Book Name |
: Ganit Prakash |
Subject |
: Mathematics (Maths) |
Class |
: 9 (Madhyamik/WB) |
Publisher |
: Prof. Nabanita Chatterjee |
Chapter Name |
: Theorems On Area (12th Chapter) |
Let I do – 12.1
1. If any triangle and rectangle are on the same base and between the same parallel then let us prove logically that the area of triangle is half of the area in the shape of rectangular region.
Solution: Given:
△ABC and Rectangle ABCD are on the same base AB and between the same parallel lines AB and CD.
R.T.P : △ABC=21 Rectangle ABCD
Construction :- Joint A and C.
Proof: By construction ABCD is a rectangle and AC is one of its diagonal.
∴△ABC=221 Rectangle ABCD[∵AB∥CD AD∥BC](Proved)
2. If any triangle and any parallelogram are on the same base and between same parallels. Let us prove logically that the area of the triangular field is half of the area of in the shape of parallelogram field.
Solution: Given △ABC and parallelogram ABDE are on the same base AB and between the same parallels AB and ED.
R.T.P : ∴△ABC=21 ParallelogramABDE
Construction: The straight line through the point A drawn parallel to BC intersects DC produced at the point F.
Proof : By construction ABCF is a parallelogram and AC is one of its diagonals.
∴△ABC=21 Parallelogram ABDE
∴△ABC=21 Parallelogram ABDE (Proved)
∴ The two parallelograms are on the same base and between the same parallels.
Let’s prove – 12
1. P and Q are the mid points of sides AB and DC of parallelogram ABCD; Let’s prove that the area of the quadrilateral field =21× area of parallelogram field.
Solution: Given:
ABCD is a parallelogram in which P and Q are the mid points of sides AB and DC.
P, Q is drawn, Join A, Q and P, C
R.T.P : Area of quadrilateral field APCQ=21× area of Parallelogram ABCD
Proof : ∵AB∥CD BC∥AD
∴BC∥AD∥PQ[∵P Qarethemid−pointsofABandCD]
∴APQDandBCQP are two parallelograms.
In △APQ and Parallelogram APQD are on the same parallel line AP and QD.
∴△APQ=21 Parallelogram APQD
Similarly, △CPQ=21ParallelogramBCQP
∴△APQ+△CPQ=
1 Parallelogram APQP +21 Parallelogram BCQP
Quadrilateral APCQ=21Parallelogram(APQP+BCQP)
∴ Quadrilateral APCQ=21 Prallelogram ABCD (Proved)
2. The distance between two sides AB and DC of a rhombus ABCD is PQ and distance between sides AD and BC is RS; Let’s prove that PQ = RS.
Solution : Given:
ABCD is a rhombus.
The distance between two sides AB and DC is PQ and distance between sides AD and BC is R S respectively. PQ and RS are drawn.
R.T.P : PQ = RS.
Construction: Join A,P ; P,B; A,S and S,D
Proof: In △APBandrhombusABCD.
AB is same base and between the same parallel lines AB and CD.
∴△APB=21 Rhombus ABCD
Again,
In △ASD and rhombus ABCD are on the same base AD and between the same parallel lines AD and BC.
∴△ASD=21 Rhombus ABCD
∴△ Area of APB= Area of △ASD
or 21×AB×PQ=21×AD×RS
∴PQ=RS[∵AB=AD] (Proved)
3. P and Q are the mid-points of sides AB and DC of parallelogram ABCD respectively. Let’s prove that PBQD is a parallelogram and △PBC=21 parallelogram PQBD.
Solution : Given:
ABCD is a parallelogram in which P and Q are mid-points of AB and DC respectively. PD, BQ and PC are drawn.
R.T.P : (i) PQBD is parallelogram.
(ii) △PBC=21 parallelogram PQBD
Proof: ∴ ABCD is a parallelogram
∴AB∥DC
∴PB∥DQ
Again,
∴ ABCD is a parallelogram.
∴AB=DC
∴21AB=21DC
∴PB=DQ[∴P and Q are mid-points of A B and D C]
Now, PB∥DQ and PB = DQ
∴ PBQD is parallelogram (i) (Proved)
∴△ PBC and parallelogram PBQD are on the same base PB between the same parallels PB and DC.
