Given:-
- HM = 2x+5
- FM = X+8
- HF = 3x+2
- perimeter (∆HFM) = 51 units
To find:-
- whether ∆ HFM is congruent to ∆FMK
Solution:-
:[tex]\implies[/tex] perimeter (∆HFM) = HM+FM+HF
:[tex]\implies[/tex] 51 = 2x+5+x+8+3x+2
:[tex]\implies[/tex] 51 = 6x+15
:[tex]\implies[/tex] 36 = 6x
:[tex]\implies[/tex] x = 6 units.
in ∆HFM,
- HM = 2x+5 = 2(6)+5 = 12+5 = 17units
- FM = X+8 = 6+8 = 14 units
- HF = 3x+2 = 3(6)+2 = 20 units
in ∆KFM,
- FM = 14 units
- KF = 4x-3 = 4(6)-3 = 21 units
- MK = 3x-1 = 3(6)-1 = 17 units
Since, all the sides of one triangle does not correspond to the other, these traingles are not congruent.
Hope it helps ⭐⭐⭐
