The length of AB is 20cm and the length of BC is 10cm.
Let's break down the information given:
PQ=8cm
QR=2cm
The perimeter of the rectangle ABCD is 60cm
To find the length of AB and BC, let's solve this step by step.
First, the perimeter of a rectangle is given by the sum of all its sides. For rectangle ABCD, the perimeter is twice the sum of its length and width.
Let the length of AB be l and the width (or BC) be w.
So, the perimeter of ABCD is:
Perimeter=2×(l+w)
Given that the perimeter is 60cm, we can express this as:
60=2×(l+w)
30=l+w (Equation 1)
Now, let's utilize the information about PQ and QR.
Given that PQ=8cm and QR=2cm, and O is the center of the rectangles:
PR=PQ+QR=8+2=10cm
From the problem statement, AC and BD are lines passing through O, so PR is a diagonal of the rectangle.
In a rectangle, the diagonal PR can be related to its sides using the Pythagorean theorem.
PR^2 =AB^2 +BC^2
10^2 =l^2 +w^2
We also know the relationship between l and w from Equation 1:
30=l+w
l=30−w
Now, let's substitute this into the Pythagorean equation:
10^2 =(30−w)^2
100=900−60w+w
100=900−60w+2w
2w^2 −60w+800=0
w^2 −30w+400=0
This is a quadratic equation that can be factored or solved using the quadratic formula. Factoring or using the quadratic formula gives:
w=20 or w=10
If w=20, then l=30−20=10.
This doesn't satisfy the condition for the perimeter (l+w=30).
Thus,w=10 and l=30−10=20.
Therefore, the length of AB is 20cm and the length of BC is 10cm.