# Class 9 SELINA Solutions Maths Chapter 16 - Area Theorems [Proof and Use]

## Area Theorems [Proof and Use] Exercise Ex. 16(A)

### Solution 1

(i)and parallelogram ABED are on the same base AB and between the same parallels DE//AB, so area of the triangle is half the area of parallelogram ABED.

Area of ABED = 2 (Area of ADE) = 120 cm^{2}

(ii)Area of parallelogram is equal to the area of rectangle on the same base and of the same altitude i.e, between the same parallels

Area of ABCF = Area of ABED = 120 cm^{2}

(iii)We know that area of triangles on the same base and between same parallel lines are equal

Area of ABE=Area of ADE =60 cm^{2}

### Solution 2

After drawing the opposite sides of AB, we get

Since from the figure, we get CD//FE therefore FC must parallel to DE. Therefore it is proved that the quadrilateral CDEF is a parallelogram.

Area of parallelogram on same base and between same parallel lines is always equal and area of parallelogram is equal to the area of rectangle on the same base and of the same altitude i.e, between same parallel lines.

So Area of CDEF= Area of ABDC + Area of ABEF

Hence Proved

### Solution 3

(i)

Since POS and parallelogram PMLS are on the same base PS and between the same parallels i.e. SP//LM.

As O is the center of LM and Ratio of area of triangles with same vertex and bases along the same line is equal to ratio of their respective bases.

The area of the parallelogram is twice the area of the triangle if they lie on the same base and in between the same parallels.

So 2(Area of PSO)=Area of PMLS

Hence Proved.

(ii)

Consider the expression :

LM is parallel to PS and PS is parallel to RQ, therefore, LM is

Since triangle POS lie on the base PS and in between the parallels PS and LM, we have,,

Since triangle QOR lie on the base QR and in between the parallels LM and RQ, we have,

(iii)

In a parallelogram, the diagonals bisect each other.

Therefore, OS = OQ

Consider the triangle PQS, since OS = OQ, OP is the median of the triangle PQS.

We know that median of a triangle divides it into two triangles of equal area.

Therefore,

Hence Proved.

### Solution 4

(i)

Given ABCD is a parallelogram. P and Q are any points on the sides AB and BC respectively, join diagonals AC and BD.

proof:

since triangles with same base and between same set of parallel lines have equal areas

area (CPD)=area(BCD)…… (1)

again, diagonals of the parallelogram bisects area in two equal parts

area (BCD)=(1/2) area of parallelogram ABCD…… (2)

from (1) and (2)

area(CPD)=1/2 area(ABCD)…… (3)

similarly area (AQD)=area(ABD)=1/2 area(ABCD)…… (4)

from (3) and (4)

area(CPD)=area(AQD),

hence proved.

(ii)

We know that area of triangles on the same base and between same parallel lines are equal

So Area of AQD= Area of ACD= Area of PDC = Area of BDC = Area of ABC=Area of APD + Area of BPC

Hence Proved

### Solution 5

(i)

Since triangle BEC and parallelogram ABCD are on the same base BC and between the same parallels i.e. BC//AD.

(ii)

Therefore, Parallelograms ANMD and NBCM have areas equal to triangle BEC

### Solution 6

Since DCB and DEB are on the same base DB and between the same parallels i.e. DB//CE, therefore we get

Hence proved

### Solution 7

APB and parallelogram ABCD are on the same base AB and between the same parallel lines AB and CD.

ADQ and parallelogram ABCD are on the same base AD and between the same parallel lines AD and BQ.

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

Subtracting Ar.PCQ from both sides, we get

Hence proved.

### Solution 8

Since triangle EDG and EGA are on the same base EG and between the same parallel lines EG and DA, therefore

Subtracting from both sides, we have

(i)

Similarly

(ii)

Now

Hence proved

### Solution 9

Joining PC we get

ABC and BPC are on the same base BC and between the same parallel lines AP and BC.

BPC and BQP are on the same base BP and between the same parallel lines BP and CQ.

