Area of parallelograms of Class 9
Polygon region can be expressed as the union of a finite number of triangular regions in a plane such that if two of these intersect, their intersection is either a point or a line segment. It is the shaded portion including its sides as shown in the figure.
(a) Area Axioms:
Every polygonal region R has an area, measure in square unit and denoted by ar(R).
- (i) Congruent area axiom :if R1 and R2 be two regions such that R1R2 then ar(R1) = ar (R2).
- (ii) Area monotone axiom :If R1 R2, then are (R1) ar(R2).
- (iii) Area addition axiom :If R1 are two polygonal regions, whose intersection is a finite number of points and line segments and R = R1 R2, then ar (R) = ar(R1) + ar(R2).
- (iv) Rectangular area axiom :If AB = a metre and AD = b metre then, ar (Rectangular region ABCD) = ab sq. m.
(b) Unit of Area :
There is a standard square region of side 1 metre, called a square metre, which is the unit of area measure. The area of a polygonal region is square metres (sq. m or m2) is a positive real number
AREA ADDITION AXIOM:
Total area R of the plane figure ABCD is the sum of two polygonal regions R1 and R2, i.e. ar. (R) = ar.
A diagonal of parallelogram divides it into two triangles of equal area.
Given: A parallelogram ABCD whose one of the diagonals is BD.
To prove: ar (ΔABD) = ar (ΔCDB).
Proof: In ΔABD and ΔCDB.
AB = DC [Opp. sides of a ||gm]
AD = BC [Opp. sides of a ||gm]
BD = BD [Common side]
∴ ΔABD≅ΔCDB [By SSS]
∴ ar (ΔABD) = ar(ΔCDB) [Congruent area axiom] Hence Proved.
Prove that parallelogram on the same base and between same parallel are equal in area.
Given: Parallelogram ABCD and ABEF are on the same base AB and between the same parallels AB and CF.
To Prove: ar. (ABCD) = ar. (ABEF)
Proof: In ΔADF and ΔBCE, we have
AF = BE [Opposite sides of the ||gms]
∠AFD = ∠BEC [Corresponding angles]
∠ADF = ∠BCE [Corresponding angles]
∴ ΔADF≅ΔBCE [By AAS Congruence]
⇒ ar. (ΔADF) = ar. (ΔBCE) [Congruence area axiom]
Adding ar.(quadrilateral ABED) to both sides, we get
ar. (ΔADF) + ar.(quadrilateral ABED) = ar (ΔBCE) + ar.(quadrilateral ABED)
⇒ ar. ||gmABEF = ar. ||gmABCD [Area addition axiom]
Prove that parallelogram and a rectangle on the same base and between the same parallels are equal in area.
Given :A Parallelogram ABCD and a rectangle ABML are on the same base AB and between the same parallels AB and CL.
To Prove : ar. ||gmABCD = ar. rectangle ABML.
Proof :In ΔADL and ΔBCM, we have
AL = BM [Opposite sides of a rectangle]
∠ALD =∠BMC[Corresponding angles]
∠ADL = ∠BCM [Corresponding angles]
∴ ΔADL≅ΔBCM [By AAS congruence]
⇒ ar. (ΔADL) = ar. (ΔBCM)
Adding ar. (quadrilateral ABMD) to both sides, we get
ar. ΔADL + ar. quadrilateral ABMD = ar. ΔBCM + ar. quadrilateral ABMD
or ar. of rectangle ABML = ar. of ||gmABCD.
If a triangle and a parallelogram are on the same base and between the same parallels, then prove that the area of the triangle is equal to half the area of the parallelogram.
Given: AParallelogram ABCD and a triangle APB are on the same base AB and between the same parallels AB and PC.
To Prove: ar.
Const: Through B draw BQ || AP which meets PC or PC produced at Q. Join BQ
Proof: Since the parallelogram ABQP and the parallelogram ABCD stand on the same base AB and between same parallels AB and PQ or DQ.
