SELINA Solution Class 9 Chapter 16 Area of Theorems [Proof and Use] Exercise 16C

Question 1

In the given figure, the diagonals AC and BD intersect at point O. If OB = OD and AB//DC,
show that:
(i) Area (Δ DOC) = Area (Δ AOB).
(ii) Area (Δ DCB) = Area (Δ ACB).
(iii) ABCD is a parallelogram.

Sol:

(i)  The ratio of the 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:

${\displaystyle \frac{\text{Area of}\Delta \text{DOC}}{\text{Area of} \Delta \text{BOC}}=\frac{DO}{BO}}$ = 1 .....(i)

Similarly

${\displaystyle \frac{\text{Area of}\Delta \text{DOA}}{\text{Area of}\Delta \text{BOA}}=\frac{DO}{BO}}$ = 1 ......(ii)

We know that the area of triangles on the same base and between the 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                      .....(iii)

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 the area of triangles on the same base and between the same parallel lines are equal.
Given: triangles are equal in the area on the common base, so it indicates AD || BC.
So, ABCD is a parallelogram.
Hence Proved.

Question 2

The given figure shows a parallelogram ABCD with area 324 sq. cm. P is a point in AB such that AP: PB = 1:2
Find The area of Δ APD.

Sol:

(i) The ratio of the 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

${\displaystyle \frac{\text{Area of}\Delta \text{APD}}{\text{Area of}\Delta \text{BPD}}=\frac{\text{AP}}{\text{BP}}}$ = ${\displaystyle \frac{1}{2}}$

Area of parallelogram ABCD = 324 sq.cm

The area of the triangles with the same base and between the same parallels are equal.
We know that the 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,

Area ( ΔABD ) = ${\displaystyle \frac{1}{2}}$ x Area [ || gm ABCD ]

                      =${\displaystyle \frac{324}{2}}$
                     = 162 sq.cm
From the diagram it is clear that,
Area (Δ ABD ) = Area ( ΔAPD ) + Area ( ΔBPD )
⇒ 162 = Area ( ΔAPD ) + 2Area ( ΔAPD ) 
⇒ 162 = 3Area ( ΔAPD )
⇒ Area ( ΔAPD ) =${\displaystyle \frac{162}{3}}$
⇒ Area ( ΔAPD ) = 54 sq.cm

(ii) Consider the triangle ΔAOP and ΔCOD
∠AOP = ∠COD       ....[ vertically opposite angles ]
∠CDO = ∠APD       .....[ AB and DC are parallel and DP is the transversal, alternate interior angles are equal ]

Thus, by Angle-Angle similarly,
ΔAOP ∼ ΔCOD.
Hence the corresponding sides are proportional.
${\displaystyle \frac{\text{AP}}{\text{CD}}=\frac{\text{OP}}{\text{OD}}=\frac{\text{AP}}{\text{AB}}}$

= ${\displaystyle \frac{\text{AP}}{\text{AP + PB}}}$

= ${\displaystyle \frac{\text{AP}}{\text{3AP}}}$

=${\displaystyle \frac{1}{3}}$

Question 3

In ΔABC, E and F are mid-points of sides AB and AC respectively. If BF and CE intersect each other at point O,
prove that the ΔOBC and quadrilateral AEOF are equal in area.

Sol:

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

Therefore, Ar.( ΔBEF ) = Ar.( ΔCOF )
⇒ Ar.( ΔBOE ) + Ar.( ΔEOF ) = Ar.( ΔEOF ) + Ar.( ΔCOF ) 
⇒ Ar.(ΔBOE ) = Ar.( ΔCOF )

Now BF and CE are the medians of the triangle ABC

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

Thus, we have, Ar. (ΔABF) = Ar. (ΔCBF)

Subtracting Ar. ΔBOE on both the sides, we have

Ar. (ΔABF) - Ar. (ΔBOE) = Ar. (ΔCBF) - Ar. (ΔBOE)

Since, Ar. ( ΔBOE ) = Ar. ( ΔCOF ),

Ar. (ΔABF) - Ar. (ΔBOE) = Ar. (ΔCBF) - Ar. (ΔCOF)

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

Question 4

In parallelogram ABCD, P is the mid-point of AB. CP and BD intersect each other at point O. If the area of ΔPOB = 40 cm², and OP: OC = 1:2, find:
(i) Areas of ΔBOC and ΔPBC
(ii) Areas of ΔABC and parallelogram ABCD.

