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In figure, compute the area of quadrilateral ABCD.
Given:
DC = 17 cm, AD = 9 cm and BC = 8 cm
In ΔBCD we have
CD2 = BD2 + BC2
⇒ 172 = BD2 + 82
⇒ BD2 = 289 − 64
⇒ BD = 15
In ΔABD we have
AB2 + AD2 = BD2
⇒ 152 = AB2 + 92
⇒ AB2 = 225 − 81 = 144
⇒ AB = 12
ar(quad ABCD) = ar(ΔABD) + ar(ΔBCD)
ar(quad ABCD) = 1/2(12 × 9) + 1/2(8 × 17) = 54 + 68 = 122 cm2
ar(quad ABCD) = 1/2(12 × 9) + 1/2(8 × 15) = 54 + 60 = 114 cm2
In figure, PQRS is a square and T and U are, respectively, the midpoints of PS and QR. Find the area of ΔOTS if PQ = 8 cm.
From the figure,
T and U are mid points of PS and QR respectively
∴ TU ∥ PQ
⇒ TO ∥ PQ
Thus, in ΔPQS, T is the midpoint of PS and TO ∥ PQ
∴ TO = (1/2) PQ = 4 cm
Also, TS = (1/2) PS = 4 cm
∴ ar(ΔOTS) = (1/2)(TO × TS) = (1/2)(4 × 4)cm2 = 8 cm2
Compute the area of trapezium PQRS in figure
We have,
ar(trap. PQRS) = ar(rect. PSRT) + ar(ΔQRT)
⇒ ar(trap. PQRS) = PT × RT + 1/2(QT × RT)
= 8 × RT + 1/2(8 × RT) = 12 × RT
In ΔQRT, we have
QR2 = QT2 + RT2
⇒ RT2 = QR2 − QT2
⇒ RT2 = 172 − 82 = 225
⇒ RT = 15
Hence, Area of trapezium = 12 × 15 cm2 = 180 cm2
In figure, ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of ΔAOB.
Since, the midpoint of the hypotenuse of a right triangle is equidistant from the vertices
∴ CA = CB = OC
⇒ CA = CB = 6.5 cm
⇒ AB = 13 cm
In right angled triangle OAB, we have
AB2 = OB2 + OA2
⇒ 132 = OB2 + 122
⇒ OB2 = 132 − 122 = 169 − 144 = 25
⇒ OB = 5
∴ ar(ΔAOB) = (1/2)(12 × 5) = 30 cm2
In figure, ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.
Draw AL ⊥ DC, BM ⊥ DC then,
AL = BM = 4 cm and LM = 7 cm.
In Δ ADL, we have
AD2 = AL2 + DL2
⇒ 25 = 16 + DL2
⇒ DL = 3 cm
Similarly,
∴ x = CD = CM + ML + LD = (3 + 7 + 3) cm = 13 cm
ar(trap. ABCD) = 1/2(AB + CD) × AL = 1/2(7 + 13) × 4 cm2 = 40 cm2
In figure, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If OE = 2√5 cm, find the area of the rectangle.
Given OD = 10 cm and OE = 2√5cm
By using Pythagoras theorem
∴ OD2 = OE2 + DE2
∴ Area of rectangle OCDE = OE × DE
In figure, ABCD is a trapezium in which AB ∥ DC. Prove that ar(ΔAOD) = ar(ΔBOC)
Given: ABCD is a trapezium in which AB ∥ DC
To prove: ar(ΔAOD) = ar(ΔBOC)
Proof: Since, ΔADC and ΔBDC are on the same base DC and between same parallels AB and DC
Then, ar(ΔADC) = ar(ΔBDC)
⇒ ar(ΔAOD) + ar(ΔDOC) = ar(ΔBOC) + ar(ΔDOC)
⇒ ar(ΔAOD) = ar(ΔBOC)
In figure, ABCD, ABFE and CDEF are parallelograms. Prove that ar(ΔADE) = ar(ΔBCF).
Given that
ABCD is parallelogram ⇒ AD = BC
CDEF is parallelogram ⇒ DE = CF
ABFE is parallelogram ⇒ AE = BF
Thus, in Δs ADF and BCF, we have
AD = BC, DE = CF and AE = BF
So, by SSS criterion of congruence, we have
ΔADE ≅ ΔBCF
ar(ΔADE) = ar(ΔBCF)
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that: ar(ΔAPB) × ar(ΔCPD) = ar(ΔAPD) × ar(ΔBPC).
