Question

In A ABC, D, E, and F are respectively the mid-points of BC, CA and AB. If the lengths of side AB,
BC, and CA are 7cm, 8cm, and 9cm, respectively, find the perimeter of ADEF.​

Answer 1

Appropriate Question :-

In triangle ABC, D, E, and F are respectively, the mid-points of BC, CA and AB. If the lengths of side AB, BC, and CA are 7cm, 8cm, and 9cm, respectively, find the perimeter of triangle DEF.

$\\ \\ \large\underline{\sf{Solution-}}$

Given that, In triangle ABC, D, E, and F are respectively, the mid-points of BC, CA and AB.

Also, given that

AB = 7 cm

BC = 8 cm

CA = 9 cm

As it is given that, D is the midpoint of BC and E is the midpoint of AC.

So, By using Midpoint Theorem, we have

$\bf\implies \:DE \: = \: \dfrac{1}{2}AB = \dfrac{7}{2} \: cm - - - (1)$

Also, given that D is the midpoint of BC and F is the midpoint of AB.

So, by using Midpoint Theorem, we have

$\bf\implies \:DF \: = \: \dfrac{1}{2}AC = \dfrac{9}{2} \: cm - - - (2)$

Also, given that E is the midpoint of AC and F is the midpoint of AB.

So, by using Midpoint Theorem, we have

$\bf\implies \:EF \: = \: \dfrac{1}{2}BC = \dfrac{8}{2} = 4\: cm - - - (3)$

Now, Consider

$\bf \: Perimeter_{(\triangle\:DEF)}$

$\sf \: = \: DE \: + \: EF \: + \: DF$

$\sf \: = \: \dfrac{7}{2} + 4 + \dfrac{9}{2}$

$\sf \: = \: \dfrac{7 + 8 + 9}{2}$

$\sf \: = \: \dfrac{24}{2}$

$\sf \: = \: 12 \: cm$

So,

$\bf\implies \:Perimeter_{(\triangle\:DEF)} \: = \: 12 \: cm$

$\rule{190pt}{2pt}$

Theorem Used :-

Midpoint Theorem : - This theorem states that line segment joining the midpoints of two sides of a triangle is parallel to third side and equals to half of it.

Answer 2

$\colorbox{blue}{★QUESTION★}$

In A ABC, D, E, and F are respectively the mid-points of BC, CA and AB. If the lengths of side AB,

BC, and CA are 7cm, 8cm, and 9cm, respectively, find the perimeter of ADEF.

$\colorbox{red}{☟︎︎︎ANSWER✍︎}$

Compute the figure

Given, that, in Δ ABC

D, E AND F ARE THE MIDPOINT OF BC, CA AND AB

THEREFORE,THE FIGURE IS A SHOW UP

☝︎☝︎☝︎☝︎☝︎☝︎☝︎☝︎☝︎☝︎☝︎☝︎☝︎☝︎☝︎☝︎☝︎☝︎☝︎☝︎☝︎☝︎☝︎☝︎☝︎

COMPUTE THE SIDE EF,DF AND DE.THIS IS

IN Δ ABC,IT CAN BE SEEN THAT IS,

F AND E ARE MIDPOINT OF AB AND AC

THEREFORE, BY MIDPOINT THEREM

$ef = \frac{1}{2 \: }bc$

SIMILARLY

$df = \frac{1}{2} \: ac \\ \: \: \: \: \: \:and \\ de = \frac{1}{2} \: ab$

COMPUTE THE PERIMETER OF Δ DEF THAT IS

PERIMETER OF ΔDEF = DE+EF+DF

$\frac{1}{2} ab + \frac{1}{2}bc + ef + df$

$= \frac{1}{2} \times 7 + \frac{1}{2} \times 8 + \frac{1}{2} \times 9$

$= \frac{7}{2} + 4 + \frac{9}{2} \\ = 3.5 + 4 + 4.5 \\ = 12cm$

$\huge\mathfrak\green{Be \: branily}$

$\sf \colorbox{cyan} {ANSWER BY ACHALMUCHHAL2}$