From (i) and (ii), we get

Hence proved.

### Solution 10

(i)

From (i) and (ii), we get

In EAC and BAF, we have, EA=AB

and AC=AF

EAC BAF (SAS axiom of congruency)

(ii)

### Solution 11

(i)

In ABC, D is midpoint of AB and E is the midpoint of AC.

DE is parallel to BC.

Again

From the above two equations, we have

Area (ADC) = Area(AEB).

Hence Proved

(ii)

We know that area of triangles on the same base and between same parallel lines are equal

Area(triangle DBC)= Area(triangle BCE)

Area(triangle DOB) + Area(triangle BOC) = Area(triangle BOC) + Area(triangle COE)

So Area(triangle DOB) = Area(triangle COE)

### Solution 12

(i)

Since EBC and parallelogram ABCD are on the same base BC and between the same parallels i.e. BC//AD.

(ii)

Parallelograms on same base and between same parallels are equal in area

Area of BCFE = Area of ABCD= 960 cm^{2}

(iii)

Area of triangle ACD=480 = (1/2) x 30 x Altitude

Altitude=32 cm

(iv)

The area of a triangle is half that of a parallelogram on the same base and between the same parallels.

Therefore,

### Solution 13

Here AD=DB and EC=DB, therefore EC=AD

Again, (opposite angles)

Since ED and CB are parallel lines and AC cut this line, therefore

From the above conditions, we have

Adding quadrilateral CBDF in both sides, we have

Area of // gm BDEC= Area of ABC

### Solution 14

In Parallelogram PQRS, AC // PS // QR and PQ // DB // SR.

Similarly, AQRC and APSC are also parallelograms.

Since ABC and parallelogram AQRC are on the same base AC and between the same parallels, then

Ar.(ABC)=Ar.(AQRC)......(i)

Similarly,

Ar.(ADC)=Ar.(APSC).......(ii)

Adding (i) and (ii), we get

Area of quadrilateral PQRS = 2 Area of quad. ABCD

### Solution 15

Given: ABCD is a trapezium

AB || CD, MN || AC

Join C and M

We know that area of triangles on the same base and between same parallel lines are equal.

So Area of Δ AMD = Area of Δ AMC

Similarly, consider AMNC quadrilateral where MN || AC.

Δ ACM and Δ ACN are on the same base and between the same parallel lines. So areas are equal.

So, Area of Δ ACM = Area of Δ CAN

From the above two equations, we can say

Area of Δ ADM = Area of Δ CAN

Hence Proved.

### Solution 16

We know that area of triangles on the same base and between same parallel lines are equal.

Consider ABED quadrilateral; AD||BE

With common base, BE and between AD and BE parallel lines, we have

Area of ΔABE = Area of ΔBDE

Similarly, in BEFC quadrilateral, BE||CF

With common base BC and between BE and CF parallel lines, we have

Area of ΔBEC = Area of ΔBEF

Adding both equations, we have

Area of ΔABE + Area of ΔBEC = Area of ΔBEF + Area of ΔBDE

=> Area of AEC = Area of DBF

Hence Proved

### Solution 17

Given: ABCD is a parallelogram.

We know that

Area of ΔABC = Area of ΔACD

Consider ΔABX,

Area of ΔABX = Area of ΔABC + Area of ΔACX

We also know that area of triangles on the same base and between same parallel lines are equal.

Area of ΔACX = Area of ΔCXD

From above equations, we can conclude that

Area of ΔABX = Area of ΔABC + Area of ΔACX = Area of ΔACD+ Area of ΔCXD = Area of ACXD Quadrilateral

Hence Proved

### Solution 18

Join B and R and P and R.

We know that the area of the parallelogram is equal to twice the area of the triangle, if the triangle and the parallelogram are on the same base and between the parallels

Consider ABCD parallelogram:

Since the parallelogram ABCD and the triangle ABR lie on AB and between the parallels AB and DC, we have

....(1)

We know that the area of triangles with same base and between the same parallel lines are equal.

