RS Aggarwal Solutions Class 9 Ex 10A

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.

 

C:\Users\user\Desktop\a.PNG

 

Prove that: ar.( ABP ) = ar. (quad. ABCD)

Sol:

 

C:\Users\user\Desktop\a.PNG

 

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(ΔABP)=ar(quad.ABCD)

Proof: ΔACPandΔACD have same base AC and lie between parallel lines AC and DP.

Therefore, ar(ΔACP)=ar(ΔACD)

Adding ar(ΔACP) on both sides, we get;

ar(ΔACP)+ar(ΔABC)=ar(ΔACD)+ar(ΔABC)

Therefore, ar(ΔABP)=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.

 

C:\Users\user\Desktop\B.PNG

 

Sol:

Given: Two triangles, i.e. ΔABCandΔDBC which have same base BC and points A and D lie on opposite sides of BC and ar(ΔABC)=ar(ΔDBC)

 

C:\Users\user\Desktop\B.PNG

 

To prove:  OA = OD

Construction: Draw APBCandDQBC

Proof: We have:

ar(ΔABC)=12×BC×AP and

ar(ΔBCD)=12×BC×DQ

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

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

Now, in ΔAOPandΔQOD, we have:

APO=DQO=90

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

AP = DQ [From (2)]

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

Therefore, ΔAOPΔQOD [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 ABC and P is appointed on AD.

Prove that:

(i) ar.( BDP) = ar.( CDP)

(ii)ar.( ABP = ar.( ACP)

Sol:

Given: A ΔABC in which AD is the median and P is a point on AD.

 

C:\Users\user\Desktop\c.PNG

 

To prove:

(i) ar(ΔBDP)=ar(ΔCDP)

(ii) ar(ΔABP)=ar(ΔAPC)

Proof:

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

So, ar(ΔBDP)=ar(ΔCDP) . . . . . . . . (1)

(ii) In ΔABC, AD is the median

So, ar(ΔABD)=ar(ΔADC)

But, ar(ΔBDP)=ar(ΔCDP) [From (1)]

Subtracting ar(ΔBPD) from both the sides of the equation, we have

Therefore, ar(ΔABD)ar(ΔBPD)=ar(ΔADC)ar(ΔBPD)

=ar(ΔBPD)ar(ΔCDP) [From (1)]

Therefore, ar(ΔABP)=ar(ΔACP)

 

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

 

C:\Users\user\Desktop\c1.PNG

 

If BO = OD, prove that

ar.( ABC)=ar. (ADC).

Sol:

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

 

C:\Users\user\Desktop\c1.PNG

 

To Prove: ar(ΔABC)=ar(ΔADC)

Proof: Since OB = OD [given]

So, AO is the median of ΔABD

Therefore, ar(ΔAOD)=ar(ΔAOB) . . . . . . . . (i)

As OC is the median of ΔCBD

Therefore, ar(ΔDOC)=ar(ΔBOC) . . . . . . . . . (ii)

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

ar(ΔAOD)+ar(ΔDOC)=ar(ΔAOB)+ar(ΔBOC)

Therefore, ar(ΔADC)=ar(ΔABC)

 

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

 

C:\Users\user\Desktop\c2.PNG

 

Prove that:

ar.( BED)= 14ar(ABC)

Sol:

Given: A ΔABC in which AD is a median and E is the mid-point of AD

C:\Users\user\Desktop\c2.PNG

To prove: ar(ΔBED)=14ar(ΔABC)

Proof: Since, ar(ΔABD)=ar(ΔACD)   [Since,ADisamedian]

i.e. ar(ΔABD)=12ar(ΔABC) . . . . . . . (1) [Since,ar(ΔABC)=ar(ΔABD)+ar(ΔADC)]

Now, as BE is the median of ΔABD

ar(ΔABE)=ar(ΔBED) . . . . . . . . . (2)

Since, ar(ΔABD)=ar(ΔABE)+ar(ΔBED) . . . . . . . (3)

Therefore, ar(ΔBED)=ar(ΔABE)……..[From(2)]

= 12ar(ΔABD)  [From (2) and (3)]

= 12[12ar(ΔABC)]  [From (1)]

= 14ar(ΔABC)

 

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

 

C:\Users\user\Desktop\c4.PNG

 

Prove that:

ar(BEC) = 12ar(ABC)

Sol:

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

To Prove: ar(ΔBEC)=12ar(ΔABC)

