1.the diagonal of a rectangular field is 60meter more than the shorter side. if the longer side is 30 metres more than the shorter side find the sides of the field.
2. the difference of squares of two number is 180. the square of the smaller number is 8 times the larger number.find the two numbers.
Answers
Answered by
19
hey dear here is ur answer
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1. let be shorter side =X
diagonal side=X+60
longer side= X + 30
by applying Pythagoras theorem;
(X + 60)²= (X + 30)²+x²
X² + 3600+120x=x²+ 900 + 60 x +x²
x²+3600+120x-x²-900-60x-x²=0
-x²+60x+2700=0
x²-60x+2700=0
x²-x(90-30)-2700=0
X²-90x+30x-2700=0
X(x-90)+30(x-90)=0
(x-90) (X+30)
x-90=0. X+30=0
X=90 X=-30(side can not be negative
hence,
shorter side=90m.
longer side=90+30=120m.
⭐⭐<==================>⭐
1. let be shorter side =X
diagonal side=X+60
longer side= X + 30
by applying Pythagoras theorem;
(X + 60)²= (X + 30)²+x²
X² + 3600+120x=x²+ 900 + 60 x +x²
x²+3600+120x-x²-900-60x-x²=0
-x²+60x+2700=0
x²-60x+2700=0
x²-x(90-30)-2700=0
X²-90x+30x-2700=0
X(x-90)+30(x-90)=0
(x-90) (X+30)
x-90=0. X+30=0
X=90 X=-30(side can not be negative
hence,
shorter side=90m.
longer side=90+30=120m.
ShaadabShamim:
Mate, your given answer is wrong.
Answered by
1
Answer:
90 m and 120m
Step-by-step explanation:
let be shorter side =X
diagonal side=X+60
longer side= X + 30
by applying Pythagoras theorem;
(X + 60)²= (X + 30)²+x²
X² + 3600+120x=x²+ 900 + 60 x +x²
x²+3600+120x-x²-900-60x-x²=0
-x²+60x+2700=0
x²-60x+2700=0
x²-x(90-30)-2700=0
X²-90x+30x-2700=0
X(x-90)+30(x-90)=0
(x-90) (X+30)
x-90=0. X+30=0
X=90 X=-30(side can not be negative
hence,
shorter side=90m.
longer side=90+30=120m
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