AREA OF A TRAINGLE

Area of parallelograms of Class 9

POLYGONAL REGION

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.

area of a triangle

Area Axioms:

Every polygonal region R has an area, measure in square unit and denoted by ar(R).

  • Congruent area axiom : if R1 and R2 be two regions such that R1≅R2 then ar(R1) = ar (R2).
  • Area monotone axiom :If R1area of a triangle R2, then are (R1) area of a trianglear(R2).
  • Area addition axiom : If R1 are two polygonal regions, whose intersection is a finite number of points and line segments and R = R1area of a triangle 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.

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.  (R1) +ar . (R2)

THEOREM – 1:

A diagonal of parallelogram divides it into two triangles of equal area.

Given: A parallelogram ABCD whose one of the diagonals is BD.

area of a triangle

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.

THEOREM – 2:

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

area of a triangle

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]

THEOREM – 3:

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]

area of a triangle

⇒ 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.

THEOREM – 4:

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.

area of a triangle

(i) (ii)

Given: AParallelogram ABCD and a triangle APB are on the same base AB and between the same parallels AB and PC.

To Prove: ar. area of a triangle

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

⇒ ar.area of a triangle.

THEOREM – 5:

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 area of a triangle 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)

Proof:

area of a triangle

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)

area of a triangleABD ≅ 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.

area of a triangle

∴ 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 a triangle

∴ Area of parallelogram ABCD = AD × BN = AB × DM

⇒ AD × 5 = 8 × 4

⇒ AD = area of a triangle

= 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)

area of a triangle

area of a triangle

area of a triangle

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 ΔCDBarea of a triangle

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 = CDQarea of a triangle

(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 BPDC

∴ar (parallelogram ABCD) = ar (parallelogram PBCQ).

Ex. In figure, ABCD is a parallelogram, AE area of a triangle DC and CFarea of a triangleAD. If AB = 16 cm, AE = 8 cm and CF = 10 cm find AD.

area of a triangle

Sol. We have AB = 16 cm, AE = 8 cm CF = 10 cm.

We know that are of parallelogram = Base × Height [Base = CD, height = AE]

ABCD = CD × AE = 16 × 8 = 128 cm2

Again, Area of parallelogram = Base × Height = AD × CF [Base = AD, height = CF]

128 = AD × 10

⇒ AD = 128/10=12.8 cm Ans.

Ex: In figure, ABCD is a parallelogram and EFCD is a rectangle. Also AL area of a triangle DC 

and AL = DE . Prove that ar(ABCD) = ar(EFCD)

Sol: Rectangle is also a parallelogram.

Thus, parallelogram ABCD and rectangle EFCD are on the same base CD.area of a triangle

 

Thus ar(||gm ABCD) = DC X  AL

Ar(DEFC) = DC  X  DE

= DC  X  AL

= ar(||gm ABCD).

Ex: Parallelogram ABCD and rectangle ABEF have the same base AB and also have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.

Sol: To prove:Perimeter of parallelogram ABCD > perimeter of

area of a triangle

rectangle ABEF i.e.

AB + BC + CD + AD > AB + BE + EF + AF

Since opposite sides of a parallelogram and a rectangle are equal.

∴ AB = DC = EF … (i) (Since ABCD is a parallelogram)

∴ AB + DC = AB + EF … (ii)

Since, of all the segments that can be drawn to a given line form a point not lying on it, the perpendicular segment is the shortest.

∴ BE < BC and AF < AD

⇒ BC + AD > BE + AF … (iii)

Adding (ii) and (iii), we get

AB + DC + BC + AD > AB + EF + BE + AF

∴ AB + BC + CD + DA > AB + BE + EF + FA.

AREA OF A TRAINGLE

Theorem – 5:

Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

Given: Two triangles ABC and PCs on the same base BC and between the same parallel lines BC and AP.

area of a triangle

To prove: ar(ΔABC) = ar(ΔPBC)

Construction: Through B, draw BD || CA intersecting PA produced in D and through C, draw CQ || BP, intersecting line AP in Q.

Proof :We have, 

BD || CA [By construction]

And,BC || DA [Given]

area of a triangle Quad. BCAD is a parallelogram.

Similarly, Quad. BCQP is a parallelogram.

