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

Question 1

In the given figure, if the area of triangle ADE is 60 cm², state, given reason, the area of :
(i) Parallelogram ABED;
(ii) Rectangle ABCF;
(iii) Triangle ABE.

Sol:

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

Area of ABED = 2 (Area of ADE) = 120 cm²

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

Area of ABCF = Area of ABED = 120 cm²

(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 cm²

Question 2

The given figure shows a rectangle ABDC and a parallelogram ABEF; drawn on opposite sides of AB.
Prove that: 
(i) Quadrilateral CDEF is a parallelogram;
(ii) Area of the quad. CDEF
= Area of rect. ABDC + Area of // gm. ABEF.

Sol:

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.

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

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

Question 3

In the given figure, diagonals PR and QS of the parallelogram PQRS intersect at point O and LM is parallel to PS. Show that:

(i) 2 Area (POS) = Area (// gm PMLS)
(ii) Area (POS) + Area (QOR) = Area (// gm PQRS)
(iii) Area (POS) + Area (QOR) = Area (POQ) + Area (SOR).

Sol;

(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 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.

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: Area ( ΔPOS) + Area ( QOR ):

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,

Area ( ΔPOS ) = ${\displaystyle \frac{1}{2}\text{Area}(\square }$ PSLM )

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

Area ( ΔQOR ) = ${\displaystyle \frac{1}{2}\text{Area}(\square }$ LMQR )

Area ( ΔPOS ) + Area ( ΔQOR ) = ${\displaystyle \frac{1}{2}\text{Area}(\square }$ PSLM ) + ${\displaystyle \frac{1}{2}\text{Area}(\square }$ LMQR )

= ${\displaystyle \frac{1}{2}[\text{Area (PSLM )}+\text{Area}(\square }$ LMQR )]

= ${\displaystyle \frac{1}{2}[\text{Area}(\square }$ PQRS) ]

(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 the median of a triangle divides it into two triangles of equal area.

Therefore,
Area ( ΔPOS) = Area ( ΔPOQ )              ....(1)
Similarly, since OR is the median of the triangle QRS, we have, Area ( ΔQOR ) = Area ( ΔSOR )           ....(2)

Adding equations (1) and (2), we have,
Area ( ΔPOS ) + Area( ΔQOR) = Area ( ΔPOQ ) + Area( SOR)
Hence Proved.

Question 4

In parallelogram ABCD, P is a point on side AB and Q is a point on side BC.
Prove that:
(i) ΔCPD and ΔAQD are equal in the area.
(ii) Area (ΔAQD) = Area (ΔAPD) + Area (ΔCPB)

Sol:

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

proof:
(i) since triangles with the same base and between the same set of parallel lines have equal areas

area ( CPD ) =  area( BCD )                           …… (1)

again, diagonals of the parallelogram bisect 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

Question 5

In the given figure, M and N are the mid-points of the sides DC and AB respectively of the parallelogram ABCD.

If the area of parallelogram ABCD is 48 cm²;
(i) State the area of the triangle BEC.
(ii) Name the parallelogram which is equal in area to the triangle BEC.

Sol:

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

So Area ( ΔBEC )= ${\displaystyle \frac{1}{2}\times \text{Area}(\square }$ABCD )

= ${\displaystyle \frac{1}{2}}$ x 48 = 24 cm²  

(ii) Area (${\displaystyle \square \text{ANMD})=\text{Area}(\square }$ BNMC )
= ${\displaystyle \frac{1}{2}\text{Area}(\square }$ ABCD)

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

= Area ( ΔBEC )

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

Question 6

In the following figure, CE is drawn parallel to diagonals DB of the quadrilateral ABCD which meets AB produced at point E.
Prove that ΔADE and quadrilateral ABCD are equal in area.

Sol:

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

Ar. ( ΔDCB) = Ar. ( ΔDEB )
Ar. ( ΔDCB + ΔADB ) = AR. (ΔDEB + ΔADB )
Ar. ( ABCD ) = Ar. ( ΔADE )
Hence proved.

Question 7

ABCD is a parallelogram a line through A cuts DC at point P and BC produced at Q. Prove that triangle BCP is equal in area to triangle DPQ.

Sol:

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

∴ Ar. ( ΔAPB ) = ${\displaystyle \frac{1}{2}}$ Ar.( parallelogram ABCD ) ......(i)

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

∴ Ar.( ΔADQ ) = ${\displaystyle \frac{1}{2}}$ Ar.( parallelogram ABCD ) ......(ii)

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

∴ Ar.( ΔAPB ) + Ar.( ΔADQ ) = Ar.(parallelogram ABCD)
Ar.( quad ADQB ) - Ar.(Δ BPQ ) = Ar.(parallelogram ABCD)
Ar.( quad ADQB) - Ar.( ΔBPQ ) = Ar.(quad ADQB ) -Ar.( ΔDCQ )
                            Ar. ( ΔBPQ ) = Ar. ( ΔDCQ )

Subtracting Ar.ΔPCQ from both sides, we get

Ar. ( ΔBPQ ) - Ar.(ΔPCQ ) = Ar. ( ΔDCQ ) - Ar. ( ΔPCQ)
                    Ar. ( ΔBCP ) = Ar. ( ΔDPQ )
Hence proved.

