The base of a triangle is 4cm longer than its altitude.If the area of the triangle is 48sq.cm then find its altitude& base
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let altitude be X
and base be y
therefore y=4+X •••••••(1)
area of ∆= 1/2 X × y =48
put the value of y
1/2 X × (4+X ) =48
1/2 . 4x + x² = 48
2x+ x²=48
eqⁿ will be x²+2x-48=0
by factorisation method
x²+8x-6x-48=0
x(x+8)-6(x+8)=0
(x-6) (X+8)=0
x=6
y=4+X
y= 10
and base be y
therefore y=4+X •••••••(1)
area of ∆= 1/2 X × y =48
put the value of y
1/2 X × (4+X ) =48
1/2 . 4x + x² = 48
2x+ x²=48
eqⁿ will be x²+2x-48=0
by factorisation method
x²+8x-6x-48=0
x(x+8)-6(x+8)=0
(x-6) (X+8)=0
x=6
y=4+X
y= 10
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