Today our topic of discussion is Area of a Plane Region.
Area of a Plane Region
Area Related Theorems and Constructions
We know that bounded plane figures may have different shapes. If the region is bounded by four sides, it is known as quadrilateral Quadrilaterals have classification and they are also named based on their shapes and properties.
Apart from these, there are many regions bounded by more than four sides. These are polygonal regions or simply polygons. Every closed region has a certain measurement which is called the area of the region.
For measurement of areas usually the area of a square with sides of 1 unit of length is used as the unit area and their areas are expressed in square units. For example, the area of Bangladesh is 1.47 lacs square kilometres (approximately).
In our day to day life we need to know and measure areas of polygons for meeting the necessity of life. So, it is important for the learners to have a comprehensive knowledge about areas of polygons. Areas of polygons and related theorems and constructions are presented here.
At the end of the chapter, the students will be able to –
⇒ explain the area of polygons
⇒ verify and prove theorems related to areas.
⇒ construct polygons and justify construction by using given data.
⇒ construct a quadrilateral with area equal to the area of a triangle.
⇒ construct a triangle with area equal to the area of a quadrilateral.
Area of a Plane Region
Every closed plane region has definite area. In order to measure such area, usually the area of a square having sides of unit length is taken as the unit. For example, the area of a square with a side of length 1 cm is 1 square centimere
We know that,
1) If the length of the rectangular region ABCD AB = a units (say, metre), breadth BC = b units (say, metre), the area of the region ABCD = ab square units (say, square metres).
2) If the length of a side of the square region ABCD, AB = a * units (say, metre), the area of the region ABCD = a² square units (say, square metres).
When the area of two regions are equal, the sign’ is used between them. For example, in the figure the area of the rectangular region ABCD area of the triangular region AED, where AB = BE
It should be mentioned that, if ΔABC and ΔDEF are congruent, we write ΔABC ≅ ΔDEF . In this case, the area of the triangular region ΔABC = area of the triangular region ΔDEF .
But, two triangles are not necessarily congruent when they have equal areas. For example, in the figure, area of ΔABC = area of ΔDBC but ΔABC and ΔDBC are not congruent
Theorem 36. The areas of all triangular regions having same base and lying between the same pair of parallel lines are equal to one another.
Let, the triangular regions ABC and DBC stand on the same base BC and lie between the pair of parallel lines BC and AD. It is required to prove that,Δregion ABC = Δ region DBC.
Drawing: At the points B and C of the line segment BC, draw perpendiculars BE and CF respectively. They intersect the line AD or extended AD at the points E and F respectively. As a result a rectangular region EBCF is formed.
Proof: According to the construction, EBCF is a rectangular region. In triangle ABC, the base and height are BC and BE respectively.
Hence, Δregion ABC =1/2 x BCX BE… (1)
In triangle DBC, the base and height are BC and CF respectively.
∴ Δ region DBC=1/2xBCxCF=1/2xBCxBE…[: EBCF rectangule]
Comparing (1) and (ii)
∴ Δregion ABC = Δregion DBC (proved)
Corollary 1. If the areas of triangular regions lying on the same base and same side are equal, the triangles He between the same pair of parallel lines.
Corollary 2. Area of a triangle is exactly half of the area of a parallelogram lying on the same base and between the same pair of parallel lines.
Hints: In the figure, ABCD is a parallelogram with AC as diagonal.
∴ ΔABC≅ΔADC
ΔABC=1/2 x Area of parallelogram ABCD,
Theorem 37. Parallelograms lying on the same base and between the same pair of parallel lines are of equal area.
Let the parallelograms regions ABCD and ABEF stand on the same base AB and lie between the pair of parallel lines AB and FC. It is required to prove that, area of the parallelogram ABCD = area of the parallelogram ABEF.
Drawing: Join A, C and A, E. From the points C and E, draw perpendiculars EK and CL to the base AB and extended AB respectively.
Proof: The area of ΔABC = 1/2 * AB x CL and
The area of ΔABE = 1/2 * ABxEK
CL = EK [by construction AL ||FC]
∴ The area of ΔABC = The area of ΔABE
Right arrow 1/2 1 area of the parallelogram ABCD = area of the parallelogram ABEF.
∴ Area of the parallelogram ABCD = area of the parallelogram ABEF. (proved)
Theorem 38. Pythagoras Theorem.
