9类NCERT解决方案-第9章平行四边形和三角形的区域–练习9.4

问题1.平行四边形ABCD和矩形ABEF位于相同的基础AB上，并具有相等的面积。显示平行四边形的周长大于矩形的周长。

解决方案：

Given: ||gm ABCD and rectangle ABEF lie on the same base AB.

ar(ABCD)=a(ABEF)

to prove: Perimeter of ||gm ABCD> perimeter of rectangle ABEF

AB+BC+CD+DA>AB+BE+EF+FA

Proof:    

AB=CD            —-[opposite sides of ||gm and rectangle are equal]  

AB=EF             —-[opposite sides of ||gm and rectangle are equal]

→CD=EF  

Adding AB to both sides  

CD+AB=EF+AB        ———1

Here  BC>BF           ———[hypotenuse is always greater than other sides]

AD>AF                   ———[hypotenuse is always greater than other sides]

Adding them:

BC+AD>BE+AF                —————2

Adding 1 and 2

CD+AB+BC+AD > EF+AB+BE+AF

BC+AD>BE+AF

问题2。在图中，D和E是BC上的两个点，因此BD = DE = EC。证明ar(ABD)= ar(ADE)= ar(AEC)。

解决方案：

Given: In ∆ABC,D and F are two points on BC such that BD=DE=EC

To prove:  

Construction: Draw AM ⊥ BC

ar.(ABD)=1/2BD*AM           ———1

ar.(ADE)=1/2*DE*AM

=1/2*BD*AM           ——-2

Ar.(AEC)=1/2*EC*MA

=1/2*BD*AM           ————3

From 1 ,2 and 3

ar.(ABD)=ar.(ADE)=ar.(AEC)

问题3.在图中，ABCD，DCFE和ABFE是平行四边形。证明ar(ADE)= ar(BCF)

解决方案：

Given: ||gm ABCD, DCFE, and ABFE

To prove: ar.(ADE)=ar.(BCF)

Proof: In ∆ADE and ∆BCF

AD = BC               ——[opposite sides of ||gm ABCD]

DE = CF               ——[opposite sides of ||gm DCFE]

AE = BF               ——[opposite sides of ||gm ABFE]

∴∆ADE ≅ ∆BCF              (S.S.S congruence)

→ar.(ADE) = ar.(BCF)   ——-[congruent triangles have equal in area]

问题4。在图中，ABCD是一个平行四边形，并且BC产生到点Q，使得AD = CQ。如果AQ在P处与DC相交，则表明ar(BPC)= ar(DPQ)。 [提示：加入AC。]

解决方案：

Given: In fig. ABCD is a ||gm AD=CQ

To prove: ar.(BPC)=ar.()

Construction: Join AC.

Proof: ACQD will be ||gm            (∴AD=CQ, AD ||gm CQ)

In ∆APC and ∆QPD

AC=QD         [opposite sides of ||gm]

∠ 3=∠ 4            [altitude interior angles]

∴∆APC≅∆QPD                      [A.S.A congruence]  

→ar.(APC)=ar.(BPC)      [because both lie on the same base PC and between same ||lines PC and AB]

From 1 and 2

ar.(BPC)=ar.(DPQ)

问题5。在图中，ABC和BDE是两个等边三角形，使得D是BC的中点。如果AE在F处与BC相交，则表明

(i)ar(BDE)= 14 ar(ABC)

(ii)ar(BDE)= 12 ar(BAE)

(iii)ar(ABC)= 2 ar(BEC)

(iv)ar(BFE)= ar(AFD)

(v)ar(BFE)= 2 ar(FED)

(vi)ar(FED)= 18 ar(AFC)

[提示：加入EC和AD。显示BE || AC和DE || AB等]

解决方案：

i) Given: ABC and BDE are equilateral triangles D is mid point of BC i.e. BD=DC

To prove:  

Proof: Let AB=x=AC=BC

Then BD=x/2=BE=DE

ar.(ABC)=√3/4×2

ar.(BDE)=√3/4*(2/2)2          

=√3/4*x2/4  

=1/4*√3/4*x2

ar.(BDE)=1/4 ar.(ABC)

ii) to prove; ar.(BDE)=1/2 ar.(BAE)

Construction: Join EC

ED is median of ∆BFC  

ar.(BDE)=1/2 ar.(BEC)    ———–1

 ∠1=∠2=60°             [altitude interior angles]

→BE||AC  

AR.(BEC)=ar.(BEA) ————2[both lie on the same base BE and AC between same||lines BE and AC]

Taking 1

ar.(BDE)=1/2 ar.(BAE)     [from 2]

iii) To prove: ar.(ABC)=2ar.(BEC)

From i

ar.(BDE)=1/4 ar.(ABC)

from ii

ar.(BDE)=1/2 ar.(BAE)

From I and ii

¼ ar(ABC)=1/2ar.(BAE)

Ar.(ABC)=1/24 ar.(BAE)

=2ar.(BAE)

Ar.(ABC)=2ar.(BFC)     [ar.(BAE)=ar.(BEC) from ii]

iv) To show ar (BFE) = ar (AFD)

Construction: Join AD.