Now, parallelogram BCQP and BCAD are on the same base BC, and between the same parallels.

area of a triangle ar(||gm BCQP) = ar(||gmBCAD) ....(i)

We know that the diagonals of a parallelogram divides it into two triangles of equal area.

area of a triangle  ar(ΔPBC=(||gm BCQP) ....(ii)

And ar(ΔABC) = ar(||gm BCAD) ....(iii)

Now, ar(||gm  BCQP) = ar(||gm BCAD) [From (i)]

ar(||gm BCAD) = ar(||gm BCQP)

Hence, ar(ΔABC) = ar(ΔPBC) [Using (ii) and (iii)]

Hence Proved.

THEOREM – 6:

The area of a trapezium is half the product of its height and the sum of the parallel sides.

35.jpg

Given: Trapezium ABCD in which AB || DC, ALarea of a triangle DC, CNarea of a triangleAB and AL = CN = h (say)

AB = a, DC = b.

To prove: ar(trap. ABCD) = h × (a + b).

Construction: Join AC.

Proof: AC is a diagonal of quad. ABCD.

∴ ar(trap. ABCD) = ar(ΔABC) + ar(ΔACD) = h × a +h × b=h(a + b). Hence Proved.

THEOREM – 7:

Triangles having equal areas and having one side of the triangle equal to corresponding side of the other, have their corresponding altitudes equal/

Given: Two triangles ABC and PQR such that (i) ar (ΔABC) = ar(ΔPQR) and (ii) AB = PQ.

CN and RT and the altitude corresponding to AB and PQ respectively of the two triangles.

To prove :CR = RT.

Proof :In ΔABC, CN is the altitude corresponding to the side AB.

ar(ΔABC) = AB × CN ....(i)

area of a triangle

Similarly, ar(ΔPQR) = PQ × RT ....(ii)

Since ar(ΔABC) =ar(ΔPQR) [Given]

AB × CN = PQ × RT

Also, AB = PQ [Given]

CN = RT Hence Proved.

Ex: Prove that the area of a triangle is half the product of any of its side and the corresponding altitude.

Sol: Given: Aarea of a triangleABC in which AL is the altitude to the side BC.

To prove: ar(ΔABC) = (BC area of a triangle AL)

Construction: Through C and A draw CD || BA and AD || BC respectively intersecting each other at D

Proof: We have BA || CD (by construction)area of a triangle

and AD || BC (by construction)

∴ BCDA is a parallelogram.

Since AC is a diagonal of ||gm BCDA. Therefore,

ar(ΔABC) = ar (||gm BCDA)

⇒ar(ΔABC) = (BC area of a triangle AL) ( BC is the base and AL is the corresponding altitude of ||gm BCDA)

Ex: AD is one of the medians of aΔ ABC. X is any point on AD. Show that

ar(Δ ABX) = ar(Δ ACX)

Sol: In ΔABC, AD is a medianarea of a triangle

∴ar(ΔABD) = ar(ΔACD) … (1)

(median divides a Δ into two Δ’s of equal area)

again, in ΔXBC, XD is median 

∴ar(ΔXBD) = ar(ΔXCD) … (2)

Subtracting (2) from (1), we get

ar(ΔABD) – ar(ΔXBD)

= ar(ΔACD) – ar(ΔXCD)

⇒ar(ΔABX) = ar(ΔACX).

Ex. Triangles ABC and DBC are on the same base BC; with A, D on opposite sides of the line BC, such that ar(ΔABC) = ar(ΔDBC). Show that BC bisects AD.

area of a triangle

Sol. Construction :Draw AL area of a triangle BC and DM area of a triangle BC.

Proof :ar(ΔABC) = ar(ΔDBC) [Given] 

area of a triangle

⇒ AL = DM ....(i)

Now in Δs OAL and OMD

AL = DM [From (i)]

⇒ ∠ALO = ∠DMO [Each = 90o]

⇒ ∠AOL = ∠MOD [Vert. opp. ∠s]

⇒ ∠OAL = ∠ODM [Third angles of the triangles]

area of a triangle ΔOALarea of a triangleΔOMD [By ASA]

area of a triangle OA = OD [By cpctc]

i.e., BC bisects AD. Hence Proved.

