NCERT Solutions for Class 9 Maths Chapter 9: Areas of Parallelograms and Triangles - Exercise 9.3

Exercise (9.3)

1. In the given figure, $\text{E}$ is any point on median $\text{AD}$ of a $\text{ }\!\!\Delta\!\!\text{ ABC}$. Show that $\text{ar}\left( \text{ABE} \right)\text{ = ar}\left( \text{ACE} \right)$

Ans: It is given that there is a point $\text{E}$ on a median $\text{AD}$ of the $\text{ }\!\!\Delta\!\!\text{ ABC}$.

We know the median divides the triangle into two equal area triangles.

So, median $\text{AD}$ will divide $\text{ }\!\!\Delta\!\!\text{ ABC}$ into two triangles of equal area.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABD} \right)\text{ = ar}\left( \text{ACD} \right)$               …$\left( \text{1} \right)$

The median $\text{ED}$ will divide $\text{ }\!\!\Delta\!\!\text{ EBC}$ into two triangles of equal area.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ EBD} \right)\text{ = ar}\left( \text{ECD} \right)$                …$\left( \text{2} \right)$

Now we will subtract the equation $\left( \text{2} \right)$ from $\left( \text{1} \right)$.

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABD} \right)\text{ - ar}\left( \text{ }\!\!\Delta\!\!\text{ EBD} \right)\text{ = ar}\left( \text{ACD} \right)\text{ - ar}\left( \text{ECD} \right)$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABE} \right)\text{ = ar}\left( \text{ACE} \right)$

Therefore, we have shown that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABE} \right)\text{ = ar}\left( \text{ACE} \right)$.

2. In a triangle $\text{ABC}$, $\text{E}$ is the mid-point of median $\text{AD}$. Show that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ BED} \right)\text{ = }\frac{\text{1}}{\text{4}}\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right)$.

Ans: It is given that point $\text{E}$ is a mid-point on a median $\text{AD}$ of the $\text{ }\!\!\Delta\!\!\text{ ABC}$.

We know the median divides the triangle into two equal area triangles.

So, median $\text{AD}$ will divide $\text{ }\!\!\Delta\!\!\text{ ABC}$ into two triangles of equal area.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABD} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ ACD} \right)$

Since the two triangles are of equal area, we can now write $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABD} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right)$

Similarly, median $\text{BE}$ will divide $\text{ }\!\!\Delta\!\!\text{ ABD}$ into two triangles of equal area.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ BED} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BEA} \right)$

Since the two triangles are of equal area, we can now write $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ BED} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \Delta \text{ABD} \right)$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ BED} \right)\text{ = }\frac{\text{1}}{\text{2}}\left[ \frac{\text{1}}{\text{2}}\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right) \right]$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ BED} \right)\text{ = }\frac{\text{1}}{\text{4}}\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right)$

Therefore, we have shown that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ BED} \right)\text{ = }\frac{\text{1}}{\text{4}}\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right)$

3. Show that the diagonals of a parallelogram divide it into four triangles of equal area.

Ans: Let us assume that $\text{ABCD}$ be a parallelogram.

If we join the diagonals $\text{AC}$ and $\text{BD}$, let $\text{E}$ be their intersection.

We know that the diagonals of a parallelogram bisect each other.

So, we can say that $\text{E}$ is the mid-point of both the diagonals.

We know the median divides the triangle into two equal area triangles.

So, median $\text{DE}$ will divide $\text{ }\!\!\Delta\!\!\text{ ADC}$ into two triangles of equal area.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ CED} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ DEA} \right)$                                       …$\left( \text{1} \right)$

The median $\text{CE}$ will divide $\text{ }\!\!\Delta\!\!\text{ DCB}$ into two triangles of equal area.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ BEC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ CED} \right)$                                        …$\left( \text{2} \right)$  

Similarly we can write $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ AEB} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BEC} \right)$      …$\left( \text{3} \right)$

From equations $\left( \text{1} \right)$, $\left( \text{2} \right)$ and $\left( \text{3} \right)$, we can write

$\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DEA} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ CED} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BEC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ AEB} \right)$

Therefore, we have shown that the diagonals of a parallelogram divide it into four triangles of equal area.

