# 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.

(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 $(\Delta APB)$