∴△PBC=21 Parallelogram PBQD (ii) (Proved)
4. In an isoceles triangle ABC, AB = AC and P is any point on produced side BC. PQ and PR are perpendicular on sides AB and AC from the point P respectively. BS is perpendicular on side AC from point B; Let’s prove that PQ – PR = BS
Solution : Given:
ABC is an isosceles triangle, AB = AC and P is any point on produced side BC. PQ and PR are perpendicular on sides AB and AC from the point P respectively. BS is perpendicular on side AC from point B.
R.T.P: PQ – PR = BS
Construction: A & P are joined.
Proof: In △ABC,
area of △ABC=21×AC×BS
In △ACP
area of △ACP=21×AC×PR
In △ABP,
area of △ABP=21×AB×PQ
∴ area of △ABP= area of △ABC+ area of △ACP
or, 21×AB×PQ=21×AC×BS+21×AC×PR
or, 21×AB×PQ=21×AC(BS+PR)
or, 21×AB×PQ=21×AB(BS+PR)[∵AB=AC]
or, PQ=BS+SR
∴PQ−PR=BS
∴ (Proved)
5. O is any point out side the equilateral triangle ABC and within the angular region on ABC; OP, OQ and OR are the perpendicular distance of AB, BC and CA respectively from the point O. Let us prove that the altitude of the triangle = OP + OQ – OR.
Solution: Given:
ABC is an equilateral triangle. O is a point out side the triangle and within the angular region ABC.OP,OQandOR are perpendicular distances of
AB, BC and CA respectively from the point O. Perpedicular AS on BC is drawn. Thus AS is the altitude of △ABC
R.T.P: AS=OP+OQ−OR.
Proof : OA,OB and OC are drawn.
∵△ABC= Quadrilateral ABCD−△AOC
or, △ABC=△ABO+△OBC−△AOC
∴21BC×AS=21×AB×OP+21BC×OQ−21AC×OR
=21×BC×OP+21BC×OQ−21AC×OR
[∵AB=BC,AC=BC]
=21BC(OP+OQ−OR)
∴AS=OP+OQ−OR (Proved)
6. A straight line parallel to AB of parallelogram ABCD intersects AD, AC and BC or their produced parts at the points E, F and G respectively. Let’s prove that △AEG=△AFD
Solution: Given,
ABCD is a parallelogram, and a straight line parallel to AB intersects sides AD,AC and BC or their produced parts at the point E, F and G respectively E G is drawn.
R.T.P: △AEG=△AFD
Construction: We draw IH∥AD∥BC joined A, G.
Proof: ∵EG∥AB∥DC&AD∥IH∥BC
∴AIFE,CGFH,BIFG&DEFH are four parallelograms.
In parallelogram ABCD,△ABC=△ADC
" " " " " " " "AIFE,△AIF=△AFE
" " " " " " " "FGCH△FGC=△FHC
∴△ABC−△AIF−△FGC=△ADC−△AEF−△FHC
∴ParallelogramBIFG=ParallelogramDEFH
∴∠BIFG+∠AIFE=∠DEFH+∠AIFE
∴∠EGBA=∠ DHIA
In △ADFand∠DHIA,AD is same base and between the same parallel linesADandHI.
∴△ADF=21×∠DHIA
In △AEGand∠EGBA,EG is same base and between the same parallel lines EG and AB.
∴△AEG=21x<EGBA
∴△AEG=△AFD[∵∠DHIA=∠EGBA]
7. E is any point on side DC of parallelogram ABCD, produced AE meets produced BC at the point F. Joined D, F. Let’s prove that (i) △ADF=△ABE, (ii) △DEF=△BEC.
Solution : Given:
ABCD is a parallelogram and E is a point on DC. AE produced intersects BC produced at F. DF and BE are drawn.
R.T.P : (i)△ADF=△ABE
(ii) △DEF=△BEC
Proof: △ADF and parallelogram ABCD are on same base AD and between the same parallels AD and BF
∴△ADF=21×△ABCD
∴△ABE and ∠ABCD are on the same base A B and between the same parallels AB and DC.
∴△ABE=21×∠ABCD
From (i) and (ii) We get.