4. In the given figure, $\text{ABC}$ and $\text{ABD}$ are two triangles on the same base $\text{AB}$. If line-segment $\text{CD}$ is bisected by $\text{AB}$ at $\text{O}$, show that $\text{ar}\left( \text{ABC} \right)\text{ = ar}\left( \text{ABD} \right)$

Ans: It is given that $\text{CD}$ is bisected by $\text{AB}$ at $\text{O}$.

Let us first consider $\text{ }\!\!\Delta\!\!\text{ ACD}$.

We are given that $\text{CO = DO}$.

Since a median divides the triangle into two equal area triangles, we can say that $\text{AO}$ is the median of $\text{ }\!\!\Delta\!\!\text{ ACD}$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ AOC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ AOD} \right)$                             …$\left( \text{1} \right)$

Similarly $\text{BO}$ is the median of $\text{ }\!\!\Delta\!\!\text{ BCD}$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ BOC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BOD} \right)$                              …$\left( \text{2} \right)$

We will now add equations $\left( \text{1} \right)$ and $\left( \text{2} \right)$:

$\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ AOC} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ BOC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BOD} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ AOD} \right)$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ ABD} \right)$

Therefore, we have shown that $\text{ar}\left( \text{ABC} \right)\text{ = ar}\left( \text{ABD} \right)$.

5. $\text{D}$, $\text{E}$ and $\text{F}$ are respectively the midpoints of the sides $\text{BC}$, $\text{CA}$ and $\text{AB}$ of a $\text{ }\!\!\Delta\!\!\text{ ABC}$. Show that:

i. $\text{BDEF}$ is a parallelogram

Ans: It is given that $\text{F}$ and $\text{E}$ are midpoints of $\text{AB}$ and $\text{AC}$ respectively.

We know that the line joining the mid-points of two sides of a triangle is parallel to the third side and also half of its length.

$\therefore \text{FE}\parallel \text{BC}$ and $\text{FE = }\frac{\text{1}}{\text{2}}\text{BC}$

Since, $\text{D}$ is the mid-point of $\text{BC}$ we know $\text{BD = }\frac{\text{1}}{\text{2}}\text{BC}$.

$\therefore \text{FE = BD}$ and $\text{FE}\parallel \text{BD}$                                          …$\left( \text{1} \right)$

It is given that $\text{D}$ and $\text{E}$ are midpoints of $\text{BC}$ and $\text{AC}$ respectively.

We know that the line joining the mid-points of two sides of a triangle is parallel to the third side and also half of its length.

$\therefore \text{DE}\parallel A\text{B}$ and $\text{DE = }\frac{\text{1}}{\text{2}}\text{AB}$

Since, $\text{F}$ is the mid-point of $\text{AB}$ we know $\text{BF = }\frac{\text{1}}{\text{2}}\text{AB}$.

$\therefore \text{DE = BF}$ and $\text{DE}\parallel \text{BF}$                                           …$\left( \text{2} \right)$        

So, from equations $\left( \text{1} \right)$ and $\left( \text{2} \right)$, we obtain:

$\text{FE = BD}$ and $\text{FE}\parallel \text{BD}$

$\text{DE = BF}$ and $\text{DE}\parallel \text{BF}$

Therefore, we have shown that $\text{BDEF}$ is a parallelogram.

ii. $\text{ar}\left( \text{DEF} \right)\text{ = }\frac{\text{1}}{\text{4}}\text{ar}\left( \text{ABC} \right)$

Ans: We have found out that $\text{BDEF}$ is a parallelogram.

A diagonal divides the parallelogram into two equal area triangles.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ BDF} \right)\text{ = ar}\left( \text{DEF} \right)$                              …$\left( \text{1} \right)$

Similarly, we can say that $\text{CDFE}$ is also a parallelogram.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ CDE} \right)\text{ = ar}\left( \text{DEF} \right)$                              …$\left( \text{2} \right)$

Similarly, we can say that $\text{AEDF}$ is also a parallelogram.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ AEF} \right)\text{ = ar}\left( \text{DEF} \right)$                               …$\left( \text{3} \right)$

From equations $\left( \text{1} \right)$, $\left( \text{2} \right)$ and $\left( \text{3} \right)$, we obtain

$\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BDF} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ CDE} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ AEF} \right)$

But we know that $ar\left( \Delta ABC \right)=\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ BDF} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ CDE} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ AEF} \right)$

$\Rightarrow ar\left( \Delta ABC \right)=\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right)\text{ = 4  }\!\!\times\!\!\text{  ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)\text{ }$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)\text{ = }\frac{\text{1}}{\text{4}}\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right)$

Therefore, we have shown that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)\text{ = }\frac{\text{1}}{\text{4}}\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right)$.

iii. $\text{ar}\left( \text{BDEF} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{ABC} \right)$

Ans: We know $\text{ar}\left( \text{BDEF} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BDF} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)$

But $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BDF} \right)$.