Construction: - Draw BQ ⊥ AC and DR ⊥ AC
Proof:-
L.H.S
= ar(ΔAPB) × ar(ΔCDP)
= (1/2) [(AP × BQ)] × (1/2 × PC × DR)
= (1/2 × PC × BQ) × (1/2 × AP × DR)
= ar(ΔAPD) × ar(ΔBPC).
= R.H.S
Hence proved.
In figure, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, show that ar(ΔABC) = ar(ΔABD).
Given that CD is bisected by AB at O
To prove: ar(ΔABC) = ar(ΔABD).
Construction: Draw CP ⊥ AB and DQ ⊥ AB.
Proof:
ar(ΔABC) = 1/2 × AB × CP⋅⋅⋅⋅⋅ (1)
ar(ΔABD) = 1/2 × AB × DQ ⋅⋅⋅⋅ (2)
In ΔCPO and ΔDQO
∠CPO = ∠DQO [Each 90°]
Given that, CO = OD
∠COP = ∠DOQ [Vertically opposite angles are equal]
Then, ΔCPO ≅ ΔDQO [By AAS condition]
∴ CP = DQ (3) [C.P.C.T]
Compare equation (1), (2) and (3)
∴ ar(ΔABC) = ar(ΔABD).
If P is any point in the interior of a parallelogram ABCD, then prove that area of the triangle APB is less than half the area of parallelogram.
Draw DN ⊥ AB and PM ⊥ AB
Now,
ar (∥gm ABCD) = AB × DN, ar(ΔAPB) = (1/2) (AB × PM)
Now, PM < DN
⇒ AB × PM < AB × DN
⇒ (1/2)(AB × PM) < (1/2)(AB × DN)
⇒ ar(ΔAPB) < 1/2 ar(∥ gm ABCD)
If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in area. If G is the mid-point of the median AD, prove that ar(ΔBGC) = 2ar(ΔAGC).
Draw AM ⊥ BC
Since, AD is the median of ΔABC
∴ BD = DC
⇒ BD = AM = DC × AM
⇒ (1/2)(BD × AM) = (1/2)(DC × AM)
⇒ ar(ΔABD) = ar(ΔACD) ⋅⋅⋅⋅⋅⋅ (1)
In ΔBGC, GD is the median
⇒ ar(ΔBGD) = ar(ΔCGD) ⋅⋅⋅⋅⋅⋅ (2)
In ΔACD, CG is the median
⇒ ar(ΔAGC) = ar(ΔCGD) ⋅⋅⋅⋅⋅ (3)
From (2) and (3) we have,
ar(ΔBGD) = ar(ΔAGC)
But, ar(ΔBGC) = 2ar(ΔBGD)
⇒ ar(ΔBGC) = 2ar(ΔAGC)
A point D is taken on the side BC of a ΔABC, such that BD = 2DC. Prove that ar(ΔABD) = 2ar(ΔADC).
Given that,
In ΔABC, BD = 2DC
To prove: ar(ΔABD) = 2ar(ΔADC).
Construction:
Take a point E on BD such that BE = ED
Proof: Since, BE = ED and BD = 2 DC
Then, BE = ED = DC
We know that median of triangle divides it into two equal triangles.
∴ In ΔABD, AE is the median.
Then, ar(ΔABD) = 2ar(ΔAED) ⋅⋅⋅⋅ (1)
In ΔAEC, AD is the median.
Then, ar(ΔAED) = 2ar(ΔADC) ⋅⋅⋅ (2)
Compare equation 1 and 2
ar(ΔABD) = 2ar(ΔADC).
ABCD is a parallelogram whose diagonals intersect at O .If P is any point on BO, prove that:
(i) ar(ΔADO) = ar(ΔCDO).
(ii) ar(ΔABP) = 2ar(ΔCBP).
Given that ABCD is the parallelogram
To Prove:
we know that diagonals of parallelogram bisect each other
∴ AO = OC and BO = OD
(i) In ΔDAC, since DO is a median.
Then ar(ΔADO) = ar(ΔCDO).
(ii) In ΔBAC, since BO is a median.
Then ar(ΔBAO) = ar(ΔBCO) ⋅⋅⋅⋅ (1)
In ΔPAC, since PO is a median.
