Question

In Figure 6.7, OA = 1 cm,

ZAOB = ZBOC = ZCOD = 30° and
ZOAB = ZOBC = ZOCD = 90°.
O
D
30°
30°
30°
1 cm
Figure 6.7
с
B
A
(i) Find the length of OD giving your
answer in the form a√√3.
(ii) Show that the perimeter of
5
OABCD is (1 + √3) cm.

Answer 1

i).The length of OD in the form a√3 is (8/9)√3.

ii). The perimeter of the figure OABCD is actually (5/3)(1 + √3)cm.

Using the trigonometric ratios of tangent and sine of angle 30° for the right triangles OAB, OBC and OCD, we can derive the length of OD and show that the perimeter of OABCD is actually 5/3(1 + √3) as follows:

tan 30° = √3/3 and sin 30° = 1/2

For ∆OAB;

tan 30 = AB/OA {opposite/adjacent}

√3/3 = AB/1

AB = √3/3

sin 30 = AB/OB {opposite/hypotenuse}

1/2 = √3/3 ÷ OB

OB = √3/3 × 2

OB = 2(√3/3)

For ∆OBC;

tan 30 = BC/OB {opposite/adjacent}

√3/3 = BC ÷ 2(√3/3)

BC = √3/3 × 2(√3/3)

BC = 2/3

sin 30 = BC/OC {opposite/hypotenuse}

1/2 = 2/3 ÷ OC

OC = 2/3 × 2/1

OC = 4/3

For ∆OCD;

tan 30 = CD/OC {opposite/adjacent}

√3/3 = CD ÷ 4/3

CD = √3/3 × 4/3

CD = 4√3/9

sin 30 = CD/OD {opposite/hypotenuse}

1/2 = 4√3/9 ÷ OD

OD = 4√3/9 × 2/1

OD = (8/9)√3

To show that the perimeter of OABCD is (5/3)(1 + √3), the sum of OA, AB, BC, CD and OD must be (5/3)(1 + √3) so;

1 + √3/3 + 2/3 + 4√3/9 + (8/9)√3 = (9 + 3√3 + 6 + 4√3 + 8√3)/9

1 + √3/3 + 2/3 + 4√3/9 + (8/9)√3 = (15 + 15√3)/9

1 + √3/3 + 2/3 + 4√3/9 + (8/9)√3 = (15/9)(1 + √3)

1 + √3/3 + 2/3 + 4√3/9 + (8/9)√3 = (5/3)(1 + √3)

Therefore, the length of OD in the form a√3 is (8/9)√3 and the perimeter of OABCD is actually (5/3)(1 + √3)cm