## RS Aggarwal Class 9 Ex 10A Chapter 10

## Exercise 10.1: Area

**Q.1: In the adjoining figure, ABCD is a quadrilateral. A line through D, parallel to AC, meets BC produced in P.**

**Prove that:Â ar.( â–³ABP ) = ar. (quad. ABCD)**

**Sol:**

**Given:** ABCD is a quadrilateral in which through D, a line is drawn parallel to AC which meets BC produced in P.

**To prove:** ar(Î”ABP)=ar(quad.ABCD)

**Proof:** Î”ACPandÎ”ACD have same base AC and lie between parallel lines AC and DP.

Therefore, ar(Î”ACP)=ar(Î”ACD)

Adding ar(Î”ACP) on both sides, we get;

ar(Î”ACP)+ar(Î”ABC)=ar(Î”ACD)+ar(Î”ABC)

Therefore, ar(Î”ABP)=ar(quad.ABCD)

**Q.2: In the adjoining figure, AD and BC are 2 diagonals of the parallelogram ABCD. The point at which the diagonals intersect is O. Prove that OA = OD.**

**Sol:**

**Given:** Two triangles, i.e. Î”ABCandÎ”DBC which have same base BC and points A and D lie on opposite sides of BC and ar(Î”ABC)=ar(Î”DBC)

**To prove:** Â OA = OD

**Construction:** Draw APâŠ¥BCandDQâŠ¥BC

**Proof:** We have:

ar(Î”ABC)=12Ã—BCÃ—AP and

ar(Î”BCD)=12Ã—BCÃ—DQ

So, 1/2 x BC x AP = 1/2 x BC x DQ [From (1)]

Therefore, AP = DQ . . . . . . . . . (2)

Now, in Î”AOPandÎ”QOD, we have:

âˆ APO=âˆ DQO=90âˆ˜

And, âˆ APO=âˆ DOQ=90âˆ˜Â [vertically opp. Angles]

AP = DQ [From (2)]

Thus, by Angle-Angle-side criterion of congruence, we have:

Therefore, Î”AOPâ‰…Î”QOD [AAS]

The corresponding parts of the congruent triangles are equal.

Therefore, OA = OD [C.P.C.T]

**Q.3: In the adjoining figure, AD is one of the medians of a â–³ABC and P is appointed on AD.**

**Prove that:**

**(i) ar.(** **â–³BDP) = ar.(** **â–³CDP)**

**(ii)ar.(** **â–³ABP = ar.( â–³ACP)**

**Sol:**

**Given:** A Î”ABC in which AD is the median and P is a point on AD.

**To prove:**

**(i)** ar(Î”BDP)=ar(Î”CDP)

**(ii)** ar(Î”ABP)=ar(Î”APC)

**Proof:**

**(i)** In Î”BPC, PD is the median. Since median of a triangle divides the triangle into two triangles of equal areas.

So, ar(Î”BDP)=ar(Î”CDP) . . . . . . . . (1)

**(ii)** In Î”ABC, AD is the median

So, ar(Î”ABD)=ar(Î”ADC)

But, ar(Î”BDP)=ar(Î”CDP) [From (1)]

Subtracting ar(Î”BPD) from both the sides of the equation, we have

Therefore, ar(Î”ABD)âˆ’ar(Î”BPD)=ar(Î”ADC)âˆ’ar(Î”BPD)

=ar(Î”BPD)âˆ’ar(Î”CDP)Â [From (1)]

Therefore, ar(Î”ABP)=ar(Î”ACP)

**Q.4: In the adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O.**

**If BO = OD, prove that**

**ar.(** **â–³ABC)=ar. (â–³ADC).**

**Sol:**

**Given:** A quadrilateral ABCD in which diagonals AC and BD intersect at O and BO = OD

**To Prove:** ar(Î”ABC)=ar(Î”ADC)

**Proof:** Since OB = OD [given]

So, AO is the median of Î”ABD

Therefore, ar(Î”AOD)=ar(Î”AOB) . . . . . . . . (i)

As OC is the median of Î”CBD

Therefore, ar(Î”DOC)=ar(Î”BOC) . . . . . . . . . (ii)

Adding both sides of (i) and (ii), we get:

ar(Î”AOD)+ar(Î”DOC)=ar(Î”AOB)+ar(Î”BOC)

Therefore, ar(Î”ADC)=ar(Î”ABC)

**Q.5: ABC is a triangle in which D is the midpoint of BC and E is the midpoint of AD.**

**Prove that:**

**ar.( â–³BED)= 14ar(â–³ABC) **

**Sol:**

**Given:** A Î”ABC in which AD is a median and E is the mid-point of AD

**To prove:** ar(Î”BED)=14ar(Î”ABC)

**Proof:** Since, ar(Î”ABD)=ar(Î”ACD) Â Â [Since,ADisamedian]

i.e. ar(Î”ABD)=12ar(Î”ABC) . . . . . . . (1) [Since,ar(Î”ABC)=ar(Î”ABD)+ar(Î”ADC)]