Sol:

(i) Joining AC we have the following figure

Consider the triangles ΔPOB and ΔCOD
∠POB = ∠DOC             ...[ vertically opposite angles ]
∠OPB = ∠ODC             ...[ AB and DC are parallel, CP and BD are the transversals, alternate interior angles are equal ]

Therefore, by Angle-Angle similarly criterion of congruence, ΔPOB ∼ ΔCOD
Since P is the mid-point AP = BP, and AB = CD, we have CD = 2 BP
Therefore, We have,

${\displaystyle \frac{\text{BP}}{\text{CD}}=\frac{\text{OP}}{\text{OC}}=\frac{\text{OB}}{\text{OD}}=\frac{1}{2}}$

⇒ OP : OC = 1: 2

(ii) Since from part ( i ), we have

${\displaystyle \frac{\text{BP}}{\text{CD}}=\frac{\text{OP}}{\text{OC}}=\frac{\text{OB}}{\text{OD}}=\frac{1}{2}}$ ,

The ratio between the areas of two similar triangles is equal to the ratio between the square of the corresponding sides.
Here, ΔDOC and ΔPOB are similar triangles.
Thus, we have , 

${\displaystyle \frac{\text{Ar.( ΔDOC)}}{\text{Ar.( ΔPOB )}}=\frac{{\text{DC}}^2}{{\text{PB}}^2}}$

⇒ ${\displaystyle \frac{\text{Ar.(ΔDOC )}}{\text{Ar.( ΔPOB )}}=\frac{{\text{(2PB)}}^2}{{\text{PB}}^2}}$

⇒ ${\displaystyle \frac{\text{Ar.( ΔDOC )}}{\text{Ar.( ΔPOB )}}=\frac{{\text{4PB}}^2}{{\text{PB}}^2}}$

⇒ ${\displaystyle \frac{\text{Ar.( ΔDOC )}}{\text{Ar.( ΔPOB )}}}$ = 4

⇒  Ar.( ΔDOC ) = 4Ar, ( ΔPOB )
                      =  4 x 40 
                      = 160 cm²  

Now consider Ar. ( ΔDBC ) = Ar. ( ΔDOC ) + Ar. (Δ BOC )
                                        = 160 + 80
                                       = 160 cm²  

Two triangles are equal in the area if they are on equal bases and between the same parallels.

Therefore, Ar. ( ΔDBC ) = Ar. ( ΔABC ) = 240 cm² 
The median divides the triangles into areas of two equal triangles.
Thus, CP is the median of the triangle ABC.
Hence, Ar. ( ΔABC ) = 2 Ar. ( ΔPBC )
Ar. ( ΔPBC ) = ${\displaystyle \frac{\text{Ar.( ΔABC )}}{2}}$
Ar. ( ΔPBC ) = 120 cm² 

( iii ) From part (ii) we have,
Ar. ( ΔABC ) = 2Ar. ( PBC ) = 240 cm² 
The area of a triangle is half the area of the parallelogram if both are on equal bases and between the same parallels.

Thus, Ar. ( ΔABC ) = ${\displaystyle \frac{1}{2}}$ Ar. [ || gm ABCD ]

AR. [ || gm ABCD ] = 2 Ar. ( ΔABC )
AR. [ || gm ABCD ] = 2 x 240
AR. [ || gm ABCD ] = 480 cm² 

Question 5

The medians of a triangle ABC intersect each other at point G. If one of its medians is AD,
prove that:
(i) Area ( ΔABD ) = 3 x Area ( ΔBGD )
(ii) Area ( ΔACD ) = 3 x Area ( ΔCGD )
(iii) Area ( ΔBGC ) = ${\displaystyle \frac{1}{3}}$ x Area ( ΔABC ).

Sol:

(i) The figure is shown below

(i) Medians intersect at centroid.
Given that C is the point of intersection of medians and hence G is the centroid of the triangle ABC.

Centroid divides the medians in the ratio 2: 1
That is AG: GD = 2: 1.

Since BG divides AD in the ratio 2: 1, we have,
${\displaystyle \frac{\text{A( ΔAGB )}}{\text{A( ΔBGD)}}=\frac{2}{1}}$

⇒ Area( ΔAGB ) = 2Area( ΔBGD )

From the figure, it is clear that,
Area( ΔABD ) = Area( ΔAGB ) + Area( ΔBGD )
⇒ Area( ΔABD ) = 2Area( ΔBGD ) + Area( ΔBGD )
⇒ Area( ΔABD ) = 3Area( ΔBGD )        ......(1)

(ii) Medians intersect at centroid.
Given that G is the point of intersection of medians and hence G is the centroid of the triangle ABC.

