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 ).