- Median; $R-1$; $S-1$
  2. $P-$ Altitude; $Q$- Median; $R-3$; $S-3$
  3. $P-$ Median; $Q$- Altitude; $R-3$; $S-3$
  4. $P-$ Median; $Q$- Altitude; $R-2$; $S-3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Medians : The line segment from any vertex of a triangle to the midpoint of its opposite side is called medians of triangle. 

Altitude : The perpendicular drawn from a vertex to opposite side is called as altitude. 
A triangle has $3$ altitudes & $3$ medians.

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.