Question

pleas help me with this question​

Answer 1

Answer:

Given:

AE//DB

BE//AD

D is the midpoint of BC

. : BD=DC

∆ADC=2cm*2

Step-by-step explanation:

∆ABD=∆AEB (diagonal divides the //gm into two equal halves)

∆ABD=∆ADC (triangles btw same parallel and having same base)

if ∆ADC = 2cm*2

Then //gm AEBD = 4cm*2

Answer 2

Given area of triangle ADC=2cm^2    ,     d is mdpt of BC

to find =ar //gm AEBD

construction= join A to D

proof= in triangle ABC, AD is the median,

              so, ar triangle ADC= ar triangle ABD=2 cm^2

in //gm AEBD , diagonal AB divides the //gm into 2 congruent triangles

                           since congruent triangles have equal areas,

                                              area of triangle ABD = area of triangle ABE

                                                =2 CM^2

                      so area of

                        ABD + ABE = AEBD

                            2   +    2   =    4 CM^2

                                           THEREFORE AREA OF //GM

                                                         AEBD IS 4 cm^2