△ADF=△ABE
Now,
△ADE+△BEC=∠ABCD−△ABE
△ABCD−21∠ABCD
[∵△ABE=21∠ABCD]
=21∠ABCD
∴△ADF[∵△ADF=21∠ABCD]
=△ADE+△DEF
∴△ADE+△BEC=△ADE+△DEF
∴△BEC=△DEF
∴△BEC=△DEF(ii) (Proved)
8. Two triangles ABC and ABD with equal area and stand on the opposite side of AB, Let’s prove that A B bisects CD.
Solution:
Given: △ABC and △ABD are equal in area and they are on the opposite sides of common base AB . CD is drawn. Let CD intersects AB at the point O.
R.T.P: AB bisects CD.
Construction: Perpendiculars CP and DQ are drawn on AB from C and D respectively.
Proof: CP∥DQ
Now,
△ABC=△ABD
or, =21×AB×CP=21AB×DQ
∴CP=DQ
∵CP∥DQ and CP=DQ
∴CQDP is parallelogram, diagonals bisect each other.
∴QP bisects CD. (Proved)
∴ CQDP is parallelogram, diagonals bisect each other.
∴ QP bisects CD. (Proved)
9. D is mid-point af side BC of triangle ABC, Parallelogram CDEF stands between side BC and parallel to BC through the point A. Let’s prove that △ABC= parallelogram CDEF.
Solution :
Given: ABC is a triangle in which D is the mid-point of side BC and parallelogram CDEF stands between side BC and parallel to BC through A.
R.T.P :△ABC= parallelogram CDEF
Construction: A and D are joined.
Proof: △ADC and ∠CDEF are situated on the same base DC and lies between same parallel lines DC and AF.
∴△ADC=21×∠CDEF
again,
∵ADisthemedianof△ABC
∴△ADC=21△ABC
∴△ABC=∠CDEF (Proved)
10. P is any point on diagonal BD of parallelogram ABCD. Let’s prove that △APD=△ABP
Solution: Given:
ABCD is a parallelogram. P is any point on diagonal BD.
R.T.P : △APD=△ABP
Construction: Perpendiculars are drawn on BD from the point A and C which intersectsBD at E and Frespectively.
Proof: In △ADE and △CBF,
<ADE= Alternate <CBF and AD=BC[∵AD∥BC]
<AED=<CFB (each are right angle)
△ADE=△CBF
∴AE=CF
Now, ∵△APDand△ABP are situated on same base
BD and altitude
[∵AE=CF]
∴△APD=△ABP(Proved)
11. AD and BE are the medians of triangle ABC. Let’s prove that △ACD=△BCE
Solution : Given:
ABC is a triangle in which AD and BE are the medians of △ABC which meets at O.
R.T.P: △ACD=△BCE
Construction: Join D, E such that DE∥AB
Proof: ∵BE is the median of △ABC
∴△BCE=21×△ABC
Again,
∵CF is median
of △ABC
∴△ACD=21×△ABC
From, (i) & (ii)
∴△ACD=△BCE(Proved)
12. A line parallel to BC of triangle ABC intersects sides AB and AC at the points P and Q respectively. CQ and BQ intersect each other at the point X. Let’s prove that
(i) △BPQ=△CPQ
(ii) △BCP=△BCQ
(iii) △ACP=△ABQ
(iv) △BXP=△CXQ
Solution: Given:
ABC is a triangle and PQ parallel to BC which intersects AB and AC at P and Q respectively. CP and BQ intersect each other at X.
R.T.P : (i) △BPQ=△CPQ
(ii) △BCP=△BCQ
(iii) △ACP=△ABQ
(iv) △BXP=△CXQ
Proof: △BPQ and △CPQ are on the same base PQ and between the same parallels PQ and BC.
∴△BPQ=△CPQ………(i) (Proved)
∴△BCPand△BCQ are on the same base BC and between the same parallels BC and PQ.
∴△BCP=△BCQ
Now, △ABQ=△BPQ+△APQ
=△CPQ+△APQ[∵△BAQ=△CPQ]……….(ii) (Proved)
=△ACP
∴△BPQ=△CPQ............. (iii) (Proved)
∴△BXP+△PXQ=△CXQ+△PXQ
∴△BXP=△CXQ…………. (iv)(Proved).