Then ar(ΔPAO) = ar(ΔPCO) ⋅⋅⋅⋅ (2)
Subtract equation 2 from 1.
⇒ ar(ΔBAO) − ar(ΔPAO) = ar(ΔBCO) − ar(ΔPCO)
⇒ ar(ΔABP) = 2ar(ΔCBP).
ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F.
(i) Prove that ar(ΔADF) = ar(ΔECF).
(ii) If the area of ΔDFB = 3 cm2, find the area of ∥ gm ABCD.
In triangles ADF and ECF, we have
∠ADF = ∠ECF [Alternate interior angles, Since AD ∥ BE]
AD = EC [since AD = BC = CE]
And ∠DFA = ∠CFA [Vertically opposite angles]
So, by AAS congruence criterion, we have
ΔADF ≅ ΔECF
⇒ ar(ΔADF) = ar(ΔECF) and DF = CF .
Now, DF = CF
⇒ BF is a median in Δ BCD.
⇒ ar(ΔBCD) = 2ar(ΔBDF)
⇒ ar(ΔBCD) = 2 × 3 cm2 = 6 cm2
Hence, area of a parallelogram = 2ar(ΔBCD) = 2 × 6 cm2 = 12 cm2
ABCD is a parallelogram whose diagonals AC and BD intersect at O. A line through O intersects AB at P and DC at Q. Prove that ar(ΔPOA) = ar(ΔQOC).
In triangles POA and QOC, we have
∠AOP = ∠COQ
AO = OC
∠PAC = ∠QCA
So, by ASA congruence criterion, we have
ΔPOA ≅ ΔQOC
⇒ ar(ΔPOA) = ar(ΔQOC).
ABCD is a parallelogram. E is a point on BA such that BE = 2EA and F is point on DC such that DF = 2FC. Prove that AECF is a parallelogram whose area is one third of the area of parallelogram ABCD.
Draw FG ⊥ AB
BE = 2 EA and DF = 2FC
⇒ AB - AE = 2 AE and DC - FC = 2 FC
⇒ AB = 3 AE and DC = 3 FC
⇒ AE = (1/3) AB and FC = (1/3)DC ⋅⋅⋅⋅ (1)
But AB = DC
Then, AE = FC [opposite sides of ∥gm]
Thus, AE = FC and AE ∥ FC
Then, AECF is a parallelogram
Now, area of parallelogram AECF = AE × FG
⇒ ar(∥gm AECF) = 1/3 AB × FG from(1)
⇒ 3ar (∥gm AECF) = AB × FG ⋅⋅⋅ (2)
And ar(∥gm ABCD) = AB × FG ⋅⋅⋅ (3)
Compare equation 2 and 3
⇒ 3ar(∥gm AECF) = ar(∥gm ABCD)
⇒ ar(∥gm AECF) = 1/3 ar(∥gm ABCD)
In a triangle ABC, P and Q are respectively the mid points of AB and BC and R is the midpoint of AP. Prove that:
(i) ar(ΔPBQ) = ar(ΔARC).
(ii) ar(ΔPRQ) = 1/2 ar(ΔARC).
(iii) ar(ΔRQC) = 3/8 ar(ΔABC).
We know that each median of a triangle divides it into two triangles of equal area.
(i) Since CR is the median of ΔCAP
∴ ar(ΔCRA) = (1/2) ar(ΔCAP) ⋅⋅⋅⋅⋅ (1)
Also, CP is the median of a Δ CAB
∴ ar(ΔCAP) = ar(ΔCPB) ⋅⋅⋅⋅ (2)
From 1 and 2, we get
∴ ar(ΔARC) = (1/2) ar(ΔCPB) ⋅⋅⋅⋅⋅ (3)
PQ is the median of a ΔPBC
∴ ar(ΔCPB) = 2ar(ΔPBQ) ⋅⋅⋅ (4)
From 3 and 4, we get
∴ ar(ΔARC) = ar(ΔPBQ) ⋅⋅⋅⋅ (5)
(ii) Since QP and QR medians of triangles QAB and QAP respectively
∴ ar(ΔQAP) = ar(ΔQBP) ⋅⋅⋅⋅ (6)
And ar(ΔQAP) = 2ar(ΔQRP) ⋅⋅⋅⋅ (7)
From 6 and 7, we get
ar(ΔPRQ) = (1/2) ar(ΔPBQ) ⋅⋅⋅⋅⋅ (8)
From 5 and 8, we get
ar(ΔPRQ) = (1/2) ar(ΔARC)
(iii) Since, LR is a median of ΔCAP
∴ ar(ΔARC) = (1/2) ar(ΔCAD)
= 1/2 × (1/2) ar(ΔABC)
= (1/4) ar(ΔABC)
Since RQ is the median of ΔRBC.