Proof: ∠3= ∠4=60°           ——–[angles of equilateral tringles]

Also they are alternate interior angle

∴AE||ED  

ar(EDB)=ar(EDA)        ——-[both lie on the same base ED  and Between same || lines ED and AB]

Subtracting ar(EDF) from both sides  

ar(EDB)-ar(EDF)=ar.(EDA)-ar.(EDF)

v) To show :ar.(BFE)=2ar.(FED)

Solution: In ∆ADB

AD²=AB²+BD²

=x²-(x/2)²

=x²-(x²/4)

=3x²/4

AD=√3x²/4

=√3x/4

ar.(AFD)=1/2*ED*AD

=1/2*ED*(√3X)/2         ————-1

Construction: Draw EL⊥ BD

→angle is midpoint of BD  

In ∆ELD

EL²=ED²-LD²

=(x/2)²-(x/4)²

=x²/4-x²/16

=4x²-x²/16

=3x²/16

EL=√3/10 X²

=√3/4 x

In ∆ADB  

ar(FED)=1/2*FD*EL

=1/2*FD*√3/4 x

=1/2*FD*√3/xx *1/2

ar.(FED)=ar.(AFD)

2ar.(FED)=ar.(BFE)        ——–[in part (iv) we prove ar(AFD=ar.(BFE)]

(vi) ar.(FED=1/8ar.(AFC))

Solution: ar(BDE)=1/4ar.(ABC)

ar.(BFE)+ar.(FED)=1/4ar.2*(ADC)

2ar.(FED)+ar(FED)=1/2ar.(AFC-1/2ar.(AFD))

3ar.(FED)=1/2ar.(AFC-AFD)

3ar.(FED)=1/2ar(AFC)-1/2ar.(AED)

3ar.(FED)=1/2ar.(AFC)-1/2 ar.*2(FED)    ————[FROM (iv) and (v)]

3ar(FED)=1/2 ar.(AFC)-ar.(FED)

3ar.(FED)+ar.(FED)=1/2ar.(AFC)

4ar.(FED)=1/2ar.(AFC)

ar.(FED)=1/2*4 ar.(AFC)

ar.(FED)=1/8 ar.(AFC)

问题6.四边形ABCD的对角AC和BD在P处相交。证明ar(APB)×ar(CPD)= ar(APD)×ar(BPC)。到BD。]

解决方案：

Given: ABCD is quadrilateral. Diagonals  AC and BD intersects at P.

To show: ar (APB) × ar (CPD) = ar (APD) × ar (BPC)

Construction: Draw AM ⊥BD &CN ⊥BD

L.H.S.  ar.(APB)*(LPD)

½*PB*AM*1/2*DP*CN      ————-1

¼*PB*AM*DP*LN  

R.H.S.  ar.(APD)*ar.(BPC)

½*DP*AM*1/2*BP*CN

¼*DP*AM*BP*CN

¼*PB*AM*DP*CN            ————-2

From 1 and 2

L.H.S=R.H.S

∴ar.(APB)*ar.(CPD)=ar.(APD)*ar.(BPC)  

问题7. P和Q分别是三角形ABC的边AB和BC的中点，R是AP的中点，表明

(i)ar(PRQ)= 12 ar(ARC)

(ii)ar(RQC)= 38 ar(ABC)

(iii)ar(PBQ)= ar(ARC)

解决方案：

i) Given: In ∆ABC, P is midpoint of AB, Q is midpoint of BC, R is midpoint of AP  

To show:  ar (PRQ) = 12 ar (ARC)

Construction: Join AQ and PC

Solution : ar.(APQ)  ———[∴QR is median of ∆APQ]

=½*1/2ar.(ABQ)   ———[∴QP is median of ∆ABQ]

=1/2*1/2*1/2ar.(ABC)      ———[∴AQ is median of ∆ABC]

=1/8ar.(ABC)     ———-1

ar.(ARC)=1/2(APC)   ——–[∴CR is median of ∆APC]

=1/2*1/2ar.(ABC)     ———-[∴CP is median of ∆ABC]

=1/4ar.(ABC)      ————2

Taking 1

ar.(PRQ)=1/8ar.(ABC)

=1/2*1/4ar.(ABC)   ———-[from 2]

ii) ar.(RQC)=3/8ar.(ABC)

solution : ar(ABC)+ar.(ARQ)-ar(ARC)   ——-1

ar(AQC)=1/2(ABC)       [∴AQ is median of ∆ABC]

ar.(ABC)=1/2ar.(APQ)  [∴QR is median of ∆APQ]

=1/2ar.(ABQ) [∴QP is median of ∆ABQ]

=1/2*1/2*1/2ar.(ABC)       [∴AQ is median of ∆ABC]

=1/8ar.(ABC)

ar.(ARC)=1/2ar.(APC)  [∴CR is median of ∆APC]

