the sides of a right angled triangle are 5x and (3x-1) . what is the length of the hypotenuse if the area is 60 cm^2
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Answered by
2
Let hypotenuse be ycm
so
5x^2+(3x-1)^2=y^2. consider it as eq 1
Now,area=60
1/2*5x*(3x-1)=60
1/2*(15x^2-5x)=60
15x^2-5x=120
divide it by 5
3x^2-x-24=0
3x^2-9x+8x-24=0
3x(x-3)+8(x-3)=0
(3x+8)(x-3)=0
x=-8/3,x=3 but it should be positive
so x=3
now sub x value in eq 1
5(3)^2+(3*3-1)^2=y^2
5(9)+8^2=y^2
45+64=y^2
y^2=109
y=✓109
y=13.7 cm
Hope it is useful!!!
so
5x^2+(3x-1)^2=y^2. consider it as eq 1
Now,area=60
1/2*5x*(3x-1)=60
1/2*(15x^2-5x)=60
15x^2-5x=120
divide it by 5
3x^2-x-24=0
3x^2-9x+8x-24=0
3x(x-3)+8(x-3)=0
(3x+8)(x-3)=0
x=-8/3,x=3 but it should be positive
so x=3
now sub x value in eq 1
5(3)^2+(3*3-1)^2=y^2
5(9)+8^2=y^2
45+64=y^2
y^2=109
y=✓109
y=13.7 cm
Hope it is useful!!!
saksham99999:
It is wrong the hypotenuse should be 17cm
how??
First substitute the values in each side and then apply pythagoras theorem
H^2 = 15^2 + 8^2
i agreee
but how about the question?
given area
Answered by
1
Answer:
Step-by-step explanation:
Hi!
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