Areas of Parallelograms and Triangles - Areas of Parallelograms and Triangles, Class 9, Mathematics PDF Download

INTRODUCTION

In earlier class, we have learnt some formulae for finding the areas of different plane figures such as triangle, rectangle, parallelogram, square etc. In this chapter, we will consolidate the knowledge about these formulae by studying some relationship between the area of these figures under the condition when they lie on the same base and between the same parallels.

PARALLEOGRAMS ON THE SAME BASE AND BETWEEN THE SAME PARALLELS

Theorem-1 : Parallelograms on the same base and between the same parallels are equal in area.

Given : Two parallelograms ABCD and ABEF on the same base AB and between the same parallels AB and FC.
To prove : ar(║gm ABCD) = ar (║gm ABEF).    
Proof :

STATEMENT		REASON
1.	In ΔBCE and ΔADF, we have: (i) BC = AD (ii) ∠BCE = ∠ADF (iii) ∠BEC = ∠AFD	Opposite sides of a ║ gm are equal. Corres. ∠s are equal, as AD ⊥ BC and FC is the transversal.
2.	ΔBCE ΔADF AAS-axiom of congruence.	Corres. ∠s are equal, as BE ⊥ AF and FC is the transversal.
3.	ar(ΔBCE) = ar(ΔADF)	Congruent figures are equal in area.
4.	ar (quad. ABED) + ar (ΔBCE) = ar (quad. ABED) + ar (ΔADF)	Adding same area on both sides of 3.
5.	ar (║ gm ABCD) = ar (║ gm ABEF)	ar (R1) + ar (R2) = ar (R1 R2).

Hence, proved.
 

Corollary 1: In parallelogram ABCD, AB ⊥ CD and BC ⊥​ AD. If AL ⊥​ BC and L is the foot of the perpendicular, then ar (ABCD) = BC × AL. 
Proof. In fig, ABCD is a parallelogram. AL ⊥​ BC. Now, we draw line through A and D, BE ⊥​ and CF ⊥​ . Then BEFC becomes a rectangle, Here, the parallelogram ABCD and the rectangle BEFC (also BEFC is a parallelogram) have same base and both are between the same parallels, Thus, we have

Area of parallelogram ABCD = Area of rectangle BEFC = BC × BE = BC × AL (∵  AL = BE)

Corollary 2: If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram.

Given : A ΔABC and a || gm BCDE on the same base BC and between the same parallels BC and AD.
To Prove :

Construction : Draw AL ⊥​ BC and DM ⊥​ BC (produced).

Proof :

STATEMENT	REASON
1. AL = DM	Perpendiculars to the same line and between the same parallels are equal.
2. ar(ΔABC) =  x BC x AL =  x BC x DM =  ar (|| gm BCDE)	Area of a Δ =   × Base × Height AL = DM (from 1) ar (|| gm BCDE)  = Base × Height = BC × DM.
3. ar(ΔABC) =  ar (|| gm BCDE)	Hence Proved

Ex.1 In fig, ABCD is a parallelogram, AL ⊥ BC, AM ⊥​ CD, AL = 4 cm and AM = 5 cm. If BC = 6.5 cm, then find CD.

Sol. We have, BC × AL = CD × AM (Each equal to area of the parallelogram ABCD)

Ex.2 In the given fig, ABCD is a parallelogram whose diagonal intersect at O. A line segment through O meets AB at P and DC at Q. Prove that : ar (quad. APQD)

Sol. Proof :

STATEMENT	REASON
1. ar(ΔACD) =  ar (|| gm ABCD)	Diagonal AC divides || gm ABCD into two Δs of equal area
2. In ΔOAP and ΔOCQ, we have : (i) OA = OC (ii) ∠OAP = ∠OCQ (iii) ∠AOP =∠COQ	Diagonal of a|| gm bisect each other. Alt. int. ∠s, DC ║ AB and CA is the transversal. Vert. opp ∠s.
3. ΔOAP ≌ ΔOCQ	AAS-axiom of congruence.
4. ar (ΔOAP) = ar (ΔOCQ)	Congruent Δs are equal in area
5. ar (ΔOAP) + ar (quad. AOQD) = ar (ΔOCQ) + ar (quad. AOQD)	Adding ar (quad. AOQD) on both sides of 4.
6. ar (quad. APQD) = ar (ΔACD)
7. ar (quad. APQD) =   x ar (|| gm ABCD)	From 1 to 6.

