Question

find the actual length of sides is triangle OTP in the following figure​

Answer 1

d2=t2 + r2

(x+1)^2 = (x-3)^2+(x-1)^2

x^+2x+1=x^2-6x +9+X^2-2x + I

- x^2+10x-9=0

x^2-10x+9=0

factorise

(x-9) (x-1)=0

x=9

sides of Rt. triangle OTPare

OP =x + I=9+I=10

TP =x-3=9-3=6

OT =x-l=9-1=8

Answer 2

Given: The following figure​.

To find: The actual length of sides is triangle OTP.

Solution: The actual length of sides is triangle OTP is 8, 6 and 10 units.

The line OT is from the centre of the circle to the point where the tangent touches the circle. Such a line is perpendicular to the tangent. Hence, the angle OTP is 90°. The line PT is the tangent to the circle. Thus, OTP is a right-angled triangle.

Also, the length of the three sides of the triangle is given in terms of x. Now, these lengths can be equated to one another by using the Pythagoras theorem.

$OP^{2} = OT^{2} + TP^{2}$

$(x+1)^{2} = (x-1)^{2} + (x-3)^{2}$

The above equation can be solved further to form a quadratic equation as stated below.

$x^{2} -10x+9 = 0$

$x^{2} -x-9x+9=0$

$x(x-1)-9(x-1)=0$

$(x-1)(x-9)=0$

Thus, the value of x is 1 or 9. The value cannot be x because the length of the line OT would become zero. Hence, x is 9. Now, the lengths of the sides are

$x-1 = 9-1$

         $=8$

$x-3 = 9-3$

         $=6$

$x+1=9+1$

         $=10$

Therefore, the actual length of sides is triangle OTP is 8, 6 and 10 units.