Question 8

The given figure shows a pentagon ABCDE. EG drawn parallel to DA meets BA produced at G and CF draw parallel to DB meets AB produced at F.
Prove that the area of pentagon ABCDE is equal to the area of triangle GDF.

Sol:

Since triangle EDG and EGA are on the same base EG and between the same parallel lines EG and DA.
Therefore,
A( ΔEDG ) = A( ΔEGA )

Subtracting ΔEOG from both sides, we have
A( ΔEOG ) = A( ΔGOA )                      ......(i)

Similarly,
A( ΔDPC ) = A( ΔBPF )                       ........(ii)
Now
A( ΔGDF ) = A( ΔGOA ) + A( ΔBPF ) + A( pen. ABPDO ) 
                 = A( ΔEOD ) + A( ΔDPC ) + A( pen.ABPDO )
                 = A( pen. ABCDE )
Hence proved.

Question 9

In the given figure, AP is parallel to BC, BP is parallel to CQ.
Prove that the area of triangles ABC and BQP are equal.

SoL:

Joining PC we get,

ΔABC and ΔBPC are on the same base BC and between the same parallel lines AP and BC.
∴ A( ΔABC ) = A( ΔBPC )          ....(i)

ΔBPC and ΔBQP are on the same base BP and between the same parallel lines BP and CQ.
∴ A( ΔBPC ) = A( ΔBQP )          ....(ii)

From (i) and (ii), we get
∴A( ΔABC ) = A( ΔBQP ) 
Hence proved.

Question 10

In the figure given alongside, squares ABDE and AFGC are drawn on the side AB and the hypotenuse AC of the right triangle ABC.

If BH is perpendicular to FG
prove that:
(i) ΔEAC ≅ ΔBAF.
(ii) Area of the square ABDE
⇒ Area of the rectangle ARHF.

SOl:

(i) ∠EAC = ∠EAB + ∠BAC
∠EAC = 90° + ∠BAC            ....(i)
∠BAF = ∠FAC + ∠BAC
∠BAF = 90° + ∠BAC            .....(ii)

From (i) and (ii), we get
∠EAC = ∠BAF
In ΔEAC and ΔBAF, we have, EA = AB
∠EAC = ∠BAF and AC = AF
∴ ΔEAC ≅ ΔBAF                ....( SAS axiom of congruency )

(ii) Since ΔABC is a right triangle, We have,
AC² = AB² + BC²              ....( Using pythagoras theorm in ΔABC )
⇒ AB² = AC² - BC² 
⇒ AB² = ( AR + RC )² - ( BR² + RC² )  ....( Since AC = AR + RC and Using Pythagoras Theorem in ΔBRC )
⇒ AB² = AR² + 2AR x RC + RC² - ( BR² + RC² ) ....( Using the identity ) 
⇒ AB² = AR² + 2AR x RC + RC² - ( AB² - AR² + RC² )  ...( Using Pythagoras Theorem in ΔABR )

⇒ 2AB² = 2AR² + 2AR x RC
⇒ AB² = AR( AR + RC )
⇒ AB² =  AR + AC
⇒ AB² = AR x AF
⇒ Area (${\displaystyle \square }$ABDE ) = Area( rectangle ARHF ).

Question 11

In the following figure, DE is parallel to BC.
Show that: 
(i) Area ( ΔADC ) = Area( ΔAEB ).
(ii) Area ( ΔBOD ) = Area( ΔCOE ).

SoL:

(i) In ΔABC, D is the midpoint of AB and E is the midpoint of AC.
${\displaystyle \frac{\text{AD}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}}$

DE is parallel to BC.
∴ A( ΔADC ) = A( ΔBDC ) = ${\displaystyle \frac{1}{2}}$ A( ΔABC )
Again,
∴ A( ΔAEB ) = A( ΔBEC ) = ${\displaystyle \frac{1}{2}}$ A( ΔABC )

From the above two equations, we have
Area( ΔADC ) = Area( ΔAEB ).
Hence Proved.

(ii) We know that the area of triangles on the same base and between the same parallel lines are equal.
Area( ΔDBC )= Area( ΔBCE )
Area( ΔDOB ) + Area( ΔBOC ) = Area( ΔBOC ) + Area( ΔCOE )
So, Area( ΔDOB ) = Area( ΔCOE ).

