An altitude of a triangle is five thirds of the length of its corresponding base. If the altitude is increased by 4 CM and the base is decreased by 2 CM, but the area of triangle remains the same. Find the base and altitude of the triangle respectively.
give the whole answer with all steps.
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For the first case,
Let the base be b cm
therefore, altitude = 5/3 b cm ( given)
therefore ,
area , A1 = 1/2× base ×altitude
= 1/2× b × 5/3b
= (5/6 b^2) cm^2
Now for 2nd case
base= (b -2)cm
altitude= (5/3b +4) cm
therefore ,
area , A2= 1/2× (b-2) × (5/3b +4)
= 1/2×( 5/3b^2 + 4b - 10/3b -8)
= 1/2×(5/3b^2 + 2/3b - 8)
= (5/6b^2 +1/3b - 4) cm^2
Now area A1 = area A2
therefore,
5/6b^2 = 5/6b^2 + 1/3b - 4
=>1/3b = 4
=> b = 12 cm
therefore , altitude =5/3 × 12 = 20cm
Let the base be b cm
therefore, altitude = 5/3 b cm ( given)
therefore ,
area , A1 = 1/2× base ×altitude
= 1/2× b × 5/3b
= (5/6 b^2) cm^2
Now for 2nd case
base= (b -2)cm
altitude= (5/3b +4) cm
therefore ,
area , A2= 1/2× (b-2) × (5/3b +4)
= 1/2×( 5/3b^2 + 4b - 10/3b -8)
= 1/2×(5/3b^2 + 2/3b - 8)
= (5/6b^2 +1/3b - 4) cm^2
Now area A1 = area A2
therefore,
5/6b^2 = 5/6b^2 + 1/3b - 4
=>1/3b = 4
=> b = 12 cm
therefore , altitude =5/3 × 12 = 20cm
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