So, $\text{ar}\left( \text{BDEF} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)$

$\Rightarrow \text{ar}\left( \text{BDEF} \right)\text{ = 2  }\!\!\times\!\!\text{  ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)$

$\Rightarrow \text{ar}\left( \text{BDEF} \right)\text{ = 2  }\!\!\times\!\!\text{  }\left[ \frac{\text{1}}{\text{4}}\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right) \right]$

$\Rightarrow \text{ar}\left( \text{BDEF} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right)$

Therefore, we have shown that $\text{ar}\left( \text{BDEF} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right)$.

6. In the given figure, diagonals $\text{AC}$ and $\text{BD}$ of quadrilateral $\text{ABCD}$ intersect at $\text{O}$ such that $\text{OB = OD}$. If $\text{AB = CD}$, then show that:

i. $\text{ar}\left( \text{DOC} \right)\text{ = ar}\left( \text{AOB} \right)$

(Hint: From $\text{D}$ and $\text{B}$, draw perpendiculars to $\text{AC}$)

Ans: In the figure, let us draw $\text{BP}\bot \text{AC}$ and $\text{DQ}\bot \text{AC}$.

Let us consider in $\text{ }\!\!\Delta\!\!\text{ DOQ}$ and $\text{ }\!\!\Delta\!\!\text{ BOP}$:

$\angle \text{DQO = }\angle \text{BPO = 9}{{\text{0}}^{\text{o}}}$ (By drawing perpendicular lines)

$\angle \text{DOQ = }\angle \text{BOP}$ (They are vertically opposite angles)

$\text{OB = OD}$ (This is given)

Therefore, by $\text{AAS}$ congruence rule, $\text{ }\!\!\Delta\!\!\text{ DOQ }\cong \text{  }\!\!\Delta\!\!\text{ BOP}$

So, we can write $\text{DQ = BP}$                                             …$\left( \text{1} \right)$

The congruent triangles also have equal areas.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DOQ} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BOP} \right)$                                             …$\left( \text{2} \right)$

Let us consider in $\text{ }\!\!\Delta\!\!\text{ DQC}$ and $\text{ }\!\!\Delta\!\!\text{ BPA}$:

$\angle \text{DQC = }\angle \text{BPA = 9}{{\text{0}}^{\text{o}}}$ (By drawing perpendicular lines)

$\text{CD = AB}$ (This is given)

$\text{DQ = BP}$ [From equation $\left( \text{1} \right)$]

Therefore, by $\text{RHS}$ congruence rule, $\text{ }\!\!\Delta\!\!\text{ DQC }\cong \text{  }\!\!\Delta\!\!\text{ BPA}$

The congruent triangles also have equal areas.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DQC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BPA} \right)$                                             …$\left( \text{3} \right)$

We will now add equations $\left( \text{2} \right)$ and $\left( \text{3} \right)$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DOQ} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ DQC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BOP} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ BPA} \right)$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DOC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ AOB} \right)$

Therefore, we have shown that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DOC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ AOB} \right)$.

ii. $\text{ar}\left( \text{DCB} \right)\text{ = ar}\left( \text{ACB} \right)$

Ans: From the above part, we obtained that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DOC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ AOB} \right)$.

We will add $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ OCB} \right)$ on both sides.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DOC} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ OCB} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ AOB} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ OCB} \right)$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DCB} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ ACB} \right)$

Therefore, we have shown that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DCB} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ ACB} \right)$.

iii. $\text{DA}\parallel \text{CB}$ or $\text{ABCD}$ is a parallelogram.

Ans: From the previous part, we obtained that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DCB} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ ACB} \right)$.

We know that if two triangles have equal areas and the same base, then they will both lie between two parallel lines.

$\therefore \text{CB }\parallel \text{ DA}$

It is given that $\text{AB = CD}$.

So, In quadrilateral $\text{ABCD}$, a pair of opposite sides is equal $\left( \text{AB = CD} \right)$ and the other pair of opposite sides is parallel $\left( \text{CB }\parallel \text{ DA} \right)$.

Hence, we say that $\text{ABCD}$ is a parallelogram.