∴ ar(ΔRQC) = (1/2) ar(ΔRBC)
= (1/2) {ar(ΔABC) − ar(ΔARC)}
= (1/2) {ar(ΔABC) – (1/4) ar(ΔABC)}
= (3/8) ar(ΔABC)
ABCD is a parallelogram. G is a point on AB such that AG = 2GB and E is point on DC such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:
(i) ar(ADEG) = ar(GBCE)
(ii) ar(ΔEGB) = (1/6) ar(ABCD)
(iii) ar(ΔEFC) = (1/2) ar(ΔEBF)
(iv) ar(ΔEGB) = 3/2 × ar(ΔEFC)
(v) Find what portion of the area of parallelogram is the area of ΔEFG.
Given: ABCD is a parallelogram
AG = 2 GB, CE = 2 DE and BF = 2 FC
To prove:
(i) ar(ADEG) = ar(GBCE).
(ii) ar(ΔEGB) = (1/6) ar(ABCD).
(iii) ar(ΔEFC) = (1/2) ar(ΔEBF).
(iv) ar(ΔEGB) = (3/2) × ar(ΔEFC)
Construction: Draw a parallel line to AB through point F and a perpendicular line to AB from C
(i) Since ABCD is a parallelogram
So, AB = CD and AD = BC
Consider the two trapezium s ADEG and GBCE
Since AB = DC, EC = 2DE, AG = 2GB
⇒ ED = (1/3) CD = (1/3) AB and EC = (2/3) CD = (2/3) AB
⇒ AG = (2/3) AB and BG = (1/3) AB
So, DE + AG = (1/3) AB + (2/3) AB = AB and EC + BG = (2/3) AB + (1/3) AB = AB
Since the two trapezium ADEG and GBCE have same height and their sum of two parallel sides are equal
Since
So, ar(ADEG) = ar(GBCE).
(ii) Since we know from above that
BG = (1/2) AB So
ar(ΔEGB) = (1/2) × GB × Height
ar(ΔEGB) = (1/2) × (1/3) × AB × Height
ar(ΔEGB) = (1/6) × AB × Height
ar(ΔEGB) = (1/6) ar(ABCD).
(iii) Since height if triangle EFC and EBF are equal. So
ar(ΔEFC) = (1/2) × FC × Height
ar(ΔEFC) = (1/2) × (1/2) × FB × Height
ar(ΔEFC) = (1/2) ar(EBF)
Hence, ar(ΔEFC) = (1/2) ar(ΔEBF).
(iv) Consider the trapezium in which
ar(EGBC) = ar(ΔEGB) + ar(ΔEBF) + ar(ΔEFC)
⇒ (1/2) ar(ABCD) = (1/6) ar(ABCD) + 2ar(ΔEFC) + ar(ΔEFC)
⇒ (1/3) ar(ABCD) = 3 ar(ΔEFC)
⇒ ar(ΔEFC) = (1/9) ar(ABCD)
Now from (ii) part we have
ar(ΔEGB) = (1/6) ar(ΔEFC)
ar(ΔEGB) = (3/2) × (1/9) ar(ABCD)
ar(ΔEGB) = (3/2) ar(ΔEFC)
∴ ar(ΔEGB) = (3/2) ar(ΔEFC)
(v) In the figure it is given that FB = 2CF.
Let CF = x and FB = 2x.
Now consider the two triangles CFI and CBH which are similar triangle.
