Tag: application of the mid-point theorem

Questions Related to application of the mid-point theorem

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 :

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

  1. $17 cm$

  2. $38.4 cm$

  3. $25.6 cm$

  4. $6.4 cm$

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Correct Option: D
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

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

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

  1. $1$

  2. $2$

  3. $3$

  4. $4$

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Correct Option: A

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 

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

  1. 60

  2. 205

  3. 115

  4. 405

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Correct Option: B
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 $205.$

In $\triangle ABC, D$ and $E$ are the mid point of $\bar {BC}$ and $\bar {AC}$ respectively. $\bar {AD}$ and $\bar {BE}$ intersect each other in $G.A$ line $m$ passing through $D$ and parallel to $\overleftrightarrow { BE } $ intersects $\bar {AC}$ in $K$.
then $AC=4CK$

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

  1. True

  2. False

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Correct Option: A