Class-ix mathematics formative Assignment

SECTION A (4 x 1M= 4M)

1. In the given figure fig. 9.1, ABCD is a parallelogram and AEB is a triangle. If Ar (ABCD) = 172.6cm²,

 then Ar ( ABE) =_________  (a) 86.3cm2 (b) 172.6cm2 (c) 345.2cm2 (d) cannot be determined.

2. In the given figure fig. 9.2, chord AB at a distance of 9cm from the center O of the given circle. If radius of the circle is 41cm, length of chord AB is:         

(a) 40cm (b) 80cm (c) 50cm (d) cannot be determined

3. In fig. 9.3 ABCD is a parallelogram, AP ^ DC and CQ ^ AD. If AB = 10cm, AD = 8cm and AP = 8cm then 

CQ = ----                                           

(a) 16cm (b) 10cm (c) 8 cm (d) none of these.

4. In fig. 9.4, If O is the center of the circle and AB = CD, then OL : OM = __ (a) 2:1 (b) 1:2 (c) 1:1 (d) none of these.

                                                                 

 SECTION B (3 x 2M = 6M)

5. in fig. 9.5, A and B are points on sides PQ and QR of parallelogram PQRS. Show that Ar (ARS) = Ar(PBS)                                                                                                                              

6. In fig. 9.6, AB and BC are two chords of a circle with centre O such that <ABO = <CBO.  Show that AB = BC

7. Construct the angle of measurement 22½ ° 

                                                                                

SECTION C (2 x 3M = 6M)      

                                                  

8. A line l parallel to side BC of ABC meets AB at X and AC at Y. Also lines parallel to AB through C and AC through B meets line l at E and F respectively. 

Then prove that Ar (ABF) = Ar (ACE).

9. In fig. 9.7, O is the centre of the circle. Determine ÐDAC, ÐACB, ÐADE.      

   

 SECTION D (1 x 4M = 4M)

10. Prove that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.                                                                                                      

                                           (OR),

In fig. 9.8, P is a point in the interior of a parallelogram ABCD. Show that 

1) Ar (APB) + Ar (PCD) = ½ Ar(ABCD)       

2) Ar(APD) + Ar(PBC) = Ar(APB) + Ar(PCD)

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