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AM:ME=9:8
BM:MD=25:9
Excellent question sir.. Thanks a lot u r great
AM:ME=9:8
BM:MD=25:9
Sir last me jo question diye h aap usme 🔺ABD ka median AM Ho skta h kya??
In this right angled triangle ABC line BD will be the altitude on hypotenuse AC.Using the property of altitude on hypotenuse in a right angled triangle AD:CD = 9:16.Now Solved using MPG and got AM:ME = 18:16 = 9:8 and BM:MD = 25:9.
25:9
9:8
9:8
25:9
AM:ME=9:8 AND BM:MD=25:9
AM:ME 9:8
BM:MD 25:9
AM:ME=9:8
BM:MD=25:9
Sir jyada fast ho jaa raha hai
1 9/8
2. 25/9
2 -5:3
1- 3:2
Couldnt solve it
AM/ME=9/8
BM/MD=25/9
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AM:ME=9:8
BM:MD=25:9
By Mass point theorem
AM:ME = 9:8
BM:MD =25:9
1)9/8 2)25/9
You r great sir
sir i have equated the area 5*BD/2 =4*3/2===>BD=12/5 now applying pythagorus theorem find CD=root(4^2 -(12/5)^2)=16/5 now AC=5 so AD=5-16/5=9/5 so ratio of BE/EC=1/1 and AD/DC=9/16 now let weight on point C=9 unit so weight distribution are on the points A=16 unit and B = 9 unit E=18 and D=25 and M=34 unit now required length ratios are reverse of mass distribution so AM/ME=18/16=9/8 and BM/MD=25/9
I shall try ur homwrk but ur explanation is awesome sir…