6.10 POLYGONS AREA AND PERIMETER. A POLYGON IS A FLAT CLOSED FIGURE DESCRIBED BY STRAIGHT- LINE SEGMENTS AND ANGLES. Polygons are named by the number.

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Presentation transcript:

6.10 POLYGONS AREA AND PERIMETER

A POLYGON IS A FLAT CLOSED FIGURE DESCRIBED BY STRAIGHT- LINE SEGMENTS AND ANGLES. Polygons are named by the number of sides they contain and other relations. The base of any polygon is the horizontal side or a side that would be horizontal if the polygon’s orientation is modified. The adjacent side is the side that has an endpoint in common with the base.

THE KINDS OF POLYGONS WE WILL WORK WITH WILL BE QUADRILATERALS AND TRIANGLES. WE WILL USE FORMULAS TO FIND BOTH PERIMETER AND AREA OF THESE SHAPES.

THE FOLLOWING ARE QUADRILATERALS FOR WHICH YOU NEED TO BE ABLE TO CALCULATE THE PERIMETER AND AREA: Parallelogram- opposite sides are parallel A = bh P = 2(a + b) a b h Base and height always form right angles

THE FOLLOWING ARE QUADRILATERALS FOR WHICH YOU NEED TO BE ABLE TO CALCULATE THE PERIMETER AND AREA: Rectangle- a parallelogram with 4 right angles. A = bh or lw P = 2(b + h) or 2(l + w) b h

THE FOLLOWING ARE QUADRILATERALS FOR WHICH YOU NEED TO BE ABLE TO CALCULATE THE PERIMETER AND AREA: Square- a rectangle with 4 equal sides A = P = 4b or 4s b

THE FOLLOWING ARE QUADRILATERALS FOR WHICH YOU NEED TO BE ABLE TO CALCULATE THE PERIMETER AND AREA: Rhombus- a parallelogram with 4 equal sides A= bh P = 4b b h b

THE FOLLOWING ARE QUADRILATERALS FOR WHICH YOU NEED TO BE ABLE TO CALCULATE THE PERIMETER AND AREA: Trapezoid- a quadrilateral with one pair of parallel sides. A = ½ (a + b)h P = a + b + c + d a b c d h

THE AREA OF A RECTANGLE IS 280 SQ CM. ITS WIDTH IS 14 CM. FIND ITS LENGTH.

A 150 SQ/FT ROLL OF FIBERGLASS IS 30 IN WIDE. WHAT IS ITS LENGTH IN FEET? IF THE COST IS $9.75/SQ YD, WHAT IS THE COST OF THE ROLL?

FIND THE AREA. 20 ft 12 ft 14 ft 6 ft 8 ft

THE REPLACEMENT COST FOR CONSTRUCTION OF THE BUILDING BELOW IS $75/SQ.FT. DETERMINE HOW MUCH INSURANCE SHOULD BE CARRIED FOR FULL REPLACEMENT. 75 ft 40 ft 24 ft 12 ft 16 ft 26 ft 44 ft

A PIECE OF SHEET METAL HAS A RECTANGULAR HOLE IN IT AS SHOWN. FIND (A) THE AREA OF THE PIECE PUNCHED OUT, AND (B) THE AREA OF METAL LEFT. 12 in 5.5 in 40 in 21 in 23 in

TRIANGLES

THE FORMULAS FOR PERIMETER AND AREA OF TRIANGLES ARE: P = a + b + c A = ½ bh a b c h a = h b c a c b h

IF ONLY THE SIDES (NOT THE HEIGHT) OF A TRIANGLE ARE KNOWN, THE AREA CAN BE FOUND USING A FORMULA CALLED HERON’S FORMULA GIVEN BY WHERE A, B AND C ARE THE SIDES OF THE TRIANGLE AND S = ½ (A+B+C).

FIND THE AREA OF THE TRIANGLE BELOW. 24 in25 in 40 in

A SQUARE HOLE IS CUT FROM THE EQUILATERAL TRIANGLE AS SHOWN. FIND THE AREA REMAINING IN THE TRIANGLE cm 40 cm