Ex: The diagonals of quadrilateral ABCD, AC and BD intersect in O. Prove that if BO = OD, the triangles ABC and ADC are equal in area.

Sol:Given: A quadrilateral ABCD in which its diagonals AC & BD intersect at O such that BO = OD.

To Prove: ar(ΔABC) = ar(ΔADC)

Proof: In ΔABD, we have BO = OD (Given)

area of a triangle

⇒ O is mid-point of BD

⇒ AO is the median

⇒ar (ΔAOB) = ar(ΔAOD) … (1)

(∴median divides a Δ into two Δ’s of equal area)

In ΔCBD, O is the mid-point of BD  ⇒  CO is median

⇒ar(ΔCOB) = ar(ΔCOD) … (2)

Adding (1) and (2), we get

ar(ΔAOB) + ar(ΔCOB) = ar(AOD) + ar(ΔCOD)

⇒ar(ΔABC) = ar(ΔADC)

Ex: Triangles ABC and DBC are on the same base BC with A, D on opposite side of line BC, such that ar(ΔABC) = ar(ΔDBC). Show that BC bisects AD.

Sol:Since Δ’s ABC and DBC are equal in area and have a common base BC. Therefore the altitudes corresponding to BC are equal i.e. AE = DF

area of a triangle

Now in Δ’s AEO and DFO, we have

∠1 = ∠2 (vertically opposite angles)

∠AEO = ∠DFO (each equal to 90o)

and AE = DF

so, by ASA criterion of congruence,

ΔAEO ≅ ΔDFO

⇒ AO = DO ⇒ BC bisects AD.

 

Ex:In figure, D, E are points on sides AB & AC respectively of ΔABC, such that ar(ΔBCE) = ar(ΔBCD). Show that DE || BC.

area of a triangle

Sol: Since Δs BCE and ΔBCD are equal in area and have a same base BC. Therefore altitude from E of 

ΔBCE = Altitude from D of ΔBCD ⇒Δs BCE and BCD are between the same parallel lines. 

⇒ DE || BC.

Ex: If the medians of ΔABC interests at G, show that 

ar(ΔAGB) = ar(ΔAGC) = ar(ΔBGC) = 1/3 ar (ΔABC).

Sol: Given: Aarea of a triangleABC such that its medians AD, BE & CF interest at G.

To prove: ar(ΔAGB) = ar(ΔAGC)area of a triangle

= ar(ΔBGC) = 1/3ar(ΔABC).

Proof: We know that medians of a triangle divides it into two triangles of equal area.

In ΔABC, AD is the median

⇒ar(ΔABD) = ar(ΔACD) … (1)

In area of a triangleGBC, GD is the median

⇒ar(ΔGBD) = ar(ΔGCD) … (2)

Subtracting (2) from (1), we get

ar(ΔABD) – ar(ΔGBD) = ar(ΔACD) – ar(ΔGCD)

⇒ar(ΔAGB) = ar(ΔAGC) … (3)

Similarly, ar(ΔAGB) = ar(ΔGBC) … (4)

from (3) and (4) we get ar(ΔAGB) = ar(AGC) = ar(ΔBGC)

but ar(ΔAGB) + ar(ΔAGC) + ar(ΔBGC) = ar(ΔABC)

∴ 3 ar(ΔAGB) = ar(ΔABC)  ⇒ ar(ΔAGB) = 1/3ar(ΔABC)

Hence ar(ΔAGB) = ar(ΔAGC) = ar(ΔBCG) = 1/3 ar (ΔABC).

Ex: ABC is a triangle in which D is the mid-point of BC and E is the mid-point of AD. Prove that area (ΔBED) = 1/4 are (ΔABC).

Sol: Since AD is the median of ΔABC and median divides a triangle into two triangles of equal areas.

∴area (ΔABD) = area (ΔADC)area of a triangle

⇒area (ΔABD) = area (ΔABC) … (i)

In ΔABD, BE is the median

∴area (ΔBED) = area (ΔBAE) … (ii)

⇒area (ΔBED) = area (ΔABD)

⇒area (ΔBED) = × area (ΔABC) [from (i)]

= 1/4area (ΔABC).

Talk to Our counsellor