# ICSE Solutions for Selina Concise Chapter 16 Area Theorems Class 9 Maths

**Exercise 16(A)**

**1. In the given figure, if area of triangle ADE is 60 cm**

^{2}, state, given reason, the area of :**(i) Parallelogram ABED;**

**(ii) Rectangle ABCF;**

**(iii) Triangle ABE.**

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

Area of ABED = 2 (Area of ADE) = 120 cm^{2 }.

**(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 cm^{2 }.

**(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 cm^{2 }.

**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.**

**Answer**

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.

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.

**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:**

**Answer**

**(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: Area (△POS) + 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 (△POS)= (1/2)Area (ㅁPSLM).

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

Area(△POS) = (1/2)Area(◻️LMQR)

⇒ Area(△POS) + Area(△QOR) = (1/2)Area (ㅁPSLM)+ (1/2)Area(ㅁLMQR)

⇒ Area(△POS) + Area(△QOR) = (1/2)[Area(ㅁPSLM) + Area(ㅁLMQR)

⇒ Area(△POS) + Area(△QOR) = (1/2)[Area(ㅁPQRS)]

**(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,

Area (△POS) = Area(△POQ) **...(1)**

Similarly, since OR is the median of the triangle QRS, we have,

Area(△QOR) = Area(△SOR)** ...(2)**

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

Area(△POS) + Area(△QOR) = Area(△POQ) + Area(△SOR)

Hence, Proved.

**4. In ABCD, P is a point on side AB and Q is a point on side BC.**

**Prove that:**

**(i) △COP and △AQD are equal in area.(ii) Area(△AQD) = Area(△APD) + Area(△CPB)**

**Answer**

**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.

**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 cm ^{2};**

**(i) State the area of the triangle BEC.**

**(ii) Name the parallelogram which is equal in area to the triangle BEC.**

**Answer**

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

So, Area(△BEC) = (1/2)×Area(ㅁABCD) = (1/2)×48= 24cm^{2 }.

**(ii) **Area(ㅁANMD) = Area(ㅁBNMC)

= (1/2)Area(ㅁABCD)

= (1/2)×2×Area(△BEC)

=Area(△BEC)

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

**6. In the following figure, CE is drawn parallel to diagonals DB of the quadrilateral ABCD which meets AB produced at point E.**

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

Ar.(△DCB) = Ar.(△DEB)

⇒ Ar.(△DCB + △ADB) = Ar.(△DEB+△ADB)

⇒ Ar.(ABCD) = Ar.(△ADE)

Hence proved.

**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.**

**Answer**

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

∴ Ar.(△APB) = (1/2)Ar.(parallelogram ABCD) **...(i)**

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

∴Ar(△ADQ) = (1/2)Ar.(parallelogram ABCD) **...(ii) **

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

∴ Ar.(△APB) + Ar.(△ADQ) = Ar.(parallelogram ABCD)

⇒ Ar.(quad ADQB) - Ar.(△BPQ) = Ar.(parallelogram ABCD)

⇒ Ar.(quad ADQB)- Ar.(△BPQ) = Ar.(quad ADQB)- Ar.(△DCQ)

⇒ Ar.(△BPQ) = Ar.(△DCQ)

Subtracting Ar.(△PCQ) from both sides, we get

Ar.(△BPQ) - Ar.(△PCQ) = Ar.(△DCQ) - Ar.(△PCQ)

⇒ Ar.(△BCP) = Ar.(△DPQ)

Hence proved.

,

**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.**

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

Ar.(△EDG) = Ar.(△EGA)

Subtracting △EOG from both sides, we have

Ar.(△EOD) = Ar.(△GOA)

**...(i)**

Similarly,

Ar.(△DPC) = Ar.(△BPF)

**...(ii)**

Now,

Ar.(△GDF) = Ar.(△GOA) + Ar.(△BPF) +Ar.(pen ABPDO)

= Ar.(△EOD)+ Ar.(△DPC) + Ar. (pen ABPDO)

= Ar.(pen ABCDE)

Hence, proved.

**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.**

△ABC and △BPC are on the same base BC and between the same parallel lines AP and BC.

∴ Ar(△ABC) = Ar.(△BPC) **...(i)**

△BPC ad △BQP are on the same base BP and between the same parallel lines BP and CQ.

