Areas of Parallelograms and Triangles

Class 9 NCERT Maths

NCERT

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

Solution :

(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 $BC$. 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 $BQ$.$\\$ (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.

2   In the given figure, $ABCD$ is parallelogram, $AE \perp DC$ and $CF \perp AD$. If $AB = 16 cm, AE = 8 cm $ and $CF = 10 cm,$ find $AD$.

Solution :

In parallelogram $ABCD, CD = AB = 16 cm$$\\$ [Opposite sides of a parallelogram are equal] We know that$\\$ Area of a parallelogram = Base * Corresponding altitude$\\$ Area of parallelogram $ABCD = CD * AE = AD * CF$$\\$ $16 cm * 8 cm = AD * 10 cm$$\\$ $AD=\dfrac{16*8}{10} cm =12.8 cm$$\\$ Thus, the length of $AD$ is $12.8 cm.$

3   If $E, F, G$ and $H$ are respectively the mid-points of the sides of a parallelogram $ABCD$ show that ar $(EFGH)=\dfrac{1}{2} ar (ABCD)$

Solution :

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

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

Solution :

It can be observed that $\Delta BQC$ and parallelogram $ABCD$ lie on the same base $BC$ and these are between the same parallel lines $AD$ and $BC$.$\\$ $\therefore $ Area$(\Delta BQC)=\dfrac{1}{2}$Area$(ABCD)....(1)$$\\$ Similarly, $\Delta APB$ and parallelogram $ABCD$ lie on the same base $AB$ and between the same parallel lines $AB$ and $DC$.$\\$ $\therefore $ Area $(\Delta APB)=\dfrac{1}{2}$Area $(ABCD)...(2)$$\\$ From Equations (1) and (2), we obtain$\\$ Area $(\Delta BQC)$ = Area $(\Delta APB)$

5   In the given figure, $P$ is a point in the interior of a parallelogram $ABCD$. Show that$\\$ (i)$ ar (APB) + ar (PCD) = \dfrac{1}{2}ar (ABCD)$$\\$ (ii)$ar(APD)+ar(PBC)=ar(APB)+ar(PCD)$[Hint: Through. $P$, draw a line parallel to $AB$]

Solution :

(i) Let us draw a line segment $EF$, passing through point $P$ and parallel to line segment $AB.$$\\$ In parallelogram $ABCD,$ $AB \parallel EF$ (By construction) ... (1)$\\$ $ABCD$ is a parallelogram.$\\$ $\therefore AD \parallel BC $(Opposite sides of a parallelogram)$\\$ $\Rightarrow AE \parallel BF ... (2)$$\\$ From Equations (1) and (2), we obtain$\\$ $AB \parallel EF$ and $AE \parallel BF$$\\$ Therefore, quadrilateral $ABFE$ is a parallelogram.$\\$ It can be observed that $\Delta APB$ and parallelogram $ABFE$ are lying on the same base $AB$ and between the same parallel lines $AB$ and $EF.$$\\$ $\therefore $ Area $(\Delta APB) = \dfrac{1}{2} $ Area $(ABFE).....(3)$$\\$ Similarly, for $\Delta PCD$ and parallelogram $EFCD,$$\\$ Area$(\Delta PCD)=\dfrac{1}{2}$Area $(EFCD)....(4)$$\\$ Adding Equations (3) and (4), we obtain$\\$ Area $(\Delta APB)$+Area($\Delta PCD )=\dfrac{1}{2}$[Area $(ABFE)$+Area $(EFCD)]$$\\$ Area $(\Delta APB)$+Area $(\Delta PCD)=\dfrac{1}{2}$Area($ABCD)....(5)$

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