Now, as BE is the median of Î”ABD

ar(Î”ABE)=ar(Î”BED) . . . . . . . . . (2)

Since, ar(Î”ABD)=ar(Î”ABE)+ar(Î”BED) . . . . . . . (3)

Therefore,Â ar(Î”BED)=ar(Î”ABE)â€¦â€¦..[From(2)]

= 12ar(Î”ABD) Â [From (2) and (3)]

= 12[12ar(Î”ABC)] Â [From (1)]

= 14ar(Î”ABC)

**Q.6: The vertex A of â–³ABC is joined to a point D on the side BC. The midpoint AD is E.**

**Prove that:**

**ar(â–³BEC) = 12ar(â–³ABC)**

**Sol: **

**Given:** A Î”ABC in which E is the mid-point of the line segment AD where D is a point on BC.

**To Prove:** ar(Î”BEC)=12ar(Î”ABC)

**Proof:** Since, BE is a median of Î”ABD

So, ar(Î”BDE)=ar(Î”ABE)

i.e. ar(Î”BDE)=12ar(Î”ABD) . . . . . . . (1)

As, CE is the median of Î”ADC

So, ar(Î”CDE)=12ar(Î”ACD) . . . . . . . (2)

**Adding (i) and (ii), we get:**

ar(Î”BDE)+ar(Î”CDE)=12ar(Î”ABD)+12ar(Î”ACD)

ar(Î”BDE)=12[ar(Î”ABD)+ar(Î”ACD)]

= 12ar(Î”ABC)

**Q.7: D is the midpoint of side BC of â–³ABC and E is the midpoint of BD. if O is the midpoint of AE.**

**Prove that:**

**ar.(** **â–³BOE ) = 18ar(â–³ABC)**

**Sol:**

**Given:** A Î”ABC in which AD is the medium and E is the mid-point of BD. O is the mid-point of AE.

**To prove:** ar(Î”BOE)=18ar(Î”ABC)

**Proof:** Since O is the mid-point of AE.

So, BO is the medium of Î”BAE

Therefore, ar(Î”BOE)=12ar(Î”ABE) . . . . (1)

Now, E is the mid-point of BD

So AE divides Î”ABD into two triangles of equal area.

Therefore, ar(Î”ABE)=12ar(Î”ABD) . . . .(2)

As D is the mid-point of BC

So, ar(Î”ABD)=12ar(Î”ABC) . . . . . (3)

Therefore, ar(Î”BOE)=12ar(Î”ABE) [From (1)]

= 12[12ar(Î”ABCD)] [From (2)]

= 14ar(Î”ABD)

= 14Ã—12ar(Î”ABC) [From (3)]

= 18ar(Î”ABC)

**Q.8: In the adjoining figure, ABCD parallelogram and O is any point on the diagonal AC.**

**Show that: ar.( â–³AOB )=ar.( â–³AOD)**

**Sol:**

**Given:** A parallelogram ABCD in which O is any point on the diagonal AC

**To Prove:** ar(Î”AOB)=ar(Î”AOD).

**Construction:** Join BD which intersects AC at P.

**Proof:** As diagonals of a parallelogram bisect each other therefore, OP is the median of Î”ODB

Therefore, ar(Î”ODP)=ar(Î”OBP)

Also, AP is the median of Î”ABD

Therefore, ar(Î”ODP)=ar(Î”OBP)

Adding both sides, we get:

ar(Î”ODP)+ar(Î”ADP)=ar(Î”OBP)+ar(Î”ABP)

Therefore, ar(Î”AOB)=ar(Î”AOD).

**Q.9: P, Q, R, S are respectively the midpoints of the sides AB, BC, CD and DA of parallelogram ABCD. Show that PQRS is parallelogram and also show that**

**ar.(parallelogram PQRS)= 12 * ar.(parallelogram ABCD).**

**Sol:**

**Given:** ABCD is a parallelogram and P,Q,R and S are the midpoints of AB,BC,CD,DA respectively.

**To prove: **PQRS is a parallelogram and parallelogram PQRS= 12 Parallelogram ABCD

**Construction: **Join AC,BD and SQ.

**Proof: **As S and R are the midpoints of AD and CD, so in â–³ADC

SRâˆ¥AC ( By midpoint theorem)

Also, as P and Q are the midpoints of AB and BC. So, in â–³ABC,

PQâˆ¥AC

Therefore, PQâˆ¥ACâˆ¥SR

So, PQâˆ¥SR

Similarly, we can prove SPâˆ¥RQ.

Thus PQRS is a parallelogram as its opposite sides are parallel.

Since the diagonals of a parallelogram bisect each other.

So in â–³ABD,

O is the midpoint of AC and S is the midpoint of AD.

Therefore, OSâˆ¥AB (by midpoint theorem )

Similarly in â–³ABC we can prove that,

OQâˆ¥AB and SQâˆ¥AB

Thus, ABQS is a parallelogram.

