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

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

Question 1

In the given figure, if area of triangle ADE is 60 cm2, state, given reason, the area of :

(i) Parallelogram ABED;

(ii) Rectangle ABCF;

(iii) Triangle ABE.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Solution 1

(i)Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Useand parallelogram ABED are on the same base AB and between the same parallels DE//AB, so area of the triangle Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Useis half the area of parallelogram ABED.

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

(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 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 quad. CDEF

= Area of rect. ABDC

+ Area of // gm. ABEF.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Solution 2

After drawing the opposite sides of AB, we get

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

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

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:

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

(i) 2 Area (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UsePOS) = Area (// gm PMLS)

(ii) Area (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UsePOS) + Area (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseQOR)

= Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea (// gm PQRS)

(iii) Area (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UsePOS) + Area (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseQOR)

= Area (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UsePOQ) + Area (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseSOR).

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 Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use:

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,Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use,

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

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use


(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,

 Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

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) Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseCPD and Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseAQD are equal in area.

(ii) Area (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseAQD)

= Area (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseAPD) + Area (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseCPB)

Solution 4

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

 

 

 

 

 

 

 

 

 (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

Question 5

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

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

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.

Solution 5

(i)

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

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

(ii)

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

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 Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseADE and quadrilateral ABCD are equal in area.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Solution 6

Since Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseDCB and Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseDEB are on the same base DB and between the same parallels i.e. DB//CE, therefore we get

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

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.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Solution 7

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseAPB and parallelogram ABCD are on the same base AB and between the same parallel lines AB and CD.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseADQ and parallelogram ABCD are on the same base AD and between the same parallel lines AD and BQ.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

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

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Subtracting Ar.Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UsePCQ from both sides, we get

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

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.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Solution 8

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

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

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Subtracting Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Usefrom both sides, we have

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use  (i)

Similarly

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use  (ii)

Now

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

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.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Solution 9

Joining PC we get

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

 

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC and Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseBPC are on the same base BC and between the same parallel lines AP and BC.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseBPC and Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseBQP are on the same base BP and between the same parallel lines BP and CQ.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

From (i) and (ii), we get

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

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.

 

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

If BH is perpendicular to FG prove that:

(i) Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseEAC Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseSelina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseBAF.

(ii) Area of the square ABDE

= Area of the rectangle ARHF.

Solution 10

(i)

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

From (i) and (ii), we get

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

In Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseEAC and Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseBAF, we have, EA=AB

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Useand AC=AF

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseSelina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseEAC Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseSelina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseBAF (SAS axiom of congruency)

(ii)

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Question 11

In the following figure, DE is parallel to BC. Show that:

(i) Area (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseADC) = Area(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseAEB).

(ii) Area (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseBOD) = Area (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseCOE).

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Solution 11

(i)

In Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC, D is midpoint of AB and E is the midpoint of AC.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

DE is parallel to BC.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Again

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

From the above two equations, we have

Area (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseADC) = Area(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseAEB).

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)


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.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Solution 12

(i)

Since Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseEBC and parallelogram ABCD are on the same base BC and between the same parallels i.e. BC//AD.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

(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 = (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,

 Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Question 13

In the given figure, D is mid-point of side AB of Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC and BDEC is a parallelogram.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Prove that:

Area of Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC = Area of // gm BDEC.

Solution 13

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

Again, Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use(opposite angles)

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

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

From the above conditions, we have

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Adding quadrilateral CBDF in both sides, we have

Area of // gm BDEC= Area of Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC

Question 14

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

Prove that:

Area of quadrilateral PQRS = 2 Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea of quad. ABCD.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Solution 14

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

Similarly, AQRC and APSC are also parallelograms.

Since Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC and parallelogram AQRC are on the same base AC and between the same parallels, then

Ar.(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC)=Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseAr.(AQRC)......(i)

Similarly,

Ar.(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseADC)=Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseAr.(APSC).......(ii)

Adding (i) and (ii), we get

Area of quadrilateral PQRS = 2 Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea of quad. ABCD

Question 15

ABCD is 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.

 

Solution 15

Given: ABCD is a trapezium

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

 

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.

 

 

 

 

Question 16

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

 

 

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use 

 

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

 

Question 17

In the given figure, ABCD is a parallelogram; BC is produced to point X. Prove that: area (Δ ABX) = area (quad. ACXD)

 

 

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use 

 

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

 

 

 

 

Question 18

The given figure shows parallelograms ABCD and APQR. Show that these parallelograms are equal in area.

[Join B and R]

 

 

 Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use 

 

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

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use....(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,

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use ....(2)

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

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

 

 

 

 

 

Hence Proved

 

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

Question 1

Show that:

(i) A diagonal divides a parallelogram into two triangles of equal area.

(ii) The ratio of the areas of two triangles of the same height is equal to the ratio of their bases.

(iii) The ratio of the areas of two triangles on the same base is equal to the ratio of their heights.

Solution 1

(i) Suppose ABCD is a parallelogram (given)

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Area of congruent triangles are equal.

Therefore, Area of ABC = Area of ADC

 

(ii) Consider the following figure:

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Here Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Since Ar.(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use)=Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

And, Ar.(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use)=Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseSelina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use,

hence proved

(iii) Consider the following figure:

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Here

Ar.(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use)=Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

And, Ar.(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use)=Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseSelina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use,

hence proved

Question 2

In the given figure; AD is median of Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC and E is any point on median AD. Prove that Area (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABE) = Area (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseACE).

