# ML Aggarwal Solutions for Chapter 14 Theorems on Area Class 9 Maths ICSE

Here, we are providing the solutions for Chapter 14 Theorems on Area from ML Aggarwal Textbook for Class 9 ICSE Mathematics. Solutions of the fourteen chapter has been provided in detail. This will help the students in understanding the chapter more clearly. Class 9 Chapter 14 Theorems on Area ML Aggarwal Solutions for ICSE is one of the most important chapter for the board exams which is based on line segments and mid-points and area of parallelograms, triangles and different quadrilaterals.

### Exercise 14

1. Prove that the line segment joining the mid-points of a pair of opposite sides of a parallelogram divides it into two equal parallelograms.

Solution

Let us consider ABCD be a parallelogram in which E and F are mid-points of AB and CD. Join EF.

To prove: ar (|| AEFD) = ar (|| EBCF)

Let us construct DG ⊥ AG and let DG = h where, h is the altitude on side AB.

Proof:

ar (|| ABCD) = AB × h

ar (|| AEFD) = AE × h

= ½ AB × h ...(1) [Since, E is the mid-point of AB]

ar (|| EBCF) = EF × h

= ½ AB × h …(2) [Since, E is the mid-point of AB]

From (1) and (2)

ar (|| ABFD) = ar (|| EBCF)

Hence, proved.

2. Prove that the diagonals of a parallelogram divide it into four triangles of equal area.

Solution

Let us consider in a parallelogram ABCD the diagonals AC and BD are cut at point O.

To prove: ar (∆AOB) = ar (∆BOC) = ar (∆COD) = ar (∆AOD)

Proof:

In parallelogram ABCD the diagonals bisect each other.

AO = OC

In ∆ACD, O is the mid-point of AC. DO is the median.

ar (∆AOD) = ar (COD) …(1) [Median of ∆ divides it into two triangles of equal areas]

Similarly in ∆ ABC,

ar (∆AOB) = ar (∆COB) …(2)

ar (∆AOD) = ar (∆AOB) …(3)

In ∆CDB

ar (∆COD) = ar (∆COB) …(4)

From (1), (2), (3) and (4)

ar (∆AOB) = ar (∆BOC) = ar (∆COD) = ar (∆AOD)

Hence, proved.

3. (a) In the figure (1) given below, AD is median of ∆ABC and P is any point on AD. Prove that

(i) Area of ∆PBD = area of ∆PDC.

(ii) Area of ∆ABP = area of ∆ACP.

(b) In the figure (2) given below, DE || BC. Prove that

(i) area of ∆ACD = area of ∆ ABE.

(ii) Area of ∆OBD = area of ∆OCE.

Solution

(a) Given:

∆ABC in which AD is the median. P is any point on AD. Join PB and PC.

To prove:

(i) Area of ∆PBD = area of ∆PDC.

(ii) Area of ∆ABP = area of ∆ACP.

Proof:

From fig (1)

AD is a median of ∆ABC

So, ar (∆ABD) = ar (∆ADC) …(1)

Also, PD is the median of ∆BPD

Similarly, ar (∆PBD) = ar (∆PDC) …(2)

Now, let us subtract (2) from (1), we get

ar (∆ABD) – ar (∆PBD) = ar (∆ADC) – ar (∆PDC)

Or, ar (∆ABP) = ar (∆ACP)

Hence, proved.

(b) Given:

∆ABC in which DE || BC

To prove:

(i) area of ∆ACD = area of ∆ ABE.

(ii) Area of ∆OBD = area of ∆OCE.

Proof:

From fig (2)

∆DEC and ∆BDE are on the same base DE and between the same || line DE and BE.

ar (∆DEC) = ar (∆BDE)

Now add ar (ADE) on both sides, we get

ar (∆DEC) + ar (∆ADE) = ar (∆BDE) + ar (∆ADE)

⇒ ar (∆ACD) = ar (∆ABE)

Hence, proved.

Similarly, ar (∆DEC) = ar (∆BDE)

Subtract ar (∆DOE) from both sides, we get

ar (∆DEC) – ar (∆DOE) = ar (∆BDE) – ar (∆DOE)

⇒ ar (∆OBD) = ar (∆OCE)

Hence, proved.

