Pythagoras Theorem

Question: The longer leg of right angle triangle is 2 more than the shorter leg, and the hypotenuse is 2 less than twice the shorter leg. find the sides of the triangle.

Solution:  

Let x be the shorter leg
longer leg=x+2
hypotenuse=2x-2

apply Pythagoras theorem

a²+b²=c²

^{x^2+(x+2)^2=(2x-2)^2}

^{x^2+x^2+4+4x=4x^2+4-8x}

^{2x^2+4x=4x^2-8x}

2x^2-12x=0

x^2-6x=0

x(x-6)=0

x=0 or x-6=0

x=6 shorter leg

x+2=8 Longer leg

2x-2=10 hypotenuse

Check:

6^2+8^2=10^2

36+64=100

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Question: The diagonal of a rectangle is 3 less than 4 times the width  of the rectangle. The length of the rectangle is 1 less than the diagonal. Find the length of the diagonal of the rectangle.

Solution:

Let width of the rectangle=x

diagonal of the rectangle=4x-3

Length of rectangle =  (4x-3)-1=4x-4

Length of the diagonal of rectangle=?

Applying Pythagoras theorem

x^2+(4x-3)^2=(4x-4)^2

x^2+16x^2+9-24x=16x^2+16-32x

x^2-24x=7-32x

x^2-8x+7=0

(x-1)(x-7)=0

x=7 width of the rectangle

4x-3=24 diagonal of the rectangle

4x-4=25 length of the rectangle.

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Question: The longer leg of the right angle triangle is 1 less than twice the shorter leg, and the hypotenuse is 1 more than twice the shorter leg. Find the sides of the triangle.

Solution: 

Let Shorter leg =x
Longer leg= 2x-1
hypotenuse=2x+1

apply Pythagoras theorem

a²+b²=c²

x^2+(2x-1)^2=(2x+1)^2
x^2+4x^2+1-4x=4x^2+1+4x
x^2-8x=0
x(x-8)=0
x=8 Shorter leg
2x-1=16-1=15 Longer leg
2x+1=16+1=17 Hypotenuse.

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Question: The shorter leg of a right angle triangle is 1 less than the longer leg, the hypotenuse is 1 less than the twice the shorter leg. Find the sides of the triangle.

Solution: 

Let Longer leg =x
shorter leg =x-1
Hypotenuse=2(x-1)-1=2x-3
apply Pythagoras theorem

a²+b²=c²

^{(x-1)^2+x^2=(2x-3)^2}

^{x^2+1-2x+x^2=4x^2+9-12x}

^{2x^2-10x+8=0}

^{x^2-5x+4=0}

^{x^2-4x-x+4=0}

^{x(x-4)-1(x-4)=0}

^(x-1)(x-4)=0

^x-4=0

^{x=4 longer leg}

^{Shorter leg=x-1=4-1=3}

^{Hypotenuse=2x-3=8-3=5}

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^{Question: The longer leg of a right angle triangle is 4 less than the hypotenuse. The longer leg is 2 less than twice the shortest leg. Find the sides of the triangle. [Hint: let x=shorter leg, and paraphrase the first sentence].}

^Solution: 

Let Shorter leg=x
given longer leg =2x-2
and  longer leg=y-4 let y be the hypotenuse
2x-2=y-4
y=2x+2

apply Pythagoras theorem

a²+b²=c²

x^2+(2x-2)^2=(2x+2)^2
x^2+4x^2+4-8x=4x^2+4+8x
x^2-16x=0
x(x-16)=0
x=0 or x-16=0
x=16 shorter leg
Longer leg=2x-2=30
hypotenuse y=2x+2=34


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Question: The hypotenuse of a right angle triangle is 4 less than 3 times the shortest side, and the longer leg is 4 more than twice the shortest side.find the sides of the triangle.

Solution:

Let shortest side=x
longest leg=2x+4
Hypotenuse=3x-4

apply Pythagoras theorem

a²+b²=c²

x^2+(2x+4)^2=(3x-4)^2
x^2+4x^2+16+16x=9x^2+16-24x
5x^2+16x+16=9x^2-24x+16
4x^2-40x=0
x^2-10x=0
x(x-10)=0
x=10 shorter side
longer leg 2x+4=24
Hypotenuse=3x-4=26.

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