Centroid divides the medians in the ratio 2: 1
That is AG: GD = 2: 1.

Since CG divides AD in the ratio 2: 1, we have,
${\displaystyle \frac{\text{A( ΔAGC )}}{\text{A( ΔCGD)}}=\frac{2}{1}}$

⇒ Area( ΔAGC ) = 2Area( ΔCGD )

From the figure, it is clear that,
Area( ΔACD ) = Area( ΔAGC ) + Area( ΔCGD )
⇒ Area( ΔACD ) = 2Area( ΔCGD ) + Area( ΔCGD )
⇒ Area( ΔACD ) = 3Area( ΔCGD )        ......(2)

(iii) Adding equations (1) and (2), We have,
Area( ΔABD ) + Area( ΔACD ) = 3Area( ΔBGD ) + 3Area( ΔCGD ) 
⇒ Area( ΔABC ) = 3[ Area( ΔBGD ) + Area( ΔCGD ) ]
⇒ Area( ΔABC ) = 3[ Area( ΔBGC ) ]

⇒ ${\displaystyle \frac{\text{Area( ΔABC )}}{3}=\left[Area(\Delta BGC)\right]}$

⇒ Area( ΔBGC ) = ${\displaystyle \frac{1}{3}}$ Area( ΔABC )

Question 6

The perimeter of a triangle ABC is 37 cm and the ratio between the lengths of its altitudes be 6: 5: 4. Find the lengths of its sides.
Let the sides be x cm, y cm, and (37 - x - y) cm. Also, let the lengths of altitudes be 6a cm, 5a cm, and 4a cm.

Sol:

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 = ${\displaystyle \frac{1}{2}}$ x base x altitude

∴ ${\displaystyle \frac{1}{2}\times x\times 6a=\frac{1}{2}\times y\times 5a=\frac{1}{2}\times (37-x-y)\times 4a}$

6x = 5y = 148 - 4x - 4y

6x = 5y and 6x = 148 - 4x - 4y

6x - 5y = 0 and 10x + 4y = 148

Solving both the equations, we have
X = 10 cm, y = 12 cm and ( 37 - x - y ) cm = 15cm.

Question 7

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 = ${\displaystyle \frac{1}{2}}$ x base x altitude

∴ ${\displaystyle \frac{1}{2}\times x\times 6a=\frac{1}{2}\times y\times 5a=\frac{1}{2}\times (37-x-y)\times 4a}$

6x = 5y = 148 - 4x - 4y

6x = 5y and 6x = 148 - 4x - 4y

6x - 5y = 0 and 10x + 4y = 148

Solving both the equations, we have
X = 10 cm, y = 12 cm and ( 37 - x - y ) cm = 15cm.

Sol:

ΔADF and ΔAFE have the same vertex A and their bases are on the same straight line DE.
∴ ${\displaystyle \frac{\text{A(ΔADF)}}{\text{A(ΔAFE)}}=\frac{\text{DF}}{\text{FE}}}$

⇒ ${\displaystyle \frac{60}{\text{A(ΔAFE)}}=\frac{5}{3}}$

⇒ A(ΔAFE) = ${\displaystyle \frac{60\times 3}{5}}$ = 36cm².

Now, A(ΔADE) = A(ΔADF) + A(ΔAFE) = 60 + 36 = 96 cm².

ΔADE and ΔEDB have the same vertex D and their bases are on the same straight line AB.\

∴ ${\displaystyle \frac{\text{A( ΔADE )}}{\text{A( ΔEDB )}}=\frac{\text{AE}}{\text{EB}}}$

⇒ ${\displaystyle \frac{96}{\text{A( ΔEDB )}}=\frac{4}{5}}$

⇒ A( ΔEDB ) = ${\displaystyle \frac{96\times 5}{4}}$ = 120 cm² .

Now, A( ΔADB ) and ||^m ABCD are on the same base AB and between the same parallels AB and DC.

∴ A( ΔADB ) = ${\displaystyle \frac{1}{2}}$ A( ||^m ABCD ) 

⇒ 216 = ${\displaystyle \frac{1}{2}}$ A( ||^m ABCD ) 

⇒ A( ||^m ABCD ) = 2 x 216 = 432 cm² .

