The base of a right angle triangle is 2cm more than its perpendicular its area is 24cm ^. Find its perimeter.
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let the perpendicular and base of the right angle∆ be x and y respectively.
ATQ,
y - 2 = x
Area = 24 cm²
=> 1/2 * b * h = 24
=> x * y = 48
=> y(y - 2) = 48
=> y² - 2y = 48
=> y² - 8y + 6y - 48 = 0
=> y( y - 8 ) +6( y - 8 ) = 0
=> (y - 8)(y + 6) = 0
Since sides are positive
=> y = 8cm
=> x = 6cm
Hypotenuse = 10 cm ( Pythagoras theorem)
=> Perimeter = 6 + 8 + 10
=> Perimeter = 24 cm
Hope this helps
let the perpendicular and base of the right angle∆ be x and y respectively.
ATQ,
y - 2 = x
Area = 24 cm²
=> 1/2 * b * h = 24
=> x * y = 48
=> y(y - 2) = 48
=> y² - 2y = 48
=> y² - 8y + 6y - 48 = 0
=> y( y - 8 ) +6( y - 8 ) = 0
=> (y - 8)(y + 6) = 0
Since sides are positive
=> y = 8cm
=> x = 6cm
Hypotenuse = 10 cm ( Pythagoras theorem)
=> Perimeter = 6 + 8 + 10
=> Perimeter = 24 cm
Hope this helps
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