7. $\text{D}$ and $\text{E}$ are points on sides $\text{AB}$ and $\text{AC}$ respectively of $\text{ }\!\!\Delta\!\!\text{ ABC}$ such that $\text{ar}\left( \text{DBC} \right)\text{ = ar}\left( \text{EBC} \right)$. Prove that $\text{DE}\parallel \text{BC}$.

Ans: It is given that $\text{ar}\left( \text{DBC} \right)\text{ = ar}\left( \text{EBC} \right)$.

We know that if two triangles have equal areas and same base, then they will both lie between two parallel lines.

$\therefore \text{DE}\parallel \text{BC}$

Hence, we have proved that $\text{DE}\parallel \text{BC}$.

8. $\text{XY}$ is a line parallel to side $\text{BC}$ of a triangle $\text{ABC}$. If $\text{BE}\parallel \text{AC}$ and $\text{CF}\parallel \text{AB}$ meet $\text{XY}$ at $\text{E}$ and $\text{F}$ respectively, show that $\text{ar}\left( \text{ABE} \right)\text{ = ar}\left( \text{ACF} \right)$.

Ans: It is given that $\text{XY}$ is a line parallel to side $\text{BC}$ of a triangle $\text{ABC}$.

$\therefore \text{XY}\parallel \text{BC}$

Since $\text{E}$ is an extension of line $\text{XY}$, we can also write $\text{EY}\parallel \text{BC}$

It is given that $\text{BE}\parallel \text{AC}$.

Since $\text{Y}$ is a point of line $\text{AC}$, we can also write $\text{BE}\parallel CY$.

So, in quadrilateral $\text{EBCY}$, we have found out that the pair of opposite sides are parallel.

Hence, $\text{EBCY}$ is a parallelogram.

Similarly, we know that $\text{XY}\parallel \text{BC}$.

Since $\text{F}$ is an extension of line $\text{XY}$, we can also write $\text{XF}\parallel \text{BC}$.

It is given that $\text{CF}\parallel \text{AB}$.

Since $\text{X}$ is a point of line $\text{AB}$, we can also write $\text{CF}\parallel BX$.

So, in quadrilateral $\text{FCBX}$, we have found out that the pair of opposite sides are parallel.

Hence, $\text{FCBX}$ is a parallelogram.

We know that if two parallelograms have same base and they lie between two parallel lines, then they have the same areas.

The parallelograms $\text{EBCY}$ and $\text{FCBX}$ lie on the same base $\text{BC}$ and between the parallel lines $\text{BC}$ and $\text{EF}$.

$\therefore \text{ar}\left( \text{EBCY} \right)\text{ = ar}\left( \text{FCBX} \right)$                                              …$\left( \text{1} \right)$

Now let us consider parallelogram $\text{EBCY}$ and $\text{ }\!\!\Delta\!\!\text{ ABE}$.

They both lie on the same base $\text{BE}$ and between two parallel lines $\text{AC}$ and $\text{EB}$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABE} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{EBCY} \right)$                                           …$\left( \text{2} \right)$ 

Similarly, consider parallelogram $\text{FCBX}$ and $\text{ }\!\!\Delta\!\!\text{ ACF}$.

They both lie on the same base $\text{FC}$ and between two parallel lines $\text{AB}$ and $\text{FC}$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ACF} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{FCBX} \right)$                                            …$\left( \text{3} \right)$

From equations  $\left( \text{1} \right)$, $\left( \text{2} \right)$ and $\left( \text{3} \right)$, we can write:

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABE} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ ACF} \right)$

Therefore, we have shown that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABE} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ ACF} \right)$.

9. The side $\text{AB}$ of a parallelogram $\text{ABCD}$ is produced to any point$\text{P}$. A line through $\text{A}$ and parallel to $\text{CP}$ meets $\text{CB}$ produced at $\text{Q}$ and then parallelogram $\text{PBQR}$ is completed (see the following figure). Show that $\text{ar}\left( \text{ABCD} \right)\text{ = ar}\left( \text{PBQR} \right)$.

Ans: Let us join the lines $\text{AC}$ and $\text{PQ}$.

We know that if two triangles have the same base and they lie between two parallel lines, then they have the same areas.

