RRB NTPC 29 April 2016 Paper

Question : 99 of 100

Marks: +1, -0

In a right-angled triangle, the hypotenuse is 2 cm longer than the perpendicular which is 2 cm longer than the base.

Solution:

(1) H = x + 4, P = x + 2, B = x

\⇒ \ { H } ^ { 2 } = { B } ^ { 2 } + { P } ^ { 2 }

\⇒ ( \ { x } + 4 ) ^ { 2 } = ( { x } + 2 ) ^ { 2 } + { x } ^ { 2 }

\⇒( \ { x } + 4 ) ^ { 2 } - ( { x } + 2 ) ^ { 2 } = { x } ^ { 2 }

\⇒ ( x + 4 + \ { x } + 2 ) ( { x } + 4 - { x } - 2 ) = { x } ^ { 2 }

\⇒ ( 2 { x } + 6 ) ( 2 ) = { x } ^ { 2 }

\⇒ 4 x + 12 = x ^ { 2 }

\⇒ x ^ { 2 } - 4 x - 12 = 0

x ^ { 2 } - 6 x + 2 x - 12 = 0

x ( x - 1 ) + 2 ( x - 6 ) = 0

( x + 2 ) ( x - 6 ) = 0

x \≠ 2 x = 6

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