Selina Concise Class 9 Maths Chapter 16 Area Theorems Exercise 16A Solutions
EXERCISE – 16A
(Q1) In the given figure, if area of triangle ADE is 60cm2, state, given reason, the area of!
(i) Parallelogram ABED
(ii) Rectangle ABCF
(iii) Triangle ABE
Solution:
Given, Area (△ADE) = 60cm2
Find: (i) Area (ABED)
(ii) Rectangle ABCF
(iii) Area (Triangle ABE)
(i) In Parallelogram ABED,
Area (△ADE) = 1/2 area (ABED)
60 = 1/2 area (ABED)
60 × 2 = area (ABED)
120 = Area (ABED)
(ii) area (ABCF) = area (ABED)
Parallelogram are on the same base and between the parallel lines arc equal in area.
Area (ABCF) = 120 cm2
(iii) area (ABE) = 1/2 area (ABED)
= 1/2 × 120
= 60 cm2
(Q2) 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 quadrilateral CDEF = area of rectangle ABDC + Area of parallelogram ABEF
To Prove:
(i) CDEF is a parallelogram
(ii) Area (CDEF) = area ABCD + area ABEF
Solution:
(i) CD parallel AB parallel EFC
(CD // AB // EF)
Also AB = CD = EF
Hence, CDEF is a parallelogram.
(iii) Area (CDNM) = area (CDBA) —- (i)
Area (EFMN) = area (EFAB) —– (ii)
Adding equation (i) and (ii),
Area (CDEF) = area (CDBA) + area (ABEF) (Hence Proved)
(Q3) In the given figure, diagonals PR and QS of the parallelogram PQRS intersect at point O and LM is parallel to PS.
Show that:
(i) 2 Area (△POS) = Area (Parallelogram PMLS)
(ii) Area (△POS) + Area (△QOR) = 1/2 Area (Parallelogram PQRS)
(iii) Area (△POS) + Area (△QOR) = Area (△POQ) + Area (△SOR)
Solution:
Given: PQRS is a parallelogram.
LM parallel PS parallel RQ (LM//PS//RQ)
To Prove:
(i) 2 area (△POS) = area (Parallelogram PMLS)
(ii) area (△POS) + area (△QOR) = 1/2 area (parallelogram PQRS)
(iii) Area (△POS) + are (△QOR) = area (△POQ) + area (△SOR)
Proof: In parallelogram PMLS,
(i) Area (△POS) = 1/2 area (PMLS)
2 area (△POS) = area (PMLS) —— (i)
(ii) Similarly, in parallelogram MQRL,
area (△QOR) = 1/2 area (QMLR)
2 area (△QOR) = area (QMLR) —- (ii)
Adding equation (i) and (ii),
2 area (△POS) + 2 area (△QOR) = area (PMLS) + area (QMLR)
2 [area (△POS) + area (△QOR)] = area (PQRS)
Area (△POS) + area (△QOR) = 1/2 area (PQRS) —– (iii)
Similarly, area (△POQ) + area (△ROS) = 1/2 area (PQRS) —— (iv)
From equation (iii) and (iv),
Area (POS) + area (QOR) = area (POQ) + area (ROS) (Hence Proved)
(Q4) In Parallelogram ABCD, P is a point on side AB and is a point on side BC.
Prove that:
(i) △CPD and △AQD are equal in area.
(ii) Area (△AQD) = Area (△APD) + Area (△CPB)
Solution:
Given: In parallelogram ABCD, P is a point on side AB and Q is a point on side BC.
To prove:
(i) Area (△CPD) = area (△AQD)
(ii) Area (△AQD) = area (△APD) + area (△CPB)
Proof: area (△CPD) = 1/2 area (ABCD) —- (i)
Also, area (△AQD) = 1/2 area (ABCD) —– (ii)
From equation (i) and (ii),
Area (△CPD) = area (△AQD) (Hence Proved)
(ii) Area (△AQD) = area (△CPD)
Area (△AQD) = area (△APD) + area (△CPB) (Hence Proved)
(Q5) 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 cm2.
(i) State the area of the triangle BEC
(ii) Name the parallelogram which is equal in area to the triangle BEC.
Given: area (ABCD) = 48cm2
To find: area (△BEC) =?
Solution:
(i) area (△BEC) = 1/2 area (ABCD)
Area (△BEC) = 1/2 × 48
Area (△BEC) = 24cm2
(ii) area (ANMB) = area (BNMC)
Area (BEC) = area (ANM) = area (BNMC) (Hence Proved)
(Q6) In the following figure, CE is drawn parallel to diagonals DB of the quadrilateral ABCD which meets AB produced at point E prove that, △ADE and quadrilateral ABCD are equal in area.
