Question

can any body answer question in the pic question 19 20 21 if it will be correct i will mark as brainliest

Answer 1

(19.)this is something similar to your question

in picture.




for QNo.21

1. Take half of angle ACD=x. hence angle ABD=2x .

2. Take angle BAD=DAC=y
3. given AB=CD
now, 2x+y+y+x=180 (sum of all angles of triangle is 180)
3x+2y=180
BY sine rule sin(x)/AD=sin(y)/CD and sin(2x+y)/AB=sin(2x)/AD
simplifying the above equation will give sin(x)*sin(2x+y)=sin(y)*sin(2x)
on further simplification you get sin(x-y)=0 hence x=y=36
Therefore angle BAC=2y=72



Q20.(ii)

ABCD is a quad.Its diagonals are AC and BD.
In triangle ACB, AB + BC > AC - ----------1
In triangle BDC, BC + CD > BD - ----------2
In triangle ACD, AD + DC > AC ------------3
In triangle BAD, AB + AD > BD -------------4
Adding 1,2,3 and 4,
AB + BC + BC + CD + AD + DC + AB + AD > AC + BD + AC + BD
2AB + 2BC + 2CD + 2AD > 2AC + 2BD
AB + BC + CD + AD > AC + BD.------------5
(i) now AC= BD
then from eqn 5:-
AB+BC+CD+AD>AC+BD
AB+BC+CD+AD>BD+BD
AB+BC+CD+AD>2BD



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