Triangles - Test Papers

 CBSE Test Paper 01

Chapter 6 Triangles

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1. In an isosceles triangle ABC if AC = BC and AB2 = 2AC2 then the measure of ∠C$\mathrm{\angle }C$ is (1)
   1. 90o
   2. 45o
   3. 60o
   4. 300
2. In the given figure XY || BC. If AX = 3cm, XB = 1.5cm and BC = 6cm, then XY is equal to (1)
   
   1. 6 cm.
   2. 4.5 cm
   3. 3 cm.
   4. 4 cm.
3. What will be the length of the hypotenuse of an isosceles right triangle whose one side is 42–√cm$4\sqrt{2}cm$ (1)
   1. 122–√cm.$12\sqrt{2}cm.$
   2. 12 cm.
   3. 8 cm.
   4. 82–√cm.$8\sqrt{2}cm.$
4. In the given figure, if ar(ΔALM)ar(trapeziumLMCB)=916,$\frac{ar(\mathrm{\Delta }ALM)}{ar(trapeziumLMCB)}=\frac{9}{16},$ and LM||BC, Then AL:LB is equal to (1)
   
   1. 3 : 5
   2. 4 : 1
   3. 3 : 4
   4. 2 : 3
5. In the follwoing figure AD : DB = 1 : 3, AE : EC = 1 : 3 and BF : FC = 1 : 4, then (1)
   
   1. AD||FC.$AD||FC.$
   2. AD||FE.$AD||FE.$
   3. DE||BC.$DE||BC.$
   4. AE||DF.$AE||DF.$

6. In the given figure, ST || RQ, PS = 3 cm and SR = 4 cm. Find the ratio of the area of △$\mathrm{△}$PST to the area of △$\mathrm{△}$PRQ. (1)
   

7. If D and E are points on the sides AB  and AC respectively of △ABC$\mathrm{△}ABC$ such that AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm, show that DE ||BC. (1)
   

8. A ladder is placed in such a way that its foot is at a distance of 5 m from a wall and its tip reaches a window 12 m above the ground. Determine the length of the ladder. (1)

9. Triangles ABC and DEF are similar. If AC = 19 cm and DF = 8 cm, find the ratio of the area of two triangles. (1)

10. In the given figure, DE ∥$\Vert$ BC.
    
    Find AD. (1)

11. In △$\mathrm{△}$ABC, X is any point on AC. If Y, Z, U and V are the middle points on AX, XC, AB and BC respectively, then prove that UY || VZ and UV || YZ.
    

12. If the angles of one triangle are respectively equal to the angles of another triangle, Prove that the ratio of their corresponding sides is the same as the ratio of their corresponding angle bisectors. (2)

13. In a Δ$\mathrm{\Delta }$ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = x, DB = x-2, AE = x + 2 and EC = x - 1, find the value of x. (2)

14. A man goes 10m due south and then 24m due west. How far is he from the starting point? (3)

15. In the given figure A, B and C are points on OP, OQ and OR respectively such that AB ∥$\parallel$ PQ and AC ∥$\parallel$ PR. Prove that BC ∥$\parallel$ QR.
    

16. In a Δ$\mathrm{\Delta }$ABC, D and E are points on the sides AB and AC respectively such that DE∥BC.$DE\Vert BC.$ If AD = 8x - 7, DB = 5x - 3, AE = 4x - 3 and EC = (3x -1), find the value of x. (3)

17. In Fig. find ∠$\mathrm{\angle }$F. (3)
    
    

18. Prove that ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. (4)

19. In a trapezium ABCD, AB | | DC and DC = 2AB. EF | | AB, where E and F lie on BC and AD respectively such that BEEC=43$\frac{BE}{EC}=\frac{4}{3}$. Diagonal DB intersects EF at G. Prove that, 7EF = 11AB. (4)

