Chapter 15 Areas of Parallelograms and Triangles RD Sharma Solutions Exercise 15.3 Class 9 Maths
Chapter Name  RD Sharma Chapter 15 Areas of Parallelograms and Triangles Exercise 15.3 
Book Name  RD Sharma Mathematics for Class 10 
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Related Study  NCERT Solutions for Class 10 Maths 
Exercise 15.3 Solutions
1. In the below figure, compute the area of quadrilateral ABCD.
Given that,
DC = 17 cm
AD = 9cm and BC = 8cm
In Î”BCD we have
2. In the below figure, PQRS is a square and T and U are respectively, the mid  points of PS and QR. Find the area of Î”OTS if PQ = 8 cm.
Solution
From the figure
T and U are the midpoints of PS and QR respectively.
∴ TU  PQ
⇒ TO  PQ
Thus, in Î”PQS, T is the midpoint of PS and TO  PQ
∴ TU  PQ
⇒ TO  PQ
Thus, in Î”PQS, T is the midpoint of PS and TO  PQ
3. Compute the area of trapezium PQRS is Fig. below.
Solution
We have
4. In the below fig. ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of Î”AOB.
Solution
Since, the midpoint of the hypotenuse of a right triangle is equidistant from the vertices
∴ CA = CB = OC
⇒ CA = CB = 6.5 cm
∴ CA = CB = OC
⇒ CA = CB = 6.5 cm
⇒ AB = 13 cm
In a right angle triangle OAB, we have
and distance between AB and DC is 4cm. Find the value of x and area of trapezium ABCD.
Solution
6. In the below fig. OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If OE = 2√5, find the area of the rectangle.
Solution
Given OD = 10 cm and OE = 2√5 cm
By using Pythagoras theorem
Solution
Given : ABCD is a trapezium with AB  DC
To prove : ar(Î”AOD) = ar (BOC)
Proof :
To prove : ar(Î”AOD) = ar (BOC)
Proof :
Since Î”ADC and Î”BDC are on the same base DC and between same parallels AB and DC
8. In the given below fig. ABCD, ABFE and CDEF are parallelograms. Prove that ar(Î”ADE) = ar(Î”BCF) .
Solution
Given that,
ABCD is a parallelogram ⇒ AD = BC
CDEF is a parallelogram ⇒ DE = CF
ABFE is a parallelogram ⇒ AE = BF
CDEF is a parallelogram ⇒ DE = CF
ABFE is a parallelogram ⇒ AE = BF
Thus, in Î”s ADE and BCF, we have
AD = BC, DE = CF and AE = BF
So, by SSS criterion of congruence, we have
Î”ADE ≅ Î”ABCF
Î”ADE ≅ Î”ABCF
∴ ar(Î”ADE) = ar(BCF)
9. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:
ar(∆APB) × ar(∆CPD) = ar(∆APD) × ar (∆BPC)
Solution
10. In the below Fig, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, show that ar (Î”ABC) = ar (Î”ABD)
Solution
Given that CD is bisected at O by AB
To prove : ar(Î”ABC) = ar(Î”ABD)
Construction : Draw CP ⊥ AB and DQ ⊥ AB
Construction : Draw CP ⊥ AB and DQ ⊥ AB
Proof :
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.
Solution
12. 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 median AD, prove that ar(Î”BGC) = 2 ar(Î”AGC).
Solution
Solution
Given that,
In Î”ABC, BD = 2DC
To prove : ar(Î”ABD) = 2ar (Î”ADC)
Construction : Take a point E on BD such that BE = ED
Proof : Since, BE = ED and BD = 2DC
Then, BE = ED = DC
To prove : ar(Î”ABD) = 2ar (Î”ADC)
Construction : Take a point E on BD such that BE = ED
Proof : Since, BE = ED and BD = 2DC
Then, BE = ED = DC
We know that median of Î”^{le} divides it into equal Î”^{les} .
∴ In Î”ABD, AE is a median
∴ In Î”ABD, AE is a median
Then, area (Î”ABD) = 2ar(Î”AED) ...(i)
In Î”AEC, AD is a median Then area (Î”AED) = area (Î”ADC) ...(ii)
Compare equation (i) and (ii)
Area (Î”ABD) = 2ar(Î”ADC).
Compare equation (i) and (ii)
Area (Î”ABD) = 2ar(Î”ADC).
14. ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, Prove that:
(i) ar (Î”ADO) = ar(Î”CDO)
(ii) ar(Î”ABP) = ar(Î”CBP)
(ii) ar(Î”ABP) = ar(Î”CBP)
Solution
Given that ABCD is a parallelogram
To prove : (i) ar(Î”ADO) = ar(Î”CDO)
(ii) ar(Î”ABP) = ar(Î”CBP)
Proof: We know that, diagonals of a parallelogram bisect each other
(ii) ar(Î”ABP) = ar(Î”CBP)
Proof: We know that, diagonals of a parallelogram bisect each other
∴ AO = OC and BO = OD
(i) In Î”DAC, since DO is a median
Then area (Î”ADO) = area (Î”CDO)
(ii) In Î”BAC, Since BO is a median
Then; area (Î”BAO) = area (Î”BCO) ...(1)
In a Î”PAC, Since PO is a median
In a Î”PAC, Since PO is a median
Then, area (Î”PAO) = area (Î”PCO) ...(2)
Subtract equation (2) from equation (1)
⇒ area(Î”BAO)  ar(Î”PAO) = ar(Î”BCO)  area(Î”PCO) ⇒ Area (Î”ABP) = Area of Î”CBP
15.ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F.
(i) Prove that ar(Î”ADF) = ar(Î”ECF)
(ii) If the area of Î”DFB = 3 cm^{2}, find the area of ^{gm} ABCD.
(ii) If the area of Î”DFB = 3 cm^{2}, find the area of ^{gm} ABCD.
Solution
In triangles ADF and ECF, we have
∠ADF = ∠ECF [Alternative interior angles, Since AD  BE]
AD = EC [Since AD = BC = CE]
And ∠DFA = ∠CFA [vertically opposite angles]
AD = EC [Since AD = BC = CE]
And ∠DFA = ∠CFA [vertically opposite angles]
So, by AAS congruence criterion, we have
Î”ADF ≅ Î”ECF
⇒ area (Î”ADF) = area (Î”ECF) and DF = CF.
Now, DF = CF
⇒ BF is a median in Î”BCD
⇒ area (Î”BCD) = 2ar(Î”BDF)
⇒ area (Î”BCD) = 2×3 cm^{2} = 6cm^{2}
⇒ area (Î”BCD) = 2×3 cm^{2} = 6cm^{2}
Hence, ar(^{gm} ABCD) = 2ar(Î”BCD) = 2 × 6 cm^{2} = 12cm^{2}
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(Î”POA) = ar(Î”QOC).
Solution
17. ABCD is a parallelogram. E is a point on BA such that BE = 2 EA and F is a point on DC such that DF = 2 FC. Prove that AE CF is a parallelogram whose area is one third of the area of parallelogram ABCD.
Solution
18. In a Î”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(Î”PBQ) = ar (Î”ARC)
(ii) ar (Î”PRQ) = 1/2 ar(Î”ARC)
(iii) ar(Î”RQC) = 3/8 ar (Î”ABC).
(ii) ar (Î”PRQ) = 1/2 ar(Î”ARC)
(iii) ar(Î”RQC) = 3/8 ar (Î”ABC).
Solution
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(Î”EGB) = 1/6 ar(ABCD)
(iii) ar(Î”EFC) = 1/2 ar(Î”EBF)
(iv) ar(Î”EBG) = ar(Î”EFC)
(v) Find what portion of the area of parallelogram is the area of Î”EFG.
(i) ar(ADEG) = ar(GBCE)
(ii) ar(Î”EGB) = 1/6 ar(ABCD)
(iii) ar(Î”EFC) = 1/2 ar(Î”EBF)
(iv) ar(Î”EBG) = ar(Î”EFC)
(v) Find what portion of the area of parallelogram is the area of Î”EFG.
