मराठी

In a ΔABC, D, E, F are the mid-points of sides BC, CA and AB respectively. If ar (ΔABC) = 16cm2, then ar (trapezium FBCE) =

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प्रश्न

In a ΔABC, D, E, F are the mid-points of sides BC, CA and AB respectively. If ar (ΔABC) = 16cm2, then ar (trapezium FBCE) =

पर्याय

  •  4 cm2

  •  8 cm2

  • 12 cm2

  • 10 cm2

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उत्तर

Given: In ΔABC

(1) D is the midpoint of BC

(2) E is the midpoint of CA

(3) F is the midpoint of AB

(4) Area of ΔABC = 16 cm2

To find: The area of Trapezium FBCE

Calculation: Here we can see that in the given figure,

Area of trapezium FBCE = Area of ||gm FBDE + Area of ΔCDE

Since D and E are the midpoints of BC and AC respectively.

∴ DE || BA ⇒  DE || BF

Similarly, FE || BD. So BDEF is a parallelogram.

Now, DF is a diagonal of ||gm BDEF.

∴ Area of ΔBDF = Area of ΔDEF ……(1)

Similarly,

DE is a diagonal of ||gm DCEF

∴ Area of ΔDCE = Area of ΔDEF ……(2)

FE is the diagonal of ||gm AFDE

∴ Area of ΔAFE = Area of ΔDEF ……(3)

From (1), (2), (3) we have

Area of ΔBDF = Area of ΔDCF = Area of ΔAFE = Area of ΔDEF

But

Area of ΔBDF + Area of ΔDCE + Area of ΔAFE + Area of ΔDEF = Area of ΔABC

∴ 4 Area of ΔBDF = Area of ΔABC

Area of ΔBDF = `1/4` Area of ΔABC 

                       = `1/4 (16)`

                      = 4 cm2 

Area of ΔBDF = Area of ΔDCE = Area of ΔAFE = Area of ΔDEF = 4 cm2 …….(4)

Now

Area of trapezium FBCE = Area of || FBDE + Area of ΔCDE 

                                      =(Area of  ΔBDF + Area of ΔDEF ) + Area of ΔCDE

                                      = 4 + 4+ 4 (from 4) 

                                     = 12 cm2

Hence we get

Area of trapezium FBCE  = 12 cm2 

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पाठ 14: Areas of Parallelograms and Triangles - Exercise 14.5 [पृष्ठ ६०]

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आर.डी. शर्मा Mathematics [English] Class 9
पाठ 14 Areas of Parallelograms and Triangles
Exercise 14.5 | Q 5 | पृष्ठ ६०

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