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**Question 1. Parallelogram ABCD and rectangle ABEF are on the same base AB 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)

âˆµ AB = CD [âˆµ Opposite sides of a parallelogram are equal]

and AB = EF [âˆµ Opposite sides of a rectangle are equal]

â‡’ CD = EF â‡’ AB + CD = AB + EF â€¦(1)

âˆµ BE < BC

and AF < AD [âˆµ In a right triangle, hypotenuse is the longest side.]

âˆ´ BC > BE and AD > AF â‡’ (BC + AD) > (BE + AF) â€¦(2)

Adding (1) and (2), we have

(AB + CD) + (BC + AD) > (AB + EF) + (BE + AF)

â‡’ (AB + BC + CD + DA) > (AB + BE + EF + FA)

â‡’ [Perimeter of parallelogram ABCD] > [Perimeter of rectangle ABEF]

**Question 2. In the figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC) Can you now, answer the question that you have left in the â€˜Introductionâ€™ 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 DABC into n triangles of equal areas. Solution: **Let us draw AF, perpendicular to BC such that AF is the height of Î”ABD, Î”ADE and Î”AEC.

âˆ´

ar (Î”ABD) = (1/2) x BD x AF

[âˆµ Area of a triangle = (1/2) x base x height]

Similarly, ar (Î”ADE) = (1/2) x DE x AF

ar (Î”AEC) = (1/2) x EC x AF

Since, BD = DE = EC

âˆ´ ((1/2) x BD x AF) = ((1/2) x DE x AF) = ((1/2) x EC x AF)

â‡’ ar (Î”ABD) = ar (Î”ADE) = ar (Î”AEC) ,

Yes, the altitudes (heights) of all the triangles are same. Therefore, Budhia can use the above result in dividing her land in three equal parts.

**Question 3. In the figure, ABCD, DCFE and ABFE are parallelograms. Show that ar(ADE) = ar (BCF). Solution:** We have parallelograms ABCD, DCFE and ABFE. âˆµ ABCD is a parallelogram [given]

âˆ´ Its opposite sides are parallel and equal. â‡’ AD = BC â€¦(1)

and AB || DC â€¦(2)

Since, DCFE is a parallelogram [Given]

âˆ´ DC || EF â€¦(3)

From (2) and (3), we have AB || EF

Now, DADE and DBCF are on equal bases,

i.e. AD = BC

[from (1)] and between the same parallels,

i.e. AB || EF

âˆ´

ar (Î”ADE) = ar (Î”BCF)

**Question 4. In the figure ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ). Hint: Join AC. Solution: **We have a parallelogram ABCD and AD = CQ.

Let us join AC. We know that triangles on the same base and between the same parallels are equal in area. Since Î”QAC and Î”QDC are on the same base QC and between the same parallels AD and BQ.

âˆ´ ar (Î”QAC) = ar (Î”QDC) Subtracting ar (Î”QPC) from both sides, we have

ar (Î”QAC) âˆ ar (Î”QPC) = ar (Î”QDC) âˆ ar (Î”QPC)

â‡’ ar (Î”PAC) = ar (Î”QDP) ...(1)

Since, Î”PAC ad Î”PBC are on the same base PC and between the same parallels AB and CD.

âˆ´ ar (Î”PAC) = ar (Î”PBC) â€¦(2)

From (1) and (2), we get ar (Î”PBC) = ar (Î”QDP)

i.e. ar (Î”PBC) = ar (Î”DPQ)

** Question 5. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) x ar (CPD) = ar (APD) x ar (BPC). Hint: From A and C, draw perpendiculars to BD. Solution: We have a quadrilateral ABCD such that its diagonals AC and BD intersect at P. Let us draw AM âŠ¥ BD and CN âŠ¥ BD.**

Now, ar (Î”APB) = (1/2) x BP x AM

ar (Î”CPD) = (1/2) x DP x CN

âˆ´ ar (Î”APB) x ar (Î”CPD) = ((1/2) x BP x AM) x ((1/2) x DP x CN)

(1/4) x BP x DP x AM x CN â€¦(1)

Similarly,

ar Î”(APD) x ar (Î”BPC) = (1/4)x BP x DP x AM x CN â€¦(2)

From (1) and (2), we get ar (Î”APB) x ar (Î”CPD) = ar (Î”APD) x ar (Î”BPC).

**Question 6. P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that**

(i) ar (âˆ† PRQ) = (1/2)ar (âˆ† ARC)

(ii) ar (âˆ† RQC) = (3/8)ar (âˆ† ABC)

(iii) ar (âˆ† PBQ) = ar (âˆ† ARC) Solution: We have a DABC such that â€˜Pâ€™ is the mid-point of AB and â€˜Qâ€™ is the mid-point of BC. Also, â€˜Râ€™ is the mid-point of AP. Let us join AQ and PC.

(i) To prove that ar (Î”PRQ) = (1/2)ar(Î”ARC)

In Î”APQ, R is the mid-point of AP. [Given]

Let us join RQ.

âˆµ RQ is a median of Î”APQ.

