9th ncert solution Ch area of parallelogram and triangle optional exercise

1.Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.

Solution. IIgm ABCD and rectangle ABEF are between the same parallels AB and CF.

 AB = EF (For rectangle) and AB = CD (For parallelogram)

∴ CD = EF  ⇒ AB + CD = AB + EF ... (1)

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

∴ AF < AD

similarly we write , BE < BC          ∴ AF + BE < AD + BC ... (2)

From equations (1) and (2), we get

AB + EF + AF + BE < AD + BC + AB + CD
Perimeter of rectangle ABEF < Perimeter of parallelogram ABCD

NCERT Solutions IX Area of parallelograms and triangles

Q.1 A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
Ans: Let quadrilateral ABCD be original shape of field. We need to join diagonal BD and have to draw a line parallel to BD through point A. that meet the extended side CD at point E.

Now we join BE and AD. that intersect each other at O. 

In this way part ∆AOB can be cut from the original field to make new shape of field will be ∆ BCE.


Now we have to prove that the area of ∆AOB (portion that was cut so as to construct Health Centre) is equal to the area of the ∆DEO (portion added to the field so as to make the area of new field so formed equal to the area of original field) 

∆DEB and ∆DAB lie on same base BD and are between same parallels BD and AE.
∴ Area (∆DEB) = area (∆DAB)
⇒ Area (∆DEB) – area (∆DOB) = area (∆DAB) – area (∆DOB)
⇒ Area (∆DEO) = area (∆AOB)

2. In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. show that   (i) ar (ACB) = ar (ACF)  (ii) ar (AEDF) = ar (ABCDE) 

  

Solution : (i)  ∆ACB and ∆ ACF are on the same base AC and are between  the same parallels AC and BF   ∴ area (∆ACB) = area (∆ACF) 

(ii) Area (∆ACB) = area (∆ACF)   

⇒ Area (∆ACB) + area (ACDE) = area (ACF) + area (ACDE) 

⇒ Area (ABCDE) = area (AEDF)

3.  In the given figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:  

 (i) ar (DOC) = ar (AOB)  (ii) ar (DCB) = ar (ACB)   (iii) DA || CB or ABCD is a parallelogram.

Solution: We have to draw DN ⊥ AC and BM ⊥ AC 

(i) In ∆DON and ∆BOM

   ∠DNO = ∠BMO   (By construction) 

∠DON = ∠BOM   (Vertically opposite angles) 

OD = OB   (Given) 

By A–A–S congruence rule 

∆DON ≅ ∆BOM 

∴ DN = BM      ... (1) 

We know that congruent triangles have equal areas. 

∴ Area (∆DON) = area (∆BOM)    ... (2) 

In ∆DNC and ∆BMA 

∠DNC = ∠BMA  

CD = AB   

DN = BM   

∴ ∆DNC ≅ ∆BMA   (RHS congruency) 

⇒ area (∆DNC) = area (∆BMA)   ... (3) 

On adding equation (2) and (3), we have 

Area (∆DON) + area (∆DNC) = area (∆BOM) + area (∆BMA) 

So, area (∆DOC) = area (∆AOB) 

(ii) We have 

Area (∆DOC) = area (∆AOB) 

   ⇒ Area (∆DOC) + area (∆OCB) = area (∆AOB) + area (∆OCB) 

      (Adding area (∆OCB) to both sides) 

   ⇒ Area (∆DCB) = area (∆ACB) 

(iii) Area (∆DCB) = area (∆ACB) 

Now if two triangles are having same base and equal areas, these will be between same parallels 

∴ DA || CB      ... (4) 

For quadrilateral ABCD, we have one pair of opposite sides are equal 

(AB = CD) and other pair of opposite sides are parallel (DA || CB).  

Therefore, ABCD is parallelogram 

IXQuadrilateral CCE Test Papers

CBSE ADDA: 9th Quadrilateral CCE Test Papers

Prove that followings:

1. A diagonal of a parallelogram divides it into two congruent triangles.
2. In a parallelogram, opposite sides and angle are equal.
3. If each pair of opposite sides of quadrilateral is equal, then it is a parallelogram.
4. If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.
5. The diagonals of a parallelogram bisect each other.
6. If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
7. A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.
8. The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.
9. The line segment joining the mid- points of the two sides of a triangle is parallel to the third side.
10. Show that each angle of a rectangle is a right angle.
11. Show that the diagonal of a rhombus are perpendicular to each other.
12. Show that the bisectors of the angles of a parallelogram form a rectangle.
13. ABCD is a parallelogram (||gm) in which P and Q are mid-points of opposite side AB and CD. If AQ intersects DP at S and BQ intersects CO at R, show that (i) APCQ is ||gm(ii)DPBQ is ||gm(iii) PSQR is ||gm
14. In Triangle ABC, D, E and Fare respectively the mid points of sides AB, BC and CA. Show that triangle ABC is divided into four congruent triangle by joining D, E and F
15. If the diagonal of a parallelogram are equal, then show that it is a rectangle.
16. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
17. Show that the diagonals of a square are equal and bisect each other at right angles.
18. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
19. In Δ ABC and Δ DEF, AB=DE, AB||DE, BC=EF and BC||EF. Vertices A, B and C are joined to vertices D, E and F respectively. Show that:(i) Quadrilateral ABCD is a parallelogram.(ii) Quadrilateral BEFC is a parallelogram.(iii) AD||CF and AD=CF(iv) Quadrilaterals ACFD is a parallelogram(v) AC=DF(vi) Δ ABC @ Δ DEF.
20. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC is a diagonal. Show that:(i) SR||AC and SR =1/2 AC(ii) PQ=SR(iii) PQRS is a parallelogram.
21. ABCD is a rhombus and P, Q, R and S are the mid- point of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.
22. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
23. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
24. ABC is a triangle right angle at C. A line through the mid-points M of hypotenuse AM and parallel to BC intersects AC at D. Show that(i) D is the mid –point of AC(ii) MD ┴ AC(iii) CM=MA=1/2 AB.