Question 8

In the following figure, BD is parallel to CA, E is mid-point of CA and BD = ${\displaystyle \frac{1}{2}}$CA
Prove that: ar. ( ΔABC ) = 2 x ar.( ΔDBC )

Sol:

Here BCED is a parallelogram, Since BD = CE and BD || CE.
ar. ( ΔDBC ) = ar. ( ΔEBC )      ......( Since they have the same base and are between the same parallels )

In ΔABC,
BE is the median,
So, ar. ( ΔEBC ) = ${\displaystyle \frac{1}{2}}$ ar. ( ΔABC )
Now, ar. ( ΔABC ) = ar. ( ΔEBC ) + ar. ( ΔABE)
Also, ar. ( ΔABC ) = 2ar. ( ΔEBC )
⇒ ar. ( ΔABC ) = 2ar. ( ΔDBC )

Question 9

In the following figure, OAB is a triangle and AB || DC.

If the area of ∆ CAD = 140 cm² and the area of ∆ ODC = 172 cm²,
find : (i) the area of ∆ DBC
(ii) the area of ∆ OAC
(iii) the area of ∆ ODB.

Sol:

Given:
ΔCAD = 140 cm²
ΔODC = 172 cm²
AB || CD
As Triangle DBC and ΔCAD have same base CD and between the same parallel lines,
Hence,
Area of ΔDBC = Area of ΔCAD = 140 cm²

Area of ΔOAC = Area of ΔCAD + Area of ΔODC
= 140 cm² + 172 cm² = 312 cm²

Area of ΔODB = Area of ΔDBC + Area of ΔODC
= 140 cm² + 172 cm² = 312 cm².

Question 10

E, F, G, and H are the midpoints of the sides of a parallelogram ABCD.
Show that the area of quadrilateral EFGH is half of the area of parallelogram ABCD.

Sol:

Join HF.

Since H and F are mid-points of AD and BC respectively,
∴ AH = ${\displaystyle \frac{1}{2}\text{AD and BF}=\frac{1}{2}\text{BC}}$

Now, ABCD is a parallelogram.
⇒ AD = BC and AD ∥ BC

⇒ ${\displaystyle \frac{1}{2}\text{AD}=\frac{1}{2}}$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,
A( ΔFHE ) = ${\displaystyle \frac{1}{2}}$A ( ||^m FHAB )     .....(i)

Similarly,
A( ΔFHG ) = ${\displaystyle \frac{1}{2}}$A ( ||^m FHDC )    .......(ii)

Adding (i) and (ii), We get,
A( ΔFHE ) + A( ΔFHG ) = ${\displaystyle \frac{1}{2}\text{A}\left(\mid {\mid }^m\text{FHAB}\right)+\frac{1}{2}}$A ( ||^m FHDC ) 
⇒ A( EFGH ) = ${\displaystyle \frac{1}{2}}$[ A ( ||^m FHAB ) + A ( ||^m FHDC ) ]

⇒ A( EFGH ) = ${\displaystyle \frac{1}{2}}$A( ||^m ABCD )

Question 11

ABCD is a trapezium with AB parallel to DC. A line parallel to AC intersects AB at X and BC at Y.
Prove that the area of ∆ADX = area of ∆ACY.

Sol:

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 )

SELINA Solution Class 9 Chapter 16 Area of Theorems [Proof and Use] Exercise 16B

Question 1.1

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

Sol:

Suppose ABCD is a parallelogram    ...(given)

Consider the triangles ABC and ADC :
AB = CD           ......[ ABCD is a parallelogram ]
ADE  = BC        ......[ ABCD is a parallelogram ]
AD = AD           .....[ common ]

By Side- Side -Side criterion of congruence, we have,
ΔABC ≅ ΔADC
Area of congruent triangles are equal.

Therefore, Area of ABC = Area of ADC

Question 1.2

Show that:
The ratio of the areas of two triangles of the same height is equal to the ratio of their bases.

Sol:

Consider  the following figure :

Here AP ⊥ BC
Since Ar. ( ΔABD ) = ${\displaystyle \frac{1}{2}}$ BD x AP
And, Ar. ( ΔADC ) =${\displaystyle \frac{1}{2}}$ DC x AP

 ${\displaystyle \frac{\text{Area}(\Delta \text{ABD})}{\text{Area(Δ ADC )}}=\frac{\frac{1}{2}\text{BD}\times \text{AP}}{\frac{1}{2}\text{DC}\times \text{AP}}=\frac{\text{BC}}{\text{DC}}}$

Hence proved

Question 1.3

Show that:
The ratio of the areas of two triangles on the same base is equal to the ratio of their heights.