Hence, proved.

Ex.3 Prove that of all parallelograms of which the sides are given, the parallelogram which is rectangle has the greatest area.

Sol. Let ABCD be a parallelogram in which AB = a and AD = b. Let h be the altitude corresponding to the base
AB. Then, 

(║gm ABCD) = AB × h = ah

Since the sides a and b are given. Therefore, with the same sides a and b we can construct infinitely many parallelograms with different heights.

Now, ar (║gm ABCD) = ah

⇒ ar (║gm ABCD) is maximum or greatest when h is maximum. [∵  a is given i.e., a is constant]

But, the maximum value which h can attain is AD = b and this is possible when AD is perpendicular to AB

i.e. the ║gm ABCD becomes a rectangle.

Thus, (ar ║gm ABCD) is greatest when AD ⊥ AB i.e. when (║gm ABCD) is a rectangle.

TRIANGLES ON THE SAME BASE AND BETWEEN THE SAME PARALLELS

Theorem-2 : Triangles on the same base and between the same parallels are equal in area.

Given :Two Δs ABC and DBC on the same base BC and between the same parallels BC and AD.
To prove : ar(ΔABC) = ar (ΔDBC).

Construction : Draw BE ⊥ CA, meeting DA produced at E and draw CF ⊥ BD, meeting AD produced at F.
Proof :

STATEMENT		REASON
1.	BCAE is a ║gm BC ⊥ EA and BE ⊥ CA	(By construction)
2.	ar (ΔABC) = ar (║gm BCAE)	Diagonal BA divides ║gm BCAE into two Δs of equal areas. (By construction)
3.	BCFD is a ║gm	BC ⊥ DF and BD ⊥ CF
4.	ar (ΔDBC) =  ar (║gm BCFD)	Diagonal CD divides ║gm BCFD into two Δs of equal areas.
5.	ar (║gm BCAE) = ar (║gm BCFD)	║gms on same base BC and between the same parallels BC and EF are equal in area.
6.	ar (ΔABC) = ar (ΔDBC)	From 2, 4 and 5.

Hence, proved.

Corollary : Area of a triangle =  × Base × Height.

Given : A ΔABC with base BC and height AL.

To prove : ar(ΔABC) =  × BC × AL

Construction : Draw CD ⊥ BA and AD ⊥ BC, intersecting each other at D.

Proof :  

STATEMENT		REASON
1.	ABCD is a  ║ gm	BC ⊥ AD and BA ⊥ CD (By construction)
2.	ar (ΔABC) = ar (║ gm ABCD) = ×  BC × AL	Diagonal CA divides ║ gm ABCD into two Δs of equal areas.  ar (║ gm ABCD) = BC × AL

Ex.Show that a median of a triangle divides it into two triangles of equal area.

Sol. Given :ΔABC in which AD is a median.
To prove : ar(ΔABD) = ar (ΔADC).

Construction : Draw AL  ⊥ BC.
Proof : Since AD is the median ΔABC.
Therefore, D is the mid-point of BC.

⇒ BD = DC

⇒ BD X AL = DC X AL [ Multiplying both sides by AL]

⇒ 1/2 (BD X AL) = 1/2(DC X AL) [ Multiplying both sides by 1/2]

⇒ Area (ΔABD) =  Area (ΔADC)
Hence, proved.

Ex. In the given figure, ABCD is a quadrilateral in which M is the mid-point of diagonal AC.
Prove that : ar(quad. ABMD) = ar(quad. DMBC).

Sol. Given : ABCD is a quadrilateral in which M is the mid-point of diagonal AC.
To prove : ar(quad. ABMD) = ar (quad. DMBC)

Proof :

Hence, proved.