Activity related to Area of parallelogram and triangle for class 9th

Activity related to to find a relation, between the areas of two parallelograms on the same base and between the same parallels.

Let us take a graph sheet and draw two parallelograms ABCD and PQCD on it as shown in Fig

Now find the areas of these two parallelograms by counting the number of complete squares enclosed by the figure, the number of squares a having more than half their parts enclosed by the figure and the number of squares having half their parts enclosed by the figure. The squares whose less than half parts are enclosed by the figure are ignored.
You will find that areas of both the parallelograms are (approximately) 15 Sq.cm

Hence, you to conclude that parallelograms on the same base and between the same parallels are equal in area

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

Proof : Two parallelograms ABCD and EFCD, on the same base DC and between the same parallels

AF and DC are given (see Fig).
We need to prove that ar (ABCD) = ar (EFCD).
In Δ ADE and Δ BCF,
∠ DAE = ∠ CBF (Corresponding angles from AD || BC and transversal AF) (1)
∠ AED = ∠ BFC (Corresponding angles from ED || FC and transversal AF) (2)
Therefore, ∠ ADE = ∠ BCF (Angle sum property of a triangle) (3)
Also, AD = BC (Opposite sides of the parallelogram ABCD) (4)
So, Δ ADE ≅ Δ BCF [By ASA rule, using (1), (3), and (4)]
Therefore, ar (ADE) = ar (BCF) (Congruent figures have equal areas) (5)
Now,adding ar (EDCB) both the sides
 ar (ADE) + ar (EDCB) = ar (BCF)+ ar (EDCB)
                 ar (ABCD) = ar (EFCD)
So, parallelograms ABCD and EFCD are equal in area.

Problem: 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.
Solution : Let Δ ABP and parallelogram ABCD be on the same base AB and between the same parallels
AB and PC (see Fig).

To prove :  ar (PAB) =1/2 ar (ABCD)
Construction: Draw BQ || AP to obtain another parallelogram ABQP. So that parallelograms ABQP
and ABCD will be on the same base AB and between the same parallels AB and PC.
Therefore, ar (ABQP) = ar (ABCD) (By Theorem that parallelograms on the same base and between the same parallels are equal in area) --------(1)

But Δ PAB ≅ Δ BQP (As diagonal PB divides parallelogram ABQP into two congruent triangles.)

So, ar (PAB) = ar (BQP) -------------(2)

Therefore, ar (PAB) =1/2ar (ABQP) [From (2)]--------------------- (3)

This gives ar (PAB) =1/2 ar (ABCD) [From (1) and (3)]

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

Let two triangles ABC and PBC on the same base BC and between the same parallels BC and AP

Construction : Draw CD || BA and CR || BP such that D and R lie on line AP(see Fig) to get two parallelograms PBCR and ABCD on the same base BC and between the same parallels BC and AR.

Now, Parallelograms PBCR and ABCD on the same base BC and between the same parallels BC and AR. Therefore, ar (ABCD) = ar (PBCR) ----------(1)

Now Δ ABC ≅ Δ CDA and Δ PBC ≅ Δ CRP ( The diagonals of a parallelogram divides it  into two congruent triangles)

So, ar(Δ ABC) = ar(Δ CDA) and ---------(ii)

ar (ABCD) = ar(Δ ABC) + ar(Δ CDA) --------------(iii)

ar (PBCR)  = ar(Δ PBC) = ar(ΔCRP) ----------------(iv)

From (1),(ii),(iii) and (iv) , we have 

So, ar (ABC) =1/2 ar (ABCD)

and ar (PBC) =1/2 ar (PBCR)

Therefore, ar (ABC) = ar (PBC)

Problem: 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 fields 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?

When she took any point A on RS and joined it to points P and Q

A divides the field into three parts. These parts are triangular in shape –> ∆PSA, ∆PAQ and ∆QRA 

So, Area of   IIgm  PQRS =  Area of ∆PSA + Area of ∆PAQ + Area of ∆QRA  ... (1)  

∴ Area (∆PAQ) = 1/2 area (PQRS)      ... (2) [a parallelogram and triangle are on the same base and 

between the same parallels, the area of triangle is half the area of the parallelogram]

From equations (1) and (2), we get

Area (∆PSA) + area (∆QRA) + 1/2 area (PQRS) = area (PQRS)   

Area (∆PSA) + area (∆QRA)=  area (PQRS) -1/2 area (PQRS) 

Area (∆PSA) + area (∆QRA)=  1/2 area (PQRS) 

Area (∆PSA) + area (∆QRA)= Area (∆PAQ)

Clearly, farmer must sow wheat in triangular part PAQ and pulses in other two triangular parts PSA and QRA or wheat in triangular part PSA and QRA and pulses in triangular parts PAQ.

Class-IX. Math. Chapter : Area of Parallelogram and Triangles. NCERT Solutions Topperlearning Download
9th Area of Parallelogram and Triangle Test paper