13. D is the mid point of BC of triangle ABC and P is any point on BC. Join P, A – through the point D a straight line parallel to line segment PA meet AB at point Q. Let’s prove that,
(i) △ADQ=△PDQ
(ii) △BPQ=21△△ABC
Solution: Given:
ABC is a triangle in which D is the mid-point of BC and P is any point on BC. Join P, A; A, D and P, Q and AP∥QD
R.T.P: (i) △ADQ=△PDQ
(ii) △BPQ=21△ABC
Proof: △ADQ and △PDQ are on the same base QD and between the same parallel line QD and AP.
∴△ADQ=△PDQ………(i) (Proved)
We have,
△BPQ=△BDQ+△PDQ
=△BDQ+△ADQ[∵△ADQ=△PDQ]
=△ABD
∴△BPQ=△ABD=21△ABC
∴△BPQ=21△ABC (Proved)
14. In triangle ABC of which AB = AC; perpendiculars through the points B and C on sides AC and A B intersed sides AC and AB at the points E and F. Let’s prove that FE∥BC.
Solution: Given: ABC is a triangle such that AB = AC, B and CFare perpendiculars on AC and AB respectively. FE is drawn.
R.T.P : FE∥BC
Proof : In △ABC,AB=AC
∴<ACB=<ABC
i.e, ∠ECB=∠FBC
Now, In △BEC and △BFC
<ECB=∠FBC
∠BEC=∠BFC(each is equal to 90∘]
BC is common
∴△BBEC≅△BFC
∴△BEC=△BFC
Since △BEC and △BFC are equal and they are on the same base BC. Hence they must be between the same parallels.
∴FE∥BC (Proved)
15. In triangle ABC < ABC = ∠ACB; bisectors of an angle ∠ABC and ∠ACB intersect the side AC and AB at the points E and F respectively. Let’s prove that, FE∥BC.
Solution :
Given: ABC is a triangle such that ∠ABC=∠ACB i.e, AB = AC, BE and CF are the bisector of ∠ABC and ∠ACB respectively.
R.T.P: FE∥BC
Proof: In ABC,∠ACB=∠ABCi.e;∠ECB=∠FBC
∴AB=AC
Now, In △BEC and △BFC
<ECB=∠FBC [Given]
<EBC=∠FCB [Given]
BC is common
∴△BEC≅△BFC
∴△BEC=△BFC
Since, △BEC are equal and they are on the same base BC.
Hence, they must be between the same parallels.
∴FE∥BC (Proved)
16. The shape of two parallelograms ABCD and AEFG of which ∠A is common are equal in area and E lies on AB, Let’s prove that DE∥FC.
Solution : Given: ABCD and AEFG be two parallelograms of which ∠A is common. DE & FC are drawn such that ∠OEBC=∠ODGF
R.T.P : DE∥FC
Proof: In △DOF and ∠ODFG are on the same base OD and between the same parallel lines OD and GF.
∴△ODF=21×<ODGF
In △COE and ∠OEBC are one the same base OE and between the same parallel lines OE and BC
∴△OEC=21<OEBC=21<ODGF
∴△ODF=△OEC
∵△ODF and △OEC are equal,
∵△ODFand△OEC are equal, Hence, they must be between the parallel lines.
∴DE∥FC (Proved)
17. ABCD is a parallelogram and ABCE is a quadrilateral. Diagonal AC divides the quadrilateral field ABCE into two equal parts. Let’s prove that AC∥DE.
Solution :
Given: ABCD is a parallelogram and ABCE is a quadrilateral. The diagonal AC bisects the quadrilateral, i.e.△ABC=△AEC.DE is drawn.
R.T.P : AC∥DE
Proof : AC is a diagonal of the parallel-ogram ABCD.
∴△ADC=△AEC
But, △ABC=△ABC
∴△ADC=△AEC
∵△ADC and △AEC are equal and they are on the same base AC.
∴AC∥DE(Proved)
18. D is the mid-point of side BC of triangle ABC ; P and Q lie on sides BC and BA in such a way that △BPQ=1△ABC. Let’s prove that, DQ∥PA
Solution :
Given: ABC is a triangle in which D is the mid-point of BC. P and Q are points on the sides BC and BA respectively such that △BPQ=△ABC. DQ and PA are drawn.
