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NCERT Solutions Class 9 Maths Exercise 9.1 and 9.2 -Areas of Parallelogram and Triangles

NCERT Solutions Class 9 Maths Exercise 9.1 and 9.2 of Chapter 9-Areas of Parallelogram and Triangle is very important to study for the exam preparations. These NCERT solutions are the solutions of unsolved questions of class 9 NCERT maths textbook exercise 9.1 and 9.2 of chapter 9 .All questions are solved by the expert by a step by step method.

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NCERT Solutions class 9 maths of chapter 9-Areas of parallelogram and triangles

Exercise 9.3-Areas of Parallelogram and Triangles

Download pdf of NCERT solutions of class 9 maths chapter 9-Areas of Parallelogram and Triangle

PDF-NCERT solutions of class 9 maths chapter 9-Areas of Parallelogram

NCERT Solutions of Class 9 Science : Chapter 1 to Chapter 15

Q1. Which of the following figure lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.

Ans.

(i) In figure (i) parallelogram ABCD and ΔPCD are  between the same parallels,the common base is DC and the two parallel lines are AB ∥ DC.

(ii)In figure (ii) parallelogram PQRS and quadrilateral MNRS are not between the same parallels, although the common base is SR

(iii) In figure (iii) quadrilateral PQRS and ΔQTR are between the same parallels,the common base is QR and the two parallel lines are PS ∥ QR.

(iv) In figure (iv) parallelogram ABCD and ΔPBQ are not between the same parallels AD∥BC, but there is no common base.

(vi) In figure (vi) quadrilateral PQRS and quadrilateral ABCD are between the same parallels PQ∥ SR but they don’t have a common base.

Exercise 9.2

Q1. In the given figure ABCD is a parallelogram ,AE ⊥ DC and CF ⊥ AD.If  AB = 16 cm,AE = 8 cm and CF = 10 cm,find AD.

Ans. Area of parallelogram = 1/2 (Base × altitude)

AB = DC= 16 cm (opposite sides of parallelogram)

Area of parallelogram ABCD = 1/2(DC × AE) =1/2(16×8) =64 cm²

Q2.If E, F,G and H are respectively are the mid points of the sides of a parallelogram ABCD ,show that

$\boldsymbol{ar\left ( EFGH \right )=\frac{1}{2}\left ( ABCD \right )}$

Ans.

GIVEN: ABCD is a parallelogram

E, F, G, and H are midpoints of AB, BC, CD, and AD respectively

CONSTRUCTION: Joining the points H and F

TO PROVE:

$\boldsymbol{ar\left ( EFGH \right )=\frac{1}{2}\left ( ABCD \right )}$

PROOF: H and F are the mid points of AD and BC respectively

∴HF ∥ DC and HF =DC

HFCD will be a parallelogram

Since the area of a triangle is half of the area of a parallelogram between the same parallels and on the common base.

$\boldsymbol{\therefore ar\left ( HGF \right )=\frac{1}{2}\left ( HFCD \right )...\left ( i \right )}$

H and F are the midpoints of AD and BC respectively

∴HF ∥ AB and HF =AB

ABFH will be a parallelogram$\boldsymbol{\therefore ar\left ( HEF \right )=\frac{1}{2}ar\left ( ABFH \right )....\left ( ii \right )}$

Adding both equations (i) and (ii)

$\boldsymbol{ ar\left ( HEF \right )+ar\left ( HGF \right )=\frac{1}{2}ar\left ( ABFH \right )+\frac{1}{2}ar\left ( HFCD \right )}$

$\boldsymbol{ar\left ( EFGH \right )=\frac{1}{2}\left ( arABFH+arHFCD \right )}$

$\boldsymbol{ar\left ( EFGH \right )=\frac{1}{2}\left ( ABCD \right )}$

Hence proved

Q3. P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD.Show that ar(APB) = ar(BQC).

Ans.

GIVEN: ABCD is a parallelogram

P and Q are the points on the sides DC and AD respectively

TO PROVE: ar(APB) = ar(BQC).

PROOF: The area of a triangle is half of the area of a parallelogram between the same parallels and on the common base

Here parallelogram ABCD and ΔAPB are on the same base AB and between the same parallels AB ∥ DC.

$\boldsymbol{ar\left ( APB \right )=\frac{1}{2}ar\left ( ABCD \right )...\left ( i \right )}$

Parallelogram ABCD and ΔBCQ are on the same base AB and between the same parallels AD ∥ BC

$\boldsymbol{ar\left ( BCQ \right )=\frac{1}{2}ar\left ( ABCD \right )...\left ( ii \right )}$

From equation (i) and equation (ii)

ar(APB) = ar(BQC), Hence proved

Q4. In the given figure,P is a point in the interior of a parallelogram ABCD. Show that

(i) ar(APB) + ar(PCD) = 1/2 ar(ABCD)

(ii) ar(APD) + ar(PBC)= ar(APB) + ar(PCD)

(Hint: Through P draw a line parallel to AB)

Ans.