Solution
20. In Fig. below, CD  AE and CY  BA.
(i) Name a triangle equal in area of Î”CBX
(ii) Prove that ar (Î”ZDE) = ar(Î”CZA)
(iii) Prove that ar(BCZY) = ar(Î”EDZ)
Solution(iii) Prove that ar(BCZY) = ar(Î”EDZ)
Since, Î”BCA and Î”BYA are on the same base BA and between same parallels BA and CY
Then area (Î”BCA) = ar(Î”BYA)
⇒ ar(Î”CBX) + ar(Î”BXA) = ar(Î”BXA) + ar(Î”AXY)
⇒ ar(Î”CBX) = ar(Î”AXY) ...(1)
Since, Î”ACE and Î”ADE are on the same base AE and between same parallels CD and AE
Then, ar(Î”ACE) = ar(Î”ADE)
⇒ ar(Î”CLA) + ar(Î”AZE) = ar(Î”AZE) + ar(Î”DZE)
⇒ ar(Î”CBX) + ar(Î”BXA) = ar(Î”BXA) + ar(Î”AXY)
⇒ ar(Î”CBX) = ar(Î”AXY) ...(1)
Since, Î”ACE and Î”ADE are on the same base AE and between same parallels CD and AE
Then, ar(Î”ACE) = ar(Î”ADE)
⇒ ar(Î”CLA) + ar(Î”AZE) = ar(Î”AZE) + ar(Î”DZE)
⇒ ar(Î”CZA) = (Î”DZE) ...(2)
Since Î”CBY and Î”CAY are on the same base CY and between same parallels
BA and CY
Since Î”CBY and Î”CAY are on the same base CY and between same parallels
BA and CY
Then ar(Î”CBY) = ar(Î”CAY)
Adding ar(Î”CYG) on both sides, we get
Adding ar(Î”CYG) on both sides, we get
⇒ ar(Î”CBX) + ar(Î”CYZ) = ar(Î”CAY) + ar(Î”CYZ)
⇒ ar(BCZX) = ar(Î”CZA) ...(3)
Compare equation (2) and (3)
ar(BCZY) = ar(Î”DZE)
⇒ ar(BCZX) = ar(Î”CZA) ...(3)
Compare equation (2) and (3)
ar(BCZY) = ar(Î”DZE)
21. In below fig., PSDA is a parallelogram in which PQ = QR = RS and AP  BQ  CR. Prove that ar(Î”PQE) = ar(Î”CFD).
Solution
Given that PSDA is a parallelogram
Since, AP  BQ  CR  DS and AD PS
∴ PQ = CD ...(i)
In Î”BED, C is the midpoint of BD and CF  BE
∴ PQ = CD ...(i)
In Î”BED, C is the midpoint of BD and CF  BE
∴ F is the midpoint of ED
⇒ EF = PE
Similarly
EF = PE
∴ PE = FD ...(2)
In Î”SPQE and CFD, we have
In Î”SPQE and CFD, we have
PE = FD [Alternative angles]
∠EDQ = ∠FDC,
And PQ = CD
∠EDQ = ∠FDC,
And PQ = CD
So, by SAS congruence criterion, we have Î”PQE ≅ Î”DCF.
12. In the below fig. 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) DCXY is a trapezium
(iii) ar(trap. DCYX) = 9/11 ar(trap. XYBA)
Solution
23. In Fig. below, ABC and BDE are two equilateral triangles such that D is the mid  point of BC. AE intersects BC in F. Prove that
(i) ar(Î”BDE) = 1/4 ar(Î”ABC)
(ii) ar(Î”BDE) = 1/2 ar(Î”BAE)
(iii) ar(Î”BEF) = ar(Î”AFD).
(ii) ar(Î”BDE) = 1/2 ar(Î”BAE)
(iii) ar(Î”BEF) = ar(Î”AFD).
(iv) ar(Î”ABC) = ar(Î”BEC)
(v) ar(Î”FED) = 1/8 ar(Î”AFC)
(vi) ar(Î”BFE) = 2ar(Î”EFD)
(v) ar(Î”FED) = 1/8 ar(Î”AFC)
(vi) ar(Î”BFE) = 2ar(Î”EFD)
Solution
Given that,
ABC and BDE are two equilateral triangles.