4. (a) In the figure (1) given below, ABCD is a parallelogram and P is any point in BC. Prove that: Area of ∆ABP + area of ∆DPC = Area of ∆APD.

(b) In the figure (2) given below, O is any point inside a parallelogram ABCD. Prove that:

(i) area of ∆OAB + area of ∆OCD = ½ area of || gm ABCD

(ii) area of ∆ OBC + area of ∆ OAD = ½ area of || gm ABCD

Solution

(a) Given:

From fig (1)

ABCD is a parallelogram and P is any point in BC.

To prove:

Area of ∆ABP + area of ∆DPC = Area of ∆APD

Proof:

∆APD and || gm ABCD are on the same base AD and between the same || lines AD and BC,

ar (∆APD) = ½ ar (|| gm ABCD) …(1)

In parallelogram ABCD

ar(|| gm ABCD) = ar (∆ ABP) + ar (∆APD) + ar (∆DPC)

Now, divide both sides by 2, we get

½ ar(|| gm ABCD) = ½ ar (∆ ABP) + ½ ar (∆APD) + ½ ar (∆DPC) …(2)

From (1) and (2)

ar (∆APD) = ½ ar (|| gm ABCD)

Substituting (2) in (1)

ar (∆APD) = ½ ar (∆ ABP) + ½ ar (∆APD) + ½ ar (∆DPC)

⇒ ar (∆APD) – ½ ar (∆APD) = ½ ar (∆ ABP) + ½ ar (∆DPC)

⇒ ½ ar (∆APD) = ½ [ar (∆ ABP) + ar (∆DPC)]

⇒ ar (∆APD) = ar (∆ ABP) + ar (∆DPC)

Or, ar (∆ ABP) + ar (∆DPC) = ar (∆APD)

Hence, proved.

(b) Given:

From fig (2)

|| gm ABCD in which O is any point inside it.

To prove:

(i) area of ∆OAB + area of ∆OCD = ½ area of || gm ABCD

(ii) area of ∆ OBC + area of ∆ OAD = ½ area of || gm ABCD

Draw POQ || AB through O. It meets AD at P and BC at Q.

Proof:

(i) AB || PQ and AP || BQ

ABQP is a || gm

Similarly, PQCD is a || gm

Now, ∆OAB and || gm ABQP are on same base AB and between same || lines AB and PQ

ar (∆OAB) = ½ ar (||gm ABQP) …(1)

Similarly,

ar (∆OCD) = ½ ar (||gm PQCD) …(2)

Now by adding (1) and (2)

ar (∆OAB) + ar (∆OCD) = ½ ar (|| gm ABQP) + ½ ar (|| gm PQCD)

= ½ [ar (|| gm ABQP) + ar (|| gm PQCD)]

= ½ ar (|| gm ABCD)

⇒ ar (∆OAB) + ar (∆OCD) = ½ ar (|| gm ABCD)

Hence, proved.

(ii) we know that,

ar (∆OAB) + ar (∆ OBC) + ar (∆OCD) + ar (∆OAD) = ar (|| gm ABCD)

⇒ [ar (∆OAB) + ar (∆OCD)] + [ar (∆ OBC) + ar (∆OAD)] = ar (|| gm ABCD)

⇒ ½ ar (|| gm ABCD) + ar (∆OBC) + ar (∆OAD) = ar (|| gm ABCD)

⇒ ar (∆OBC) + ar (∆OAD) = ar (|| gm ABCD) – ½ ar (|| gm ABCD)

⇒ ar (∆OBC) + ar (∆OAD) = ½ ar (|| gm ABCD)

Hence, proved.

5. If E, F, G and H are mid-points of the sides AB, BC, CD and DA respectively of a parallelogram ABCD, prove that area of quad. EFGH = 1/2 area of || gm ABCD.

Solution

Given:

In parallelogram ABCD, E, F, G, H are the mid-points of its sides AB, BC, CD and DA.

Join EF, FG, GH and HE.

To prove:

area of quad. EFGH = ½ area of || gm ABCD

Proof:

Let us join EG.

We know that, E and G are mid-points of AB and CD.

EG || AD || BC

AEGD and EBCG are parallelogram

Now, || gm AEGD and ∆EHG are on the same base and between the parallel lines.

ar ∆EHG = ½ ar || gm AEGD …(1)

Similarly,

ar ∆EFG = ½ ar || gm EBCG …(2)

Now by adding (1) and (2)

ar ∆EHG + ar ∆EFG = ½ ar || gm AEGD + ½ ar || gm EBCG

⇒ area quad. EFGH = ½ ar || gm ABCD

Hence, proved.