20. In a triangle, if the square of one side is equal to the sum of the squares on the other two sides. Prove that the angle opposite to the first side is a right angle. Use the above theorem to find the measure of ∠$\mathrm{\angle }$ PKR in the figure given below. (4)
    

CBSE Test Paper 01
Chapter 6 Triangles

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Solution

1. 1. 90o
      Explanation: Given: AB2=2AC2$A{B}^2=2A{C}^2$
      ⇒$\Rightarrow$AB2=AC2+AC2$A{B}^2=A{C}^2+A{C}^2$
      ⇒$\Rightarrow$AB2=AC2+BC2$A{B}^2=A{C}^2+B{C}^2$ [Given: AC = BC] ∴$\therefore$ΔABC$\mathrm{\Delta }\text{ABC}$is a right angled triangle, by converse of Pythagoras theorem
      Now, since ΔABC$\mathrm{\Delta }\text{ABC}$ is an isosceles triangle also.
      Therefore, its two sides are equal i.e., AC = BC Therefore, AB is hypotenuse.
      ∴$\therefore$∠$\mathrm{\angle }$C is a right angle i.e., 90∘${90}^{\circ }$
2. 4. 4 cm.
      Explanation: Since XY||BC,$XY||BC,$ then using Thales theorem,
      ⇒$\Rightarrow$AXAB=XYBC$\frac{\text{AX}}{\text{AB}}=\frac{\text{XY}}{\text{BC}}$
      ⇒$\Rightarrow$34.5=XY6$\frac{3}{4.5}=\frac{\text{XY}}{6}$
      ⇒$\Rightarrow$XY = 4 cm
3. 3. 8 cm.
      Explanation: Let AC be hypotenuse. Its equal sides are AB and BC and AB = BC = 42–√$4\sqrt{2}$cm.

      Using Pythagoras Theorem,
      AC2 = AB2 + BC2$\text{A}{\text{C}}^2\text{ = A}{\text{B}}^2\text{ + B}{\text{C}}^2$
      ⇒$\Rightarrow$AC2 = (42–√)2 + (42–√)2$\text{A}{\text{C}}^2\text{ = }{\left(4\sqrt{2}\right)}^2\text{ + }{\left(4\sqrt{2}\right)}^2$ = 32 + 32 = 64 cm2
      ⇒$\Rightarrow$AC = 8 cm
      

4. 2. 4 : 1
      Explanation: In ΔALM$\mathrm{\Delta }\text{ALM}$ and ΔABC,$\mathrm{\Delta }\text{ABC},$ ∠A = ∠A$\mathrm{\angle }\text{A = }\mathrm{\angle }\text{A}$ [Common] ∠ALM = ∠ABC$\mathrm{\angle }\text{ALM = }\mathrm{\angle }\text{ABC}$ [Corresponding angles as LM∥$\parallel$BC]
      ∴$\therefore$ΔALM$\mathrm{\Delta }\text{ALM}$∼$\sim$ΔABC$\mathrm{\Delta }\text{ABC}$ [AA similarity]
      ∴$\therefore$ar(ΔALM)ar(ΔABC)=AL2AB2$\frac{\text{ar}\left(\mathrm{\Delta }\text{ALM}\right)}{\text{ar}\left(\mathrm{\Delta }\text{ABC}\right)}=\frac{\text{A}{\text{L}}^2}{\text{A}{\text{B}}^2}$ Now, ar(trap.LMCB)ar(ΔALM)=916$\frac{\text{ar}\left(\text{trap}\text{.LMCB}\right)}{\text{ar}\left(\mathrm{\Delta }\text{ALM}\right)}=\frac{9}{16}$
      ⇒$\Rightarrow$ar(ΔABC)−ar(ΔALM)ar(ΔALM)=916$\frac{\text{ar}\left(\mathrm{\Delta }\text{ABC}\right)-\text{ar}\left(\mathrm{\Delta }\text{ALM}\right)}{\text{ar}\left(\mathrm{\Delta }\text{ALM}\right)}=\frac{9}{16}$
      ⇒$\Rightarrow$ar(ΔABC)ar(ΔALM)−1=916$\frac{\text{ar}\left(\mathrm{\Delta }\text{ABC}\right)}{\text{ar}\left(\mathrm{\Delta }\text{ALM}\right)}-1=\frac{9}{16}$
      ⇒$\Rightarrow$ar(ΔABC)ar(ΔALM)=916+1$\frac{\text{ar}\left(\mathrm{\Delta }\text{ABC}\right)}{\text{ar}\left(\mathrm{\Delta }\text{ALM}\right)}=\frac{9}{16}+1$
      ⇒$\Rightarrow$ar(ΔABC)ar(ΔALM)=2516$\frac{\text{ar}\left(\mathrm{\Delta }\text{ABC}\right)}{\text{ar}\left(\mathrm{\Delta }\text{ALM}\right)}=\frac{25}{16}$
      ⇒$\Rightarrow$AB2AL2=2516$\frac{\text{A}{\text{B}}^2}{\text{A}{\text{L}}^2}=\frac{25}{16}$
      ⇒$\Rightarrow$ABAL=54$\frac{\text{AB}}{\text{AL}}=\frac{5}{4}$
      Let AB = 5x and AL = 4x then LB = AB - AL = 5x - 4x = 1x
      ∴$\therefore$ ALLB=4x1x=41$\frac{\text{AL}}{\text{LB}}=\frac{4x}{1x}=\frac{4}{1}$
      ⇒$\Rightarrow$ AL : LB = 4 : 1

