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Triangles on Same Base and Between Same Parallels

Areas Of Parallelograms And Triangles II - Concepts

Median and Altitude of Triangle

Triangles on Same Base and Between Same Parallels

Relation between Area of Parallelogram and Triangle

Class - 9th IMO Subjects

Concept Explanation

Triangles on Same Base and Between Same Parallels

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

Given Two triangles ABC and PBC on the same base BC and between the same parallel line BC and AP.

To prove $ar (Delta ABC)=ar(Delta PBC)$

Construction Through $B,$ draw $BD parallel CA$ intersecting $PA$ produced in $D$ and through $C$ , draw $CQparallel BP$ , intersecting line $AP;in;Q.$

Proof we have, $BDparallel CA$ [By construction]

and, $BCparallel DA$ [Given AD is the extension of line AP]

$therefore$ $BCDA$ is a parallelogram.

Similarly, $BCQP$ is a parallelogram.

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

$therefore$ $ar(parallel ^{gm}BCQP)=ar(parallel ^{gm}BCAD)$ ...(i)

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

$therefore$ $ar(Delta PBC) =frac{1}{2}ar(parallel ^{gm}BCQP)$ ...(ii)

and, $ar(Delta ABC) =frac{1}{2}ar(parallel ^{gm}BCAD)$ ...(iii)

Now, $ar(parallel ^{gm}BCQP)=ar(parallel ^{gm}BCAD)$ [From (i)]

$Rightarrow$ $frac{1}{2}ar(parallel ^{gm}BCAD)=frac{1}{2}ar(parallel ^{gm}BCQP)$

$Rightarrow$ $ar(Delta ABC)=ar(Delta PBC)$ [Using (ii)and(iii)]

Hence, $ar(Delta ABC)=ar(Delta PBC)$

ILLUSTRATION: The side AB of a parallelogram ABCD is produced to any point P. A line through A parallel to CP meets CB produced at Q then parallelogram PBQR is constructed. Show that ar( ABCD)= ar (PBQR)

Solution: $large Delta$ ACQ and $large Delta$ APQ stand on the same base AQ and they are between same parallel AQ || CP.

ar ( $large Delta$ ACQ)= ar ( $large Delta$ APQ) [Triangles on the same base and beteween same parallel have equal area]

Subtracting common ar (ABQ) from both sides we get

ar ( $large Delta$ ACQ) - ar (ABQ) = ar ( $large Delta$ APQ) - ar (ABQ)

ar ( $large Delta$ ABC)= ar ( $large Delta$ ABP) ................(1)

We know that the diagonal divides a parallelogram in two congruent triangles

$large ar ( Delta ABC)= frac{1}{2}ar (ABCD)$ ..................(2)

$large ar ( Delta PBQ)= frac{1}{2}ar (PBQR)$ ..................(3)

From Eq (1), (2) and (3) we get

$large frac{1}{2}ar (ABCD)= frac{1}{2}ar (PBQR)$

ar( ABCD)= ar (PBQR)

Hence Proved.

Sample Questions

(More Questions for each concept available in Login)

Question : 1

A, B, D, C, E, F are points on three lines I, m and n.

If ar(ABC) = ar(ADC) = ar(AFC) = ar(AEC), then

$BD parallel EF$

$BD$ is not parallel to EF

$AD parallel CF$

$AD parallel CE$

Right Option : A

View Explanation

Question : 2

In the given figure $ABCD$ is a parallelogram, with height of parallelogram is 4 cm, then area of triangle $AFB$ is ____________________.
A. $16;cm^{2}$	B. 34	C. $8;cm^{2}$	D. $6;cm^{2}$
Right Option : C
View Explanation

Question : 3

In the given figure, ABCD is a parallelogram. If P and Q are the mid-points of the sides AB and BC, respectively, then which of the following relationships is definitely true ?

$ar(ABQ)=ar(ADP)$

$ar(APD)=frac {1}{2}ar(ABCD)$

$ar(ABQ)=2;ar(APD)$

None of the above

Right Option : A

View Explanation

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Language - English

Chapters

Polynomials II

Direct and Inverse Variation

Linear Equations in One Variable

Introduction to Trigonometry

Introduction to Euclids Geometry

Herons Formula Complete

Polynomial I

Surface Area and Volume II

Surface Area and Volume I

Coordinate Geometry

Constructions

Areas Of Parallelograms And Triangles II

Areas Of Parallelograms And Triangles I

Linear Equations in Two Variables

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Triangles on Same Base and Between Same Parallels

Triangles on Same Base and Between Same Parallels