R.T.P : DQ∥PA
Proof : D is the mid-point of BC.
∴AD is median of △ABC
∴△ABD=21△ABC
But △BPQ=21△ABC
∴△ABD=△BPQ
or, △BQD+△AQD=△BQD+△PQD
∴△AQD=△BPQ
Since △AQD and △PQD are equal and they are on the same base DQ. Hence, they must be between same parallels.
∴DQ∥PA (Proved)
19. Parallelogram ABCD of which mid-points are E, F, G and H of sides AB, BC, CD and D A respectively. Let’s prove that
(i) EFGH is a parallelogram
(ii) Area of the shape of parallelogram EFGH is half of area of the shape of parallelogram ABCD.
Solution : (i) Given:
ABCD is a parallelogram of which mid-points are E, F, G & H of sides AB, BC, CD and DA respectively. EF, FG, GH. & HE are drawn.
R.T.P : EFGH is a parallelogram.
Construction: Join A and C.
Proof: G and H are the mid-points of CD and DA of △ACD
∴GH∥AC and GH21AC
Again,
E and F are the mid-points of AB & BC of △ABC
∴EF∥AC and EF21AC
∵GH∥AC and EF∥AC
∴GH∥EF and GH21AC=EF
∴GH∥EF and GH=EF
Hence, EFGH is a parallelogram. (Proved)
(ii) Given: ABCD is a parallelogram of which E, F, G & H are the mid points of AB, BC, CD & AD respectively such that EFGH is a parallelogram.
Construction: Join E & G
Proof: In △HEG and ∠AEGD are on the same base GE and between the same parallel lines GE and AD.
∴△HEG=21<AEGD
Again, In EFG and EBCG are on the same base GE and between the same parallel lines GE and BC
∴△EFG=21<BEGC
∴△HEG+△EFG=21<AEGD+=21<BEGC
or ∠EFGH=21(∠AEGD+∠BEGC)
∴∠EFGH=21∠ABCD (Proved).
20. AB∥DC of a trapezium ABCD and E is mid-point of BC. Let’s prove that area of tringular field AED=1× 2 area of the shape of trapezium field ABCD
Solution :
Given: ABCD is a trapezium and E is mid-point of BC. AE and DE are drawn.
R.T.P : Area of triangular field AED =21 x area of the trapezium field ABCD
Constructuion: We draw EF parallel to AB
Proof : In △EFD and trapezium CEFD are on the same base FE and between the same parallel lines FE and CD.
∴△AEF=21× trapazium CEFD
Again,
In △AEFand trapezium ABEF are on the same base EF and between the same parallel lines FE and AB
∴△AEF=21× trapazium AFEB
∴△EFD+△AEF=21 Trapezium CEFD + Trapezium AFEB)
∴△AE=21× Trapeziun ARCO
21. M.C.Q :
(i) D, E and F are mid-points of sides BC, CA and AB respectively of a triangle ABC. If △ABC=16sq . cm; then the area of the shape of trapezium FBCE is
(a) 40sq.cm.
(b) 8 sq. cm.
(c) 12sq⋅cm
(d) 100sq⋅cm.
Solution :∵△ABC=16sq.cm
In the figure, We see that,
Four triangles are formed with equal areas since D, E and F are the mid-points of BC, CA and A B respectively.
∴ Area △AFE=416Sq.cm.=4sq⋅cm.
∴ Area of trapezium FBCE=4×3sq.cm.(Ans.)
∴ (c) is correct option
(ii) A, B, C, D are the mid-points of sides PQ, QR, RS and SP respectively. of parallelogram PQRS. I area of the shape of parallelogram PQ S = 36 sq. cm. then area of ABCD field is
(a) 24 sq. cm.
(b) 18 sq. cm.
(c) 30 sq. cm.
(d) 36 sq. cm.
Solution : We have,
Area of parallelogram \ABCD
Area of Parallelogram =21× area of parallelogram PQRS
ABCD
=21×36Sq.cm
=18sqcm.
∴ (b) is correct option.
(iii) O is any a point inside parallelogram ABCD.
If △AOB+△COD=16 sq.cm., then area of the shape of parallelogram ABCD is
(a) 8 sq. cm.
(b) 4 sq. cm.
(c) 32 sq. cm.
(d) 64 sq. cm.