Sol:

Consider the following figure :

Here
Ar. ( ΔABC ) = ${\displaystyle \frac{1}{2}}$ BM x AC
and, Ar. ( ΔADC ) = ${\displaystyle \frac{1}{2}}$ DN x AC

${\displaystyle \frac{\text{Area}(\Delta \text{ABD})}{\text{Area(Δ ADC )}}=\frac{\frac{1}{2}\text{BM}\times \text{AC}}{\frac{1}{2}\text{DN}\times \text{AC}}=\frac{\text{BM}}{\text{DN}}}$

hence proved

Question 2

In the given figure; AD is median of ΔABC and E is any point on median AD.
Prove that Area (ΔABE) = Area (ΔACE).

Sol:

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

∴ Area (ΔABD)= Area (ΔACD)           ...(i)

ED is the median of ΔEBC
∴Area (ΔEBD)= Area (ΔECD)            ...(ii)

Subtracting equation (ii) from (i), we obtain
Area (ΔABD)- Area (ΔEBD) = Area (ΔACD)- Area (ΔECD)
Area (ΔABE) = Area (ΔACE).
Hence proved

Question 3

In the figure of question 2, if E is the mid- point of median AD, then
prove that:
Area  ( ΔABE ) = ${\displaystyle \frac{1}{4}}$ Area ( ΔABC ).

Sol:

AD is the median of ΔABC. Therefore it will divide ΔABC into two triangles of equal areas.
∴ Area( ΔABD ) =  Area( ΔACD )

Area ( ΔABD ) = ${\displaystyle \frac{1}{2}}$Area( ΔABC)       ...(i)
In ΔABD, E is the mid-point of AD. Therefore BE is the median.

∴ Area( ΔBED ) = Area( ΔABE )

Area( ΔBED ) = ${\displaystyle \frac{1}{2}}$ Area( ΔABD )

Area( ΔBED ) = ${\displaystyle \frac{1}{2}\times \frac{1}{2}}$Area( ΔABC )...[from equation (i)]

Area( ΔBED ) = ${\displaystyle \frac{1}{4}}$ Area( ΔABC )

Question 4

ABCD is a parallelogram. P and Q are the mid-points of sides AB and AD respectively.
Prove that area of triangle APQ = ${\displaystyle \frac{1}{8}}$ of the area of parallelogram ABCD.

Sol:

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( ΔABD )= Area ( ΔDBC )

=${\displaystyle \frac{1}{2}}$ Area ( parallelogram ABCD)       ...(i)

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

∴ Area( ΔAPD )= Area ( ΔDPB )

= ${\displaystyle \frac{1}{2}}$ Area ( ΔABD )

= ${\displaystyle \frac{1}{2}\times \frac{1}{2}}$ Area (parallelogram ABCD) ...[from equation (i)]

= ${\displaystyle \frac{1}{4}}$ Area (parallelogram ABCD)     ...(ii)

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

∴ Area(ΔAPQ)= Area (ΔDPQ)

=  ${\displaystyle \frac{1}{2}}$ Area (ΔAPD)

= ${\displaystyle \frac{1}{2}\times \frac{1}{4}}$ Area (parallelogram ABCD)...[from equation (ii)]

Area (ΔAPQ)= ${\displaystyle \frac{1}{8}}$ Area (parallelogram ABCD),
hence proved

Question 5

The base BC of triangle ABC is divided at D so that BD = ${\displaystyle \frac{1}{2}}$DC.
Prove that area of ΔABD = ${\displaystyle \frac{1}{3}}$ of the area of ΔABC.

Sol:

In ΔABC, ∵ BD = ${\displaystyle \frac{1}{2}\text{DC}\Rightarrow \frac{\text{BD}}{\text{DC}}=\frac{1}{2}}$

∴ Ar.( ΔABD ) : Ar.( ΔADC ) = 1:2

But Ar.( ΔABD ) + Ar.( ΔADC ) = Ar.( ΔABC )

Ar.( ΔABD ) + 2Ar.( ΔABD ) = Ar.( ΔABC )

3 Ar.( ΔABD ) = Ar.( ΔABC )

Ar.( ΔABD ) = ${\displaystyle \frac{1}{3}}$Ar.( ΔABC )

Question 6

In a parallelogram ABCD, point P lies in DC such that DP: PC = 3:2. If the area of ΔDPB = 30 sq. cm.
find the area of the parallelogram ABCD.

