Question

the area of 11 gm abcd is​

Answer 1

Answer:

Step-by-step explanation:

Given If ABCD is parallelogram and E is the mid point of AB And DE meets diagonal AC At point F

Then Side DE divided in ratio 1 : 2 by AC

So DF=2 FE

In both triangle ADF and ADE the height AF is common

Then DE=

hen area of triangle ADF=  

2

1

​  

DF×AF=60

AND   area of triangle ADE=  

2

1

​  

DE×AF=  

2

3

​  

(  

2

1

​  

DF×AF)=  

2

3

​  

×60=90 sq cm  

In  parallelogram ABCD  

Triangle ADE and triangle ADB height  is common because AB II DC  

AND E is the mid point of AB Then AB=2 AE

So area of triangle ADB=  

2

1

​  

×AB×height=  

2

1

​  

(2AE)×height=2×areaoftriangleADE=2×90=180 sq cm    

In figure diagonal AB divided parallelogram ABCD in two equal triangle ADB and BDC

Then area of parallelogram  ABCD=2×areaoftriangleADB=2×180=360 sq cm  

DF