*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

And

*Q.17.**From the following figures, find out which figures lie between the same parallels and same base. If the case is found, then write the common base and two parallels.*

**Solution:**

(I)The figures (quadrilateral APCD and quadrilateral ABCD), and ( quadrilateral PBCD and quadrilateral ABCD), lie between the same parallels DC and AB and lie on the same base DC.

(II)The figures (

(III)Quads APCD and ABQD lie on the same base AD, and between the same parallels AD and BQ.