NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles

Ans : (i) 

Yes. It can be observed  that trapezium ABCD and triangle PCD have a common base CD and these are lying between the same parallel lines AB and CD.
(ii) 

No. It can be observed that parallelogram PQRS and trapezium MNRS have a 
common base RS. However, their vertices, (i.e., opposite to the common base) P, Q 
of parallelogram and M, N of trapezium, are not lying on the same line. 
(iii) 

Yes. It can be observed that parallelogram PQRS and triangle TQR have a common 
base QR and they are lying between the same parallel lines PS and QR. 
(iv) 

No. It can be observed that parallelogram ABCD and triangle PQR are lying between 
same parallel lines AD and ac. However, these do not have any common base. 
(v) 

Yes. It can be observed that parallelogram ABCD and parallelogram APQD have a 
common base AD and these are lying between the same parallel lines AD and 3Q. 
(vi) 

No. It can be observed that parallelogram PBCS and PQRS are lying on the same 
base PS. However, these do not lie between the same parallel lines. 

Ans : In parallelogram ABCD, CD = AB = 16 cm
[Opposite sides of a parallelogram are equal] 
We know that 
Area of a parallelogram = Base x Corresponding altitude 
Area of parallelogram ABCD = CD x AE = AD x CF 
Area of parallelogram = Base x Corresponding altitude 
Area of parallelogram ABCD = CD x AE = AD x CF 
\( 16 \mathrm{cm} \times 8 \mathrm{cm}=\mathrm{AD} \times 10 \mathrm{cm} \) 
\( \mathrm{AD}=\frac{16 \times 8}{10} \mathrm{cm}=12.8 \mathrm{cm} \) 
Thus, the length of AD is 12.8 cm. 

Ans : 
Let us join HF.
In parallelogram ABCD, 
AD = BC and \( \mathrm{AD}\|\mathrm{BC} \) ( Opposite sides of a parallelogram are equal and parallel)
AB = CD ( Opposite sides of a parallelogram are equal)
\( \Rightarrow \frac{1}{2} \mathrm{AD}=\frac{1}{2} \mathrm{BC} \) and 
\( \mathrm{AH}\|\mathrm{BF} \) 
Therefore, ABFH is a parallelogram.
Since \( \Delta \mathrm{HEF} \) and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF,
\( \therefore \text { Area }(\Delta \mathrm{HEF})=\frac{1}{2} \text { Area (ABFH) } \ldots \) (1)
Similarly, it can be proved that 
Area \( (\Delta H G F)=\frac{1}{2} \text { Area (HDCF) } \ldots \) (2)
On adding equations (1) and (2), we obtain
Area \( (\Delta \mathrm{HEF})+\text { Area }(\Delta \mathrm{HGF})=\frac{1}{2} \text { Area }(\mathrm{ABFH})+\frac{1}{2} \text { Area }(\mathrm{HDCF}) \) 
\( =\frac{1}{2}[\text { Area }(\mathrm{ABFH})+\text { Area }(\mathrm{HDCF})] \) 
\( \Rightarrow \text { Area }(\mathrm{EFGH})=\frac{1}{2} \text { Area }(\mathrm{ABCD}) \) 

Ans : 
It can be observed that \( \triangle B Q C \) and paralleogram ABCD lie on the same base BC and these are between the same parallel lines AD and BC.
\( \therefore \text { Area }(\Delta B Q C)=\frac{1}{2} \text { Area (ABCD) } \dots \) (1)
Similarly, \( \triangle \mathrm{APB} \) and parallelogram ABCD lie on the same base AB and between the same parallel lines AB and DC.
\( \therefore \text { Area }(\Delta \mathrm{APB})=\frac{1}{2} \text { Area (ABCD) } \ldots \) (2) 
From equation (1) and (2), we obtain
Area \( (\Delta B Q C)=\text { Area }(\Delta A P B) \) 

Ans : 
(i) Let us draw a line segment EF, passing through point P and parallel to line segment AB.
In parallelogram ABCD,
\( \mathrm{AB}\|\mathrm{EF}(\mathrm{By} \text { construction }) \ldots(1) \) 
\( \begin{array}{l}{\text { ABCD is a parallelogram. }} \\ {\therefore \text { AD }\|\text { BC (Opposite sides of a parallelogram) }}\end{array} \) 
\( \Rightarrow \mathrm{AE}\|\mathrm{BF} \ldots(2) \) 
From equations (1) and (2), we obtain
\( \mathrm{AB}\|\mathrm{EF} \text { and } \mathrm{AE}\| \mathrm{BF} \)
Therefore, quadrilateral ABFE is a parallelogram 
It can be observed that \( \triangle \mathrm{APB} \) and parallelogram ABFE are lying on the same base AB and between the same parallel lines AB and EF.
\( \therefore \text { Area }(\Delta \mathrm{APB})=\frac{1}{2} \text { Area (ABFE) } \ldots \) (3)
Similarly, for \( \Delta \mathrm{PCD} \) and parallelogram EFCD,
Area \( (\triangle \mathrm{PCD}) \) = \( \frac{1}{2}{\text { Area }(E F C D) \ldots(4)} \) 
Adding equation (3) and (4), we obtain
\( (\Delta \mathrm{APB})+\text { Area }(\Delta \mathrm{PCD})=\frac{1}{2}[\text { Area }(\mathrm{ABFE})+\text { Area }(\mathrm{EFCD})] \) 
\( (\Delta \mathrm{APB})+\text { Area }(\Delta \mathrm{PCD})=\frac{1}{2} \text {Area }(\mathrm{ABCD}) \) …(5)
(ii)  

Let us draw a line segment MN, passing through point P and parallel to line segment AD.
In parallelogram ABCD,
MN || AD (By construction)...(6)
ABCD is a parallelogram.
\( \therefore \text { AB }\|\text { DC (Opposite sides of a parallelogram) } \) 
\( \Rightarrow \mathrm{AM}\|\mathrm{DN} \dots(7) \) 
From equations (6) (7), we obtain
\( \mathrm{MN}\|\mathrm{AD} \text { and } \mathrm{AM}\| \mathrm{DN} \) 
Therefore, quadrilateral AMND is a parallelogram.
It can be observed that \( \triangle A P D \) and parallelogram AMND are lying on the same base
AD and between the same parallel lines AD and MN.
\( \therefore \text { Area }(\Delta A P D)=\frac{1}{2} \text { Area (AMND) } \dots \) (8)
Similarly, for \( \triangle \mathrm{PCB} \) and parallelogram MNCB,
Area \( (\Delta \mathrm{PCB})=\frac{1}{2} \text { Area (MNCB) } \ldots \) (9)
Adding equations (8) and (9), we obtain
\( (\Delta \mathrm{APD})+\text { Area }(\Delta \mathrm{PCB})=\frac{1}{2}[\text { Area }(\mathrm{AMND})+\text { Area }(\mathrm{MNCB})] \) 
\( (\Delta \mathrm{APD})+\text { Area }(\Delta \mathrm{PCB}) \) = \( \frac{1}{2} \operatorname{Area}(\mathrm{ABCD}) \quad \ldots(10) \) 
On comparing equations (5) and (10), we obtain
\( (\Delta \mathrm{APD})+\text { Area }(\Delta \mathrm{PBC})=\text { Area }(\Delta \mathrm{APB})+\text { Area }(\Delta \mathrm{PCD}) \)