6. (a) In the figure (1) given below, ABCD is a parallelogram. P, Q are any two points on the sides AB and BC respectively. Prove that, area of ∆ CPD = area of ∆ AQD.

(b) In the figure (2) given below, PQRS and ABRS are parallelograms and X is any point on the side BR. Show that area of ∆ AXS = ½ area of ||gm PQRS.

Solution

(a) Given:

From fig (1)

||gm ABCD in which P is a point on AB and Q is a point on BC.

To prove:

area of ∆ CPD = area of ∆ AQD.

Proof:

∆ CPD and ||gm ABCD are on the same base CD and between the same parallels AB and CD.

ar (∆ CPD) = ½ ar (||gm ABCD) …(1)

∆ AQD and ||gm ABCD are on the same base AD and between the same parallels AD and BC.

ar (∆AQD) = ½ ar (||gm ABCD) …(2)

from (1) and (2)

ar (∆ CPD) = ar (∆AQD)

Hence, proved.

(b) From fig (2)

Given:

PQRS and ABRS are parallelograms on the same base SR. X is any point on the side BR.

Join AX and SX.

To prove:

area of ∆ AXS = ½ area of ||gm PQRS

we know that, || gm PQRS and ABRS are on the same base SR and between the same parallels.

So, ar ||gm PQRS = ar ||gm ABRS …(1)

we know that, ∆ AXS and || gm ABRS are on the same base AS and between the same parallels.

So, ar ∆ AXS = ½ ar ||gm ABRS

= ½ ar ||gm PQRS [From (1)]

Hence, proved.

7. D, E and F are mid-point of the sides BC, CA and AB respectively of a ∆ ABC. Prove that

(i) FDCE is a parallelogram

(ii) area of ∆ DEF = ¼ area of ∆ ABC

(iii) area of || gm FDCE = ½ area of ∆ ABC

Solution

Given:

D, E and F are mid-point of the sides BC, CA and AB respectively of a ∆ ABC.

To prove:

(i) FDCE is a parallelogram

(ii) area of ∆ DEF = ¼ area of ∆ ABC

(iii) area of || gm FDCE = ½ area of ∆ ABC

Proof:

(i) F and E are mid-points of AB and AC.

So, FE || BC and FE = ½ BC …(1)

Also, D is mid-point of BC

CD = ½ BC ...(2)

From (1) and (2)

FE || BC and FE = CD

⇒ FE || CD and FE = CD ...(3)

Similarly,

D and F are mid-points of BC and AB.

So, DF || EC is a parallelogram.

Hence proved.

(ii) we know that, FDCE is a parallelogram.

And DE is a diagonal of ||gm FDCE

So, ar (∆ DEF) = ar (∆DEC) …(4)

Similarly, we know BDEF and DEAF are ||gm

So, ar (∆ DEF) = ar (∆ BDF) = ar (∆ AFE) …(5)

From (4) and (5)

ar (∆ DEF) = ar (∆DEC) = ar (∆ BDF) = ar (∆ AFE)

Now,

ar (∆ ABC) = ar (∆ DEF) + ar (∆ DEF) + ar (∆ DEF) + ar (∆ DEF)

= 4 ar (∆ DEF)

⇒ ar (∆ DEF) = ¼ ar (∆ ABC) …(6)

Hence proved.

(iii) ar of || gm FDCE = ar (∆ DEF) + ar (∆ DEC)

= ar (∆ DEF) + ar (∆ DEF)

= 2 ar (∆ DEF) [From (4)]

= 2 [¼ ar (∆ ABC)] [From (6)]

⇒ ar of || gm FDCE = ½ ar of ∆ ABC

Hence, proved.

8. In the given figure, D, E and F are mid points of the sides BC, CA and AB respectively of ∆ ABC. Prove that BCEF is a trapezium and area of trap. BCEF = ¾ area of ∆ ABC.

Solution

Given:

In ABC, D, E and F are mid points of the sides BC, CA and AB.

To prove:

area of trap. BCEF = ¾ area of ∆ ABC

Proof:

We know that D and E are the mid-points of BC and CA.