∴ ar. ||gmABQP = ar. ||gmABCD [By theorem 1] …(i)
Since the diagonal of a ||gmABQP divides it into two Δs of equal area
∴ ar. ||gmABQP = 2 ar. (ΔPAB) …(ii)
From (i) and (ii), we have
2 ar. (ΔPAB) = ar. ||gmABCD
The arc of parallelogram is the product of its base and the corresponding altitude.
Given: A||gm ABCD in which AB is the base and AL is the corresponding height.
To prove: Area (||gmABCD) = AB × AL.
Construction: Draw BM DC so that rectangle ABML is formed.
Proof: ||gm ABCD and rectangle ABML are on the same base AB and between the same parallel lines AB and LC.
∴ ar(||gmABCD) = ar(rectangle ABML) = AB × AL.
∴ area of a ||gm = base × height.
Ex: Prove that a diagonal of a parallelogram divides it into two triangles of equal area.
Solution: Given: A parallelogram ABCD in which BD is one of the diagonals.
To prove → ar((ΔABD) = ar((ΔCDB)
In triangles ABD and CDB, we have
AB = CD (in a parallelogram opposite sides are of equal length)
AD = CB
and BD = DB (common side)
⇒ΔABD (ΔCDB (by SSS)
Hence by congruent area axiom ar((ΔABD) = ar((ΔCDB).
Ex. A parallelogram ABCD and a rectangle ABPQ are on the same base AB and between the same parallels AB and CQ. If AB = 8 cm and AQ = 6 cm, find the area of ||gmABCD.
Sol. A rectangle is a parallelogram in which each angle is 90° and parallelogram on the same base and between same parallels are equal in area.
∴ ar. (||gmABCD) = ar. (rectangle ABPQ)
But ar. (||gmABPQ) = AB×AQ
= (8 × 6) cm2 = 48 cm2
∴ Area of parallelogram ABCD = 48 cm2.
Ex. In a parallelogram ABCD, AB = 8 cm. The altitudes corresponding to sides AB and AD are respectively 4 m and 5 cm. Find AD.
Sol. We know that, Area of a parallelogram = Base × Corresponding altitude
∴ Area of parallelogram ABCD = AD × BN = AB × DM
⇒ AD × 5 = 8 × 4
⇒ AD =
= 6.4 cm. Ans.
Ex. In figure, ABCD is a parallelogram, AE⊥CD and CF⊥AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.
Sol. AB = 16 cm, AE = 8 cm and CF = 10 cm, AD= ?
ar. ||gmABCD (on base AD) = ar. ||gmABCD (on base CD)
Ex: ABCD is a quadrilateral and BD is one its diagonals as shown in figure. Show that ABCD is a parallelogram and find its area.
Sol: In ΔABD and ΔCDB
∠ABD = ∠CBD = 90o (alternate interior angles are equal)
∠ADB = ∠CBD
∠BDA = ∠BDC
Hence, ABCD is a parallelogram.
Now, area of parallelogram ABCD = (base ×corresponding altitude)
⇒area of parallelogram ABCD = AB × BD = 5 ×12 sq. units = 60 sq. cms.
Ex: Prove that parallelograms on the same base and between the same parallel lines are equal in area.
Sol: Given: Two parallelogram ABCD and PBCQ having same base BC and between the same parallel lines BC and AQ.
To Prove: ar(parallelogram ABCD)= ar (parallelogram PBCQ)
Proof: In triangles ABP and DCQ,
∠BAP = ∠CDQ
(Corresponding angles when AQ intersects parallel lines AB and DC).
∠BPA = ∠CDQ
(Corresponding angles when AQ intersects parallel lines BP and CQ).
AB = DC (Opposite sides of a parallelogram)
∴ΔABP ≅ ΔDCQ (AAS)
Hence area ΔABP = area ΔDCQ
or, area ΔABP + area BPDC = area ΔDCQ + area Δ BPDC
∴ar (parallelogram ABCD) = ar (parallelogram PBCQ).