       

5. 3. DE||BC.$DE||BC.$
      Explanation: Given: ADDB=13$\frac{\text{AD}}{\text{DB}}=\frac{1}{3}$ and AEEC=13$\frac{\text{AE}}{\text{EC}}=\frac{1}{3}$
      Therefore, in △ABC,$\mathrm{△}\text{ABC},$ADDB$\frac{\text{AD}}{\text{DB}}$=AEEC$=\frac{\text{AE}}{\text{EC}}$
      ∴$\therefore$ DE∥$\parallel$BC [Using Thales Theorem]
      Here we are not considering BF : FC =1 : 4.

6. PS = 3 cm, SR = 4 cm and ST || RQ.
   
   PR = PS + SR
   = 3 + 4 = 7 cm
   In △$\mathrm{△}$PST and △$\mathrm{△}$PRQ
   ∠$\mathrm{\angle }$SPT ≅$\cong$ ∠$\mathrm{\angle }$RPQ (common angle)
   ∠$\mathrm{\angle }$PST ≅$\cong$ ∠$\mathrm{\angle }$PRQ (Alternate angle)
   △$\mathrm{△}$PST ∼$\sim$ △$\mathrm{△}$PRQ (AA configuration)
    ar ΔPST ar ΔPQR=PS2PR2$\frac{\text{ ar }\mathrm{\Delta }PST}{\text{ ar }\mathrm{\Delta }PQR}=\frac{P{S}^2}{P{R}^2}$=3272=949$=\frac{3^2}{7^2}=\frac{9}{49}$
   Hence required ratio = 9 : 49.

7. Given: AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm
   ∴ADAB=1.45.6=14$\therefore \frac{AD}{AB}=\frac{1.4}{5.6}=\frac{1}{4}$ and AEAC=1.87.2=14$\frac{AE}{AC}=\frac{1.8}{7.2}=\frac{1}{4}$
   ⇒ADAB=AEAC$\Rightarrow \frac{AD}{AB}=\frac{AE}{AC}$
   Hence, by the converse of Thales' theorem, DE∥BC$DE\Vert BC$.

