Area of triangle ABC=24cm2. F, E and D are the midpoints of sides AB, AC, BC respectively. Find the area of triangle EFD and of parallelogram BDEF.
Answers
The area of triangle EFD is 6 cm² and of parallelogram BDEF is 12 cm².
Step-by-step explanation:
Step 1:
It is given that E and F are the midpoints of AC and AB.
We know that 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.
So, BC//FE & FE = ½ BC = BD
⇒ BD // FE & BD = FE and similarly BF//DE & BF = DE
⇒ BDEF is a parallelogram …… [since in a parallelogram pair of opposite sides are equal and parallel]
Similarly, we can also prove that quadrilaterals FDCE & AFDE are also parallelograms.
Step 2:
We know that the diagonal of a parallelogram divides it into two triangles of equal areas.
Also, we have BDEF as a parallelogram, so its diagonal FD divides it into two triangles of equal areas.
∴ ar(ΔBFD) = ar(ΔEFD) …….. (i)
Similarly, in parallelogram AFDE and FDCE
ar(ΔAEF) = ar(ΔEFD) …….. [here EF is a diagonal] …….. (ii)
and,
ar(ΔCED) = ar(ΔEFD) …….. [here DE is a diagonal] …….. (iii)
From (i), (ii) and (iii)
ar(ΔBFD) = ar(ΔAEF) = ar(ΔCED) = ar(ΔEFD) ..... (iv)
Step 3:
From the figure attached below, we can write
ar(ΔABC) = ar(ΔBFD) + ar(ΔAFE) + ar(ΔCDE) + ar(ΔEFD)
⇒ ar(ΔABC) = 4 ar(ΔEFD) ……… [From eq (iv)]
⇒ ar(∆EFD) = (1/4) * ar(∆ABC)
since ar(∆ABC) is given as 24 cm²
⇒ ar(∆EFD) = (1/4) * 24
⇒ ar(∆EFD) = 6 cm²
Step 4:
From the figure attached below, we can write
Area (parallelogram BDEF) = ar(ΔEFD) + ar(ΔBFD)
⇒ ar(parallelogram BDEF) = ar(ΔEFD) + ar(ΔEFD) …….. [from eq. (iv)]
⇒ ar(parallelogram BDEF) = 2 * ar(ΔEFD)
⇒ ar(parallelogram BDEF) = 2 * 6
⇒ ar(parallelogram BDEF) = 12 cm²
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Also View:
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