∴Ar.(△BPC) = Ar.(△BQP) **...(ii)**

From (i) and (ii), we get

∴Ar.(△ABC) =Ar. (△BQP)

Hence, proved.

**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) ****△EAC ≅△BAF**

**(ii) Area of the square ABDE ****= Area of the rectangle ARHF.**

**Answer**

**(i) ∠EAC = ∠EAB + ∠BAC**

∠EAC = 90°

∠EAC = 90

**+ ∠BAC ...(i)**

**∠BAF = ∠FAC + ∠BAC**

∠BAF = 90°

∠BAF = 90

**+ ∠BAC ...(ii)**

**From (i) and (ii), we get**

∠EAC = ∠BAF

In △EAC and △BAF, we have, EA = AB

∠EAC = ∠BAF and AC = AF

∴ △EAC ≅ △BAF (SAS axiom of congruency)

(ii) Since △ABC is a right triangle, we have,

AC

⇒ AB

⇒ AB

⇒ AB

⇒ AB

⇒ 2AB

⇒ AB

⇒ AB

⇒ AB

⇒ Area(□ABDE) = Area(rectangle ARHF)

∠EAC = ∠BAF

In △EAC and △BAF, we have, EA = AB

∠EAC = ∠BAF and AC = AF

∴ △EAC ≅ △BAF (SAS axiom of congruency)

(ii) Since △ABC is a right triangle, we have,

AC

^{2}= AB^{2}+ BC^{2}[Using Pythagoras Theorem in △ABC]⇒ AB

^{2}= AC^{2}- BC^{2}⇒ AB

^{2}= (AR + RC)^{2}- (BR^{2}+ RC^{2}) [Since AC = AR + RC and Using Pythagoras Theorem in △BRC]⇒ AB

^{2}= AR^{2}+ 2AR×RC + RC^{2}- (BR^{2}+ RC^{2}) [Using the identity]⇒ AB

^{2}= AR^{2}+ 2AR×RC + RC^{2}- (AB^{2}- AR^{2}+ RC^{2}) [Using Pythagoras Theorem in △ABR]⇒ 2AB

^{2}= 2AR^{2}+ 2AR×RC⇒ AB

^{2}= AR(AR+ RC)⇒ AB

^{2}= AR×AC⇒ AB

^{2}= AR×AF⇒ Area(□ABDE) = Area(rectangle ARHF)

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

**(i) Area (ΔADC) = Area(ΔAEB).**

**(ii) Area (ΔBOD) = Area (ΔCOE).**

**Answer**

**(i)** In △ABC, D is midpoint of AB and E is the midpoint of AC.

AD/AB =AE/AC

DE is parallel to BC.

∴ Ar.(△ADC) = Ar.(△BDC) = (1/2)Ar.(△ABC)

Again,

∴ Ar.(△AEB)= Ar.(△BEC) = (1/2)Ar.(△ABC)

From the above two equations, we have

Area(△ADC) = Area(△AEB).

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)

**12. ABCD and BCFE are parallelograms. If area of triangle EBC = 480 cm ^{2}; 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.**

**Answer**

**(i)**Since, △EBC and parallelogram ABCD are on the same base BC and between the same parallels i.e. BC॥AD.

∴ Ar.(△EBC) = (1/2)×Ar.(parallelogram ABCD)

⇒ (parallelogram ABCD) = 2×Ar.(△EBC)

= 2×480 cm

^{2}

= 960 cm

^{2}

**(ii)**Parallelograms on same base and between same parallels are equal in area

Area of BCFE = Area of ABCD = 960cm

^{2}

**(iii)**Area of triangle ACD = 480 = (1/2)×30×Altitude

Altitude = 32cm

**(iv)**The area of a triangle is half that of a parallelogram on the same base and between the same parallels.

Therefore,

Area(△ECF) = (1/2)Area (▢CBEF)

Similarly, Area(△BCE) = (1/2)Area(◻️CBEF)

⇒ Area (△ECF) = Area(△BCE) = 480 cm

^{2}

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

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

△EFC = △AFD

Adding quadrilateral CBDF in both sides, we have

Area of ॥gm BDEC= Area of ΔABC.