Solution:
Area (△ADE) = area (ABCD)
Proof:
△ADE = △ADB + △BDE
△ADE = △ADB + △BCD
△ADE = ABCD
Area (△ADE) = area (ABCD) (Hence Proved)
(Q7) 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.
Solution:
Area (△BCP) = area (△DPQ)
Proof: area (△APB) = 1/2 area (ABCD) —– (i)
Area (△AQD) = 1/2 area (ABCD) —– (ii)
Adding equation (i) and (ii)
Area equation (i) and (ii)
Area (△APB) + area (△AQD) = 1/2 area (ABCD) + 1/2 area (ABCD)
Area (△APB) + area (AQD) = 1/2 [area (ABCD) + area (ABCD)]
Area (△APB) + area (△AQD) = 1/2 × 2 area (ABCD)
Area (△APB) + area (△AQD) = area (ABCD).
Area (ABQD) – area (△BPQ) = area (ABQD – area (QCD)
Area (△BPQ) = area (△QCD)
Subtracting on both side by △CPQ,
Area (△BPQ) – area (CPQ) = area (△QCD) – area (△CPQ)
Area (△BPC) = area (△DPQ) (Hence proved)
(Q8) 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.
Solution:
To prove:
Area (ABCDE) = area (GDF)
Proof: Area (△AED) = area (AGD) —– (i)
Area (BCD) = area (BFD) —- (ii)
Adding equation (i) and (ii),
Area (AED) + area (BCD) = area (AGD) + area (BFD)
Adding on both side by area (△ADB)
Area (AED) + area (BCD) + area (ADB) =
Area (AGD) + area (BFD) + area (ADB)
Area (ABCDE) = area (GDF) (Hence Proved)
(Q9) 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.
Solution:
To prove:
Area (ABC) = area (BQP)
Proof:
Area (BAC) = area (BPC) —- (i)
Area (BPC) = area (BQP) —— (ii)
From equation (i) and (ii),
Area (BAC) = area (BQP) (Hence proved)
(Q10) In the figure given alongside, squares ABDE and AFGC are drawn on the side AB and the hypotenuse AC of the right triangle ABC.
Solution:
ABDE and AFGC are squares BH Perpendicular FG (BH ⊥ FG)
To prove:
(i) △EAC ≅ △BAF
(ii) Area (ABDE) = area (rectangle ARHF)
(i) Proof: In △EAC and △BAF
EA = AB
∠EAB = ∠FAR
∠EAB + ∠BAR = ∠FAR + ∠BAR
∠EAC = ∠BAF
AC = AF
∴ △EAC ≅ △BAF (By side-angle-side (SAS)
(ii) In △ABC, By Pythagoras theorem,
AC2 = AB2 + BC2
AB2 = AC2 – BC2
AB2 = (AR + RC)2 – (BR2 + RC2)
AB2 = AR2 + RC2 + 2 (AR) (RC) – BR2 – RC2
AB2 = AR2 + 2 (AR) (RC) – (AB2 – AR2)
AB2 = AR2 + 2AR.RC – AB2 + AR2
AB2 + AB2 = AR2 + 2AR.RC + AR2
2AB2 = 2AR2 + 2AR.RC
2AB2 = 2 (AR2 + AR.RC)
AB2 = AR (AR + RC)
AB2 = AR.AC
AB2 = AR.AF
∴ Area (Square ABDE) = area (rectangle ARHF) (Hence Proved)
(Q11) In the following figure, DF is parallel to BC. Show that:
(i) Area (△ADC) = Area (△AEB)
(ii) Area (△BOD) = Area (△COE)
Solution:
Given: DE is parallel to BC
To prove:
(i) Area (△ADC) = area (△AEB)
(ii) area (△BOD) = area (△COE)
Proof:
Area (△DBE) + area (△DCE) —- (i)
Adding on both side by area (△ADE)
Area (△DBE) + area (△ADE) = area (△DCE) + area (△ADE)
(i) area (△AEB) = area (△ADC)
(ii) Subtracting on both side by area (△DOE) in equation (i)
Area (△DBE) – area (△DOE) = area (△DCE) – area (△DOE)
Area (△BOD) = area (△COE) (Hence Proved)
(Q12) 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.
Solution:
Given:
Area (△EBC) = 480 cm2
AB = 30 cm
BC = 40 cm
Find (i) Area (ABCD) =?