Solution : We draw a parallel line EF Such that AB∥CD∥EF
∴△AOB=21 area of parallelogram ABEF
∴△COD=21 area of parallelogram CDEF
∴ area of parallelogram ABCD=2(△AOB+△COD)=2×16sq.cm.=32sq.cm
∴ (c) is correct option.
(iv) D is the mid-point of side BC of triangle ABC. E is the mid-point of side BD and O is the mid-point of A E; area of triangular field BOE is
(a) 31× Area of△ABC
(b) 1 × Area of△ABC
(c) 61×Area of△ABC
(d) 8 × Area of△ABC
Solution :
Area of △BOE=21△ABE
=21×21△ABD
=21×21×21△ABC
=81△ABC
∴(d) is correct option.
(v) A parallelogram, a rectangle and a triangle stand on same base and between same parallel and if their area are P, Q and T respectively then
(a) P=R=2 T
(b) P=R=2T
(c) 2P = 2R = T
(d) P=R=T
Solution : We know that,
Area of a triangle =21 Area of parallelogram.
∴ area of parallelogram = 2 × area of triangle
∴ (a) is correct option.
22. Short answer type:
(i) DE is perpendicular on side AB from the point D of parallelogram ABCD and B F is perpendicular on side AD from the point B; If AB = 10 cm, AD = 8 cm and DE = 6 cm, let us write how much length of BF is
Solution : Given,
AB=CD=10 cm
AD=BC=8 cm
DE=6 cm
Area of △ABD=21AB×DE
Also, Area of △ADB=21AD×BF
∴21AD×BF=21AB×DE
or, 8×BF=10×6
or, BF=810×6×215
∴BF=7.5 cm (Ans.)
(ii) The area of the shape of parallelogram ABCD is 100 sq. units. P is mid-point of side BC, let us write how much area of triangular field ABP is
Solution :
We have,
Area of △ABP
=21 Area of △ABC
=21×21 Parallelogram ABCD
=21×100 sq.cm.
=25 sq. units. (Ans.)
(iii) AD is the median of triangle ABC and P is any point on side AC in such a way that area of △ADP: area of △ABD=2:3 Let us write the area of △PDC: area of △ABC
Solution:
We have,
area of △ADP : area of △ABD=2:3
or, area of △ABD area of △ADP=32
Area of △ABD=23 Area of △ADP
∴D is the median of △ABC
△ABD=21△ABC
△ADC=21△ABC
∴△ABD=△ADC
Now,
△ADC=21△ABC
or, △ADP+△PDC=21△ABC
or, △ADP+△PDC=21(△ABD+△ADC)
or, △ADP+△PDC=21×2△ABD[∴△ABD=△ADC]
or, △ADP+△PDC=23×△ADP
or, △PDC=(23−1)△ADP
or, △PDC=21△ADP
∴△ADP=2△PDC
∴ area of △ABC area of △PDC=2× area of △ACD area of △PDC
=2×(△ADP+△PDC) area of △PDC=2×(2△ACD+△PDC) area of △PDC
=2×3 area of △PDC area of △PDC=61
∴ Area of PDC : Area of ABC=1:6 (Ans.)
(iv) ABDE is a parallelogram. F is mid-point of side ED. If area of triangular field ABD is 20sq. unit, then let us write how much area of triangular field AEF is.
Solution : ∵△ABD= 20 sq. unit.
∴△ABD=△ADE=20 sq. unit.
∴△AEF=21△ADE
=21×20 sq.unit
=10sq.unit.(Ans.)
(v) PQRS is a parallelogram X and Y are the mid-points of side PQ and SR respectively. Joint diagonal SQ, Let us write the area of the shape of parallelogram field XQRY: area of triangular field QSER
Solution:
∵ Parallelogram XQRY and Parallelogram PQRS are on the same base PQ and between the same parallel lines PQ and SR.
∴ Parallelogram XQRY =21× Parallelogram PQRS………(i)
Again,
△QSR and Parallelogram PQRS are on the same base PQ and between the same parallel lines PQ and SR.
∴△QRS=21 Parallelogram PQRS……. (ii)
∴ from (i) and (ii)
Parallelogram XQRY = △QSR
∴ Parallelogram XQRY: △QSR=1:1 (Ans)