NCERT solutions class 9 maths chapter 9 areas of parallelograms and triangle is given so that students of the 9th standard can prepare mathematics in a better way. The NCERT solutions for class 9 maths chapter 9 areas of parallelograms and triangle is one of the most crucial section in class 9 examination. Students must practice the NCERT solutions class 9 maths chapter 9 regularly to understand the concepts of areas of parallelograms and triangle in details. Check the NCERT solutions for class 9 maths chapter 9 pdf given below.

### NCERT Solutions For Class 9 Maths Chapter 9 Exercises

- NCERT Solutions For Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.1
- NCERT Solutions For Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.2

**Exercise-9.1**

*Q.1.In the given figure, PQRS is a parallelogram, PE⊥SRandRF⊥PS..If PQ = 16 cm, PE = 8 cm & RF = 10 cm. Calculate AD.*

**Solution:**

Area of the parallelogram

Now, area of parallelogram PQRS=

From equation (1) and (2), we get

*Q.2.If E, F, G and H are the mid-points respectively, of the sides of a parallelogram ABCD show that Area(EFGH)=12Area(ABCD)*

**Solution:**

Lets join HF.

In the parallelogram, i.e ABCD, AD = BC and AD || BC (because in a parallelogram the opposite sides are equal and parallel)

Since

Similarly, we can prove that,

Add Equation (1) and Equation (2), we obtain

*Q.3. DC and AD are two sides on which P and Q are two points lying respectively of a parallelogram ABCD. Prove that area(APB)=area(BQC)*

**Solution:**

It is observed that

Similarly,we can say that ∆APB and parallelogram ABCD lie between the same lines AB and DC that are parallel and on the same base AB .

Equating both the equations, i.e

Hence, proved.

*Q.4. In the given figure, intheinteriorofaparallelogramABCD,thereexistapointP.Show that*

*(i) area(APB)+area(PCD)=12area(ABCD)*

*(ii) area(APD)+area(PBC)=area(APB)+area(PCD)*

*[Hint: Draw a line i.e. parallel to AB, through P]*

Solution:

- A line segment EF is drawn, parallel to line segment AB and passing through point P.

In the parallelogram ABCD,

By equating , Equation (1) and Equation(2) , we get,

Therefore, quad ABFE is a parallelogram.

Similarly

Add equation(3) and equation(4), we get,

(ii)

A line segment MN is drawn

In the parallelogram ABCD,

By equating , Equation (6) and Equation(7) , we get,

Therefore

It can be said that ∆APD and parallelogram AMND are between

Similarly

Add equation(8) and equation(9), we get,

Now compare equation(5) with equation(10), we get

*Q.5. In the figure given below X is a point on the side BR and ABRS and PQRS are parallelograms. Prove that*

*(i) area (PQRS) = area (ABRS)*

*(ii) area(ΔPXS)=12area(PQRS)*

**Solution:**

(i)It can be said that

(ii)Consider

As these lie on the same base and are between the same parallel lines AS and BR

By equating , equation (1) and equation(2)

*Q.6. A farmer had a field and that was in a parallelogram shape PQRS .He took a point A on RS and he joined it to points P and Q. In how many parts the field is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately.How should he do it ?*

**Solution:**

From the figure

The parts which are triangular in shape are –

We know that if a parallelogram and a triangle are between the same parallels and on the same base

By equation , equation (i) and (ii)

Clearly

**Exercise – 9.2**

*Q.1.In the given figure , E is any point on median AD of a ΔABC. Prove that*

**Solution:**

**Given:** AD is the median of

**To Prove:**

**Proof:** In

AD is a median.

In

ED is a median.

Subtracting equation (ii) from equation (i)

*Q.2.In a ΔABC *

, E is the mid-point of median AD. Prove that**Solution:**

**Given:** E is the mid-point of median AD in

**To Prove:**

**Proof: **In

AD is a median.

In

BE is a median.

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

**Solution:**

We know that diagonals of parallelogram bisect each other.

BO is the median in ∆ABC. Therefore, it will divide it into two triangles of equal areas.

In

Similarly

By Equating equation(i)

Therefore

*Q.4.**In figure , ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O , show that Area(ΔABC)=Area(ΔABD).*

**Solution:**

**Given:**

**To Prove:**

**Proof:** Line-segment CD is bisected by AB at O.

BO is the median of

Adding (i) and (ii), we get

*Q.5.**D , E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC.*

Prove that

(i) BDEF is a parallelogramProve that

(i) BDEF is a parallelogram

*(ii) *

*(iii) *

**Solution:**

**Given:** D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC.

**To Prove:** (i) BDEF is a parallelogram

(ii)

(iii)

**Proof:** In ΔABC

F is the mid-point of side AB and E is the mid-point of side AC.

