Question

if two side of the right triangle are (x+3)and (x+1)and it's hypotenuse is 100 find the value of x​

Answer 1

Correct Question:-

If two side of the right triangle are (x+3)and (x+1)and it's hypotenuse is 10 find the value of x

Given :-

• (x + 3) , (x + 1) are the sides of right angle triangle
• Hypotenuse is 10

To find:-

• $\sf{Value\:of \:x }$

Solution :-

$\sf{As\: we \:know\: that\: from \:Pythagoras\: Theorem , }$

$\sf{\:In \:a \:right \:angle\: triangle\: ,}$

$\sf (side)^2 + (side)^2 = (hypotenuse)^2$

$\sf (x+3)^2+(x+1)^2 =(10)^2$

$\sf (x)^2+2(x)(3) +(3)^2+ (x)^2+2(x)(1) +(1)^2= 10\times10$

$\sf x^2+6x+9 +x^2+2x+1 =100$

$\sf x^2+x^2+6x+2x +9+1 = 100$

$\sf 2x^2+8x+10=100$

$\sf 2x^2 +8x = 100-10$

$\sf 2x^2+8x = 90$

$\sf 2x^2+8x -90=0$

$\sf Take\: common\: "2"$

$\sf2(x^2+4x-45)=0$

$\sf x^2+4x-45=0$

$\sf Solving \: the \:Quadratic \: equation\: by \: Quadratic\: formula$

$\sf x = \sf\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}$

$\sf a= 1\\b =4\\c = -45$

$\sf x=\dfrac{-4\pm\sqrt{(4)^2-4(1)(-45)} }{2(1)}$

$\sf x =\dfrac{-4\pm\sqrt{16+180} }{2}$

$\sf x = \dfrac{-4\pm\sqrt{196} }{2}$

$\sf x= \dfrac{-4\pm 14}{2}$

$\sf x= \dfrac{-4+14}{2} , \dfrac{-4-14}{2}$

$\sf x= \dfrac{10}{2} , \dfrac{-18}{2}$

$\sf x = 5, -9$

Verification :-

$\sf We \: got\: the\: value\: of\: x \: Since\: , It\: should\: satisfies \: the \:Pythagoras\:theorem$

$\sf At \: x = 5$

$\sf (Side)^2+(Side)^2 = (hypotenuse)^2$

$\sf(x+3)^2+(x+1)^2 = (10)^2$

$\sf(5+3)^2+(5+1)^2= 100$

$\sf(8)^2+(6)^2= 100$

$\sf 64+36 =100$

$\sf 100=100$

$\sf1st\: Condition \: Verified!$

$\sf At\: x= -9$

$\sf (Side)^2+(Side)^2 = (hypotenuse)^2$

$\sf(x+3)^2+(x+1)^2 = (10)^2$

$\sf(-9+3)^2+(-9+1)^2 =1 00$

$\sf (-6)^2+(-8)^2= 100$

$\sf 64+36 =100$

$\sf 100=100$

$\sf 2nd \: Condition \: also \: verified!$