So, DE || AB and ½ AB

Similarly,

EF || BC and ½ BC

And FD || AC and ½ AC

∴ BDEF, CDFE, AFDE are parallelograms which are equal in area.

ED, DF, EF are diagonals of these ||gm which divides the corresponding parallelogram into two triangles equal in area.

Hence, BCEF is a trapezium.

area of trap. BCEF = ¾ area of ∆ ABC

9. (a) In the figure (1) given below, the point D divides the side BC of ∆ABC in the ratio m: n. Prove that area of ∆ ABD: area of ∆ ADC = m: n.

(b) In the figure (2) given below, P is a point on the side BC of ∆ABC such that PC = 2BP, and Q is a point on AP such that QA = 5 PQ, find area of ∆AQC: area of ∆ABC.

(c) In the figure (3) given below, AD is a median of ∆ABC and P is a point in AC such that area of ∆ADP: area of ∆ABD = 2:3. Find
(i) AP: PC

(ii) area of ∆PDC: area of ∆ABC.

Solution

(a) Given:

From fig (1)

In ∆ABC, the point D divides the side BC in the ratio m: n.

BD: DC = m: n

To prove:

area of ∆ ABD: area of ∆ ADC = m: n

Proof:

area of ∆ ABD = ½ × base ×height

ar (∆ ABD) = ½ × BD ×AE …(1)

ar (∆ ACD) = ½ × DC ×AE …(2)

let us divide (1) by (2)

[ar(∆ABD) = ½×BD×AE]/[ar(∆ACD) = ½×DC×AE]

⇒ [ar (∆ ABD)]/[ar (∆ ACD)] = BD/DC

⇒ m/n [it is given that, BD: DC = m: n]

Hence, proved.

(b) Given:

From fig (2)

In ∆ABC, P is a point on the side BC such that PC = 2BP, and Q is a point on AP such that QA = 5 PQ.

To Find:

area of ∆AQC: area of ∆ABC

Now,

It is given that: PC = 2BP

PC/2 = BP

We know that, BC = BP + PC

Now substitute the values, we get

BC = BP + PC

= PC/2 + PC

= (PC + 2PC)/2

= 3PC/2

2BC/3 = PC

ar (∆APC) = 2/3 ar (∆ABC) …(1)

It is given that, QA = 5PQ

QA/5 = PQ

We know that, QA= QA + PQ

So, QA = 5/6 AP

ar (∆AQC) = 5/6 ar (∆APC) = 5/6 (2/3 ar(∆ABC)) [From (1)]

⇒ ar (∆AQC) = 5/9 ar (∆ABC)

⇒ ar (∆AQC)/ ar (∆AQC) = 5/9

Hence proved.

(c) Given:

From fig (3)

AD is a median of ∆ABC and P is a point in AC such that area of ∆ADP: area of ∆ABD = 2:3

To Find:

(i) AP: PC

(ii) area of ∆PDC: area of ∆ABC

Now,

(i) we know that AD is the median of ∆ABC

ar (∆ABD) = ar (∆ADC) = ½ ar (∆ABC) …(1)

It is given that,

ar (∆ADP): ar (∆ABD) = 2: 3

AP: AC = 2: 3

AP/AC = 2/3

AP = 2/3 AC

Now,

PC = AC – AP

= AC – 2/3 AC

= (3AC-2AC)/3

= AC/3 …(2)

So,

AP/PC = (2/3 AC) / (AC/3)

= 2/1

AP: PC = 2:1

(ii) we know that from (2)

PC = AC/3

⇒ PC/AC = 1/3

So,

ar (∆PDC)/ar (∆ADC) = PC/AC = 1/3

⇒ ar (∆PDC)/1/2 ar (∆ABC) = 1/3

⇒  ar (∆PDC)/ar (∆ABC) = 1/3 × ½ = 1/6

⇒ ar (∆PDC): ar (∆ABC) = 1: 6

Hence, proved.

10. (a) In the figure (1) given below, area of parallelogram ABCD is 29 cm2. Calculate the height of parallelogram ABEF if AB = 5.8 cm

(b) In the figure (2) given below, area of ∆ABD is 24 sq. units. If AB = 8 units, find the height of ABC.