Proof: Since, BE is a median of ΔABD

So, ar(ΔBDE)=ar(ΔABE)

i.e. ar(ΔBDE)=12ar(ΔABD) . . . . . . . (1)

As, CE is the median of ΔADC

So, ar(ΔCDE)=12ar(ΔACD) . . . . . . . (2)

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

ar(ΔBDE)+ar(ΔCDE)=12ar(ΔABD)+12ar(ΔACD)
ar(ΔBDE)=12[ar(ΔABD)+ar(ΔACD)]

= 12ar(ΔABC)

 

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

 

C:\Users\user\Desktop\D1.PNG

 

Prove that:

ar.( BOE ) = 18ar(ABC)

Sol:

Given: A ΔABC in which AD is the medium and E is the mid-point of BD. O is the mid-point of AE.

 

 

To prove: ar(ΔBOE)=18ar(ΔABC)

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

So, BO is the medium of ΔBAE

Therefore, ar(ΔBOE)=12ar(ΔABE) . . . . (1)

Now, E is the mid-point of BD

So AE divides ΔABD into two triangles of equal area.

Therefore, ar(ΔABE)=12ar(ΔABD) . . . .(2)

As D is the mid-point of BC

So, ar(ΔABD)=12ar(ΔABC) . . . . . (3)

Therefore, ar(ΔBOE)=12ar(ΔABE) [From (1)]

= 12[12ar(ΔABCD)] [From (2)]

= 14ar(ΔABD)

= 14×12ar(ΔABC) [From (3)]

= 18ar(ΔABC)

 

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

 

C:\Users\user\Desktop\g2.PNG

 

Show that: ar.( AOB )=ar.( AOD)

Sol:

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

C:\Users\user\Desktop\g2.PNG

To Prove: ar(ΔAOB)=ar(ΔAOD).

Construction: Join BD which intersects AC at P.

Proof: As diagonals of a parallelogram bisect each other therefore, OP is the median of ΔODB

Therefore, ar(ΔODP)=ar(ΔOBP)

Also, AP is the median of ΔABD

Therefore, ar(ΔODP)=ar(ΔOBP)

Adding both sides, we get:

ar(ΔODP)+ar(ΔADP)=ar(ΔOBP)+ar(ΔABP)

Therefore, ar(ΔAOB)=ar(ΔAOD).

 

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).

 

https://lh3.googleusercontent.com/8Xo8Dg5N6Mr-f7jH_i70-xN2yAi-cv-nvsq6ph4o2KtvAL5Y-nB9hBtNEzdZwv50YqH0gpdxieurszEcbxCBeY9H9uP48Jwc-_EiiC8igh1c3s3oD3CJga-G4M8A6yb4l-KB1JGXiD4SNZzLWg

 

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 ADC

SRAC ( By midpoint theorem)

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

PQAC

Therefore, PQACSR

So, PQSR

Similarly, we can prove SPRQ.

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

Since the diagonals of a parallelogram bisect each other.

So in ABD,

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

Therefore, OSAB (by midpoint theorem )

Similarly in ABC we can prove that,

OQAB and SQAB

Thus, ABQS is a parallelogram.

Now, SPQ=12(parallelogramABQS) . . . . . . . (i)

Similarly, we can prove that:

SPQ=12(parallelogramSQCD) . . . . . . . (ii)

Adding (i) and (ii) we get,

SPQ+SPQ=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.( DGF).

 

https://lh5.googleusercontent.com/r1vjYT1HUTv4Tn1hfj8Y7QM3fAaNa2bwekD8zhGBQ9XwAJ7D8NodPuCIPFe-tSD13BG0NVyNpADHofyodH1yIrpKZRb8zhrm6VsQfV-v_y-FWq9sSkYqmWfw2P8PDYuWl2b17aHv

 

Sol:

Given: ABCDE is a pentagon, EDDA and CFDB.