Since the triangles ABR and APR lie on the same base AR and between the parallels AR and QP, we have,

....(2)

From equations (1) and (2), we have,

Hence Proved

## Area Theorems [Proof and Use] Exercise Ex. 16(B)

### Solution 1

(i) Suppose ABCD is a parallelogram (given)

Area of congruent triangles are equal.

Therefore, Area of ABC = Area of ADC

(ii) Consider the following figure:

Here _{}

Since Ar.(_{})=_{}

And, Ar.(_{})=_{}

_{},

hence proved

(iii) Consider the following figure:

Here

Ar.(_{})=_{}

And, Ar.(_{})=_{}

_{},

hence proved

### Solution 2

AD is the median of ABC. Therefore it will divide ABC into two triangles of equal areas.

Area(ABD)= Area(ACD) (i)

ED is the median of EBC

Area(EBD)= Area(ECD) (ii)

Subtracting equation (ii) from (i), we obtain

Area(ABD)- Area(EBD)= Area(ACD)- Area(ECD)

Area (ABE) = Area (ACE). Hence proved

### Solution 4

We have to join PD and BD.

BD is the diagonal of the parallelogram ABCD. Therefore it divides the parallelogram into two equal parts.

Area(ABD)= Area(DBC)

=Area (parallelogram ABCD) (i)

DP is the median of ABD. Therefore it will divide ABD into two triangles of equal areas.

Area(APD)= Area(DPB)

= Area (ABD)

= Area(parallelogram ABCD)[from equation (i)]

= Area (parallelogram ABCD) (ii)

In APD, Q is the mid-point of AD. Therefore PQ is the median.

Area(APQ)= Area(DPQ)

= Area(APD)

= Area (parallelogram ABCD) [from equation (ii)]

Area (APQ)= Area (parallelogram ABCD),hence proved

### Solution 5

In ABC, BD = DC

Ar.(ABD):Ar.(ADC)=1:2

But Ar.(ABD)+Ar.(ADC)=Ar.(ABC)

Ar.(ABD)+2Ar.(ABD)=Ar.(ABC)

3 Ar.(ABD)= Ar.(ABC)

Ar.(ABD)= Ar.(ABC)

### Solution 6

Ratio of area of triangles with same vertex and bases along the same line is equal to ratio of their respective bases. So, we have

_{}

Given: Area of ΔDPB = 30 sq. cm

So area of ΔPCB = 20 sq. cm

Consider the following figure.

From the diagram, it is clear that,

Diagonal of the parallelogram divides it into two triangles of equal area.

### Solution 7

### Solution 8

### Solution 3

AD is the median of ABC. Therefore it will divide ABC into two triangles of equal areas.

Area(ABD)= Area(ACD)

Area (ABD)= Area(ABC) (i)

In ABD, E is the mid-point of AD. Therefore BE is the median.

Area(BED)= Area(ABE)

Area(ABE)= Area(ABD)

Area(ABE)= Area(ABC)[from equation (i)]

Area(ABE)= Area(ABC)

## Area Theorems [Proof and Use] Exercise Ex. 16(C)

### Solution 1

(i)

Ratio of area of triangles with same vertex and bases along the same line is equal to the ratio of their respective bases. So, we have:

_{} ----1

Similarly

_{}------2

We know that area of triangles on the same base and between same parallel lines are equal.

Area of Δ ACD = Area of Δ BCD

Area of Δ AOD + Area of Δ DOC = Area of Δ DOC + Area of Δ BOC

=> Area of Δ AOD = Area of Δ BOC ------3

From 1, 2 and 3 we have

Area (Δ DOC) = Area (Δ AOB)

Hence Proved.

(ii)

Similarly, from 1, 2 and 3, we also have

Area of Δ DCB = Area of Δ DOC + Area of Δ BOC = Area of Δ AOB + Area of Δ BOC = Area of Δ ABC

So Area of Δ DCB = Area of Δ ABC

Hence Proved.

(iii)

We know that area of triangles on the same base and between same parallel lines are equal.