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Solution 2

AD is the median of Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC. Therefore it will divide Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC into two triangles of equal areas.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABD)= Area(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseACD)   (i)

ED is the median of Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseEBC

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseEBD)= Area(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseECD)  (ii)

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

Area(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABD)- Area(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseEBD)= Area(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseACD)- Area(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseECD)

Area (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABE) = Area (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseACE). Hence proved

Question 3

In the figure of question 2, if E is the mid point of median AD, then prove that:

Area (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABE) = Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC).

Solution 3

AD is the median of Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC. Therefore it will divide Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC into two triangles of equal areas.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABD)= Area(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseACD)

Area (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABD)= Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC) (i)

In Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABD, E is the mid-point of AD. Therefore BE is the median.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseBED)= Area(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABE)

Area(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseBED)= Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABD)

Area(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseBED)= Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseSelina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC)[from equation (i)]

Area(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseBED)= Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC)

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 = Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Useof the area of parallelogram ABCD.

Solution 4

We have to join PD and BD.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

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

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABD)= Area(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseDBC)

=Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea (parallelogram ABCD) (i)

DP is the median of Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABD. Therefore it will divide Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABD into two triangles of equal areas.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseAPD)= Area(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseDPB)

= Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABD)

= Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseSelina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea(parallelogram ABCD)[from equation (i)]

= Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea (parallelogram ABCD) (ii)

In Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseAPD, Q is the mid-point of AD. Therefore PQ is the median.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseAPQ)= Area(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseDPQ)

= Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseAPD)

= Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseSelina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseSelina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea (parallelogram ABCD) [from equation (ii)]

Area (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseAPQ)= Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea (parallelogram ABCD),hence proved

Question 5

The base BC of triangle ABC is divided at D so that BD = Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseDC.

Prove that area of Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABD = Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Useof the area of Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC.

Solution 5

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

In Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC, Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseBD = Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseDCSelina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseSelina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseAr.(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABD):Ar.(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseADC)=1:2

But Ar.(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABD)+Ar.(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseADC)=Ar.(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC)

Ar.(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABD)+2Ar.(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABD)=Ar.(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC)

3 Ar.(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABD)= Ar.(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC)

Ar.(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABD)= Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseAr.(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC)

Question 6

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

 

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


Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use 


Given: Area of ΔDPB = 30 sq. cm

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

So area of ΔPCB = 20 sq. cm


Consider the following figure.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use


From the diagram, it is clear that,


Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

 

Diagonal of the parallelogram divides it into two triangles Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use of equal area.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

 

 

 


 

Question 7

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

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use 

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

Solution 7

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

 

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

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)

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use 

Solution 8

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

 

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

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.

 

 

 Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use 

 

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:

 

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use ----1


Similarly


 Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use------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

 

 

 

 

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.

 Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use 

 

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


Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use 


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,

 Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

(ii)


Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

 

 

 

 

 

 

 

 

 

 

 

 


 

 

Hence OP:OD = 1:3

 

 

 

 

Question 3

In Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC, 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 Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseOBC and quadrilateral AEOF are equal in area.

Solution 3

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

Consider the following figure.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

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

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

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.Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABF=Ar.Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseCBF

Subtracting Ar.Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseBOE on the both the sides, we have

Ar.Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABF - Ar.Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseBOE=Ar.Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseCBF - Ar.Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseBOE

Since, Ar.(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseBOE)= Ar.(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseCOF),

Ar.Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABF- Ar.Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseBOE=Ar.Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseCBF- Ar.Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseCOF

Ar. (quad. AEOF)=Ar.(Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseOBC), hence proved

Question 4

In parallelogram ABCD, P is mid-point of AB. CP and BD intersect each other at point O. If area of Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UsePOB = 40 cm2, and OP : OC = 1:2, find:

(i) Areas of Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseBOC and Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UsePBC

(ii) Areas of Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC and parallelogram ABCD.

Solution 4

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

 

 

 

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 (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABD) = 3 Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseBGD)

(ii) Area (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseACD) = 3 Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea (CGD)

(iii) Area (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseBGC) = Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea (Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseABC)

Solution 5

(i) The figure is shown below

 

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

(ii)

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

(iii)

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

 

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.

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.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseArea of a triangle=Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UsebaseSelina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Usealtitude

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseSelina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseandSelina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseandSelina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Solving both the equations, we have

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

Question 7

In parallelogram ABCD, E is a point in AB and DE meets diagonal AC at point F. If DF: FE = 5:3 and area of Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use ADF is

60 cm2 ; find

(i) area of Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And UseADE

(ii) if AE :EB = 4:5, find the area  of Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use ADB

(iii) aslo, find area of parallelogram ABCD

Solution 7

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Question 8

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use 

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use 

Solution 8

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Question 9

In the following figure, OAB is a triangle and ABDC.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use 

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.

Solution 9

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Question 10

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

Solution 10

Join HF.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use 

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

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

Now, ABCD is a parallelogram.

AD = BC and AD BC

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use

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,

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use 

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 area of ∆ADX = area of ∆ACY.

Solution 11

Join CX, DX and AY.

Selina Solutions Icse Class 9 Mathematics Chapter - Area Theorems Proof And Use 

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)