R,O,C,A are the midpoints of the sides of the rhombus DJSB. if DJ= 16 sq rt of 2 cm and d=120 degrees, what is the perimeter of ROCA?
Review your question if it is correctly asked
In triangle DRA, draw an altitude DK where K is on RA, to get a 30-60-90° triangle
and you should know that the corresponding sides are in the ratio 1 : √3 : 2
So RK/(6√2) = √3/2
x = 3√6
then RA = 6√6
and the perimeter is 24√6
To find the perimeter of the quadrilateral ROCA, we need to know the lengths of its sides.
Since R, O, C, and A are midpoints of the sides of the rhombus DJSB, we can consider the rhombus as two congruent triangles, DJR and DBO.
We have the following information:
- DJ = 16√2 cm
- d = 120 degrees
To find the side lengths of DJR, we need to use trigonometry.
Let's find the length of the side DR:
Since DJ is the hypotenuse of the right triangle DJR, we can use trigonometry to find the length of the side DR. We know that the sine of an angle is the ratio of the opposite side to the hypotenuse.
sin(d) = DR/DJ
Rearranging the formula, we have:
DR = DJ * sin(d)
DR = 16√2 cm * sin(120 degrees)
Using the sine of 120 degrees (which is √3/2), we have:
DR = 16√2 cm * (√3/2)
DR = 8√6 cm
Similarly, considering the triangle DBO, we can find the length of side DO:
DO = DJ * sin(180 degrees - d)
DO = 16√2 cm * sin(180 degrees - 120 degrees)
DO = 16√2 cm * sin(60 degrees)
Using the sine of 60 degrees (which is √3/2), we have:
DO = 16√2 cm * (√3/2)
DO = 8√6 cm
Now we can find the perimeter of quadrilateral ROCA:
Perimeter = RO + OC + CA + AR
Perimeter = (DR + DO + DJ + DJ)
Substituting the values we found earlier:
Perimeter = (8√6 cm + 8√6 cm + 16√2 cm + 16√2 cm)
Perimeter = 32√2 cm + 16√6 cm
Simplifying, we get:
Perimeter = 16(2√2 cm + √6 cm)
Therefore, the perimeter of ROCA is 16(2√2 cm + √6 cm).