## Trigonometry 7th Edition

(a) Diagonal of any face = $x\sqrt 2$ (b) Diagonal that passes through th center of the cube = $x\sqrt 3$
(a) Given length of each edge is $'x'$ Each face of a cube is a square and a diagonal divides it into two 45°–45°–90° triangles. Therefore- In 45°–45°–90° triangle Hypotenuse = $\sqrt 2 \times$ Shorter Side Therefore- Diagonal of a face = $\sqrt 2 \times$ EDGE Diagonal of a face = $\sqrt 2 \times x$ ($x$ is the edge of cube ) Diagonal of a face = $x\sqrt 2$ (b) Diagonal that passes through the center of the cube, diagonal of face and respective edge of the cube make a right triangle together . Applying Pythagorean theorem- $Center Diagonal^{2}$ = $Face Digonal^{2}$ + $Edge^{2}$ $Center Diagonal^{2}$ = $(x\sqrt 2)^{2}$ + $x^{2}$ [Substituting values of Face Diagonal and Edge] $Center Diagonal^{2}$ = $2x^{2} + x^{2}$ = $3x^{2}$ Therefore- Center Diagonal = $\sqrt (3x^{2})$ Or Center Diagonal = $x\sqrt 3$