Given: triangles are equal in area on the common base, so it indicates AD|| BC.

So, ABCD is a parallelogram.

Hence Proved

### Solution 2

Ratio of area of triangles with the same vertex and bases along the same line is equal to the ratio of their respective bases.

So, we have

_{}

Area of parallelogram ABCD = 324 sq.cm

Area of the triangles with the same base and between the same parallels are equal.

We know that area of the triangle is half the area of the parallelogram if they lie on the same base and between the

parallels.

Therefore, we have,

(ii)

Hence OP:OD = 1:3

### Solution 3

E and F are the midpoints of the sides AB and AC.

Consider the following figure.

Therefore, by midpoint theorem, we have, EF || BC

Triangles BEF and CEF lie on the common base EF and between the parallels, EF and BC

Now BF and CE are the medians of the triangle ABC

Medians of the triangle divides it into two equal areas of triangles.

Thus, we have, Ar.ABF=Ar.CBF

Subtracting Ar.BOE on the both the sides, we have

Ar.ABF - Ar.BOE=Ar.CBF - Ar.BOE

Since, Ar.(BOE)= Ar.(COF),

Ar.ABF- Ar.BOE=Ar.CBF- Ar.COF

Ar. (quad. AEOF)=Ar.(OBC), hence proved

### Solution 5

(i) The figure is shown below

(ii)

(iii)

### Solution 6

Consider that the sides be x cm, y cm and (37-x-y) cm. also, consider that the lengths of altitudes be 6a cm, 5a cm and 4a cm.

Area of a triangle=basealtitude

and

and

Solving both the equations, we have

X=10 cm, y=12 cm and (37-x-y)cm=15 cm

### Solution 7

### Solution 8

### Solution 9

### Solution 10

Join HF.

Since H and F are mid-points of AD and BC respectively,

Now, ABCD is a parallelogram.

⇒ AD = BC and AD ∥ BC

⇒ AH = BF and AH ∥ BF

⇒ ABFH is a parallelogram.

Since parallelogram FHAB and ΔFHE are on the same base FH and between the same parallels HF and AB,

### Solution 11

Join CX, DX and AY.

Now, triangles ADX and ACX are on the same base AX and between the parallels AB and DC.

∴ A(ΔADX) = A(ΔACX) ….(i)

Also, triangles ACX and ACY are on the same base AC and between the parallels AC and XY.

∴ A(ΔACX) = A(ΔACY) ….(ii)

From (i) and (ii), we get

A(ΔADX) = A(ΔACY)

### Solution 4

In ∆DOC and ∆BOP

∠POB = ∠DOC … (vertically opp. Angles)

Now, ▭ABCD is parallelogram,

Hence AB||DC and BD transversal

So, ∠OBP = ∠ODC … (alternate angles)

Hence,

∆DOC ~ ∆BOP …(AA test)

Hence, by area of similar triangles property

Now join DP

Now,

Triangles having equal base and between two same parallel lines have equal area.

Hence,

∆DPC = ΔCBD

∆DOC + ∆POD = ∆DOC + ∆BOC

Thus, ∆POD = ∆BOC …(1)

Also, we know that in a parallelogram the diagonal divides the parallelogram into two equal triangles.

Hence,

∆CBD = ∆ABD

∆DOC + ∆BOC = ∆ADP + ΔPOD + ΔPOB

From (1)

∆DOC = ∆ADP + ΔPOB

∆ADP = 160 - 40 =
120 cm^{2}

Now, in ΔADB, AP is the median, hence

∆ADP = ∆DPB = 120

∆DPB = ΔPOD + ΔPOB

Hence ΔPOD = 120
- 40 = 80 cm^{2}

From (1)

∆BOC = 80 cm^{2}

ΔPBC = ΔPOB +
ΔBOC = 120 cm^{2}

Area of Parallelogram ABCD = 2ΔABD

= 2(120+40+80) =
480 cm^{2}

ΔABC = ½ Area of
Parallelogram ABCD = ½ (480) = 240 cm^{2}