ICSE Solutions for Selina Concise Chapter 16 Area Theorems Class 9 Maths
Exercise 16(A)
(i) Parallelogram ABED;
(ii) Rectangle ABCF;
(iii) Triangle ABE.
Answer
(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 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 .
(i) Quadrilateral CDEF is a parallelogram;
(ii) Area of quad. CDEF = Area of rect. ABDC + Area of ||gm. ABEF.
AnswerAfter 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.
(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.
Prove that:
(i) △COP and △AQD are equal in area.
(ii) Area(△AQD) = Area(△APD) + Area(△CPB)
(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.
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.
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= 24cm2 .
(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.
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.
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
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.
Joining PC we get
△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.
(i) △EAC ≅△BAF
(ii) Area of the square ABDE = Area of the rectangle ARHF.
Answer∠EAC = 90° + ∠BAC ...(i)
∠BAF = 90° + ∠BAC ...(ii)
∠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,
AC2 = AB2 + BC2 [Using Pythagoras Theorem in △ABC]
⇒ AB2 = AC2 - BC2
⇒ AB2 = (AR + RC)2 - (BR2 + RC2) [Since AC = AR + RC and Using Pythagoras Theorem in △BRC]
⇒ AB2 = AR2 + 2AR×RC + RC2 - (BR2 + RC2) [Using the identity]
⇒ AB2 = AR2 + 2AR×RC + RC2 - (AB2 - AR2 + RC2) [Using Pythagoras Theorem in △ABR]
⇒ 2AB2 = 2AR2 + 2AR×RC
⇒ AB2 = AR(AR+ RC)
⇒ AB2 = AR×AC
⇒ AB2 = 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).
(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 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.
Answer∴ Ar.(△EBC) = (1/2)×Ar.(parallelogram ABCD)
⇒ (parallelogram ABCD) = 2×Ar.(△EBC)
= 2×480 cm2
= 960 cm2
(ii) Parallelograms on same base and between same parallels are equal in area
Area of BCFE = Area of ABCD = 960cm2
(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 cm2
Prove that:
Area of ΔABC = Area of ॥gm BDEC.
AnswerHere, 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.
Prove that:
Area of quadrilateral PQRS = 2× Area of quad. ABCD.
AnswerIn 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.
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.
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
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.
[Join B and R]
AnswerJoin 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)
(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
Here AP ⊥ BC
Since,
And, Ar.(△ADC) = (1/2)DC×AP
∴ [Area(△ABD)]/[Area (△ADC)]= [(1/2)BD×AP]/[(1/2)DC×AP] = BD/DC,
Hence, proved.
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).
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.
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)
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.
Answer
In △ABC,
∴ 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)
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 cm2; find the area of parallelogram
Answer
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) = 30cm2
⇒ Area(△BCD) = Area(△BDF) + Area(△BCF) = 30+ 30 = 60cm2
Hence, Area of parallelogram ABCD = 2×Area(△BCD) = 2×60 = 120cm2
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)
AnswerR and Q are the mid- points of AC and BC respectively.
⇒ RQ ॥ AB
that is RQ॥PB
So, area(∆PBQ) = area(∆APR) ...(i)
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)
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 (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)
(i) Area (Δ DOC) = Area (Δ AOB).
(ii) Area (Δ DCB) = Area (Δ ACB).
(iii) ABCD is a parallelogram.
(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.
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)
E and F are the midpoints of the sides AB and AC.
Consider the following figure.
Triangles 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.
(i) Areas of ΔBOC and ΔPBC
(ii) Areas of ΔABC and parallelogram ABCD
Answer
∠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
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)] = DC2/PB2
⇒ Ar.( ΔDOC ) = 4Ar, ( ΔPOB )
= 4 × 40
= 160 cm2
Now consider Ar. ( ΔDBC ) = Ar. ( ΔDOC ) + Ar. (Δ BOC )
= 160 + 80
= 160 cm2
Two triangles are equal in the area if they are on equal bases and between the same parallels.
Therefore, ar. (ΔDBC ) = Ar. (ΔABC) = 240 cm2
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 cm2
(ii) From part (ii) we have,
Ar. (ΔABC) = 2Ar. (PBC) = 240 cm2
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 cm2
(i) Area (ΔABD) = 3 Area (ΔBGD)
(ii) Area (ΔACD) = 3 Area (CGD)
(iii) Area (ΔBGC) = 1/3 Area (ΔABC)
Answer
(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)
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∴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.
(i) area of ΔADE.
(ii) if AE: EB = 4:5, find the area of ΔADB.
(iii) also, find the area of parallelogram ABCD.
∴[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)
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)
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.
Answer△CAD = 140 cm2
△ODC = 172 cm2
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 cm2
⇒ Area of △OAC = Area of △CAD = Area of △ODC = 140 cm2 + 172cm2 = 312 cm2
⇒ Area of △ODB = Area of △DBC + Area of △ODC = 140 cm2 + 172cm2 = 312 cm2
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)
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)