RD Chapter 15 Areas of Parallelograms and Triangles Ex 15.3 Solutions
Question - 11 : - 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.
Answer - 11 : -
Given: Here from the question we get
(1) ABCD is a parallelogram
(2) P is any point in the interior of parallelogram ABCD
To prove:
Construction: Draw DN perpendicular to AB and PM perpendicular AB
Proof: Area of triangle =
× base× heightArea of
ÄAPB = .AB .PM …… (1)Also we know that: Area of parallelogram = base× height
Area of parallelogram ABCD = AB DN …… (2)
Now PM < DN (Since P is a point inside the parallelogram ABCD)
Hence it is proved that
Answer - 12 : -
Given:
(1) ABC is a triangle
(2) AD is the median of ΔABC
(3) G is the midpoint of the median AD
To prove:
(a) Area of Δ ADB = Area of Δ ADC
(b) Area of Δ BGC = 2 Area of Δ AGC
Construction: Draw a line AM perpendicular to AC
Proof: Since AD is the median of ΔABC.
Therefore BD = DC
So multiplying by AM on both sides we get
In ΔBGC, GD is the median
Since the median divides a triangle in to two triangles ofequal area. So
Area of ÄBDG = Area of ÄGCD
⇒ Areaof ÄBGC =2(Area of ÄBGD)
Similarly In ΔACD, CG is the median
⇒ Areaof ÄAGC =Area of ÄGCD
From the above calculation we have
Area of ÄBGD = Area of ÄAGC
But Area of ÄBGC = 2(Area of ÄBGD)
So we have
Area of ÄBGC = 2(Area of ÄAGC)
Hence it is proved that
(1) 
(2) 
Answer - 13 : -
Given:
(1) ABC is a triangle
(2) D is a point on BC such that BD = 2DC
To prove: Area of ΔABD = 2 Area of ΔAGC
Proof:
In ΔABC, BD = 2DC
Let E is the midpoint of BD. Then,
BE = ED = DC
Since AE and AD are the medians of ΔABD and ΔAEC respectively
and The median divides a triangle in to two triangles of equal area. So
Hence it is proved that
ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that:
(i) ar (Δ ADO) = ar(Δ CDO)
(ii) ar (Δ ABP) = ar (Δ CBP).
Answer - 14 : -
Given: Here from the given figure we get
(1) ABCD is a parallelogram
(2) BD and CA are the diagonals intersecting at O.
(3) P is any point on BO
To prove:
(a) Area of ΔADO = Area ofΔ CDO
(b) Area of ΔAPB = Area ofΔ CBP
Proof: We know that diagonals of a parallelogram bisect each other.
O is the midpoint of AC and BD.
Since medians divide the triangle into two equal areas
In ΔACD, DO is the median
Area of ΔADO = Area ofΔ CDO
Again O is the midpoint of AC.
In ΔAPC, OP is the median
⇒ Areaof ÄAOP =Area of ÄCOP …… (1)
Similarly O is the midpoint of AC.
In ΔABC, OB is the median
⇒ Areaof ÄAOB =Area of ÄCOB …… (2)
Subtracting (1) from (2) we get,
Area of ΔAOB − Area of ΔAOP = Area of ΔCOB − Area of ΔCOP
⇒ Areaof ÄABP =Area of ÄCBP
Hence it is proved that
(a) 
(b) 
ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F.
(i) Prove that ar (Δ ADF) = ar (Δ ECF)
(ii) If the area of Δ DFB = 3 cm2, find the area of ||gm ABCD.
Answer - 15 : -
Given: Here from the given figure we get
(1) ABCD is a parallelogram with base AB,
(2) BC is produced to E such that CE = BC
(3) AE intersects CD at F
(4) Area of ΔDFB = 3 cm
To find:
(a) Area of ΔADF = Area of ΔECF
(b) Area of parallelogram ABCD
Proof: Δ ADF and ΔECF, we can see that
∠ADF = ∠ECF (Alternateangles formed by parallel sides AD and CE)
AD = EC
∠DFA = ∠CFA (Verticallyopposite angles)
(ASA condition of congruence)
As 
DF = CF
Since DF = CF. So BF is a median in ΔBCD
Since median divides the triangle in to two equaltriangles. So
Since 
.So
Hence Area of parallelogram ABCD
Hence we get the result
(a)
(b)
Question - 16 : - 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 (Δ POA) = ar (Δ QOC).
Answer - 16 : -
Given:
(1) Diagonals AC and BD of a parallelogram ABCD intersect at point O.
(2) A line through O intersects AB at P point.
(3) A line through O intersects DC at Q point.