Now, â–³SPQ=12(parallelogramABQS) . . . . . . . (i)

Similarly, we can prove that:

â–³SPQ=12(parallelogramSQCD) . . . . . . . (ii)

Adding (i) and (ii) we get,

â–³SPQ+â–³SPQ=12(parallelogramABQS)+12(parallelogramSQCD)

Therefore, (parallelogramPQRS)=12(parallelogramABCD)

**Q.10: The given figure shows a pentagon ABCDE. EG, drawn parallel to DA, meets BA produced at G, and CF, drawn parallel to DB, meets AB produced at F. Show that ar.(pentagon ABCDE) = ar.(** **â–³DGF).**

**Sol:**

**Given: **ABCDE is a pentagon, EDâˆ¥DA and CFâˆ¥DB.

**To prove:** ar(pentagon ABCDE) = ar(â–³DGF)

**Proof:**

ar(pentagon ABCDE) = ar(â–³DBC) +ar(â–³ADE)+ar(â–³ABD) . . . . . . . . (i)

Also, ar(â–³DGF)= ar(â–³DBF)+ar(â–³ADG)+ â–³ABD . . . . . (ii)

Now, â–³DBC and â–³DBF lie on the same base and between the same parallel lines.

Therefore, ar(â–³DBC)=ar(â–³DBF) . . . . . . (iii)

Similarly, â–³ADE and â–³ADG lie on the same base and between the same parallel lines.

Therefore, ar(â–³ADE)=ar(â–³ADG) . . . . . . . (iv)

**From (iii) and (iv) we have:**

ar(â–³DBC)+ar(â–³ADE)=ar(â–³DBF)+ar(â–³ADG)

Adding ar(â–³ABD) on both sides

ar(â–³DBC)+ar(â–³ADE)+ar(â–³ABD)=ar(â–³DBF)+ar(â–³ADG)+ar(â–³ABD)

By substituting the values from (i) and (ii) we get:

ar(pentagon ABCDE) = ar(â–³DGF)

**Q.11: Prove that a median divides a triangle into two triangles of equal area.**

**Sol:**

Let, AD is a median of â–³ABC and D is the midpoint of BC. AD divides â–³ABC in two triangles :â–³ABDandâ–³ADC

**To prove:** ar(â–³ABD)=ar(â–³ADC)

**Construction:** Draw ALâŠ¥BC

**Proof:**

Since, D is the midpoint of BC, we have

BD=DC

Multiplying with 12AL on both sides, we have

12Ã—BDÃ—AL=12Ã—DCÃ—AL

Therefore, ar(â–³ABD)=ar(â–³ADC)

**Q.12: Show that a diagonal divides a parallelogram into two triangles of equal area.**

**Sol:**

Let, ABCD be a parallelogram and BD be its diagonal.

**To prove:** ar(â–³ABD)=ar(â–³CDB)

Proof:

In â–³ABDandâ–³CDB we have,

AB = CD (opposite sides of a parallelogram)

AD = CD (opposite sides of a parallelogram)

BD = DB (common side)

Therefore, â–³ABDâ‰…â–³CDBÂ (SSS criteria )

ar(â–³ABD)=ar(â–³CDB)

**Q.13: The base BC ofÂ â–³ABC is divided at D such that BD = 12DC . Prove that**

**ar(Â â–³ABD) = 13Ã—ar(â–³ABC)**

**Sol:**

**Given:** D is a point on BC of â–³ABC, such that BD=12DC

**To prove:** ar(â–³ABD)=13ar(â–³ABC)

**Construction:** Draw ALâŠ¥BC

**Proof:**

In â–³ABC, we have :

BC = BD + DC

â‡’BD+2BD=3Ã—BD

Now we have:

ar(â–³ABD)=12Ã—BDÃ—AL

ar(â–³ABC)=12Ã—BCÃ—AL

â‡’ar(â–³ABC)=12Ã—3BDÃ—AL=3(12Ã—BDÃ—AL)

â‡’ar(â–³ABC)=3Ã—ar(â–³ABD)

Therefore, ar(â–³ABD)=13ar(â–³ABC)

**Q.14: In the adjoining figure, the point D divides the side BC ofÂ ****â–³ABC in the ratio m : n.**

**Prove that:**

**ar(â–³ABD):ar(â–³ADC) = m:n.**

**Sol:**

**Given:** D is a point on BC of â–³ABC such that BD:DC=m:n

**To prove:** ar(â–³ABD):ar(â–³ADC)=m:n

**Construction :** Draw ALâŠ¥BC

**Proof:**

ar(â–³ABD)=12Ã—BDÃ—AL . . . . . . (i)

ar(â–³ABD)=12Ã—DCÃ—AL . . . . . . (ii)

**Dividing (i) and (ii) we have:**

ar(â–³ABD)ar(â–³ABC)=12Ã—BDÃ—AL12Ã—DCÃ—AL

=BDDC = mn

Therefore, ar(â–³ABD):ar(â–³ADC)=m:n