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How would you find the area of a right angle triangle when its hypotenuse is greater than its height by 2.2 cm and greater than its base by 4.4 cm?
Wiki User
∙ 12y ago
Let say
A is the length of the height
B the length of the base
H the length of hypotenuse
Then A=H-2.2 and B=H-4.4
Using Pythagore one gets H2=A2+B2
Then H2=(H-2.2)2+(H-4.4)2 or H2=H2 -4.4 H +4.84 +H2 -8.8H +19.36
H2=2H2 -13.2H +24.2
H2 -13.2H +24.2 = 0
H = (13.2-sqrt(77.44)) / 2 = 2.2 or H = (13.2+sqrt(77.44)) / 2 =11
The first solution is impossible because B=H-4.4
So H=11, A=8.8 and B=6.6
The area of the triangle is Area=(AxB)/2 = (8.8 x 6.6)/2 = 29.04
Wiki User
∙ 12y ago
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