Sol:

The ratio of the 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

${\displaystyle \frac{\text{Area of DPB}}{\text{Area of PCB}}=\frac{\text{DP}}{\text{PC}}=\frac{3}{2}}$

Given: Area of ΔDPB = 30 sq. cm
Let 'x' be the area of the triangle PCB
Therefore, We have,
⇒ ${\displaystyle \frac{30}{x}=\frac{3}{2}}$
⇒ x = ${\displaystyle \frac{30}{3}\times 2}$ = 20 sq.cm.

So area of ΔPCB = 20 sq. cm
Consider the following figure.

From the diagram, it is clear that,
Area( ΔCDB ) = Area( ΔDPB ) + Area( ΔCDB )
                      = 30 + 20 = 50 sq.cm.
The diagonal of the parallelogram divides it into two triangles ΔADB and ΔCDB of equal area.
Therefore,
Area( parallelogram ABCD ) = 2 x ΔCDB = 2 x 50 = 100 sq.cm.

Question 7

ABCD is a parallelogram in which BC is produced to E such that CE = BC and AE intersects CD at F.

If ar.(∆DFB) = 30 cm²; find the area of parallelogram.

Sol:

BC = CE                     .....( given )
Also, in parallelogram ABCD, BC = AD
⇒ AD = CE
Now, in ΔADF and ΔECF, We have
AD = CE
∠ADF = ∠ECF           .....( Alternate angles )
∠DAF = ∠CEF           ......( Alternate angles )
∴ ΔADF ≅ ΔECF       ......( ASA Criterion )
⇒ Area( ΔADF ) = Area( ΔECF )     ....(1)

Also, in ΔFBE, FC is the median     ....( Since BC = CE )
⇒ Area( ΔBCF ) = Area( ΔECF )      .....(2)

From (1) and (2)
Area( ΔADF ) = Area( ΔBCF )         ......(3)
Again, ΔADF and ΔBDF are on the base DF and between parallels DF and AB.
⇒ Area( ΔBDF ) = Area( ΔADF )    ........(4)

From (3) and (4),
Area( ΔBDF ) = Area( ΔBCF ) = 30 cm²
Area( ΔBCD ) = Area( ΔBDF ) + Area( ΔBCF ) = 30 + 30 = 60 cm²
Hence, Area of parallelogram ABCD = 2 x Area( ΔBCD ) = 2 x 60 = 120cm^2.

Question 8

The following figure shows a triangle ABC in which P, Q, and R are mid-points of sides AB, BC and CA respectively. S is mid-point of PQ:
Prove that: ar. ( ∆ ABC ) = 8 × ar. ( ∆ QSB )

Sol:

In ΔABC,
R and Q are the mid-points of AC and BC respectively.
⇒ RQ || AB
that is RQ || PB

So, area ( ΔPBQ ) = area( ΔAPR )    ....(i)( Since AP = PB and triangles on the same base and between the same parallels are equal in area. )

Since P and R are the mid-points of AB and AC respectively.
⇒ PR || BC
that is PR || BQ
So, quadrilateral PMQR is a parallelogram.
Also, area( ΔPBQ ) = area( ΔPQR )    ....(ii)( diagonal of a parallelogram divide the parallelogram into two triangles with the equal area ) 

From (i) and (ii)
area( ΔPQR ) = area ( ΔPBQ ) = area( ΔAPR )   ....(iii)
Similarly, P and Q are the mid-points of AB and BC respectively.
⇒ PQ || AC
that is PQ || RC
So, quadrilateral PQRC is a parallelogram.
Also, area( ΔRQC ) = area( ΔPQR )       .....(iv)( diagonal of a parallelogram divide the parallelogram into two triangles with the equal area )

From (iii) and (iv),
area( ΔPQR ) = area( ΔPBQ ) = area( ΔRQC ) = area( ΔAPR )
So, area( ΔPBQ ) = ${\displaystyle \frac{1}{4}}$ area( ΔABC )     ....(v)

Also, since S is the mid-point of PQ,
BS is the median of ΔPBQ
SO, area( ΔQSB ) = ${\displaystyle \frac{1}{2}}$area( ΔPBQ )

From (v),
area( ΔQSB ) = ${\displaystyle \frac{1}{2}\times \frac{1}{4}}$ area( ΔABC )

⇒ area( ΔABC ) = 8 area( ΔQSB ).