^{rd} side.

Similarly

From (1) and (2)

BDEF is a parallelogram.

(ii)As in (i), we can prove that

AFCE and AFDE are parallelograms.

FD is diagonal of the parallelogram BDEF.

Similarly

From equation(3),(4) and (5), we have

=

(iii)

=

*Q.6. **In the given figure , diagonals AC and BD of a quadrilateral ABCD intersect at O such that OB = OD. If AB = CD,then show that:*

*(i) *

*(ii) *

*(iii) *

*[Hint: From D and B ,draw perpendiculars to AC]*

**Solution: **

**Given:** Diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD.

**To Prove:** If AB = CD, then

(i)

(ii)

(iii)

**Construction:** Draw

**Proof:** In

OD = OB (Given)

By AAS congruence rule

We know that congruent triangles have equal areas.

In

Add equation(2) and (3)

(ii)We got

Now, add

(iii)We got

If two triangles have the same base and equal areas

In quadrilateral ABCD

*Q.7. **D and E are points on sides AB and AC respectively of ∆ABC such that Area(ΔDBC)=Area(ΔEBC) Prove that DE || BC.*

**Solution:**

**Given :** D and E are points on sides AB and AC respectively of ∆ABC such that

**To Prove:** DE || BC

**Proof:** As

*Q.8. **XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and E respectively , show that Area(ΔABE)=Area(ΔACF)*

**Solution:**

**Given:** XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and E respectively.

**To Prove:**

**Proof:** XY||BC(given)

And CF||BX (

BC= XF

BC=XY+YF …..(1)

Again

XY||BC (

And BE||CY

From (1) and (2)

XY+YF=XY+XE

Add the corresponding sides of (4) and (5), we get

*Q.9. **The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see the following figure). Show that Area(ParallelogramABCD)=Area(ParallelogramPBQR). *

*[Hint: Join AC and PQ. Now compare Area(ΔACQ)andArea(ΔAPQ)]*

**Solution:**

**Given:** The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed.

**To Prove:** Area(Parallelogram ABCD) = Area(Parallelogram PBQR).

**Construction:** Join AC and PQ.

**Proof:** AC is a diagonal of parallelogram ABCD

PQ is a diagonal of parallelogram BQRP

Now

*Q.10. **Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that Area(ΔAOD)=Area(ΔBOC).*

**Solution:**

**Given:** Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O.

**To Prove: **

**Proof:**

Now

*Q.11. **In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that *

*(i) Area(ΔACB)=Area(ΔACF)*

* (ii) Area(AEDF)=Area(ABCDE)*

**Solution:**

**Given:** ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F.

**To Prove:** (i)

(ii)

**Proof:**

** **(i)

(ii)

Add

*Q.12. **A villager Ram has a plot of land in 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. Ram agrees to the above proposal with the condition that he should be given an 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.*

**Solution:**

Let ABCD be the plot of land in the shape of a quadrilateral.

Let the portion ADE be taken over by the Gram Panchayat of the village from one corner D to construct a Health Centre.

Join AC. Draw a line through D parallel to AC to meet BC produced in P.

Then Ram must given the land ECP adjoining his plot so as to form a triangular plot ABP as then.

Proof:

Substract

Now

*Q.13. **ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that Area(ΔADX)=Area(ΔACY). [Hint: Join CX.]*

**Solution:**

**Given: **ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y.

**To Prove:**

**Construction: **Join CX

**Proof:**

From equation(1) and (2)

*Q.14.**In the given figure, AP || BQ || CR. Prove that Area(ΔAQC)=Area(ΔPBR).*

**Solution:**

**To Prove:**

**Proof:**

Now

*Q.15. **Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that Area(ΔAOD)=Area(ΔBOC).Prove that ABCD is a trapezium.*

**Solution:**

**Given: **Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that

**To Prove: **ABCD is a trapezium.

**Proof:**

Now

But

and

*Q.16. **In the given figure , Area(ΔDRC)=Area(ΔDPC)andArea(ΔBDP)=Area(ΔARC).Show that both the quadrilaterals ABCD and DCPR are trapeziums.*

**Solution:**

**Given:**

**To Prove: **both the quadrilaterals ABCD and DCPR are trapeziums.

**Proof: **

But

And

Again

But