Question

In the given figure, ABCD is a rhombus. If OB = 3 units, OC = 4 units and AE = 6.4 units. Find the length of CE. (CE is perpendicular to AB)

• A. √26.04 units
• B. √25 units
• C. √23.04 units
• D. √20.04 units

Answer 1

The correct option is C

√23.04 units

Since, diagonals of a rhombus intersect at, ${72}^{\circ }$BOC is a right angled triangle. So, using Pythagoras theorem,

$O{B}^2$ + $O{C}^2$ = $B{C}^2$

BC = $\sqrt{(3^2+4^2)}$ = $\sqrt{(9+16)}$ = $\sqrt{25}$ = 5 units

Since all the sides of rhombus are equal, AB = BC = 5 units

AE = AB + BE

BE = AE - AB = 6.4 - 5 = 1.4 units

Again applying Pythagoras theorem on DBEC

$C{E}^2$ + $B{E}^2$ = $B{C}^2$

CE = $\sqrt{(B{C}^2-B{E}^2)}$ = $\sqrt{(5^2-{1.4}^2)}$ = $\sqrt{(25-1.96)}$ = $\sqrt{\left(23.04\right)}$ units