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

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7

step-by-step explanation:

the answer would be 7