GIVEN:P is a point in the interior of a parallelogram ABCD

CONSTRUCTION: Drawing a line FE through P such that FE∥AB∥ DC.

TO PROVE:

(i) ar(APB) + ar(PCD) = 1/2 ar(ABCD)

(ii) ar(APD) + ar(PBC)= ar(APB) + ar(PCD)

PROOF:

(i) AB ∥ FE (constructed)

ABEF will be a parallelogram

ΔAPB and ABEF are the triangle and parallelogram on the same base AB and between the same parallels AB ∥ FE.

$\boldsymbol{\therefore ar\left ( APB \right )=\frac{1}{2}ar\left ( ABEF\right )...\left ( i \right )}$

ΔPCD and FECD are the triangle and parallelogram on the same base DC and between the same parallels DC ∥ FE

$\boldsymbol{\therefore ar\left ( PCD \right )=\frac{1}{2}ar\left ( FECD\right )...\left ( ii \right )}$

Adding both equation (i) and (ii)

$\boldsymbol{ar\left ( APB \right )+ar\left ( PCD \right )=\frac{1}{2}ar\left ( FECD\right )+\frac{1}{2}ar\left ( ABEF \right )}$

$\boldsymbol{ar\left ( APB \right )+ar\left ( PCD \right )=\frac{1}{2}ar\left ( ABCD \right )}$

(ii)

Drawing a line EF such that EF∥AD∥BC

AFED and ΔAPD are the parallelogram and triangle on the same base AD and between the same parallels AD ∥ FE

$\boldsymbol{ar\left ( APD \right )=\frac{1}{2}ar\left (ADEF \right )...\left ( i \right )}$

FBCE and ΔPBC are the parallelogram and triangle on the same base BC and between the same parallels BC ∥ FE

$\boldsymbol{ar\left ( PBC \right )=\frac{1}{2}ar\left (FBCE \right )...\left ( i i\right )}$

Adding both equation (i) and (ii)

$\boldsymbol{ar\left ( APD \right )+ar\left ( PBC \right )=\frac{1}{2}ar\left ( ADEF \right )+\frac{1}{2}ar\left (FBCE \right )}$

$\boldsymbol{ar\left ( APD \right )+ar\left ( PBC \right )=\frac{1}{2}ar\left ( ABCD \right )}$

Since, we already have proved in (i)

$\boldsymbol{ar\left ( APB \right )+ar\left ( PCD \right )=\frac{1}{2}ar\left ( ABCD \right )}$

Therefore

ar(APD) + ar(PBC)= ar(APB) + ar(PCD),Hence proved

Q5.In the given figure ,PQRS and ABRS are parallelograms and X is any point on side BR .Show that

(i) ar(PQRS) = ar(ABRS)

(ii) ar(AXS) = 1/2 ar( PQRS)

Ans.(i) Parallelograms PQRS and ABRS are on the same base SR and between the same parallels PB ∥ SR

∴ ar(PQRS) = ar(ABRS)

(ii) Tringle AXS and Parallelogram ABRS are on the same base AS and between the same parallels AS∥ BR.

ar(AXS) = 1/2 ar(ABRS)…..(i)

ar(PQRS) = ar(ABRS)[proved above]…….(ii)

From equation (i) and (ii)

ar(AXS) = 1/2 ar(PQRS), Hence proved

Q6.A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?

Ans.

Ans. PQRS is a field in the form of a parallelogram

A is a point on RS that is joined to P and Q

The field is divided into three triangular parts ΔPAS, ΔPQA and ΔQAR

ar(ΔPAS) +ar(ΔPQA) + ar(ΔQAR) = ar(PQRS)

PQRS and ΔPQA are the parallelogram and triangle on the same base PQ and between the same parallels PQ ∥ SR

ar(ΔPQA) = 1/2 ar(PQRS)….(i)

ar(ΔPAS) + ar(ΔQAR)= 1/2 ar(PQRS)

ar(ΔPAS )+ ar(ΔQAR)+ 1/2 ar(PQRS)= ar(PQRS)

ar(ΔPAS) + ar(ΔQAR) = ar(PQRS) – 1/2 ar(PQRS)

ar(ΔPAS) + ar(ΔQAR) =  1/2 ar(PQRS)….(ii)

From equation (i) and (ii)

ar(ΔPAS) + ar(ΔQAR) = ar(ΔPQA)

The relationship between the triangular parts ΔPAS ,ΔPQA and ΔQAR shows that the farmer will sow wheat in ΔPQA and pulses in ΔQAR and ΔPAS or its vise versa.

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