8. Let AC be the ladder, AB be the wall and BC be the distance of ladder from the foot of the wall.
   In right △$\mathrm{△}$ABC,
   
   AC2 = AB2 + BC2{ using Pythagoras theorm for right-angled triangle}
   ⇒$\Rightarrow$ AC2 = (12)2 + 52
   ⇒$\Rightarrow$ AC2 = 144 + 25
   ⇒$\Rightarrow$ AC = 13 m

9. We have,
   Δ$\mathrm{\Delta }$ABC ~ Δ$\mathrm{\Delta }$DEF
   AC = 19 cm and DF = 8 cm
   By area of similar triangle theorem
   Area(ΔABC)Area(ΔDEF)=AC2DF2$\frac{Area(\mathrm{\Delta }ABC)}{Area(\mathrm{\Delta }DEF)}=\frac{A{C}^2}{D{F}^2}$=(19)282=36164$=\frac{{(19)}^2}{8^2}=\frac{361}{64}$

10. ∵$\because$ DE ∥$\Vert$ BC
    ∴ADBD=AECE$\therefore \frac{\mathrm{A}\mathrm{D}}{\mathrm{B}\mathrm{D}}=\frac{\mathrm{A}\mathrm{E}}{\mathrm{C}\mathrm{E}}$ (from BPT)
    ⇒AD7.2=1.85.4⇒$\Rightarrow \frac{\mathrm{A}\mathrm{D}}{7.2}=\frac{1.8}{5.4}\Rightarrow$ AD = 2.4 cm

11. Join BX
    In ABX, U is midpoint of AB and Y is mid-point AX (given)
    ∴$\therefore$ UY ∥$\Vert$ BX (using mid-point theorem) .....(i)
    
    In BCX, v is mid-point of BC and z is mid-point of XC 
    VZ ∥$\Vert$ BX ..(ii)
    from (i) and (ii)
    UY ∥$\Vert$ VZ
    In ABC, U is mid-point of AB and V is mid-point of BE.
    ∴$\therefore$ UV ∥$\Vert$ AC
    ⇒$\Rightarrow$ UV ∥$\Vert$ YZ Hence proved.

12. 
    Given: Two triangles ABC and DEF in which ∠$\mathrm{\angle }$ A = ∠$\mathrm{\angle }$D, ∠$\mathrm{\angle }$B = ∠$\mathrm{\angle }$E and ∠$\mathrm{\angle }$C = ∠$\mathrm{\angle }$F, AL and DM are angle bisectors of ∠$\mathrm{\angle }$A
    and ∠$\mathrm{\angle }$D respectively
    To prove: BCEF=ALDM$\frac{BC}{EF}=\frac{AL}{DM}$
    Proof: Triangle ABC and DEF are Similar.
    ⇒$\Rightarrow$ ABDE=BCEF$\frac{AB}{DE}=\frac{BC}{EF}$ ......(i)
    In ​△$\mathrm{△}$​ ABL and ​△$\mathrm{△}$​ DEM, we have
    ∠$\mathrm{\angle }$B= ∠$\mathrm{\angle }$E [Given]
    ∠$\mathrm{\angle }$ BAL= ∠$\mathrm{\angle }$ EDM [ ∵$\because$ ∠$\mathrm{\angle }$ A= ∠$\mathrm{\angle }$ D ⇒$\Rightarrow$ 12∠A=12∠D$\frac{1}{2}\mathrm{\angle }A=\frac{1}{2}\mathrm{\angle }D$]
    ⇒$\Rightarrow$ ​△$\mathrm{△}$​ ABL ∼$\sim$ ​△$\mathrm{△}$​ DEM [AA similarity]
    ⇒$\Rightarrow$ ABDE=ALDM$\frac{AB}{DE}=\frac{AL}{DM}$ .......(ii)
    From (i) and (ii) we have
    BCEF=ALDM$\frac{BC}{EF}=\frac{AL}{DM}$

13. We have,
    
    DE || BC
    Therefore, by basic proportionality theorem,
    We have,
    ADDB=AEEC$\frac{AD}{DB}=\frac{AE}{EC}$
    ⇒xx−2=x+2x−1$\Rightarrow \frac{x}{x-2}=\frac{x+2}{x-1}$
    ⇒$\Rightarrow$ x(x - 1) = (x + 2)(x - 2)
    ⇒$\Rightarrow$ x2 - x = x2 - (2)2 [∵$\because$ (a - b)(a + b) = a2 - b2]
    ⇒$\Rightarrow$ -x = -4
    ⇒$\Rightarrow$ x = 4 cm.