So by the property of similar triangle CI = k and IH = 2k
Now consider the triangle EGF in which
ar(ΔEFG) = ar(ΔESF) + ar(ΔSGF)
ar(ΔEFG) = (1/2) SF × k + (1/2) SF × 2k
ar(ΔEFG) = (3/2) SF × k ⋅⋅⋅ (i)
ar(ΔEGBC) = ar(SGBF) + ar(ESFC)
ar(ΔEGBC) = (1/2)(SF + GB) × 2k + (1/2)(SF + EC) × k
ar(ΔEGBC) = (3/2) k × SF + (GB + (1/2)EC) × k
ar(ΔEGBC) = (3/2) k × SF + (1/3 AB + (1/2) × (2/3) AB) × k
(1/2) ar(ΔABCD) = (3/2) k × SF + (2/3) AB × k
⇒ ar(ΔABCD) = 3k × SF + (4/3) AB × k [Multiply both sides by 2]
⇒ ar(ΔABCD) = 3k × SF + 4/9 ar(ABCD)
⇒ k × SF = (5/27 )ar(ABCD) ⋅⋅⋅⋅ (2)
From 1 and 2 we have,
ar(ΔEFG) = (3/2) × (5/27) ar(ABCD)
ar(ΔEFG) = (5/18) ar(ABCD)
In figure, CD ∥ AE and CY ∥ BA.
(i) Name a triangle equal in area of ΔCBX
(ii) Prove that ar(ΔZDE) = ar(ΔCZA)
(iii) Prove that ar(BCZY) = ar(ΔEDZ)
Since, triangle BCA and triangle BYA are on the same base BA and between same parallel s BA and CY.
Then ar(ΔBCA) = ar(ΔBYA)
⇒ ar(ΔCBX) + ar(ΔBXA) = ar(ΔBXA) + ar(ΔAXY)
⇒ ar(ΔCBX) = ar(ΔAXY) ⋅⋅⋅⋅ (1)
Since, triangles ACE and ADE are on the same base AE and between same parallels CD and AE
Then, ar(ΔACE) = ar(ΔADE)
ar(ΔCZA) + ar(ΔAZE) = ar(ΔAZE) + ar(ΔDZE)
ar(ΔCZA) = ar(ΔDZE) ⋅⋅⋅ (2)
Adding ar(ΔCYG) on both sides , we get
⇒ ar(ΔCBX) + ar(ΔCYZ) = ar(ΔCAY) + ar(ΔCYZ)
⇒ ar(BCZY) = ar(ΔCZA) ⋅⋅⋅ (3)
⇒ ar(BCZY) = ar(ΔDZE)
In figure, PSDA is a parallelogram in which PQ = QR = RS and AP ∥ BQ ∥ CR. Prove that ar (ΔPQE) = ar(ΔCFD).
Given that PSDA is a parallelogram
Since, AP ∥ BQ ∥ CR ∥ DS and AD ∥ PS
Therefore, PQ = CD (equ. 1)
In triangle BED, C is the midpoint of BD and CF ∥ BE
Therefore, F is the midpoint of ED
⇒ EF = PE
EF = PE
Therefore, PE = FD (equ. 2)
In triangles PQE and CFD, we have
PE = FD
Therefore, ∠ EPQ = ∠FDC [Alternate angles]
So, by SAS criterion, we have
ΔPQE ≅ ΔDCF
⇒ ar(ΔPQE) = ar(ΔDCF)
In figure, ABCD is a trapezium in which AB || DC and DC = 40 cm and AB = 60 cm .If X and Y are, respectively, the mid points of AD and BC, prove that:
(i) XY = 50 cm
(ii) DCYX is a trapezium
(iii) ar(trap. DCYX) = (9/11) ar(XYBA).
(i) Join DY and produce it to meet AB produced at P.
In triangles BYP and CYD we have,
∠BYP = ∠ CYD [Vertically opposite angles]
∠DCY = ∠ PBY [Since, DC ∥ AP]
And BY = CY
(ΔBYP) ≅ (ΔCYD)
⇒ DY = Yp and DC = BP
⇒ Y is the midpoint of DP
Also, x is the midpoint of AD
Therefore, XY ∥ AP and XY ∥ (1/2) AP
⇒ XY = (1/2) (AB + BP)
⇒ XY = (1/2) (AB + DC)
⇒ XY = (1/2) (60 + 40)
= 50 cm
(ii) We have, XY ∥ AP
⇒ XY ∥ AB and AB ∥ DC
⇒ XY ∥ DC
⇒ DCYX is a trapezium
(iii) Since x and y are the mid points of Ad and BC respectively.
