Question

Triangle ABC is a right triangle with sides of length 4, 6 and x. If 4 < x < 6, what is the approximate value of x ?
A. 4
B. 4.47
C. 5.21
D. 5.63
E. 7.21

Answer 1

Hint: We have to only use the property of triangles that the length of the side opposite to the longest angle of the triangle is greatest. And had to use the Pythagorean theorem \[\left[ {{{\left( {{\text{Hypotenuse}}} \right)}^2} = {{\left( {{\text{Perpendicular}}} \right)}^2} + {{\left( {{\text{Base}}} \right)}^2}} \right]\] to find the value of x.

Complete step-by-step answer:
As we know that ABC is a right-angled triangle.
And the maximum angle of the right-angled triangle is 90 degrees. And the side which is opposite to the 90 degrees angle is the hypotenuse of that triangle.
So, from the property of the triangle which states that the side opposite to the greatest angle of the triangle Is greatest.
Greatest side of the right-angled triangle is the hypotenuse of the triangle.
As we know that sides of the triangle are 4, 5 and x. where 4 < x < 6.
So, the length of the hypotenuse of the triangle ABC will be equal to 6.
Now as we know that according to the Pythagorean theorem if XYZ is a Pythagoras triangle, right angled at Y. Then the square of its hypotenuse (XZ) is equal to the sum of squares of its other two sides (XY and YZ).

So, applying Pythagorean theorem in triangle ABC. We get,
\[
  A{C^2} = A{B^2} + B{C^2} \\
  {\left( 6 \right)^2} = {x^2} + {\left( 4 \right)^2} \\
  36 = {x^2} + 16 \\
\]
Now subtracting 16 from both sides of the above equation. We get,
\[
  {x^2} = 20 \\
  x = \sqrt {20} \approx 4.47 \\
 \]
x is positive length is always positive.
Hence, the correct option will be B.

Note: Whenever we come up with this type of problem then we use the property of right-angled triangle that the greatest side is the hypotenuse of the triangle to find the hypotenuse of the triangle. After that we will apply the Pythagorean theorem to find the third side of the triangle. At last we can approximate our answer up to two decimal places.