Tag: mid-point theorem and its converse

Questions Related to mid-point theorem and its converse

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 $205.$