To prove: ar(pentagon ABCDE) = ar(DGF)

 

https://lh5.googleusercontent.com/r1vjYT1HUTv4Tn1hfj8Y7QM3fAaNa2bwekD8zhGBQ9XwAJ7D8NodPuCIPFe-tSD13BG0NVyNpADHofyodH1yIrpKZRb8zhrm6VsQfV-v_y-FWq9sSkYqmWfw2P8PDYuWl2b17aHv

 

Proof:

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

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

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

Therefore, ar(DBC)=ar(DBF) . . . . . . (iii)

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

Therefore, ar(ADE)=ar(ADG) . . . . . . . (iv)

From (iii) and (iv) we have:

ar(DBC)+ar(ADE)=ar(DBF)+ar(ADG)

Adding ar(ABD) on both sides

ar(DBC)+ar(ADE)+ar(ABD)=ar(DBF)+ar(ADG)+ar(ABD)

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

ar(pentagon ABCDE) = ar(DGF)

 

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

Sol:

Let, AD is a median of ABC and D is the midpoint of BC. AD divides ABC in two triangles :ABDandADC

https://lh5.googleusercontent.com/j5C1scmrnHALZkdpknB3oDhlIKH-qro5VaCi49al3G6jwCAhxkjzNXmD7g8XuMXEiAxOScZ5Z1C20nVzBqCFJaNHUQqttLMMjy5jj52Aws2e3rWCW8iwVcebPT_zKBwcveyiMAi5

 

To prove: ar(ABD)=ar(ADC)

Construction: Draw ALBC

Proof:

Since, D is the midpoint of BC, we have

BD=DC

Multiplying with 12AL on both sides, we have

12×BD×AL=12×DC×AL

Therefore, ar(ABD)=ar(ADC)

 

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

Sol:

 

https://lh5.googleusercontent.com/crj6hwCwQWxVJJSKBC-f8Z151wCUrnstlDK5-JUrRux75_YCrSwLzA7wjMNVgK1Sl4FPnhM8AKIaBB7kkvNgzbS2qo5-1jUPqG3yrfqA1dnpJuYlrhUpeXcCU-kX_KEOTyQQYdN_

 

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

To prove: ar(ABD)=ar(CDB)

Proof:

In ABDandCDB we have,

AB = CD (opposite sides of a parallelogram)

AD = CD (opposite sides of a parallelogram)

BD = DB (common side)

Therefore, ABDCDB (SSS criteria )

ar(ABD)=ar(CDB)

 

Q.13: The base BC of ABC is divided at D such that BD = 12DC . Prove that

ar( ABD) = 13×ar(ABC)

Sol:

 

https://lh5.googleusercontent.com/_CpOrs6Y-m0oo7N92xlNSUUGDZmpVIsakdkZ_Ds6Z3zUC5Bg28h-RxtMdSUmoZ5nPMVRQH9Mghr9ifoLuASJtAVHzeS27s1qtrD5Xv4KwN86sSEeO05RN_iirs8BKbnGiEPwSV5j

 

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

To prove: ar(ABD)=13ar(ABC)

Construction: Draw ALBC

Proof:

In ABC, we have :

BC = BD + DC

BD+2BD=3×BD

Now we have:

ar(ABD)=12×BD×AL
ar(ABC)=12×BC×AL
ar(ABC)=12×3BD×AL=3(12×BD×AL)
ar(ABC)=3×ar(ABD)

Therefore, ar(ABD)=13ar(ABC)

 

Q.14: In the adjoining figure, the point D divides the side BC of ABC in the ratio m : n.

 

https://lh4.googleusercontent.com/yNdhmqjvSyYp0aY1VHPpnBxucLG-wGwVE2Lqbpla69AsQqM9_QO7wJYsXZkhdgCGMyx0fL3gfIEwr9O_sFOVJSigBixL67Wlw_WWy-DpPBHXZzMOUYuT3xzEmXqpV-k_Fex9OVgd

 

Prove that:

ar(ABD):ar(ADC) = m:n.

Sol:

https://lh4.googleusercontent.com/yNdhmqjvSyYp0aY1VHPpnBxucLG-wGwVE2Lqbpla69AsQqM9_QO7wJYsXZkhdgCGMyx0fL3gfIEwr9O_sFOVJSigBixL67Wlw_WWy-DpPBHXZzMOUYuT3xzEmXqpV-k_Fex9OVgd

 

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

To prove: ar(ABD):ar(ADC)=m:n

Construction : Draw ALBC

Proof:

ar(ABD)=12×BD×AL . . . . . . (i)

ar(ABD)=12×DC×AL . . . . . . (ii)

Dividing (i) and (ii) we have:

ar(ABD)ar(ABC)=12×BD×AL12×DC×AL

=BDDC = mn

Therefore, ar(ABD):ar(ADC)=m:n


Practise This Question

DIRECTIONS: The following questions have four choices (a), (b), (c) and (d) out of which only one is correct. You have to choose the correct alternative.

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