In the diagram, ABCD is a part of a right angle triangle ODC. If AB = 6 cm, CD = 15 cm, BC = 8 cm angle BCD = 90 degrees and AB is parallel to DC, calculate, correct to 1 decimal place, the: (a) height (b) perimeter, of the triangle ODC

a) Height = OC = 13.3cm

b) Perimeter = 48.4cm

Step-by-step explanation:

a) Given:

ODC is a right angle triangle and

ABCD is part of it.

AB = 6 cm

CD = 15 cm

BC = 8 cm

∠BCD = 90 degrees

AB is parallel to DC

Find attached the diagram obtained from the given information.

From the diagram, ∆OAB is similar to ∆ODC.

∠OBA = ∠OCD = 90 degrees

To find the height, we would apply the similar triangles theorem.

The ratio of corresponding sides are equal and the angles are congruent.

OB/BA = OC/CD

OC = OB+BC = OB+8

OB/6 = (OB+8)/15

15OB = 6(OB+8)

15OB = 6OB + 48

9OB = 48

OB = 48/9 = 16/3

OC = 16/3 + 8

OC = 13⅓ cm = 13.3cm

Height = OC = 13.3cm

b) To get perimeter, we have to first determine OD (the hypotenuse of ∆OCD) as it is a right angled triangle

Using Pythagoras theorem

Hypotenuse ² = opposite ² + adjacent ²

OD² = OC² + CD²

OD² = (13⅓)² + 15²

OD² = 1600/9 + 225 = 3625/9

OD = √(3625/9)

OD = 20.1

Perimeter of ∆ODC= OC + CD + OD

= 13.3 + 15 + 20.1

Perimeter = 48.4cm

all the angles opposite to the angle 150 would be 150 and all the angles that are next to 150 are 30 because they equal 180 it you add it up.

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