Question

Suppose the lengths of two sides of a right triangle are represented by 2x and 3 (x + 1), and the longest side is 17 units. Find the value of x.

Answer 1

Answer:

x=4

Step-by-step explanation:

Step 1:-

given the lengths of two sides of a right angle are represented by 2x and 3(x+1) and longest side is 17 units.

AB = 2x and BC = 3(x+1) and longest side AC= 17

by using Pythagoras theorem

$AC^2 = AB^2 + BC^2$

step 2:-

The hypotenuse is longest side is AC = 17 units

(17)^2 = 4x^2 +9(X+1)^2

on simplification, we will use formula

$(a + b)^2 = a^2 +2ab+b^2$

289 = 4x^2 +9(x^2+2x+1)

$13x^2 +18x-280 = 0$

finding factors  70 X 52 = 3640

$13x^2 +70x-52x-280 = 0$

$13x^2 -52x+ 70x-280 = 0$

Taking common , we get

13x(x-4)+70(x-4)=0

x-4=0 and 13x+70=0

x=4 and $13x =-70$

x=4 and $x=\frac{-70}{13}$

we can not choose negative value so x value is 4

Final answer:- x = 4

verification:-

$AC^2 = AB^2 + BC^2$

289 = 4(4)^2+9(4+1)^2

289 = 64 +9(25)

289=289

Answer 2

Answer:

x = 4

Step-by-step explanation:

The length of two legs of a right angle triangle are 2x and 3(x+1).  

The length of longest side is 17 units. It means measure of hypotenuse is 17 units.

According to the pythagoras theorem,

$leg_1^2+leg_2^2=hypotenuse^2$

$(2x)^2+(3(x+1))^2=(17)^2$

$4x^2+9(x^2+2x+1)=289$

$4x^2+9x^2+18x+9=289$

$13x^2+18x-280=0$

On splitting the middle terms we get

$13x^2+70x-52x-280=0$

$x(13x+70)-4(13x+70)=0$

$(13x+70)(x-4)=0$

Using zero product property we get

$13x+70=0\Rightarrow x=-\dfrac{70}{13}$

$x-4=0\Rightarrow x=4$

The value of x can not be a negaive number, because for negative value of x the value of 2x is negative and side can not be negative.

Therefore, the value of x is 4.