**14. In the following, AC॥PS॥QR and PQ॥DB॥SR.**

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

Similarly, AQRC and APSC are also parallelograms.

Since ABC and parallelogram AQRC are on the same base AC and between the same parallels, then

Ar. (△ABC) = (1/2)Ar.(AQRC) **...(i)**

Similarly,

Ar.(△ADC)= (1/2)Ar. (APSC) **...(ii)**

⇒ Area of quadrilateral PQRS = 2 Area of quad. ABCD.

**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.**

**Answer**

Given: ABCD is a trapezium

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.

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

**Answer**

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

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

**Answer**

**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.

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

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(◻️ABCD) = 2×Area(△ABR) ** ...(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(△ABR) = Area(△APR) **...(2)**

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

Area(◻️ABCD) = 2×Area(△APR) **...(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(△APR) = (1/2)×Area(◻️ARQP) **...(4)**

Using (4) in equation(3), we have,

Area(◻️ABCD) = 2×(1/2)×Area(◻️ARQP)

⇒ Area(◻️ABCD) = Area(◻️ARQP)

Hence, Proved.

**Exercise 16(B) **

**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.**

**Answer**

**(i)** Suppose ABCD is a parallelogram** (given)**

AB = CD

**[ABCD is a parallelogram]**

AD = BC

**[ABCD is a parallelogram]**

AD = AD

**[common]**

By Side - Side - Side criterion of congruence, we have,

△ABC ≅ △ADC

Area of congruent triangles are equal.

Therefore, Area of ABC = Area of ADC

**(ii)**Consider the following figure:

And, Ar.(△ADC) = (1/2)DC×AP

∴ [Area(△ABD)]/[Area (△ADC)]= [(1/2)BD×AP]/[(1/2)DC×AP] = BD/DC,

Hence, proved.

**(iii)**Consider the following figure :

Here,

Ar. (△ABC) = (1/2)BM×AC

⇒ Ar.(△ADC) = (1/2)DN×AC

∴ [Area(△ABC)]/[Area(△ADC)] = [(1/2)BM×AC]/[(1/2)DN×AC] = BM/DN

Hence, proved.

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

**Answer**

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 EBC

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.

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

**Area of triangle ABE = (1/4)area of triangle ABC. **

**Answer**

∴ Area (△ABD) = Area(△ACD)

Area(△ABD) = (1/2)Area(△ABC)

**...(i)**

In △ABD, E is the mid - point of AD. Therefore BE is the median.

∴Area(△BED) = Area(△ABE)

⇒Area(△BED) = (1/2)Area(△ABD)

⇒Area(△BED) = (1/2)×(1/2)Area(△ABC)

**[From equation(i)]**

⇒Area(△BED) = (1/4)Area(△ABC)

**4. ABCD is a parallelogram. P and Q are the mid-points of sides AB and AD respectively.**

**Prove that area of triangle APQ = 1/8 of the area of parallelogram ABCD.**

**Answer**

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(△ABD) = Area (△DBC)

= (1/2)Area(parallelogram ABCD)

**...(i)**

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

∴Area(△APD) = Area(△DPB)

= (1/2)Area(△ABD)

= (1/2)×(1/2)Area(parallelogram ABCD)

**[from equation (i)]**

= (1/4)Area(parallelogram ABCD)

**...(ii)**

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

∴ Area(△APQ) = Area(△DPQ)

= (1/2)Area(△APD)

= (1/2)×(1/4)Area(parallelogram ABCD)

**[from equation(ii)]**

Area(△APQ) = (1/8)Area(parallelogram ABCD),

Hence, proved.

**5. The base BC of triangle ABC is divided at D so that BD = 1/2 DC.**

**Prove that area of ΔABD = 1/3 of the area of ΔABC.**

**Answer**

∴ Ar.(△ABD) : Ar.(△ADC) = 1:2

But, Ar. (△ABD) + Ar.(△ADC) = Ar.(△ABC)

⇒ Ar.(△ABD) + 2Ar.(△ABD) = Ar.(△ABC)

⇒ 3Ar.(△ABD) = Ar.(△ABC)

⇒ Ar.(△ABD) = (1/3)Ar.(△ABC)

**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.**

**Answer**

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

(Area of DPB)/(Area of PCB) = DP/PC = 3/2**Given:** Area of △DPB = 30sq.cm

Let 'x' bet the area of the triangle PCB

Therefore, we have,

30/x = 3/2

⇒ x = (30/3)×2 = 20 sq.cm.