(c) In the figure (3) given below, E and F are mid points of sides AB and CD respectively of parallelogram ABCD. If the area of parallelogram ABC is 36 cm2.

(i) State the area of ∆ APD.

(ii) Name the parallelogram whose area is equal to the area of ∆ APD.

Solution

(a) Given:

From fig (1)

ar ||gm ABCD = 29cm2

To find:

Height of parallelogram ABEF if AB = 5.8 cm

Now, let us find

We know that ||gm ABCD and ||gm ABEF with equal bases and between the same parallels so that there area are same.

ar (||gm ABEF) = ar (||gm ABCD)

⇒ ar (||gm ABEF) = 29cm2 …(1) [Since, ar ||gm ABCD = 29cm2]

also, ar (||gm ABEF = base × height)

29 = AB × height [From (1)]

⇒ 29 = 5.8 × height

⇒ Height = 29/5.8 = 5

∴ Height of parallelogram ABEF is 5cm

(b) Given:

From fig (2)

area of ∆ABD is 24 sq. units. AB = 8 units

To find:

Height of ABC

Now, let us find

We know that ar ∆ABD = 24 sq. units …(1)

So, ar ∆ABD = ∆ABC …(2)

From (1) and (2)

ar ∆ABC = 24 sq. units

⇒ ½ × AB × height = 24

⇒ ½ × 8 × height = 24

⇒ 4 × height = 24

⇒ Height = 24/4 = 6

∴ Height of ∆ABC = 6 sq. units

(c) Given:

From fig (3)

In ||gm ABCD, E and F are mid points of sides AB and CD respectively.

ar (||gm ABCD) = 36cm2

To find:

(i) State the area of ∆ APD.

(ii) Name the parallelogram whose area is equal to the area of ∆ APD.

Now, let us find

(i) we know that ∆ APD and ||gm ABCD are on the same base AD and between the same parallel lines AD and BC.

ar (∆ APD) = ½ ar (||gm ABCD) …(1)

⇒ ar (||gm ABCD) = 36cm2 …(2)

From (1) and (2)

ar (∆ APD) = ½ × 36 = 18 cm2

(ii) we know that E and F are mid-points of AB and CD

In ∆CPD, EF || PC

Also, EF bisects the ||gm ABCD in two eual parts.

So, EF || AD and AE || DF

AEFD is a parallelogram.

ar (||gm AEFD) = ½ ar (||gm ABCD) …(3)

From (1) and (3)

ar (∆APD) = ar (||gm AEFD)

∴ AEFD is the required parallelogram which is equal to area of ∆APD.

11. (a) In the figure (1) given below, ABCD is a parallelogram. Points P and Q on BC trisect BC into three equal parts. Prove that:

area of ∆APQ = area of ∆DPQ = 1/6 (area of ||gm ABCD)

(b) In the figure (2) given below, DE is drawn parallel to the diagonal AC of the quadrilateral ABCD to meet BC produced at the point E. Prove that area of quad. ABCD = area of ∆ABE.

(c) In the figure (3) given below, ABCD is a parallelogram. O is any point on the diagonal AC of the parallelogram. Show that the area of ∆AOB is equal to the area of ∆AOD. Solution

(a) Given:

From fig (1)

In ||gm ABCD, points P and Q trisect BC into three equal parts.

To prove:

area of ∆APQ = area of ∆DPQ = 1/6 (area of ||gm ABCD)

Firstly, let us construct: through P and Q, draw PR and QS parallel to AB and CD.

Proof:

ar (∆APD) = ar (∆AQD) [Since, ∆APD and ∆AQD lie on the same base AD and between the same parallel lines AD and BC]

⇒ ar (∆APD) – ar (∆AOD) = ar (∆AQD) – ar (∆AOD) [On subtracting ar ∆AOD on both sides]

⇒ ar (∆APO) = ar (∆OQD) ….. (1)

⇒ ar (∆APO) + ar (∆OPQ) = ar (∆OQD) + ar (∆OPQ) [On adding ar ∆OPQ on both sides]

⇒ ar (∆APQ) = ar (∆DPQ) …(2)

We know that, ∆APQ and ||gm PQSR are on the same base PQ and between same parallel lines PQ and AD.

ar (∆APQ) = ½ ar (||gm PQRS) …(3)

Now,

[ar (||gm ABCD)/ar (||gm PQRS)] = [(BC×height)/(PQ×height)] = [(3PQ×height)/(1PQ×hight)]

⇒ ar (||gm PQRS) = 1/3 ar (||gm ABCD) …(4)

by using (2), (3), (4), we get

ar (∆APQ) = ar (∆DPQ)

= ½ ar (||gm PQRS)

= ½ × 1/3 ar (||gm ABCD)

= 1/6 ar (||gm ABCD)

Hence, proved.

