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