Let AB = BC = CA = x. Then BD = x/2 = DE = BE
(ii) It is given that triangles ABC and BED are equilateral triangles
∠ACB = ∠DBE = 60°
⇒ BE  AC (Since alternative angles are equal)
Triangles BAF and BEC are on the same base
∠ACB = ∠DBE = 60°
⇒ BE  AC (Since alternative angles are equal)
Triangles BAF and BEC are on the same base
BE and between the same parallel BE and AC
∴ ar(Î”BAE) = area(Î”BEC)
⇒ ar(Î”BAE) = 2ar(Î”BDE)
[∵ED is a median of Î”EBC; ar(Î”BEC) = 2ar(Î”BDE)]
⇒ area(Î”BDE) = 1/2 ar(Î”BAE)
⇒ ar(Î”BAE) = 2ar(Î”BDE)
[∵ED is a median of Î”EBC; ar(Î”BEC) = 2ar(Î”BDE)]
⇒ area(Î”BDE) = 1/2 ar(Î”BAE)
(iii) Since Î”ABC and Î”BDE are equilateral triangles
∴ ∠ABC = 60° and ∠BDE = 60°
∠ ABC = ∠BDE
∠ ABC = ∠BDE
⇒ AB  DE (Since alternative angles are equal)
Triangles BED and AED are on the same base ED and between the same parallels AB and DE.
Triangles BED and AED are on the same base ED and between the same parallels AB and DE.
∴ ar(Î”BED) = area(Î”AED)
⇒ ar(Î”BED)  area (Î”EFD) = area (AED)  area (Î”EFD)
⇒ ar(BEF) = ar(Î”AFD).
⇒ ar(Î”BED)  area (Î”EFD) = area (AED)  area (Î”EFD)
⇒ ar(BEF) = ar(Î”AFD).
(iv) Since ED is the median of Î”BEC
∴ area(Î”BEC) = 2ar(Î”BDE)
⇒ ar(Î”BEC) = 2 × 1/4 ar(Î”ABC) [from (i)]
⇒ ar(Î”BEC) = 2 × 1/4 ar(Î”ABC) [from (i)]
⇒ ar(Î”BEC) = 1/2 area(Î”ABC)
⇒ area(Î”ABC) = 2area (Î”BEC)
⇒ area(Î”ABC) = 2area (Î”BEC)
(v) Let h be height of vertex E, corresponding to the side BD on triangle BDE
Let H be the height of the vertex A corresponding to the side BC in triangle ABC
From part (i)
From part (i)
24. D is the mid  point of side BC of Î”ABC and E is the mid  point of BD. If O is the mid  point of AE, prove that ar(Î”BOE) = 1/8 ar(Î”ABC).
Solution
25. In the below fig. X and Y are the mid  points of AC and AB respectively, QP  BC and CYQ and BXP are straight lines. Prove that ar(Î”ABP) = ar(Î”ACQ).
We observe that the quadrilateral XYAP and XY AQ are on the same base XY and between the same parallel XY and PQ .
(ii) Since, Î”PEA and Î”QFD stand on the same base PE and FQ lie between the same parallels EQ and AD
Solution
Since x and y are the midpoint AC and AB respectively
∴ XY  BC
Clearly, triangles BYC and BXC are on the same base BC and between the same parallels XY and BC
We observe that the quadrilateral XYAP and XY AQ are on the same base XY and between the same parallel XY and PQ .
∴ area (quad XY AP) = ar(quad XYPA)
Adding (i) and (ii) , we get
Adding (i) and (ii) , we get
ar(Î”BXY) + ar(quadXYAP) = ar(CXY) + ar(quad XYQA)
⇒ ar(Î”ABP) = ar(Î”ACQ)
⇒ ar(Î”ABP) = ar(Î”ACQ)
26. In the below fig. ABCD and AEFD are two parallelograms. Prove that
(i) PE = FQ
(ii) ar(Î”APE) : ar(Î”PFA) = ar(Î”QFD) : ar(Î”PFD)
(iii) ar(Î”PEA) = ar(Î”QFD)
(ii) ar(Î”APE) : ar(Î”PFA) = ar(Î”QFD) : ar(Î”PFD)
(iii) ar(Î”PEA) = ar(Î”QFD)
Solution
Given that, ABCD and AEFD are two parallelograms
To prove : (i) PE = FQ
27. In the below figure, ABCD is parallelogram. O is any point on AC. PQ  AB and LM  AD. Prove that ar(^{gm} DLOP) = ar(^{gm} BMOQ)
Solution
Since, a diagonal of a parallelogram divides it into two triangles of equal area
∴ area (Î”ADC) = area(Î”ABC)
Since, AO and OC are diagonals of parallelograms AMOP and OQCL respectively.
Since, AO and OC are diagonals of parallelograms AMOP and OQCL respectively.
