**Areas of Parallelograms and Triangles**

**Question 1.Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.**

**Solution.**

We have a parallelogram ABCD and rectangle ABEF such that

ar (parallelogram ABCD) = ar (rectangle ABEF)

**Question 2.In figure, D and E are two points on BC such that BD = DE = EC. Show thatar (Δ ABD) = ar (Δ ADE) = ar (Δ AEC).Can you now answer the question that you have left in the ‘Introduction’ of this chapter, whether the field of Budhia has been actually ‘ divided into three parts of equal area?**

[**Remark**: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide Δ ABC into n triangles of equal areas.]

**Solution.**Given : ABC is a triangle, D and E are two points on BC, such that BD = DE – EC.

To prove : ar (Δ ABD) = ar (Δ ADE) = ar (Δ AEC)

Proof: Let AO be the perpendicular to BC.

We know that,

**Question 3.In the figure, ABCD, DCFE and ABFE are parallelograms.Show that ar (Δ ADE) = ar (Δ BCF).**

**Solution.**

Given : ABCD, DCFE and ABFE are parallelograms.

To prove : ar (Δ ADE) = ar (Δ BCE)

**Question 4.In the figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersects DC at P, then show that ar (Δ BPO = ar (Δ DPQ.) [Hint: Join AC]**

**Solution.**

**Question 5.In the figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, then show that(i) ar (Δ BDE) = ar (Δ ABO)(ii) ar (Δ BDE) = ar (Δ BAE)(iii) ar (Δ ABC) = 2 ar (Δ BEO(iv) ar (Δ BFE) = ar (Δ AFD)(v) ar (Δ BFE) = 2 ar (Δ FED)(vi) ar (Δ FED) = ar (Δ AFC).**

[**Hint**: Join EC and AD. Show that BE || AC and DE || AB, etc.]

**Solution.**

Join AD and EC

Let x be the side of equilateral Δ ABC.

**Question 6.Diagonals AC and BD of a quadrilateral ABCD intersect each other at P.Show that ar (ΔAPB) x ar (Δ CPD)= ar (Δ APD) x ar (Δ BPC). [Hint: From A and C, draw perpendiculars to BD].**

**Solution.**

**Question 7.P and Q are respectively the mid-points of sides AB and BC of a Δ ABC and R is the mid-point of AP, show that(i) ar (Δ PRQ) = ar (Δ ARC)(ii) ar (Δ RQC = ar (Δ ABC)(iii) ar (Δ PBQ) = ar (Δ ARC).**

**Solution.**

**Question 8.In the figure, ABC is a right angled triangle, right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB, respectively, Line segment AX ⊥DE meets BC at Y. Show that(i) Δ MBC ≅ Δ ABD(ii) ar (BYXD) = 2 ar (Δ MBC)(iii) ar (BYXD) = ar (Δ BMN)(iv) ΔFCB ≅ AACE(v) ar (CYXE) = 2 ar (Δ FCB)(vi) ar (CYXE) = ar (ACFG)(vii) ar (BCED) = ar (ABMN) + ar (ACFG).**

[**Note**: Result (vii) is the famous Theorem of Pythagoras. You shall term a similar proof by this Theuonern is class X.]

**Solution.**Given: ABC is a right angled triangle in which ∠A = 90°. BCED, ACFG and ABMN are squares. Line segment AX ⊥ DE meets BC at Y