"Question 5 D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC. Show that (i) BDEF is a parallelogram. (ii) ar (DEF) = ar (ABC) (iii) ar (BDEF) = ar (ABC)
Class 9 - Math - Areas of Parallelograms and Triangles Page 163"
Answers
Parallelogram :
A quadrilateral in which both pairs of opposite sides are parallel is called a parallelogram.
In a parallelogram diagonal divides it into two triangles of equal areas.
Mid point theorem:
The line segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of it.
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Given:
ABC is a Triangle in which the midpoints of sides BC ,CA and AB are D, E and F.
To show:
(i) BDEF is a parallelogram.(ii) ar (DEF) = ar (ABC)
(iii) ar (BDEF) = ar (ABC)
Proof: i)
Since E and F are the midpoints of AC and AB.
BC||FE & FE= ½ BC= BD
(By mid point theorem)
BD||FE & BD= FE
Similarly, BF||DE & BF= DE
Hence, BDEF is a parallelogram.
[A pair of opposite sides are equal and parallel]
(ii) Similarly, we can prove that FDCE & AFDE are also parallelograms.
Now, BDEF is a parallelogram so its diagonal FD divides its into two Triangles of equal areas.
∴ ar(ΔBDF)
= ar(ΔDEF) — (i)
In Parallelogram AFDE
ar(ΔAFE) = ar(ΔDEF) (EF is a diagonal) — (ii)
In Parallelogram FDCE
ar(ΔCDE) = ar(ΔDEF) (DE is a diagonal) — (iii)
From (i), (ii) and (iii)
ar(ΔBDF) = ar(ΔAFE) = ar(ΔCDE) = ar(ΔDEF).....(iv)
ar(ΔBDF) + ar(ΔAFE) + ar(ΔCDE) + ar(ΔDEF) = ar(ΔABC)
4 ar(ΔDEF) = ar(ΔABC)
(From eq iv)
ar(∆DEF) = 1/4 ar(∆ABC)........(v)
(iii) Area (parallelogram BDEF) = ar(ΔDEF) + ar(ΔBDF)
ar(parallelogram BDEF) = ar(ΔDEF) + ar(ΔDEF)
ar(parallelogram BDEF) = 2× ar(ΔDEF)
(From eq iv)
ar(parallelogram BDEF) = 2× 1/4 ar(ΔABC)
(From eq v)
ar(parallelogram BDEF) = 1/2 ar(ΔABC)
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