Chapter 9 - Areas of Parallelograms and Triangles, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9 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 :

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 ℓ through A and D, BE ⊥ ℓ and CF ⊥ ℓ. 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 :  ΔABC and a || gm BCDE on the same base BC and between the same parallels BC and AD.

To Prove : ar (ΔABC) = 1/2 ar (|| gm BCDE) .

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

Proof :

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)

⇒ 6.5 x 4 = CD x 5

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 :

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,

ar (║ 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.

Ex.4 In the adjoining figure, O is any point inside a parallelogram ABCD. Prove that : ar (ΔOAB) + ar (ΔOCD) = ar (ΔOBC) + ar (ΔOAD).

 

Sol. Given : A || gm ABCD in which O is a point.

To prove : ar (ΔOAB) + ar (ΔOCD) = ar (ΔOBC) + ar (ΔOAD).

Construction : Draw EOF || AB and GOH || AD.

Proof :

Hence Proved.

Ex.5 In fig, ABCD is a parallelogram and EFCD is a rectangle. Also AL  DC. Prove that
 (i) ar (ABCD) = (EFCD) 

(ii) ar (ABCD) = DC × AL

Sol. (i) We know that a rectangle is also a parallelogram.

Thus, parallelogram ABCD and rectangle EFCD are on the same base CD and between the same parallels CD and BE.
∴ ar (||^gm ABCD) = ar (EFCD)

(ii) From (i), we have ar (ABCD) = ar (EFCD)
⇒ ar (ABCD) = CD × FC                        [∵ Area of a rectangle = Base × Height]
⇒ ar (ABCD) = CD × AL                        [∵ AL = FC as ALCF is a rectangle]
⇒ ar (ABCD) = DC × AL 

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 :

Hence, proved.

Corollary : Area of a triangle

Given : A ΔABC with base BC and height AL.

To prove : ar (ΔABC) = 1/2 x BC x AL

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

Proof :

Hence,

Ex.6 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 ⊥ ABC.

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 ]
⇒ ar (ΔABD) = ar (ΔADC)

Hence, proved.

Ex.7 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.

Ex.8 In the given figure, PQRS and PXYZ are two parallelograms of equal area. Prove that YR || QZ.

Sol. We have : ar (||gm PQRS) = ar (||gm PXYZ)

⇒ ar (||^gm PQRS) – ar (||^gm PQOZ)

= ar (||gm PXYZ) – ar (||gm PQOZ)

⇒ ar (||gm ZORS) = ar (||gm QXYO)

⇒1/2 ar (||gm ZORS) = 1/2 ar (||gm QXYO)

⇒ ar (ΔZOR) = ar (ΔOQY)

⇒ ar (ΔZOR) + ar (ΔOYR) = ar (ΔOQY) + ar (ΔOYR)

⇒ ar (ΔZYR) = ar (ΔQYR)

⇒ YR ||  QZ. [ΔZYR & ΔQYR are equal in area & have the same base. So, they must be between the same parallels]

Ex.9 Prove that the area of the quadrilateral formed by joining the mid-points of the adjacent sides of a quadrilateral is half the area of the given quadrilateral.

Sol. Given :A quadrilateral ABCD, and PQRS is the quadrilateral formed by joining mid-points of the sides AB,
BC, CD and DA respectively.

To prove : ar (quad. PQRS)

Construction : Join AC and AR.

Proof :

Ex.10 If the diagonals of a quadrilateral separate it into four triangles of equal area, show that it is a parallelogram.

Sol. Given : A quad. ABCD whose diagonals intersect at O such that :

ar (ΔAOD) = ar (ΔAOB) = ar (ΔBOC) = ar (ΔCOD)

To prove : ABCD is a parallelogram.

Proof :

ar (ΔAOD) = ar (ΔBOC)
 ⇒ ar (ΔAOD) + ar (ΔAOB) = ar (ΔBOC) + ar (ΔAOB)
 ⇒ ar (ΔABD) = ar (ΔABC)

⇒ DC || AB [Equal Δs on same base must be between the same parallels]

Similarly, AD || BC.

Hence ABCD is ||gm.