28. In a Î”ABC, if L and M are points on AB and AC respectively such that LM  BC. Prove that :
(i) ar(Î”LCM) = ar(Î”LBM)
(ii) ar(Î”LBC) = ar(Î”MBC)
(iii) ar(Î”ABM) = ar(Î”ACL)
(iv) ar(Î”LOB) = ar(Î”MOC)
(ii) ar(Î”LBC) = ar(Î”MBC)
(iii) ar(Î”ABM) = ar(Î”ACL)
(iv) ar(Î”LOB) = ar(Î”MOC)
Solution
(i) Clearly Triangles LMB and LMC are on the same base LM and between the same parallels LM and BC.
∴ ar(Î”LMB) = ar(Î”LMC) ...(i)
∴ ar(Î”LMB) = ar(Î”LMC) ...(i)
(ii) We observe that triangles LBC and MBC area on the same base BC and between the same parallels LM and BC
∴ arc Î”LBC = ar(Î”MBC) ...(ii)
ar(Î”LMB) = ar(Î”LMC) [from (i)]
⇒ ar(Î”ALM) + ar(Î”LMB) = ar(Î”ALM) + ar(Î”LMC)
⇒ ar(Î”ABM) = ar(Î”ACL)
⇒ ar(Î”ALM) + ar(Î”LMB) = ar(Î”ALM) + ar(Î”LMC)
⇒ ar(Î”ABM) = ar(Î”ACL)
(iv) We have
ar(Î”CBC) = ar(Î”MBC) ∴ [from (i)]
⇒ ar(Î”LBC) = ar(Î”BOC) = ar(Î”MBC)  ar(Î”BOC)
⇒ ar(Î”LBC) = ar(Î”BOC) = ar(Î”MBC)  ar(Î”BOC)
⇒ ar(Î”LOB) = ar(Î”MOC)
29. In the below fig. D and E are two points on BC such that BD = DE = EC. Show that ar(Î”ABD) = ar(Î”ADE) = ar(Î”AEC).
Solution
Draw a line through A parallel to BC
Given that, BD = DE = ECWe observe that the triangles ABD and AEC are on the equal bases and between the same parallels C and BC. Therefore, Their areas are equal.
Hence, ar(Î”ABD) = ar(Î”ADE) = ar(Î”ACDE)
30. If below fig. ABC is a right triangle right angled 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) Î”MBC ≅ Î”ABD
(ii) ar(BYXD) = 2 ar(Î”MBC)
(iii) ar(BYXD) = ar(Î”ABMN)
(iv) Î”FCB ≅ Î”ACE
(v) ar(CYXE) = 2 ar(Î”FCB)
(vi) ar(CYXE) = ar(ACFG)
(vii) ar (BCED) = ar(ABMN) + ar(ACFG)
Solution
(i) In Î”MBC and Î”ABD, we have
MB = AB
BC = BD
And ∠MBC = ∠ABD
[∵ ∠MBC and ∠ABC are obtained by adding ∠ABC to a right angle]
So, by SAS congruence criterion, We have
Î”MBC ≅ Î”ABD
⇒ ar(Î”MBC) = ar(Î”ABD) ...(1)
(ii) Clearly, Î”ABC and BYXD are on the same base BD and between the same parallels AX and BD
(iii) Since triangle M. BC and square MBAN are on the same Base MB and between the same parallels MB and NC
∴ 2ar(Î”MBC) = ar(Î”MBAN) ...(3)
From (2) and (3) we have
From (2) and (3) we have
ar(sq. MBAN) = ar(rect BYXD).
(iv) In triangles FCB and ACE we have
FC = AC
CB = CF
And ∠FCB = ∠ACE
[∵ ∠FCB and ∠ACE are obtained by adding ∠ACB to a right angle]
So, by SAS congruence criterion, we have
Î”FCB ≅ Î”ACE
(v) We have
Î”FCB ≅ Î”ACE
⇒ ar(Î”FCB) = ar(Î”ECA)
⇒ ar(Î”FCB) = ar(Î”ECA)
Clearly, Î”ACE and rectangle CYXE are on the same base CE and between the same parallels CE and AX
∴ 2ar (Î”ACE) = ar(CYXE)
(vi) Clearly, Î”FCB and rectangle FCAG are on the same base FC and between the same parallels FC and BG
∴ 2ar(Î”FCB) = ar(FCAG)
From (4) and (5), we get
Area (CYXE) = ar(ACFG)
(vii) Applying Pythagoras theorem in Î”ACB, we have