So, area of △PCB = 20sq.cm

Consider the following figure.

Area(△CDB) = Area(△DPB) + Area(△CPB)

= 30+20

= 50 sq.cm

Diagonal of the parallelogram divides it into two triangles △ADB and △CDB of equal area.

Therefore,

Area(∥gm ABCD) = 2×△CDB

= 2×50 = 100 sq.cm.

**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 cm ^{2}; find the area of parallelogram**

**Answer**BC = CE

**(given)**

Also, in parallelogram ABCD, BC = AD

⇒ AD = CE

Now, in △ADF and △ECF, we have

AD = CE

∠ADF = ∠ECF

**(Alternate angles)**

∠DAF = ∠CEF

**(Alternate angles)**

∴△ADF ≅ △ECF

**(ASA Criterion)**

⇒ Area(△ADF) = Area(△ECF)

**...(1)**

Also, in △FBE, FC is the median

**(Since BC = CE)**

⇒ Area(△BCF) = Area(△ECF)

**...(2)**

From (1) and (2),

Area(△ADF) = Area(△BCF)

**...(3)**

Again, △ADF and △BDF are on the base DF and between parallels DF and AB.

⇒ Area(△BDF) = Area(△ADF)

**...(4)**

From (3) and (4),

Area(△BDF) = Area(△BCF) = 30cm

^{2}

⇒ Area(△BCD) = Area(△BDF) + Area(△BCF) = 30+ 30 = 60cm

^{2}

Hence, Area of parallelogram ABCD = 2×Area(△BCD) = 2×60 = 120cm

^{2}

**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)**

**Answer**

**In ∆ABC,**

R and Q are the mid- points of AC and BC respectively.

⇒ RQ ॥ AB

that is RQ॥PB

So, area(∆PBQ) = area(∆APR) ...(i)

R and Q are the mid- points of AC and BC respectively.

⇒ RQ ॥ AB

that is RQ॥PB

So, area(∆PBQ) = area(∆APR) ...(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(∆PBQ) = area(∆PQR) ...(ii)

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(∆PBQ) = area(∆PQR) ...(ii)

**(Diagonal of a parallelogram divide the parallelogram in two triangles with equal area)**

from (i) and (ii),

area(∆PQR) = area(PBQ) = area(∆APR) ...(iii)

Similarly, P and Q are the mid-points of AB and BC respectively.

⇒ PQ ॥ AC that is PQ ॥ RC

So, quadrilateral PQCR is a parallelogram.

Also, area(∆RQC) = area(∆PQR) ...(iv)

from (i) and (ii),

area(∆PQR) = area(PBQ) = area(∆APR) ...(iii)

Similarly, P and Q are the mid-points of AB and BC respectively.

⇒ PQ ॥ AC that is PQ ॥ RC

So, quadrilateral PQCR is a parallelogram.

Also, area(∆RQC) = area(∆PQR) ...(iv)

**(Diagonal of a parallelogram divide the parallelogram in two triangles with equal area)**

From (iii) and (iv) ,

area(∆PQR) = area(∆PBQ) = (∆RQC) = area(∆APR)

So, area(∆PBQ) = (1/4)area(∆ABC) ...(v)

Also, since S is the mid - point of PQ,

BS is the median of ∆PBQ

So, area(∆QSB) = (1/2)area(∆PBQ)

From(v),

Area(∆QSB) = (1/2)×(1/4) area(∆ABC)

⇒ area(∆ABC) = 8

From (iii) and (iv) ,

area(∆PQR) = area(∆PBQ) = (∆RQC) = area(∆APR)

So, area(∆PBQ) = (1/4)area(∆ABC) ...(v)

Also, since S is the mid - point of PQ,

BS is the median of ∆PBQ

So, area(∆QSB) = (1/2)area(∆PBQ)

From(v),

Area(∆QSB) = (1/2)×(1/4) area(∆ABC)

⇒ area(∆ABC) = 8

**×**

**area(∆QSB)**

**Exercise 16(C)**

**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.**

**Answer**

**(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:

(Area of △DOC)/(Area of △BOC) = DO/BO = 1 **...(1)**

Similarly,

(Area of △DOA)/(Area of △BOA) = DO/BO = 1 **...(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.