(ii) Area (BCFE) =?
(iii) Length of altitude from A on CD
(iv) Area (△ECF)
Solution:
(i) Area (△EBC) = 1/2 area (ABCD)
480 = 1/2 area (ABCD)
480×2 = area (ABCD)
960 cm2 = area (ABCD)
Area (BCFE) = area (ABCD)
(ii) Area (BCFE) = 960 cm2
(iii) Now, area (ABCD) = CD × AM
960 = 30 × AM
960/30 = AM
32 cm = AM
(iv) area (BEC) = area (ECF) = 480 cm2
(Q13) In the given figure, D is mid-point of side AB of △ABC and BDC is a parallelogram.
Prove that: Area of △ABC = Area of parallelogram BDEC
Solution:
Given:
D is mid-point of side AB
BDEC is a parallelogram.
To Prove:
Area (△ABC) = area (Parallelogram BDEC)
Proof: AD = ED —– (i)
BD = EC —- (ii)
From equation (i) and (ii),
AD = BD
In △AFD and △CFE,
AD = BD
∠FAD = ∠FCE
∠FDA = ∠FEC
∴ △AFD ≅ △CFE (by angle-side-angle (ASA))
Area (△AFD) = area (△CFE)
Adding on both side by area (CBDE)
Area (△AFD) + area (CBDF) = area (△CFE) + area (CBDF)
Area (△ABC) = area (BDEC)
Hence proved
(Q14) In the following, AC||PS||QR and PQ||DB||SR prove that:
Area of quadrilateral PQRS = 2×Area of quadrilateral ABCD.
Solution:
AC parallel PS parallel QR (AC||PS||QR) and PQ parallel DB parallel SR (PQ||DB||SR)
To prove: area (PQRS) = 2 area (ABCD)
Proof: In △PSC
Area (△ADC) = 1/2 area (APSC) —- (i)
In △QRC,
Area (△ABC) = 1/2 area (AQRC) —- (ii)
Adding equation (i) and (ii),
Area (△ADC) + area (△ABC) = 1/2 area (APSC) + 1/2 area (AQRC)
Area (ABCD) = 1/2 [area (APSC) + area (AQRC)]
2 area (ABCD) = area (PQRS) (Hence Proved)
(Q15) 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 △ADM = area of △ACN.
Solution:
ABCD is trapezium with AB parallel DC (AB||DC)
A line parallel to AC intersects AB at point M and BC at point N.
To Prove: area (△ADM) = area (△ACN)
Proof: Area (△ADM) = area (△ACM) —- (i)
Also, area (△AMC) = area (△ANC) —– (ii)
From equation (i) and (ii),
Area (△ADM) = area (△ANC) (Hence Proved)
(Q16) In the given figure, AD||BE||CF.
Prove that area (△AFC) = area (△DBF)
Solution:
Given: AD parallel BE parallel CF.
(AD||BE||CF)
To Prove:
Area (△AEC) = area (△DBF)
Proof: In quadrilateral ABED,
Area (△ABE) = area (△BDE) —– (i)
In quadrilateral BCFE,
Area (△BCE) = area (△BFE) —– (ii)
Adding equation (i) and (ii),
Area (△ABE) + area (△BCE) = area (△BDE) + area (△BFE)
Area (△AEC) = area (△DBF) (Hence Proved)
(Q17) In the given figure, ABCD is a parallelogram. BC is produced to point X. Prove that area (△ABX) = area (Quadrilateral ACXD)
Solution:
Given, ABCD is a parallelogram.
To Prove:
Area (△ABX) = Area (Quadrilateral ACXD)
Proof: area (△ABX) = area (△ABC) + area (△CAX)
Area (△ABX) = area (△ABC) + area (△CDX)
Area (△ABX) = area (△ACD) + area (△CDX)
Area (△ABX) = area (ACXD) (Hence Proved)
(Q18) The given figure shows parallelograms ABCD and APQR. Show that these parallelograms are equal I area.
Solution:
To Prove:
Area (ABCD) = area (APQR)
Proof: In parallelogram ABCD,
Area (△ARB) = 1/2 area (ABCD) —– (i)
In parallelogram APQR,
Area (△ARB) = 1/2 area (APQR) —- (ii)
From equation (i) and (ii),
1/2 area (ABCD) = 1/2 area (APQR)
Area (ABCD) = area (APQR) (Hence Proved)
Here is your solution of Selina Concise Class 9 Maths Chapter 16 Area Theorems Exercise 16A
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