Question

In the given figure D E F are the midpoints of the sides BC CA and AB respectively
(I) if AB = 6.2 cm find the if DE
(II) DE =3.8 cm find AC
(III) if perimeter of triangle ABC is 21 cm find FE​

Answer 1

Given:

D is mid-point of side BC

E is mid-point of side CA

F is mid-point of side AB

AB = 6.2 cm

DF = 3.8 cm

The perimeter of ΔABC = 21 cm

To find:

(I) The length of DE

(II) The length of AC

(III) The length of FE​

Solution:

We know,

$\boxed{\bold{\underline{MID-POINT\:THEOREM}}}:$ This theorem states that the line segment joining the two sides of a triangle at the midpoints of those two sides is half the length of the third side.

(I). Finding the length of DE:

In the given ΔABC,

D and E are the midpoints of the side BC and CA of the triangle and the third side opposite to DE is AB whose length is 6.2 cm.

Therefore, using the mid-point theorem, we get

$DE = \frac{1}{2} \times AB = \frac{1}{2} \times 6.2 = 3.1 \:cm$

Thus, $\boxed{\bold{DE = 3.1\: cm}}$

(II). Finding the length of AC:

In the given ΔABC,

D and F are the midpoints of the side BC and AB of the triangle and the third side opposite to DF, whose length is given as 3.8 cm, is AC.

Therefore, using the mid-point theorem, we get

$AC = 2 \times DF = 2 \times 3.8 = 7.6 \:cm$

Thus, $\boxed{\bold{AC = 7.6\: cm}}$

(III). Finding the length of FE:

We know,

The perimeter of a triangle = AB + BC + CA

Substituting the values of perimeter = 21 cm, AB = 6.2 cm & AC = 7.6 cm, we get

⇒ 6.2 + BC + 7.6  = 21

⇒ 13.8 + BC = 21

⇒ BC = 21 - 13.8

⇒ BC = 7.2 cm

In the given ΔABC,

F and E are the midpoints of the side AB and CA of the triangle and the third side opposite to FE is BC whose length is 7.2 cm.

Therefore, using the mid-point theorem, we get

$FE = \frac{1}{2} \times BC = \frac{1}{2} \times 7.2 = 3.6 \:cm$

Thus, $\boxed{\bold{FE = 3.6\: cm}}$.

------------------------------------------------------------------------------------

Also View:

Prove mid-point theorem

https://brainly.in/question/2322835

triangle ABC, P, Q and R are the midpoints of AB, BC and CA respectively. AB= I2 cm, BC = 16 cm and CA = 20 cm, find the perimeter of trapezium PBCR.

https://brainly.in/question/13453555

In triangle ABC ,P and Q are the mid points of AB and AC .P and Q are joined .IfPQ=3.5cm, then find the length of BC.

https://brainly.in/question/19617376

Answer 2

Answer:

Given:

D is mid-point of side BC

E is mid-point of side CA

F is mid-point of side AB

AB = 6.2 cm

DF = 3.8 cm

The perimeter of ΔABC = 21 cm

To find:

(I) The length of DE

(II) The length of AC

(III) The length of FE

Solution:

We know,

\boxed{\bold{\underline{MID-POINT\:THEOREM}}}:

MID−POINTTHEOREM

: This theorem states that the line segment joining the two sides of a triangle at the midpoints of those two sides is half the length of the third side.

(I). Finding the length of DE:

In the given ΔABC,

D and E are the midpoints of the side BC and CA of the triangle and the third side opposite to DE is AB whose length is 6.2 cm.

Therefore, using the mid-point theorem, we get

DE = \frac{1}{2} \times AB = \frac{1}{2} \times 6.2 = 3.1 \:cmDE=

2

1

×AB=

2

1

×6.2=3.1cm

Thus, \boxed{\bold{DE = 3.1\: cm}}

DE=3.1cm

(II). Finding the length of AC:

In the given ΔABC,

D and F are the midpoints of the side BC and AB of the triangle and the third side opposite to DF, whose length is given as 3.8 cm, is AC.

Therefore, using the mid-point theorem, we get

AC = 2 \times DF = 2 \times 3.8 = 7.6 \:cmAC=2×DF=2×3.8=7.6cm

Thus, \boxed{\bold{AC = 7.6\: cm}}

AC=7.6cm

(III). Finding the length of FE:

We know,

The perimeter of a triangle = AB + BC + CA

Substituting the values of perimeter = 21 cm, AB = 6.2 cm & AC = 7.6 cm, we get

⇒ 6.2 + BC + 7.6 = 21

⇒ 13.8 + BC = 21

⇒ BC = 21 - 13.8

⇒ BC = 7.2 cm

In the given ΔABC,

F and E are the midpoints of the side AB and CA of the triangle and the third side opposite to FE is BC whose length is 7.2 cm.

Therefore, using the mid-point theorem, we get

FE = \frac{1}{2} \times BC = \frac{1}{2} \times 7.2 = 3.6 \:cmFE=

2

1

×BC=

2

1

×7.2=3.6cm

Thus, \boxed{\bold{FE = 3.6\: cm}}

FE=3.6cm