Therefore, trapezium DCYX and ABYX are of the same height say h cm
ar(trap. DCXY) = (1/2)(DC + XY) × h
⇒ ar(trap. DCXY) = (1/2) (50 + 40) × h cm2 = 45 h cm2
⇒ ar(trap. ABYX) = (1/2)(AB + XY) × h
⇒ ar(trap. ABYX) = (1/2)(60 + 50) × h cm2 = 55h cm2
ar(trap. DCYX) ar(trap. ABYX) = 45h/55h = 9/11
⇒ ar(trap. DCYX) = 9/11 ar(trap. ABYX)
In figure ABC and BDE are two equilateral triangles such that D is the midpoint of BC. AE intersects BC in F. Prove that:
(i) ar(ΔBDE) = (1/4) ar(ΔABC)
(ii) ar(ΔBDE) = (1/2) ar(ΔBAE)
(iii) ar(ΔBFE) = ar(ΔAFD)
(iv) ar(ΔABC) = 2 ar(ΔBEC)
(v) ar(ΔFED) = 1/8 ar(ΔAFC)
(vi) ar(ΔBFE) = 2 ar(ΔEFD)
Given that ABC and BDE are two equilateral triangles.
Let AB = BC = CA = x. Then, BD = x/2 = DE = BE
(i) We have,
Therefore, ar(ΔBDE) = 1/4 ar(ΔABC).
(ii) It is given that triangles, ABC and BED are equilateral triangles
∠ ACB = ∠DBE = 60°
⇒ BE ∥ AC (Since, Alternative angles are equal)
Trinagles BAF and BEC are on the same base BE and between same parallels BF and AC.
Therefore, ar(ΔBAE) = ar(ΔBEC)
⇒ ar(ΔBAE) = 2ar(ΔBDE) [Since , ED is a median of triangle EBC; ar(ΔBEC) = 2ar(ΔBDE) ]
∴ ar(ΔBDE) = (1/2) ar(ΔBAE)
(iii) Since, triangles ABC and BDE are equilateral triangles
∴ ∠ABC = 60° and ∠BDE = 60°
∠ABC = ∠BDE
⇒ AB ∥ DE (since, alternate angles are equal)
Triangles BED and AED are on the same base ED and between same parallels AB and DE.
Therefore, ar(ΔBED) = ar(ΔAED)
⇒ ar(ΔBED) − ar(ΔEFD) = ar(ΔAED) − ar(ΔEFD)
⇒ ar(ΔBEF) = ar(ΔAFD)
(iv) Since ED is the median of triangle BEC
Therefore, ar(ΔBEC) = 2ar(ΔBDE)
⇒ ar(ΔBEC) = 2 × (1/2) ar(ΔABC) [From 1, ar(ΔBDE) = (1/2) ar(ΔABC)]
⇒ ar(ΔBEC) = (1/2) ar(ΔABC)
⇒ ar(ΔABC) = 2ar(ΔBEC)
(v) ar(ΔAFC) = ar(ΔAFD) + ar(ΔADC)
⇒ ar(ΔBFE) + (1/2) ar(ΔABC) [using part (iii) , and AD is the median of triangle ABC ]
= ar(ΔBFE) + (1/2) × 4ar(ΔBDE) (using part (i))
= ar(ΔBFE) = 2ar(ΔFED) ⋅⋅⋅ (3)
ar(ΔBDE) = ar(ΔBFE) + ar(ΔFED)
⇒ 2ar(ΔFED) + ar(ΔFED)
⇒ 3ar(ΔFED) ⋅⋅⋅⋅ (4)
From 2, 3 and 4 we get,
ar(ΔAFC) = 2ar(ΔFED) + 2 × 3 ar(ΔFED) = 8 ar(ΔFED)
ar(ΔFED) = (1/8) ar(ΔAFC)
(vi) Let h be the height of vertex E, corresponding to the side BD in triangle BDE.
Let H be the height of vertex A, corresponding to the side BC in triangle ABC
From part (i)
ar(ΔBDE) = (1/4) ar(ΔABC)
⇒ (1/2) × BD × h = (1/4) (1/2 × BC × h)
⇒ BD × h = (1/4)(2BD × H)
⇒ h = (1/2) H ⋅⋅⋅ (1)
From part (iii)
ar(ΔBFE) = ar(ΔAFD)
ar(ΔBFE) = (1/2) × FD × H
ar(ΔBFE) = (1/2) × FD × 2h
ar(ΔBFE) = 2((1/2) × FD × h)
ar(ΔBFE) = 2ar(ΔEFD)
D is the midpoint of side BC of ΔABC and E is the midpoint of BD. If O is the midpoint of AE, Prove that ar(ΔBOE) = (1/8) ar(ΔABC).