âˆ´ ar (Î”PRQ) = (1/2)ar (Î”APQ) â€¦(1)

âˆµ QP is a median of Î”ABQ. [âˆµ P is the mid-point of AB]

âˆ´ ar (Î”APQ) = (1/2)ar (Î”ABQ) â€¦(2)

From (1) and (2), we get ar (Î”PRQ) = (1/2)x (1/2) ar (Î”ABQ)

= (1/4) ar (Î”ABQ)

= (1/4) x (1/2) ar (Î”ABC) [ âˆµ AQ is a median of Î”ABC]

= (1/8)(ar ABC) â€¦ (3)

Now, ar (Î”ARC) = (1/2)ar (Î”APC) [âˆµ CR is a median of Î”APC]

= (1/2)x (1/2)ar (Î”ABC) [âˆµ CP is a median of Î”ABC]

= (1/4)ar (Î”ABC) ... (4)

Now from (3) and (4), we get ar (Î”PRQ) = (1/8)ar (Î”ABC)

= (1/2) x (1/4)ar ABC)

= (1/2)ar (Î”ARC)

Thus, ar (Î”PRQ) = (1/2) ar (Î”ARC)

(ii) To prove that ar (Î”RQC) = (3/8)ar (Î”ABC)

In Î”RBC, RQ is a median. âˆ´

ar (Î”RQC) = ar (Î”RBQ) = ar (Î”PRQ) + ar (Î”BPQ)

= (1/8)ar (Î”ABC) + (1/2) ar (Î”PBC) [Proved in (i)]

= (1/8)ar (Î”ABC) + (1/2) . (1/2) (Î”ABC)]

[âˆµ P is the mid-point of AB]

= (1/8) ar (Î”ABC) + (1/4)ar (Î”ABC)]

= ( (1/4) + (1/8) ) ar (Î”ABC)

= (3/8)ar (Î”ABC)

(iii) To prove that ar (Î”PBQ)

= ar (Î”ARC)

âˆµ PQ is a median of Î”ABQ.

âˆ´ ar (Î”PBQ) = (1/2)(Î”ABQ)

= (1/2) x (1/2)ar (Î”ABC) [âˆµ AQ is a median of D ABC]

= (1/4)ar (Î”ABC)

= ar (Î”ARC) [using (4)]

Thus, ar (Î”PBQ) = ar (Î”ARC)

**Question 7. In the figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX âŠ¥ DE meets BC at Y. Show that: (i) **Î”**MBC â‰Œ **Î”**ABD (ii) ar (BYXD) = 2ar (âˆ† MBC) (iii) ar (BYXD) = ar (ABMN) (iv) **Î”**FCB â‰Œ **Î”**ACE (v) ar (CYXE) = 2ar (âˆ† FCB) (vi) ar (CYXE) = ar (ACFG) (vii) ar (BCED) = ar (ABMN) + ar (ACFG)**

**Note: Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X. Solution:** We have a right DABC such that BCED, ACFG and ABMN are squares on its sides BC, CA and AB respectively.

Line segment AX âŠ¥ DE is also drawn such that it meets BC in Y.

(i) To prove that Î”MBC â‰Œ Î”ABD In DABD and DMBC, we have

[Sides of a square]

âˆ CBD = âˆ MBA [Each = 90Âº]

âˆ´ âˆ CBD + âˆ ABC = âˆ MBA + âˆ ABC

or âˆ MBC = âˆ ABD

âˆ´ Î”MBC â‰Œ Î”ABD

(ii) To prove that: ar (BYXD) = 2 ar (âˆ† MBC)

Since parallelogram BYXD and DABD are on the same base BD and between the same parallels BD and AX.

âˆ´

ar (Î”ABD) = (1/2) ar (parallelogram BYXD)

But ar (Î”ABD) = ar (Î”MBC) [âˆµ Congruent triangles have equal areas]

â‡’ ar (Î”MBC) = (1/2) ar (parallelogram BYXD)

or 2ar (Î”MBC) = ar (BYXD) Thus, ar (BYXD)

= 2ar (Î”MBC)

(iii) To prove that ar (BYXD) = ar (ABMN)

âˆµ ar (BYXD) = 2ar (Î”MBC) [from (ii)] â€¦(1)

and ar (ABMN) = 2ar (MBC) ...(2) [âˆµ ABMN and DMBC are on the same base MB and between the same parallels MB and NC]

âˆ´ From (1) and (2), we have ar (BYXD) = ar (ABMN)

(iv) Î”FCB â‰Œ Î”ACE

In Î”FCB and Î”ACE, F C = AC [Sides of a square]

CB = CE [Sides of a square]

âˆ FCA = âˆ BCE

or âˆ FCA + âˆ ACB = âˆ BCE + âˆ ACB

â‡’âˆ FCB = âˆ ACE

â‡’Î”FCB â‰Œ Î”ACE [SAS rule]

(v) To prove that ar (CYXE) = 2ar (FCB)

Since (CYXE) and (Î”ACE) are on the same base CE and between the same parallels CE and AX.

âˆ´ ar (CYXE) = 2ar (Î”ACE)

But Î”ACE â‰Œ Î”FCB

Since, congruent triangles are equal in areas.

âˆ´ ar (CYXE) = 2ar(Î”FCB)

(vi) To prove ar(CYXE) = ar (ACFG)

Since, ar (CYXE) = 2ar (Î”FCB) [from (v)] ...(3)

Also, (ACFG) and Î”FCB are on the same base FC and between the same parallels FC and BG.

â‡’ ar (ACFG) = 2ar (Î”FCB) â€¦(4)

From (3) and (4), we get ar (CYXE) = ar (ACFG) â€¦(5)

(vii ) To prove that ar (BCED) = ar (ABMN) + ar (ACFG)

We have ar (BCED) = ar (CYXE) + ar (BYXD) = ar (CYXE) + ar (ABMN) [from (iii)]

Now, ar (BCED) = ar (ABMN) + ar (CYXE).

Thus, ar (BCED) = ar (ABMN) + ar (ACFG) [from (5)

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