In a right-angled triangle, the area of the square drawn on the hypotenuse is equal to the sum of the areas of the squares drawn on the other two sides.
Special Nomination: Let ABC be a right- angled triangle in which ∠ACB is a right angle and hypotenuse is AB. It is to be proved that, A * B² = B * C² + A * C²
Drawing: Draw three squares ABED, ACGF and BCHK on the external sides of AB, AC and BC respectively. Through C, draw the line segment CL parallel to AD or BE. Let CL intersects AB at M and DE at L respectively. Join C, D and B, F.
Proof:
Step 1. In ΔCAD and ΔBAF, CA = AF , AD = AB and included ∠CAD = ∠CAB+∠BAD = ∠CAB+∠CAF = included ∠BAF [∠BAD = ∠CAF = = 1 right angle]
Therefore, ΔCAD ≅ ΔBAF
Step 2. Triangle ΔCAD and rectangular region ADLM lie on the same base AD and between the parallel lines AD and CL. Therefore, Rectangular region ADLM-2ΔCA D [Theorem 37]
Step 3. ΔBAF and the square ACGF lie on the same base AF and between the parallel lines AF and BG.
Hence, square region ACGF=2 ΔFAB =2 ΔCAD [Theorem 37]
Step 4. Rectangular region ADLM = square region ACGF.
Step 5. Similarly joining C, E and A, K it can be proved that rectangular region BELM = square region BCHK
Step 6. Rectangular region (ADLM + BELM) square region ACGF+ square region BCHK. or, square region ABED = square region ACGF+ square region BCHK.
That is, A * B² = B * C² + A * C² (Proved)
Construction 13. Construct a parallelogram with an angle equal to a definite angle and area equal to that of a triangular region.
Let ABC be a triangular region and ∠x be a definite angle. It is required to construct a parallelogram with angle equal to ∠x and area equal to the area of the triangular region ABC.
Drawing: Bisect the line segment BC at E. At the point E of the line segment EC, draw CEF equal to ∠ x . Through A, draw AG parallel to BC which intersects the ray EF at F. Again, through C, draw the ray CG parallel to EF which intersects the ray AG at G. Hence, ECGF is the required parallelogram.
Proof: Join A, E.
Now, area of ΔABE = area of ΔAEC [ since base BE = base EC and heights of both the triangles are equal].
∴ area of ΔABC = 2 (area of ΔAEC ) .
Again, area of the parallelogram ECGF= 2 (area of ΔAEC ) [since both lie on the same base EC and EC ||AG)
∴ area of the parallelogram region ECGF= area of ΔABC
Again ∠CEF= ∠x [sin EF ||CB by construction].
So the parallelogram ECGF is the required parallelogram.
Construction 14. Construct a triangle with area of the triangular region equal to that of a quadrilateral region.
Let ABCD be a quadrilateral region. It is required to construct a triangle such that area of the triangular region is equal to that of a rectangular region ABCD.
Drawing: Join D, B. Through C, draw CE parallel to DB which intersects the side AB extended at E. Join D, E. Then, ADAE is the required triangle.
Proof: ΔBDC and ABDE lie on the same base BD and CE || DB (by construction).
∴ area of ΔBDC area of ΔBDE
∴ area of ΔBDC + area of ΔABD = area of ΔBDE + area of ΔABD
area of the quadrilateral region ΔBCD = area of ΔADE
Therefore, ΔADE is the required triangle.
Nota Bene: Applying the above mentioned method innumerable numbers of triangles can be drawn whose area is equal to the area of a given quadrilateral region.
Construction 15. Construct a parallelogram, with a given angle and the area of the bounded region equal to that of a quadrilateral region.
Let ABCD be a quadrilateral region and ∠ x be a definite angle. It is required to construct a parallelogram with angle ∠x and the area equal to area of the quadrilateral region ABCD.
Drawing: Join B, D. Through C, draw CF parallel to DB which intersects the side AB extended at F. Find the midpoint G of the line segment AF. At A of the line segment AG, draw GAK equal to angle x and draw AK ||GH through G. Again, draw KDH || AG through D and let, KDH intersects AK and GH at K and H respectively. Hence AGHK is the required parallelogram.
Proof: Join D, F. By construction AGHK is a parallelogram, where angle GAK = ∠x . Again, area of ΔDAF = area of the rectangular region ABCD and area of the parallelogram AGHK = area of the triangular region DAF. Therefore, AGHK is the required parallelogram.
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