the hypotenouse of a right angle triangle is two more than perpendicular and one more than twice the base find the sides
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consider a given right angle triangle
let whose shortest side is base(B) and third side is length(L) and H be a hypotenuse.
According to the given question,
H=2B+6......(1)
L=H−2........(2)
By equation (1),
B=
2
H−6
.......(3)
We know that,
By Pythagoras theorem,
H
2
=B
2
+L
2
⇒H
2
=(
2
H−6
)
2
+(H−2)
2
⇒H
2
=
4
H
2
+36−12H
+H
2
+4−4H
⇒4H
2
=H
2
+36−12H+4H
2
+16−16H
⇒H
2
+36−12H+16−16H=0
⇒H
2
+52−28H=0
⇒H
2
−28H+52=0
⇒H
2
−(26+2)H+52=0
⇒H
2
−26H−2H+52=0
⇒H(H−26)−2(H−26)=0
⇒(H−26)(H−2)=0
⇒H=2,26
H=2 is very small then H=26 exists.
Put the value of H in equation (2) and (3) to, we get
L=H−2
L=26−2=24
B=
2
H−6
B=
2
26−6
B=
2
20
B=10
Hence, B=10m, L=24m and H=26m
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