$\text{ }\!\!\Delta\!\!\text{ ACQ}$ and $\text{ }\!\!\Delta\!\!\text{ QPA}$ lie on the same base $\text{AQ}$ and between two parallel lines $\text{AQ}$ and $\text{CP}$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ACQ} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ QPA} \right)$

Subtracting $\text{ar}\left( \text{ABQ} \right)$ on both sides:

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ACQ} \right)\text{ - ar}\left( \text{ABQ} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ QPA} \right)\text{ - ar}\left( \text{ABQ} \right)$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ QBP} \right)$                                    …$\left( \text{1} \right)$

It is given that $\text{AC}$ and $\text{PQ}$ are diagonals of parallelograms $\text{ABCD}$ and $\text{QBPR}$ respectively. The diagonal divide the parallelogram into two triangles of equal areas.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABC} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{ABCD} \right)$                                 …$\left( \text{2} \right)$

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ QBP} \right)\text{ = }\frac{\text{1}}{\text{2}}\text{ar}\left( \text{QBPR} \right)$                                  …$\left( \text{3} \right)$

From equations $\left( \text{1} \right)$, $\left( \text{2} \right)$ and $\left( \text{3} \right)$, we will obtain $\text{ar}\left( \text{ABCD} \right)\text{ = ar}\left( \text{QBPR} \right)$

Therefore, we have shown that $\text{ar}\left( \text{ABCD} \right)\text{ = ar}\left( \text{QBPR} \right)$.

10. Diagonals $\text{AC}$ and $\text{BD}$ of a trapezium $\text{ABCD}$ with $\text{AB}\parallel \text{DC}$ intersect each other at $\text{O}$. Prove that $\text{ar}\left( \text{AOD} \right)\text{ = ar}\left( \text{BOC} \right)$.

Ans: It is given that $\text{AB}\parallel \text{DC}$.

We know that if two triangles have the same base and they lie between two parallel lines, then they have the same areas.

Here, we can see that $\text{ }\!\!\Delta\!\!\text{ ADC}$ and $\text{ }\!\!\Delta\!\!\text{ BCD}$ lie on the same base $\text{DC}$ and between two parallel lines $\text{DC}$ and $\text{AB}$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ADC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BCD} \right)$

Subtracting $\text{ar}\left( \text{DOC} \right)$ on both sides:

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ADC} \right)\text{ - ar}\left( \text{ }\!\!\Delta\!\!\text{ DOC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BCD} \right)\text{ - ar}\left( \text{ }\!\!\Delta\!\!\text{ DOC} \right)$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ AOD} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BOC} \right)$

Therefore, we have proved that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ AOD} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BOC} \right)$.

11. In the given figure, $\text{ABCDE}$ is a pentagon. A line through $\text{B}$ parallel to $\text{AC}$ meets $\text{DC}$ produced at $\text{F}$. Show that

i. $\text{ar}\left( \text{ACB} \right)\text{ = ar}\left( \text{ACF} \right)$

Ans: We know that if two triangles have the same base and they lie between two parallel lines, then they have the same areas.

Here, we can see that $\text{ }\!\!\Delta\!\!\text{ ACB}$ and $\text{ }\!\!\Delta\!\!\text{ ACF}$ lie on the same base $\text{AC}$ and between two parallel lines $\text{AC}$ and $\text{BF}$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ACB} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ ACF} \right)$

Therefore, we have shown that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ACB} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ ACF} \right)$

ii. $\text{ar}\left( \text{AEDF} \right)\text{ = ar}\left( \text{ABCDE} \right)$

Ans: From the previous part, we obtained that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ACB} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ ACF} \right)$.

We add $ar\left( \text{ACDE} \right)$ on both sides.

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ACB} \right)\text{ + ar}\left( \text{ACDE} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ ACF} \right)\text{ + ar}\left( \text{ACDE} \right)$

$\Rightarrow \text{ar}\left( \text{ABCDE} \right)\text{ = ar}\left( \text{AEDF} \right)$

Therefore, we have proved that $\text{ar}\left( \text{AEDF} \right)\text{ = ar}\left( \text{ABCDE} \right)$.

12. 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 us assume quadrilateral $\text{ABCD}$ be the original shape of the land.

Let us join the points $\text{B}$ and $\text{D}$ to form the diagonal of $\text{ABCD}$.

Now we draw a line parallel to $\text{BD}$ through point $\text{A}$. Let it meet the extending of side $\text{CD}$ at new point $\text{E}$.

Now we join $\text{BE}$ and $\text{AD}$ and let their intersection be $\text{F}$.