**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.**

**Answer**

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

(Area of △APD)/(Area of △BPD) = AP/BP = 1/2

Area of parallelogram ABCD = 324sq.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,

Area(△ABD) = (1/2) ×Area[॥gm ABCD]

⇒ Area(△ABD) = 324/2 = 162 sq.cm

From the diagram it is clear that,

Area (△ABD) = Area(△APD) + Area(△BPD)

⇒ 162 = Area(△APD) + 2Area(△APD)

⇒162 = 3Area(△APD)

⇒ Area(△APD) = 162/3

⇒ Area(△APD) = 54 sq. cm

Consider the triangles △AOP and △COD

∠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 similarity,

△AOP ~ △COD.

Hence, the corresponding sides are proportional.

AP/CD = OP/OD = AP/AB

= AP/(AP+ PB)

= AP/3AP

= (1/3)

**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.**

**Answer**

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

Consider the following figure.

Therefore, by midpoint theorem, we have, EF॥ BCTriangles BEF and CEF lie on the common base EF and between the parallels, EF and BC

Therefore, Ar.(△BEF) = Ar.(△CEF)

⇒ Ar.(△BOE) + Ar.(△EOF) = Ar.(△EOF) + Ar.(△COF)

⇒ Ar.(△BOE) = Ar.(△COF)

Now,

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

Thus, we have,

Subtracting Ar.△BOE on the both the sides, we have

Ar.△ABF - Ar.△BOE = Ar. △CBF - Ar.△BOE

Since,

⇒ Ar.△ABF - Ar.△BOE = Ar.△CBF - Ar.△COF

⇒ Ar.(quad. AEOF) = Ar.(△OBC),

Hence, proved.

**4. In parallelogram ABCD, P is mid-point of AB. CP and BD intersect each other at point O. If area of ΔPOB = 40 cm**

^{2}, and OP : OC = 1:2, find:**(i) Areas of ΔBOC and ΔPBC**

**(ii) Areas of ΔABC and parallelogram ABCD**

**Answer**Consider the triangles △POB and △COD

∠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,

Since P is the mid-point AP = BP, and AB = CD, we have,

Therefore, We have,

BP/CD = OP/OC = OB/OD = 1/2

⇒ OP : OC = 1:2

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

BP/CD = OP/OC = OB/OD = 1/2,

The ratio between the areas of two similar triangles is equal to the ratio between the square of the corresponding sides.

Here, △DOC and △POB are similar triangles.

[Ar.(△DOC)]/[Ar.(△POB)] = DC

^{2}/PB

^{2}

^{2}/PB

^{2}

^{2}/PB

^{2}

⇒ Ar.( ΔDOC ) = 4Ar, ( ΔPOB )

= 4 × 40

= 160 cm^{2 }

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

= 160 + 80

= 160 cm^{2 }

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

Therefore, ar. (ΔDBC ) = Ar. (ΔABC) = 240 cm^{2 }

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) = Ar.(ΔABC)/2

Ar. (ΔPBC) = 120 cm^{2 }**(ii)** From part (ii) we have,

Ar. (ΔABC) = 2Ar. (PBC) = 240 cm^{2 }

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) = 1/2 Ar. [||gm ABCD]

ar. [||gm ABCD] = 2 Ar. (ΔABC)

⇒ ar. [||gm ABCD] = 2 × 240

⇒ ar. [||gm ABCD] = 480 cm^{2 }

^{}

**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 Area (ΔBGD)**

**(ii) Area (ΔACD) = 3 Area (CGD)**

**(iii) Area (ΔBGC) = 1/3 Area (ΔABC)**

**Answer**

**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,

ar.(ΔAGB)/ar.( ΔBGD) = 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,

Ar.(ΔAGC)/Ar.(ΔCGD) = 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)]

⇒ [Area(ΔABC)]/3 = [Area(ΔBGC)]

⇒ Area(ΔBGC) = (1/3)Area(ΔABC)

**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.**

**Answer**

**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 = (1/2)×base × altitude

∴(1/2)×x×6a = (1/2)×y×5a = (1/2)×(37 - x- y)×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.