(b) Given:

In the figure (2) given below, DE || AC the diagonal of the quadrilateral ABCD to meet at point E on producing BC. Join AC, AE.

To prove:

area of quad. ABCD = area of ∆ABE

Proof:

We know that, ∆ACE and ∆ADE are on the same base AC and between the same parallelogram.

ar (∆ACE) = ar (∆ADC)

Now by adding ar (∆ABC) on both sides, we get

ar (∆ACE) + ar (∆ABC) = ar (∆ADC) + ar (∆ABC)

⇒ ar (∆ ABE) = ar quad. ABCD

Hence, proved.

(c) Given:

From fig (3)

In ||gm ABCD, O is any point on diagonal AC.

To prove:

area of ∆AOB is equal to the area of ∆AOD

Proof:

Let us join BD which meets AC at P.

In ∆ABD, AP is the median.

ar (∆ABP) = ar (∆ADP) …(1)

Similarly, ar (∆PBO) = ar (∆PDO) …(2)

Now add (1) and (2), we get

ar (∆ABO) = ar (∆ADO) …(3)

So, ∆AOB = ar ∆AOD

Hence proved.

12. (a) In the figure given, ABCD and AEFG are two parallelograms.

Prove that area of || gm ABCD = area of || gm AEFG.

(b) In the fig. (2) Given below, the side AB of the parallelogram ABCD is produced to E. A straight line through A is drawn parallel to CE to meet CB produced at F and parallelogram BFGE is Completed prove that area of || gm BFGE=Area of || gm ABCD.

(c) In the figure (3) given below AB || DC || EF, AD || BE and DE || AF. Prove the area of DEFH is equal to the area of ABCD.

Solution:

(a) Given:

From fig (1)

ABCD and AEFG are two parallelograms as shown in the figure.

To prove:

area of || gm ABCD = area of || gm AEFG

Proof:

let us join BG.

We know that,

ar (∆ABG) = ½ (ar ||gm ABCD) …(1)

Similarly,

ar (∆ABG) = ½ (ar ||gm AEFG) …(2)

From (1) and (2)

½ (ar ||gm ABCD) = ½ (ar ||gm AEFG)

So,

ar ||gm ABCD = ar ||gm AEFG)

Hence, proved.

(b) Given:

From fig (2)

A parallelogram ABCD in which AB is produced to E. A straight line through A is drawn parallel to CE to meet CB produced at F and parallelogram BFGE is Completed.

To prove:

area of || gm BFGE=Area of || gm ABCD

Proof:

Let us join AC and EF.

We know that,

ar (∆AFC) = ar (∆AFE) ...(1)

Now, subtract ar (∆ABF) on both sides, we get

ar (∆AFC) – ar (∆ABF) = ar (∆AFE) – ar (∆ABF)

⇒ ar (∆ABC) = ar (∆BEF)

Now multiply by 2 on both sides, we get

2× ar (∆ABC) = 2× ar (∆BEF)

⇒  ar (||gm ABCD) = ar (||gm BFGE)

Hence, proved.

(c) Given:

From fig (3)

AB || DC || EF, AD || BE and DE || AF

To prove:

area of DEFH = area of ABCD

Proof:

We know that,

DE || AF and AD || BE

It is given that ADEG is a parallelogram.

So,

ar (||gm ABCD) = ar (||gm ADEG) …(1)

Again, DEFG is a parallelogram.

ar (||gm DEFH) = ar (||gm ADEG) …(2)

From (1) and (2)

ar (||gm ABCD) = ar (||gm DEFH)

⇒  ar ABCD = ar DEFH

Hence, proved.

13. Any point D is taken on the side BC of, a ∆ ABC and AD is produced to E such that AD=DE, prove that area of ∆ BCE = area of ∆ ABC.

Solution

Given:

In ∆ABC, D is taken on the side BC.