D is the midpoint of sides BC of triangle ABC
E is the midpoint of BD and O is the midpoint of AE
Since AD and AE are the medians of triangles, ABC and ABD respectively
∴ ar(ΔABD) = (1/2) ar(ΔABC) ⋅⋅⋅⋅ (1)
∴ ar(ΔABE) = (1/2) ar(ΔABD) ⋅⋅⋅ (2)
OB is the median of triangle ABE
Therefore,
∴ ar(ΔBOE) = (1/2) ar(ΔABE)
From 1, 2 and 3, we have
∴ ar(ΔBOE) = (1/8) ar(ΔABC)
In figure, X and Y are the mid points of AC and AB respectively, QP ∥ BC and CYQ and BXP are straight lines. Prove that ar(ΔABP) = ar(ΔACQ).
Since X and Y are the mid points of AC and AB respectively.
Therefore, XY ∥ BC
Clearly, triangles BYC and BXC are on the same base BC and between the same parallels XY and BC
∴ ar(ΔBYC) = ar(ΔBXC)
⇒ ar(ΔBYC) − ar(ΔBOC) = ar(ΔBXC) − ar(ΔBOC)
⇒ ar(ΔBOY) = ar(ΔCOX)
⇒ ar(ΔBOY) + ar(ΔXOY) = ar(ΔCOX) + ar(ΔXOY)
⇒ ar(ΔBXY) = ar(ΔCXY) .... (2)
We observed that the quadrilaterals XYAP and XYAQ are on the same base XY and between same parallels XY and PQ.
∴ ar(quad. XYAP) = ar(quad XYQA) ⋅⋅⋅⋅ (2)
Adding 1 and 2, we get
∴ ar(ΔBXY) + ar(quad. XYAP) = ar(ΔCXY) + ar(quad XYQA)
⇒ ar(ΔABP) = ar(ΔACQ)
In figure, ABCD and AEFD are two parallelograms. Prove that
(i) PE = FQ
(ii) ar(ΔAPE) : ar(ΔPFA) = ar(ΔQFD): ar(ΔPFD)
(iii) ar(ΔPEA) = ar(ΔQFD)
Given that, ABCD and AEFD are two parallelograms
(i) In triangles, EPA and FQD
∠PEA = ∠QFD [corresponding angles]
∠EPA=∠FQD [corresponding angles]
PA = QD [opposite sides of parallelogram]
Then, ΔEPA ≅ ΔFQD [By AAS condition]
Therefore, EP = FQ [C.P.C.T]
(ii) Since triangles, PEA and QFD stand on equal bases PE and FQ lies between the same parallels EQ and AD
Therefore, ar(ΔPEA) = ar(ΔQFD) ... (1)
Since, triangles PEA and PFD stand on the same base PF and between same parallels PF and AD
Therefore, ar(ΔPFA) = ar(ΔPFD) .... (2)
Divide the equation 1 by equation 2
ar(ΔPEA) ar(ΔPFA) = ar(ΔQFD) ar(ΔPFD)
(iii) From part (i),
ΔEPA ≅ ΔFQD
Then, ar(ΔPEA) = ar(ΔQFD).
In figure, ABCD is a parallelogram. O is any point on AC. PQ ∥ AB and LM ∥ AD. Prove that: ar(∥ gm DLOP) = ar(∥gm BMOQ).
Since a diagonal of a parallelogram divides it into two triangles of equal area
Therefore, ar(ΔADC) = ar(ΔABC)
⇒ ar(ΔAPO) + ar(∥gm DLOP) + ar(ΔOLC)
⇒ ar(ΔAOM) + ar(∥gm BMOQ) + ar(ΔOQC) ... (1)
Since AO and Oc are diagonals of parallelograms AMOP and OQCL respectively.
∴ ar(ΔAPO) = ar(ΔAMO) .... (2)
And ar(ΔOLC) = ar(ΔOQC) .... (3)
Subtracting 2 and 3 from 1, we get
ar(∥gm DLOP) = ar(∥gm BMOQ).
