Question

plzz help me to solve this one. plzz plzz it's very important

Answer 1

LOOK ....
IN TRI. ACD....

AC=CD

SO IT IS AN ISOSCELES TRIANGLE.....

SO AC=AD=11√2....

ANG. C = 90°

BY PYTHAGORAS.....
AC2+CD2=AD2

(11√2)2+(11√2)2=AD2

121*2+121*2=AD2

242+242=AD2

484=AD2

AD = 22.....

IN TRI. ABC....

AB = BC = X

AND ANGLE B = 90°

SO BY PYTHAGORAS THM.....

AB2+BC2= AC2

AC= 11√2

SO X2+X2= (11√2)2

2X2= 121*2

2X2= 242

X2= 242/2

X2= 121

X= 11

SO AB= BC= 11

CD= 11√2

AD = 22

SO BY FINDING THE PERIMETER WE GET....

AB+BC+AD+CD= 11+11+11√2+22

SO PERIMETER= 44+11√2

SO THE PERIMETER OF GIVEN QUADRILATERAL= 44+11√2.......

I HOPE IT WILL BE RIGHT AND HELPFUL.....

THINK DIFFERENTLY......
BE BRAINLY....☺

PLZ MARK AS BRAINLIEST..... ☺

Answer 2

hope this will help you