AD produced to E such that AD = DE

To prove:

area of ∆ BCE = area of ∆ ABC

Proof:

In ∆ABE, it is given that AD = DE

So, BD is the median of ∆ABE

ar (∆ABD) = ar (∆BED) …(1)

Similarly,

In ∆ACE, CD is the median of ∆ACE

ar (∆ACD) = ar (∆CED) …(2)

By adding (1) and (2), we get

ar (∆ABD) + ar (∆ACD) = ar (∆BED) + ar (∆CED)

⇒ ar (∆ABC) = ar (∆BCE)

Hence, proved.

14. ABCD is a rectangle and P is mid-point of AB. DP is produced to meet CB at Q. Prove that area of rectangle ∆BCD = area of ∆ DQC.

Solution

Given:

ABCD is a rectangle and P is mid-point of AB. DP is produced to meet CB at Q.

To prove:

area of rectangle ∆BCD = area of ∆ DQC

Proof:

In ∆APD and ∆BQP

AP = BP [Since, D is the mid-point of AB]

∠DAP = ∠QBP [each angle is 90o]

∠APD = ∠BPQ [vertically opposite angles]

So, ∆APD ≅ ∆BQP [By using ASA postulate]

ar (∆APD) = ar (∆BQP)

Now,

ar ABCD = ar (∆APD) + ar PBCD

= ar (∆BQP) + ar PBCD

= ar (∆DQC)

Hence proved.

15. (a) In the figure (1) given below, the perimeter of parallelogram is 42 cm. Calculate the lengths of the sides of the parallelogram.

(b) In the figure (2) given below, the perimeter of ∆ ABC is 37 cm. If the lengths of the altitudes AM, BN and CL are 5x, 6x, and 4x respectively, Calculate the lengths of the sides of ∆ABC.

(c) In the fig. (3) Given below, ABCD is a parallelogram. P is a point on DC such that area of ∆DAP = 25 cm² and area of ∆BCP = 15 cm². Find

(i) area of || gm ABCD

(ii) DP: PC.

Solution

(a) Given:

The perimeter of parallelogram ABCD = 42 cm

To find:

Lengths of the sides of the parallelogram ABCD.

From fig (1)

We know that,

AB = P

Then, perimeter of ||gm ABCD = 2 (AB + BC)

42 = 2(P + BC)

⇒ 21 = P + BC

⇒ BC = 21 – P

So, ar (||gm ABCD) = AB × DM

= P ×6

= 6P …(1)

Again, ar (||gm ABCD) = BC × DN

= (21 – P)× 8

= 8(21 – P) …(2)

From (1) and (2), we get

6P = 8(21 – P)

⇒ 6P = 168 – 8P

⇒ 6P + 8P = 168

⇒ 14P = 168

⇒ P = 168/14 = 12

Hence, sides of ||gm are

AB = 12cm and BC = (21 – 12)cm = 9cm

(b) Given:

The perimeter of ∆ ABC is 37 cm. The lengths of the altitudes AM, BN and CL are 5x, 6x, and 4x respectively.

To find:

Lengths of the sides of ∆ABC. i.e., BC, CA and AB.

Let us consider BC = P and CA = Q

From fig (2),

Then, perimeter of ∆ABC = AB + BC + CA

37 = AB + P + Q

⇒ AB = 37 – P – Q

Area (∆ABC) = ½ × base × height

= ½ × BC × AM = ½ × CA × BN = ½ × AB × CL

= ½ × P × 5x = ½ × Q × 6x = ½ (37 – P – Q) × 4x

= 5P/2 = 3Q = 2(37 – P – Q)

Let us consider first two parts:

5P/2 = 3Q

⇒ 5P = 6Q

⇒ 5P – 6Q = 0 …(1)

25P – 30Q (multiplying by 5)…(2)

Let us consider second and third parts:

3Q = 2(37 – P – Q)

⇒ 3Q = 74 – 2P – 2Q

⇒ 3Q + 2Q + 2P = 74

⇒ 2P + 5Q = 74 …(3)

⇒ 12P + 30Q = 444 (multiplying by 6)…(4)

By adding (2) and (4), we get

37P = 444

⇒ P = 444/37 = 12

Now, substitute the value of P in equation (1), we get

5P – 6Q = 0

⇒ 5(12) – 6Q = 0

⇒ 60 = 6Q

⇒ Q = 60/6 = 10

Hence, BC = P = 12cm

CA = Q = 10cm

And AB = 37 – P – Q = 37 – 12 – 10 = 15cm

(c) Given:

ABCD is a parallelogram. P is a point on DC such that area of ∆DAP = 25 cm² and area of ∆BCP = 15 cm².