=1/2ar.(ABC)       [∴CP is median of ∆ABC]

ar.(RQC)=ar.(AQC)+ar.(ARQ)+ar.(ARC)

=1/2ar.(ABC)+1/8ar.(ABC)-1/4ar.(ABC)

=(1/2+1/8-1/4)ar.(ABC)

=(4+1-2/8)ar.(ABC)

=(5-2/8)ar.(ABC)

=3/8ar.(ABC)

iii) To show:  ar.(PBQ)=ar.(ARC)

ar(PBQ)=1/2ar.(ABQ)         [∴∆ABQ , QP is median]

=1/2*1/2ar.(ABC)              [∴∆ABC , AQ is median]

=1/4ar.(ABC)         ———1

ar.(ARC)=1/2ar.(APC)    [∴∆APC , CR is median]

=1/2*1/2ar.(ABC)     [∴∆ABC , CP is median]

=1/4ar.(ABC) ——–2

Form I and 2

ar.(PBQ)=ar.(ARC)

问题8.在图9.34中，ABC是与A成直角的直角三角形。BCED，ACFG和ABMN分别是BC，CA和AB侧的正方形。线段AX⊥DE在Y与BC相遇。

显示：

∆MBC ≅∆ABD

(ii)ar(BYXD)= 2 ar(MBC)

(iii)ar(BYXD)= ar(ABMN)(iv)∆FCB ≅∆ACE

(v)ar(CYXE)= 2 ar(FCB)

(vi)ar(CYXE)= ar(ACFG)

(vii)ar(BCED)= ar(ABMN)+ ar(ACFG)

注意：结果(vii)是毕达哥拉斯的著名定理。您将在X类中学习该定理的更简单证明。

解决方案：

Given: ∆ABC, right angled at A. BCED,ACFG,ABMN are squares at Y.

i). to prove: ∆MBC ≅∆ABD  

solution : In ∆MBC       and    ∆ABD

MB=AB        [∴ABMN is an square ]

BC=BD        [∴BCED is an square]

∠MBC=∠ABD     [∠MBA=∠CBO=90°    

                               ∠MBA= +∠CBD+∠ABC

                                ∴∠MBC=∠ABD]

∆MBC≅∆ABD            [S.A.S congruency rule]  

ii) To show: ar.(BYXD)=2ar.(MBC)

ar.(ABD)=1/2ar.(BYXD)       [Both on the same base BD and between same || lines BD and AX]

ar.(MBC)=1/2ar.(BYXD)       [in (i) ∆MBC≅∆ABD       ∴ar.(MBC)=ar.(ABD)  ]

2ar.(ABC)=ar.( BYXD)

ar.(MBC)=1/2ar.(ABMN)[ ∆MBC and ||gm ABMN  both lie on the same base MB and between same || lines MB and NC.]

2ar.(MBC)=ar.(ABMN)              ————–1

Also,

2ar.(MBC)=ar.(BYXD)              [from ii] —————2

From 1 and 2

ar.(BYXD)-ar.(ABMN)

(iv) To show: ∆FCB≅∆ACE

In ∆FCB  and ∆ACE

FC=AC                 [sides of square ACFG]

BC=CE                [sides of square BCED]

∠FCB=∠ACE         [∠ACE=∠BCE=90°

                                 ∠ACF+∠ACB=∠BCE+∠ACB

                                 ∴∠FCB=∠ACE]

∴∆FCB≅∆ACE        [by S.A.S    congruence ]

(v) To show : ar.(YXE)=2ar.(FCB)

ar.(ACE)=1/2ar.(CYXE)                 [∆ ACE and ||gm CYXE both lie on the same base CE and between same|| lines CE and AX]

ar. (FCB)=1/2(CYXE)          [in (iv) ∆FCB≅∆ACE               ∴ar.(FCB=ar.(ACE)]

2ar.(FCB)=ar.(CYXE)

(vi) To show : ar.(CYXE)=ar.(ACFG)[∆FCB and ||gm ACEFG both lie on the same base CF and BG Between  CF and BG]

2 ar.(FCB)=ar.(ACFG)  ———1

2ar.(FCB)=ar.(CYXE)

From 1 and 2

ar.(CYXE)=ar.(ACFG)

ar.(ABMN)+ar.(ACEFG)

ar.(BYXD)=ar.(ABMN)          [form (iii)]        ———1

ar.(CYXE)=ar.(ACFG)             [From (iv)] ———–2

Adding 1 and 2

ar.(BYXD)+ar. (CYXE)=ar.(ABMN)+ar.(ACEFG)

(vii) ar.(ABMN)+ar.(ACEFG)

ar.(BYXD)=ar.(ABMN)          [form (iii)]        ———1

ar.(CYXE)=ar.(ACFG)             [From (iv)] ———–2

adding 1 and 2

ar.(BYXD)+ar(CYXE)=ar.(ABMN)+ar.(ACEFG)

ar.(BCED)=ar.(ABMN)+ar.(ACFG)