**7 .**

(i) area of ΔADE.

(ii) if AE: EB = 4:5, find the area of ΔADB.

(iii) also, find the area of parallelogram ABCD.

**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 ΔADF is 60 cm**^{2}; find(i) area of ΔADE.

(ii) if AE: EB = 4:5, find the area of ΔADB.

(iii) also, find the area of parallelogram ABCD.

**Answer**ΔADF and ΔAFE have the same vertex A and their bases are on the same straight line DE.

∴[ar.(ΔADF)]/[ar.(ΔAFE)] = DF/FE

⇒60/[ar.(ΔAFE)] = 5/3

⇒ ar.(ΔAFE) = (60×3)/5 = 36 cm

Now, ar.(ΔADE) = ar.(ΔADF) + ar.(ΔAFE) = 60+ 36 = 96cm

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

∴ [ar.(ΔADE)]/[ar.(ΔEDB)] = AE/EB

**8. In following figure, BD is parallel to CA, E is midpoint of CA and BD = (1/2)CA.prove that : **

**Ar.(△ABC) = 2×ar.(****△DBC)**

**Answer**

Here BCED is a parallelogram, since BD = CE and BD ∥ CE.

ar.(△DBC) = ar.(△EBC) **(since they have the same base and are between the same parallels)**

In △ABC,

BE is the median,

So, ar.(△EBC) = (1/2)ar.(△ABC)

Now, ar.(△ABC) = ar.(△EBC) + ar.(△ABE)

Also, ar.(△ABC) = 2ar.(△EBC)

⇒ ar.(△ABC) = 2ar. (△DBC)

**9. In the figure, OAB is a triangle and AB∥DC.**

**If the area of ∆ CAD = 140 cm ^{2} and the area of ∆ ODC = 172 cm^{2}, find**

**(i) the area of ∆ DBC**

**(ii) the area of ∆ OAC**

**(iii) the area of ∆ ODB.**

**Answer**

△CAD = 140 cm

^{2}

△ODC = 172 cm

^{2}

AB ∥ CD

As Triangle DBC and △CAD have same base CD and between the same parallel lines Hence,

Area of △DBC = Area of △CAD = 140 cm

^{2}

⇒ Area of △OAC = Area of △CAD = Area of △ODC = 140 cm

^{2}+ 172cm

^{2}= 312 cm

^{2}

⇒ Area of △ODB = Area of △DBC + Area of △ODC = 140 cm

^{2}+ 172cm

^{2}= 312 cm

^{2}

**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.**

**Answer**

Join HF.

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

∴ AH = (1/2)AD and BF = (1/2)BC

Now, ABCD is a parallelogram.

⇒ AD = BC and AD ॥ BC

⇒ (1/2)AD = (1/2)BC and AD ॥ BC

⇒ AH = BF and AH॥ BF

⇒ ABFH is a parallelogram.

Since parallelogram FHAB and △FHE are on the same base FH and between the same parallels HF and AB,

ar.(△FHE) = (1/2)ar.(∥

^{m}FHAB)

**...(i)**

Similarly,

ar.(△FHG) = (1/2)ar.(∥

^{m}FHDC)

**...(ii)**

Adding (i) and (ii), we get

ar.(△FHE) + ar.(△FHG) = (1/2)ar.(∥

^{m}FHAB) + (1/2)ar.(∥

^{m}FHDC)

⇒ ar.(EFGH) = (1/2)[ar.(∥

^{m}FHAB) + ar.(∥

^{m}FHDC)]

⇒ ar.(EFGH) = (1/2)ar.(∥

^{m}ABCD)

**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.**

**Answer**

Join CX, DX and AY.

Now, triangles ADX and ACX are on the same base AX and between the parallels AB and DC.

∴ ar.(ΔADX) = ar.(ΔACX) **...(i)**

Also, triangles ACX and ACY are on the same base AC and between the parallels AC and XY.

∴ ar.(ΔACX) = ar.(ΔACY) **...(ii)**

From (i) and (ii), we get

ar.(ΔADX) = ar.(ΔACY)