Question

what is the largest integer value of mo for which rhombus mnop will have an obtuse angle at p?
a. 12
b. 14
c. 17
d. 19

Explanation:

Step1: Recall rhombus angle condition

In rhombus $MNOP$, all sides = 9. For $\angle P$ to be obtuse, diagonal $MO$ must satisfy the inequality from the law of cosines: $MO^2 > MP^2 + OP^2$ (since $\cos\theta < 0$ for obtuse $\theta$), and $MO < MP + OP$ (triangle inequality).

Step2: Calculate lower bound for $MO$

Substitute side length 9:
$MO^2 > 9^2 + 9^2 = 81 + 81 = 162$
$MO > \sqrt{162} \approx 12.727$

Step3: Calculate upper bound for $MO$

Triangle inequality for $\triangle MPO$:
$MO < 9 + 9 = 18$

Step4: Find largest integer $MO$

The largest integer less than 18 and greater than ~12.727 is 17.

Answer:

C. 17