The length of a hypotenuse of a right angle triangle exceeds the length of its base by 2 cm and exceeds twice the length of the altitude by 1 cm. Find the length of each side of the triangle (in cm).
A
6,8,10
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B
7,24,25
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C
8,15,17
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D
7,40,41
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Solution
The correct option is C8,15,17 Let the lenght of hypotenuse of the triangle be xcm Lenght of base of the triangle be ycm Lenght of altitude of the triangle be zcm According to given equation, x=y+2 and x=2z+1 =>y=x−2 and =>z=x−12 By Pythagoras Theorem (AC)2=(AB)2+(AC)2 x2=y2+z2 (i) putting values of y and z in equation (i) we get, x2=(x−2)2+(x−12)2 =>x2=4(x−2)2+(x−1)24 =>4x2=4(x2+4−4x)+x2+1−2x =>4x2=4x2+16−16x+x2+1−2x =>x2−18x+17=0 =>x2−17x−x+17=0 =>x(x−17)−1(x−17)=0 =>(x−17)(x−1)=0 Therefore, x=17 or x=1 x=1 is not possible as base =x−2=1−2=−1cm which cannot be in negative. Therefore,x=17 Putting the value in y y=x−2 y=17−2=15cm and z=x−12 z=17−12 z=8cm Therefore, lenght of the sides of the triangle is 17cm,15cm,8cm.