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Answer:
The area of polygon MNOPQR = Area of a rectangle that is 15 square units + Area of a rectangle that is <u>2</u> square units.
Step-by-step explanation:
We are given that a polygon MNOPQR is shown. The top vertex on the left is labeled M, and the rest of the vertices are labeled clockwise starting from the top-left vertex labeled, M. The side MN is parallel to side QR. The side MR is parallel to side PQ.
The side MN is labeled as 5 units. The side QR is labeled as 7 units. The side MR is labeled as 3 units, and the side NO is labeled as 2 units.
Firstly, we will draw a perpendicular line from point O which meets the line RQ at point T. Now, the polygon MNOPQR is divided into two rectangles MNTR and OPQT.
As we know that the area of the rectangle =
In the rectangle MNTR, the length (MN) = 5 units and the breadth (MR) = 3 units.
So, the area of the rectangle MNTR =
= = 15 square units
Now, as we know that in rectangle MNTR, the side NT = 3 units and the side NO is labeled as 2 units. This means that the side OT = NT - NO = 3 units - 2 units = 1 unit
Similarly, the side QR is labeled as 7 units and the side RT is labeled as 5 units. This means that the side TQ = QR - RT = 7 - 5 = 2 units
Now, in the rectangle OPQT, the length (TQ) = 2 unit and the breadth (OT) = 2 units.
So, the area of the rectangle OPTQ =
= = 2 square units
Hence, the area of polygon MNOPQR = Area of a rectangle that is 15 square units + Area of a rectangle that is <u>_2_</u> square units.
the idea behind the recurring decimal as a fraction, is to first off, multiply or divide by some power of 10, in order that we leave the recurring decimal to the right of the decimal point.
then we multiply by a power of 10, in order to move the repeating digits to the left of the decimal point, anyhow, let's proceed.
notice, we first divided by 10, to move the decimal point over to the right by 1 slot, then we multiplied by 100, to move it two digits over the decimal point, namely the repeating "12", thus we use 100.