So, we assume that the portion $\text{ }\!\!\Delta\!\!\text{ AFB}$ can be cut from the original field so that the villager gets the new shape of his field as $\text{ }\!\!\Delta\!\!\text{ DEF}$.

So, we will prove this assumption of ours that $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ AFB} \right)$.

We know that if two triangles have the same base and they lie between two parallel lines, then they have the same areas.

Here, we can see that $\text{ }\!\!\Delta\!\!\text{ DEB}$ and $\text{ }\!\!\Delta\!\!\text{ BAD}$ lie on the same base $\text{DB}$ and between two parallel lines $\text{DB}$ and $\text{EA}$.

$\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DEB} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BAD} \right)$

We subtract $\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DFB} \right)$ on both sides.

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DEB} \right)\text{ - ar}\left( \text{ }\!\!\Delta\!\!\text{ DFB} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BAD} \right)\text{ - ar}\left( \text{ }\!\!\Delta\!\!\text{ DFB} \right)$

$\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ DEF} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ AFB} \right)$

So, our assumption is correct.

Therefore, the land must be cut according to the figure drawn.

13. $\text{ABCD}$ is a trapezium with $\text{AB}\parallel \text{DC}$. A line parallel to $\text{AC}$ intersects $\text{AB}$ at $\text{X}$ and $\text{BC}$ at $\text{Y}$. Prove that $\text{ar}\left( \text{ADX} \right)\text{ = ar}\left( \text{ACY} \right)$. (Hint: Join $\text{CX}$)

Ans:

It is given that $\text{AB}\parallel \text{DC}$ and $\text{XY}\parallel A\text{C}$.

We know that if two triangles have the same base and they lie between two parallel lines, then they have the same areas.

Here, we can see that $\text{ }\!\!\Delta\!\!\text{ ADX}$ and $\text{ }\!\!\Delta\!\!\text{ XCA}$ lie on the same base $\text{AX}$ and between two parallel lines $\text{AX}$ and $\text{DC}$.

\[\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ADX} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ XCA} \right)\]                       …$\left( \text{1} \right)$

Similarly, $\text{ }\!\!\Delta\!\!\text{ ACY}$ and $\text{ }\!\!\Delta\!\!\text{ ACX}$ lie on the same base $\text{AC}$ and between two parallel lines $\text{AC}$ and $\text{XY}$.

\[\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ACY} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ XCA} \right)\]                       …$\left( \text{2} \right)$

Hence, from equations $\left( \text{1} \right)$ and $\left( \text{2} \right)$, we obtain:

$\text{ar}\left( \text{ADX} \right)\text{ = ar}\left( \text{ACY} \right)$

Therefore, we proved that $\text{ar}\left( \text{ADX} \right)\text{ = ar}\left( \text{ACY} \right)$.

14. In the given figure, $\text{AP}\parallel \text{BQ}\parallel \text{CR}$. Prove that $\text{ar}\left( \text{AQC} \right)\text{ = ar}\left( \text{PBR} \right)$.

Ans: We know that if two triangles have the same base and they lie between two parallel lines, then they have the same areas.

Here, we can see that $\text{ }\!\!\Delta\!\!\text{ ABQ}$ and $\text{ }\!\!\Delta\!\!\text{ PBQ}$ lie on the same base $\text{BQ}$ and between two parallel lines $\text{BQ}$ and $\text{AP}$.

\[\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABQ} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ PBQ} \right)\]                            …$\left( \text{1} \right)$

Similarly, $\text{ }\!\!\Delta\!\!\text{ BQC}$ and $\text{ }\!\!\Delta\!\!\text{ BQR}$ lie on the same base $\text{BQ}$ and between two parallel lines $\text{BQ}$ and $\text{CR}$.

\[\therefore \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ BQC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BQR} \right)\]                            …$\left( \text{2} \right)$

We will add equations $\left( \text{1} \right)$ and $\left( \text{2} \right)$.

\[\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ ABQ} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ BQC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ BQR} \right)\text{ + ar}\left( \text{ }\!\!\Delta\!\!\text{ PBQ} \right)\]

\[\Rightarrow \text{ar}\left( \text{ }\!\!\Delta\!\!\text{ AQC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ PBR} \right)\]

Therefore, we have proved that \[\text{ar}\left( \text{ }\!\!\Delta\!\!\text{ AQC} \right)\text{ = ar}\left( \text{ }\!\!\Delta\!\!\text{ PBR} \right)\].