To Find:

(i) area of || gm ABCD

(ii) DP: PC

Now let us find,

From fig (3)

(i) we know that,

ar (∆APB) = ½ ar (||gm ABCD)

Then,

½ ar (||gm ABCD) = ar (∆DAP) + ar (∆BCP)

= 25 + 15

= 40cm2

So, ar (||gm ABCD) = 2×40 = 80cm2

(ii) we know that,

∆ADP and ∆BCP are on the same base CD and between same parallel lines CD and AB.

ar (∆DAP)/ar(∆BCP) = DP/PC

⇒ 25/15 = DP/PC

⇒ 5/3 = DP/PC

So, DP: PC = 5: 3

16. In the adjoining figure, E is mid-point of the side AB of a triangle ABC and EBCF is a parallelogram. If the area of ∆ ABC is 25 sq. units, find the area of || gm EBCF.

Solution

Let us consider EF, side of ||gm BCFE meets AC at G.

We know that, E is the mid-point and EF || BC

G is the mid-point of AC.

So,

AG = GC

Now, in ∆AEG and ∆CFG,

The alternate angles are: ∠EAG, ∠GCF

Vertically opposite angles are: ∠EGA = ∠CGF

So, AG = GC

Proved.

∴ ∆AEG ≅ ∆CFG

ar (∆AEG) = ar (∆CFG)

Now,

ar (||gm EBCF) = ar BCGE + ar (∆CFG)

= ar BCGE + ar (∆AEG)

= ar (∆ABC)

We know that, ar (∆ABC) = 25sq. units

Hence, ar (||gm EBCF) = 25sq. units

17. (a) In the figure (1) given below, BC || AE and CD || BE. Prove that: area of ∆ABC= area of ∆EBD.

(b) In the figure (2) given below, ABC is right angled triangle at A. AGFB is a square on the side AB and BCDE is a square on the hypotenuse BC. If AN ⊥ ED, prove that:

(i) ∆BCF ≅ ∆ ABE.

(ii) area of square ABFG = area of rectangle BENM.

Solution

(a) Given:

From fig (1)

BC || AE and CD || BE

To prove:

area of ∆ABC= area of ∆EBD

Proof:

By joining CE.

We know that, from ∆ABC and ∆EBC

ar (∆ABC) = ar (∆EBC) …(1)

From EBC and ∆EBD

ar (∆EBC) = ar (∆EBD) ...(2)

From (1) and (2), we get

ar (∆ABC) = ar (∆EBD)

Hence, proved.

(b) Given:

ABC is right angled triangle at A. Squares AGFB and BCDE are drawn on the side AB and hypotenuse BC of ∆ABC. AN ⊥ ED which meets BC at M.

To prove:

(i) ∆BCF ≅ ∆ ABE.

(ii) area of square ABFG = area of rectangle BENM

From the figure (2)

(i) ∠FBC = ∠FBA + ∠ABC

So,

∠FBC = 90o + ∠ABC …(1)

∠ABE = ∠EAC + ∠ABC

So,

∠ABE = 90o + ∠ABC …(2)

From (1) and (2), we get

∠FBC = ∠ABE …(3)

So, BC = BE

Now, in ∆BCF and ∆ABE

BF = AB

By using SAS axiom rule of congruency,

∴ ∆BCF ≅ ∆ ABE

Hence proved.

(ii) we know that,

∆BCF ≅ ∆ ABE

So, ar (∆BCF) = ar (∆ABE) …(4)

⇒ ∠BAG + ∠BAC = 90o + 90= 180o

So, GAC is a straight line.

Now, from ∆BCF and square AGFB

ar (∆BCF) = ½ ar (square AGFB) …(5)

From ∆ABE and rectangle BENM

ar (∆ABE) = ½ ar (rectangle BENM) …(6)

From (4), (5) and (6)

½ ar (square AGFB) = ½ ar (rectangle BENM)

ar (square AGFB) = ar (rectangle BENM)

Hence, proved.