Answer:
[tex]\sf IK=IJ\sqrt{2}[/tex]
Step-by-step explanation:
The interior angles of a triangle sum to 180°. Therefore, if ΔIJK is an isosceles right triangle where m∠J = 90°:
- vertex is m∠J = 90°
- two base angles are m∠K and m∠I
To calculate the base angles:
⇒ m∠K + m∠I + m∠J = 180°
⇒ m∠K + m∠I + 90° = 180°
⇒ m∠K + m∠I = 90°
⇒ m∠K = m∠I = 45°
Therefore, IJ and JK are the legs of the right triangle and IK is the hypotenuse.
To find the length of the hypotenuse, use Pythagoras' Theorem.
Pythagoras’ Theorem: [tex]\sf a^2+b^2=c^2[/tex]
(where a and b are the legs, and c is the hypotenuse, of a right triangle)
Given:
- a = IJ = 4
- b = JK = 4
- c = IK
Substitute the given values into the formula and solve for IK:
[tex]\sf \implies IJ^2+JK^2=IK^2[/tex]
[tex]\sf \implies 4^2+4^2=IK^2[/tex]
[tex]\sf \implies IK^2=32[/tex]
[tex]\implies \sf IK=\sqrt{32}[/tex]
[tex]\implies \sf IK=\sqrt{16 \cdot 2}[/tex]
[tex]\implies \sf IK=\sqrt{16}\sqrt{2}[/tex]
[tex]\implies \sf IK=4\sqrt{2}[/tex]
As IJ = 4 then:
[tex]\implies \sf IK=IJ\sqrt{2}[/tex]