15. Diagonals $\text{AC}$ and $\text{BD}$ of a quadrilateral $\text{ABCD}$ intersect at $\text{O}$ in such a way that $\text{ar}\left( \text{AOD} \right)\text{ = ar}\left( \text{BOC} \right)$. Prove that $\text{ABCD}$ is a trapezium.

Ans: It is given that $\text{ar}\left( \text{AOD} \right)\text{ = ar}\left( \text{BOC} \right)$.

We add $\text{ar}\left( \text{AOB} \right)$ on both sides.

$\Rightarrow \text{ar}\left( \text{AOD} \right)\text{ + ar}\left( \text{AOB} \right)\text{ = ar}\left( \text{BOC} \right)\text{ + ar}\left( \text{AOB} \right)$

$\Rightarrow \text{ar}\left( \text{ABD} \right)\text{ = ar}\left( \text{ABC} \right)$

We know that if two triangles have the same base and they lie between two parallel lines, then they have the same areas.

So, these two triangles lie between two parallel lines $\text{AB}$ and $\text{DC}$.

Therefore, we can say that $\text{ABCD}$ is a trapezium.

16. In the given figure, $\text{ar}\left( \text{DRC} \right)\text{ = ar}\left( \text{DPC} \right)$ and $\text{ar}\left( \text{BDP} \right)\text{ = ar}\left( \text{ARC} \right)$. Show that both the quadrilaterals $\text{ABCD}$ and $\text{DCPR}$ are trapeziums.

Ans: It is given that $\text{ar}\left( \text{DRC} \right)\text{ = ar}\left( \text{DPC} \right)$.

We know that if two triangles have the same base and they have the same areas, then they must lie between two parallel lines.

Here, we can see that $\text{ }\!\!\Delta\!\!\text{ DRC}$ and $\text{ }\!\!\Delta\!\!\text{ DPC}$ lie on the same base $\text{DC}$. So, they must lie between two parallel lines.

$\therefore \text{DC}\parallel \text{RP}$

Hence, we can say that $\text{DCPR}$ is a trapezium.

It is also given that $\text{ar}\left( \text{ARC} \right)\text{ = ar}\left( \text{BDP} \right)$

So, we will subtract this equation from the previous equation$\text{ar}\left( \text{DRC} \right)\text{ = ar}\left( \text{DPC} \right)$

$\Rightarrow \text{ar}\left( \text{ARC} \right)\text{ - ar}\left( \text{DRC} \right)\text{ = ar}\left( \text{BDP} \right)\text{ - ar}\left( \text{DPC} \right)$

$\Rightarrow \text{ar}\left( \text{ADC} \right)\text{ = ar}\left( \text{BDC} \right)$

Here, we can see that $\text{ }\!\!\Delta\!\!\text{ ADC}$ and $\text{ }\!\!\Delta\!\!\text{ BDC}$ lie on the same base $\text{DC}$. So, they must lie between two parallel lines.

$\therefore \text{AB}\parallel \text{DC}$

Hence, we can say that $\text{ABCD}$ is a trapezium.

Therefore, we have shown that $\text{DCPR}$ and $\text{ABCD}$ are trapeziums.

Weightage of NCERT Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles

Chapter 9 of Class 9 Maths NCERT is a Geometry chapter that introduces students to concepts like how to find the areas of various figures under different conditions. The Areas of Parallelograms and Triangles chapter carries a total of 14 marks in the exam. While from the total unit of Menstruation, one can expect 2 multiple choice questions of 1 mark each, 2 short questions of 6 marks each and 1 long question of 6 marks.

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.3

Opting for the NCERT solutions for Ex 9.3 Class 9 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 9.3 Class 9 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 9 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 9 Maths Chapter 9 Exercise 9.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 9 Maths Chapter 9 Exercise 9.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

Do not delay any more. Download the NCERT solutions for Class 9 Maths Chapter 9 Exercise 9.3 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

NCERT Solutions for Class 9 Maths

• Chapter 1 - Number Systems

• Chapter 2 - Polynomials

• Chapter 3 - Coordinate Geometry

• Chapter 4 - Linear Equations in Two Variables

• Chapter 5 - Introduction to Euclids Geometry

• Chapter 6 - Lines and Angles

• Chapter 7 - Triangles

• Chapter 8 - Quadrilaterals

• Chapter 9 - Areas of Parallelograms and Triangles

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Heron's Formula

• Chapter 13 - Surface Areas and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

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