To find: Area of (ΔPOA) = Area of (ΔQOC)
Proof:
From ÄPOA and ÄQOC we get that
= 
OA = OC
= 
So, by ASA congruence criterion, we have
So
Area (ÄPOA) = Area (ÄQOC)
Hence it is proved that
Answer - 17 : -
Given:
(1) ABCD is a parallelogram.
(2) E is a point on BA such that BE = 2EA
(3) F is a point on DC such that DF = 2FC.
To find:
Area of parallelogram
Proof: We have,
BE = 2EA and DF = 2FC
AB − AE = 2AE and DC − FC = 2FC
AB = 3AE and DC = 3FC
AE =
AB and FC = DC AE = FC [since AB = DC]
Thus, AE || FC such that AE = FC
Therefore AECF is a parallelogram.
Clearly, parallelograms ABCD and AECF have the same altitude and
AE =
AB.Therefore
Hence proved that
In a Δ ABC, P and Q are respectively the mid-points of AB and BC and R is the mid-point of AP. Prove that:
(i) ar (Δ PBQ) = ar (Δ ARC)
(ii) ar (Δ PRQ) = 1212 ar (Δ ARC)
(iii) ar (Δ RQC) = 3838 ar (Δ ABC).
Answer - 18 : -
Given:
(1) In a triangle ABC, P is the mid-point of AB.
(2) Q is mid-point of BC.
(3) R is mid-point of AP.
To prove:
(a) Area of ΔPBQ = Area of ΔARC
(b) Area of ΔPRQ =
Area of ΔARC(c) Area of ΔRQC =
Area of ΔABCProof: We know that each median of a triangle divides it into two triangles of equal area.
(a) Since CR is a median of ΔCAP
Therefore
…… (1)Also, CP is a median of ΔCAB.
Therefore
…… (2)From equation (1) and (2), we get
Therefore
…… (3)PQ is a median of ΔABQ
Therefore
Since
Put this value in the above equation we get
…… (4)
From equation (3) and (4), we get
Therefore
…… (5)(b)
…… (6) 
…… (7) From equation (6) and (7)
…… (8) From equation (7) and (8)
(c)
=
…… (9)
Question - 19 : - ABCD is a parallelogram, G is the point on AB such that AG = 2 GB, E is a point of 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 (Δ EGB) = 1616 ar (ABCD)
(iii) ar (Δ EFC) = 1212 ar (Δ EBF)
(iv) ar (Δ EBG) = ar (Δ EFC)
(v) Find what portion of the area of parallelogram is the area of Δ EFG
Answer - 19 : -
Given:
ABCD is a parallelogram
G is a point such that AG = 2GB
E is a point such that CE = 2DE
F is a point such that BF = 2FC
To prove:
(i)
(ii)
(iii)
(iv)
What portion of the area of parallelogram ABCD is the area of ΔEFG
Construction: draw a parallel line to AB through point F and a perpendicular line to AB through
PROOF:
(i) Since ABCD is a parallelogram,
So AB = CD and AD = BC
Consider the two trapeziums ADEG and GBCE:
Since AB = DC, EC = 2DE, AG = 2GB
, and 
, and
So
, and 
Since the two trapeziums ADEG and GBCE have same height and their sum of two parallel sides are equal
Since
So
Hence
(ii) Since we know from above that
. So
Hence
(iii) Since height of triangle EFC and triangle EBF are equal. So
Hence
(iv) Consider the trapezium in which
(From (iii))Now from (ii) part we have
(v) In the figure it is given that FB = 2CF. Let CF = x and FB = 2x
Now consider the tow triangles CFI and CBH which are similar triangles
So by the property of similar triangle CI = k and IH = 2k
Now consider the triangle EGF in which
Now
(Multiply both sides by 2)
…… (2)
From (1) and (2) we have
In the given figure, CD || AE and CY || BA.
(i) Name a triangle equal in area of ΔCBX
(ii) Prove that ar (Δ ZDE) = ar (Δ CZA)
(iii) Prove that ar (BCZY) = ar (Δ EDZ).
Answer - 20 : -
Given:
(1) CD||AE.
(2) CY||BA.
To find:
(i) Name a triangle equal in area of CBX.
(ii)
.(iii)
.Proof:
(i) Since triangle BCY and triangle YCA are on the same base and between same parallel, so their area should be equal. Therefore
Therefore area of triangle CBX is equal to area of triangle AXY
(ii) Triangle ADE and triangle ACE are on the same base AE and between the same parallels AE and CD.
(iii) Triangle ACY and BCY are on the same base CY and between same parallels CY and BA. So we have
Now we know that