In a triangle ABC, if L and M are points on AB and AC respectively such that LM ∥ BC. Prove that:
(i) ar(ΔLCM) = ar(ΔLBM)
(ii) ar(ΔLBC) = ar(ΔMBC)
(iii) ar(ΔABM) = ar(ΔACL)
(iv) ar(ΔLOB) = ar(ΔMOC)
(i) Clearly triangles LMB and LMC are on the same base LM and between the same parallels LM and BC.
∴ ar(ΔLMB) = ar(ΔLMC) ... (1)
(ii) We observe that triangles LBC and MBC are on the same base BC and between same parallels LM and BC.
∴ ar(ΔLBC) = ar(ΔMBC) ... (2)
(iii) We have,
ar(ΔLMB) = ar(ΔLMC) [From 1]
⇒ ar(ΔALM) + ar(ΔLMB) = ar(ΔALM) + ar(ΔLMC)
⇒ ar(ΔABM) = ar(ΔACL)
(iv) we have,
ar(ΔLBC) = ar(ΔMBC) [From 1]
⇒ ar(ΔLBC) − ar(ΔBOC) = ar(ΔMBC) − ar(ΔBOC)
⇒ ar(ΔLOB) = ar(ΔMOC).
In figure, D and E are two points on BC such that BD = DE = EC. Show that ar(ΔABD) = ar(ΔADE) = ar(ΔAEC).
Draw a line l through A parallel to BC.
Given that, BD = DE = EC
We observed that the triangles ABD and AEC are on the equal bases and between the same parallels l and BC. Therefore, their areas are equal.
Hence, ar(ΔABD) = ar(ΔADE) = ar(ΔAEC).
In figure, ABC is a right angled triangle at A, BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that
(i) ΔMBC ≅ ΔABD
(ii) ar(BYXD) = 2ar(ΔMBC)
(iii) ar(BYXD) = ar(ABMN)
(iv) ΔFCB ≅ ΔACE
(v) ar(CYXE) = 2ar(ΔFCB)
(vi) ar(CYXE) = ar(ACFG)
(vii) ar(BCED) = ar(ABMN) + ar(ACFG)
(i) In ΔMBC and ΔABD, we have
MB = AB
BC = BD
And ∠MBC = ∠ABD [since, ∠MBC and ∠ABC are obtained by adding ∠ABC to a right angle.]
So, by SAS congruence criterion, we have
ΔMBC ≅ ΔABD
⇒ ar(ΔMBC) = ar(ΔABD) ⋅⋅⋅⋅⋅ (1)
(ii) Clearly, triangle ABC and rectangle BYXD are on the same base BD and between the same parallels AX and BD.
∴ ar(ΔABD) = (1/2) ar(rect BYXD)
⇒ ar (rect BYXD) = 2ar(ΔABD)
⇒ ar (rect BYXD) = 2ar(ΔMBC) ⋅⋅⋅⋅ (2) [From equ .1]
(iii) Since triangles MBC and square MBAN are on the same base Mb and between the same parallels MB and NC.
∴ 2ar(ΔMBC) = ar(MBAN) ⋅⋅⋅⋅⋅ (3)
From equ. 2 and 3, we have
ar(sq. MBAN) = ar(rect BYXD)
(iv) In triangles FCB and ACE, we have
FC = AC
CB = CE
And, ∠FCB = ∠ACE [since, ∠FCB and ∠ACE are obtained by adding ∠ACB to a right angle.]
ΔFCB ≅ ΔACE
(v) We have,
⇒ ar(ΔFCB) = ar(ΔACE)
Clearly, triangle ACE and rectangle CYXE are on the same base CE and between same parallels CE and AX.
∴ 2ar(ΔACE) = ar(CYXE)
⇒ 2ar(ΔFCB) = ar(ΔCYXE) ⋅⋅⋅ (4)
(vi) Clearly, triangle FCb and rectangle FCAG are on the same base FC and between the same parallels FC and BG.
∴ 2ar(ΔFCB) = ar(FCAG) ⋅⋅⋅⋅ (5)
From 4 and 5, we get
ar(CYXE) = ar(ACFG)
(vii) Applying Pythagoras theorem in triangle ACB, we have
BC2 = AB2 + AC2
⇒ BC × BD = AB × MB + AC × FC
⇒ ar(BCED) = ar(ABMN) + ar(ACFG)
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Chapter 15: Areas of Parallelograms And Triangles...