Tag: proving the mid-point theorem

Questions Related to proving the 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.

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
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.

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

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

  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. $AE^2+CD^2=\frac {4}{5}AC^2$

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Correct Option: B

Find the midpoint of the segment connecting the points $(a, -b)$ and $(5a, 7b)$.

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

  1. $(3a, -3b)$

  2. $(2a, -3b)$

  3. $(3a, -4b)$

  4. $(-2a, 4b)$

  5. none of these

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Correct Option: D
Explanation:

The mid point is $(\dfrac{a-5a}2,\dfrac{b+7b}2)$.i.e $(-2a,4b)$
Option D is correct.

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

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

  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; $R-2$; $S-3$

Show answer
Correct Option: C
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.

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the _______ side.

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

  1. first

  2. second

  3. third

  4. none of the above

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Correct Option: C
Explanation:

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Fill in the blank:

Line joining the mid-points of any two sides of a triangle is _____ to the third side.

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

  1. perpendicular

  2. parallel

  3. Inclined at $60^o$

  4. None of these

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Correct Option: B
Explanation:

Line joining the mid-points of any two sides of a triangle is parallel to the third side. 

This is the standard Midpoint theorem of a triangle.

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

Show answer
Correct Option: A