Question
One leg of a right angled triangle exceeds the other leg by 4 inches. The hypotenuse is 20 inches. Find the length of the shorter leg of the triangle.
Answer: Option A
:
A
Let thelength of the legs of the triangle be x and x+4inches.
Applying Pythagoras' theorem,
x2+(x+4)2=(20)2x2+x2+8x+16=4002x2+8x−384=0x2+4x−192=0
Solving the quadratic equation,x2+16x−12x−192=0x(x+16)−12(x+16)=0(x+16)(x−12)=0∴x=−16&12
Length can't be negative, hence
x=12inches
The shortest side(leg) is 12 .
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:
A
Let thelength of the legs of the triangle be x and x+4inches.
Applying Pythagoras' theorem,
x2+(x+4)2=(20)2x2+x2+8x+16=4002x2+8x−384=0x2+4x−192=0
Solving the quadratic equation,x2+16x−12x−192=0x(x+16)−12(x+16)=0(x+16)(x−12)=0∴x=−16&12
Length can't be negative, hence
x=12inches
The shortest side(leg) is 12 .
Was this answer helpful ?
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