Tag: mid-point and its converse

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Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

D,E,F are midpoints of sides BC, CA and AB of $\Delta ABC$. If perimeter of $\Delta ABC$ is 12.8 cm, then perimeter of $\Delta DEF$ is :

  1. $17 cm$
  2. $38.4 cm$
  3. $25.6 cm$
  4. $6.4 cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given in $\triangle ABC$, $D,E,F$ are the mid points of sides $AB,BC$ and $CA$ respectively

Now using mid point theorem line segment joining the mid points of two sides is parallel to third side and also half of it.
$\therefore DF=\dfrac{1}{2}BC$
$\Rightarrow \dfrac{DF}{BC}=\dfrac{1}{2}.......(i)$
Similarly $\dfrac{DE}{AC}=\dfrac{1}{2}.........(ii)$
$\dfrac{EF}{AB}=\dfrac{1}{2}...........(iii)$
Using $(i),(ii)$ and $(iii)$
$\dfrac { DF }{ BC } =\dfrac { DE }{ AC } \dfrac { EF }{ AB } =\dfrac { 1 }{ 2 } $
$\dfrac { DF }{ BC } =\dfrac { DE }{ AC } \dfrac { EF }{ AB } =\dfrac { 1 }{ 2 } \ \therefore \triangle ABC\sim \triangle DEF$
Now if triangles are similar then ratio of their perimeter is equal to ratio of  of their corresponding sides.
$\ \dfrac { Perimeter(\triangle DEF) }{ Perimeeter(\triangle ABC) } =\dfrac { 1 }{ 2 } $
$\Rightarrow $ Perimeter of $\triangle ABC=\dfrac{1}{2}\times 12.8=6.4$ cm

Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

If A, B and C are the midpoint of the sides PQ, QR and PR of $\triangle $PQR respectively, then the area of $\triangle $ABC equals if area of $\triangle PQR$ is $4$ units

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

The triangle formed by joining the midpoints of the sides of a triangle has an area equal to 1/4 of the area of the original triangle. (1/4) * 4 = 1.

Multiple choice maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

The sides $AB, BC$ and $CA$ of a triangle $ABC$ have $3, 4$ and $5$ interior points respectively on them.The number of triangles that can be constructed using these interior points as vertices is 

  1. 60

  2. 205

  3. 115

  4. 405
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

No. of ways $={^{ 3 }{ { C } _{ 1 } }}\times {^{ 4 }{ { C } _{ 1 }} }\times {^{ 5 }{ { C } _{ 1 } }}+{^{ 3 }{ { C } _{ 2 } }}\left( ^{ 4 }{ { C } _{ 1 } }+{^{ 5 }{ { C } _{ 1 } }}\right)+{^{ 4 }{ { C } _{ 2 }} } \left(^{ 3 }{ { C } _{ 1 } }+{^{ 5 }{ { C } _{ 1 }} }\right) +{^{ 5 }{ { C } _{ 2 }} }\left( ^{ 3 }{ { C } _{ 1 } }+{^{ 4 }{ { C } _{ 1 } }}\right)$

$\Rightarrow$ No of ways $=60+3\left( 4+5\right) +6\left( 3+5\right) +10 \left( 3+4\right)$
$\Rightarrow$ No of ways $=60+27+48+70=205.$
Hence, the answer is $205.$

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.