Question 12

ABCD and BCFE are parallelograms. If area of triangle EBC = 480 cm²; AB = 30 cm and BC = 40 cm.

Calculate : 
(i) Area of parallelogram ABCD;
(ii) Area of the parallelogram BCFE;
(iii) Length of altitude from A on CD;
(iv) Area of triangle ECF.

Sol:

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

∴ A( ΔEBC ) = ${\displaystyle \frac{1}{2}}$ x A( parallelogram ABCD )

parallelogram ABCD = 2 x A( ΔEBC )
                                  = 2 x 480 cm²
                                  = 960 cm²
(ii) Parallelograms on same base and between same parallels are equal in area.
Area of BCFE = Area of ABCD = 960 cm²

(iii) Area of triangle ACD=480 = ${\displaystyle \frac{1}{2}}$ 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,
Area( ΔECF ) = ${\displaystyle \frac{1}{2}}$ Area(${\displaystyle \square }$CBEF )
Similarly, Area( ΔBCE ) = ${\displaystyle \frac{1}{2}}$Area(${\displaystyle \square }$CBEF )

⇒ Area( ΔECF ) = Area( ΔBCE ) = 480 cm².

Question 13

In the given figure, D is mid-point of side AB of ΔABC and BDEC is a parallelogram.

Prove that: Area of ABC = Area of // gm BDEC.

Sol:

Here AD = DB and EC = DB, therefore EC = AD
Again, 
∠EFC = ∠AFD         .....( Opposite angles )

Since ED and CB are parallel lines and AC cut this line, therefore
∠ECF = ∠FAD 
From the above conditions, we have
ΔEFC = ΔAFD
Adding quadrilateral CBDF in both sides, we have
Area of // gm BDEC = Area of ΔABC.

Question 14

In the following, AC // PS // QR and PQ // DB // SR.

Prove that: Area of quadrilateral PQRS = 2 x Area of the quad. ABCD.

SoL:

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

Similarly, AQRC and APSC are also parallelograms.

Since ΔABC and parallelogram AQRC are on the same base AC and between the same parallels, then
A( ΔABC ) = ${\displaystyle \frac{1}{2}}$A(AQRC)  ......(i)

Similarly,
A( ΔADC ) = ${\displaystyle \frac{1}{2}}$A( APSC ) .......(ii)

Adding (i) and (ii), we get
Area of quadrilateral PQRS = 2 x Area of the quad. ABCD.

Question 15

ABCD is a trapezium with AB // DC. A line parallel to AC intersects AB at point M and BC at point N.
Prove that: area of Δ ADM = area of Δ ACN.

Sol:

Given: ABCD is a trapezium.

AB || CD, MN || AC
Join C and M

We know that the area of triangles on the same base and between the same parallel lines are equal.
So Area of ΔAMD = Area of ΔAMC

Similarly, consider the AMNC quadrilateral where MN || AC.
ΔACM and ΔACN are on the same base and between the same parallel lines. So areas are equal.

So, Area of ΔACM = Area of ΔCAN
From the above two equations, we can say
Area of ΔADM = Area of ΔCAN

Hence Proved.

Question 16

In the given figure, AD // BE // CF.
Prove that area (ΔAEC) = area (ΔDBF)

SOl:

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

Consider ABED quadrilateral; AD || BE.
With the 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.

Question 17

In the given figure, ABCD is a parallelogram; BC is produced to point X.
Prove that: area ( Δ ABX ) = area (${\displaystyle \square }$ACXD )

Sol:

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 the area of triangles on the same base and between the same parallel lines are equal.
Area of ΔACX = Area of ΔCXD
From the 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.

Question 18

The given figure shows the parallelograms ABCD and APQR.
Show that these parallelograms are equal in the area.
[ Join B and R ]

Sol:

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
Area(${\displaystyle \square }$ABCD ) = 2 x Area( ΔABR )        ....(1)

We know that the area of triangles with the 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,
Area ( ΔABR ) = Area ( ΔAPR )               ....(2)

From equations (1) and (2), we have,
Area(${\displaystyle \square }$ABCD) = 2 x Area( ΔAPR )        .....(3)

Also, the triangle APR and the parallelograms, AR and QR, lie on the same base AR and between the parallels, AR and QP,
Area( ΔAPR ) = ${\displaystyle \frac{1}{2}}$ x Area(${\displaystyle \square }$ARQP )    ....(4)

Using (4) in equation (3), We have,
Area(${\displaystyle \square }$ABCD ) = 2 x ${\displaystyle \frac{1}{2}\times \text{Area}(\square }$ARQP )

Area( ${\displaystyle \square \text{ABCD})=\text{Area}(\square }$ ARQP)
Hence Proved.