14. Starting from O, let the man goes from O to A and then A to B as shown in the figure.
    Then,
    OA = 10m, AB = 24m and ∠$\mathrm{\angle }$OAB = 90o
    
    Using Pythagoras theorem:
    OB2 = OA2 + AB2
    ⇒$\Rightarrow$ OB2 = 102 + 242
    ⇒$\Rightarrow$ OB2 = 100 + 576
    ⇒$\Rightarrow$ OB2 = 676
    ⇒$\Rightarrow$ OB =676−−−√$\sqrt{676}$ = 26m
    Hence, the man is 26m south-west from the starting position.

15. Proof : In △POQ,AB||PQ$\mathrm{△}POQ,AB||PQ$,(Given)
    AOAP=OBBQ$\frac{AO}{AP}=\frac{OB}{BQ}$.......(i) (BPT)
    In △$\mathrm{△}$OPR AC∥PR$AC\Vert PR$
    OAAP=OCCR.....(ii)$\frac{OA}{AP}=\frac{OC}{CR}.....(ii)$
    From eqn (I) and (ii)
    OBBQ=OCCR$\frac{OB}{BQ}=\frac{OC}{CR}$
    Hence BC∥QR ( By converse of BPT)$BC\Vert QR\text{ ( By converse of BPT)}$

16. We have,
    
    We are given that, DE || BC
    Therefore, by thales theorem,
    We have,
    ADDB=AEEC$\frac{AD}{DB}=\frac{AE}{EC}$
    ⇒8x−75x−3=4x−33x−1$\Rightarrow \frac{8x-7}{5x-3}=\frac{4x-3}{3x-1}$
    ⇒$\Rightarrow$ (8x - 7)(3x - 1) = (4x - 3)(5x - 3)
    ⇒$\Rightarrow$ 24x2 - 8x - 21x + 7 = 20x2 - 12x - 15x + 9
    ⇒$\Rightarrow$ 24x2 - 20x2 - 29x + 27x + 7 - 9 = 0
    ⇒$\Rightarrow$ 4x2 - 2x - 2 = 0
    ⇒$\Rightarrow$ 2[2x2 - x - 1] = 0
    ⇒$\Rightarrow$ 2x2 - x - 1 = 0
    ⇒$\Rightarrow$ 2x2 - 2x + 1x - 1 = 0
    ⇒$\Rightarrow$ 2x(x - 1) + 1(x - 1) = 0
    ⇒$\Rightarrow$ (2x + 1)(x - 1) = 0
    ⇒$\Rightarrow$2x + 1 = 0 or x - 1 = 0
    ⇒x=−12$\Rightarrow x=-\frac{1}{2}$or x = 1
    x=−12$x=-\frac{1}{2}$ is not possible.
    ∴$\therefore$ x = 1.

17. In triangles ABC and DEF, we have
    ABDF=BCFE=CAED=12$\frac{AB}{DF}=\frac{BC}{FE}=\frac{CA}{ED}=\frac{1}{2}$
    Therefore, by SSS-criterion of similarity, we have
    ΔABC∼ΔDFE$\mathrm{\Delta }ABC\sim \mathrm{\Delta }DFE$
    ⇒$\Rightarrow$ ∠$\mathrm{\angle }$A = ∠$\mathrm{\angle }$D, ∠$\mathrm{\angle }$B = ∠$\mathrm{\angle }$F and ∠$\mathrm{\angle }$C = ∠$\mathrm{\angle }$E
    ⇒$\Rightarrow$ ∠$\mathrm{\angle }$D = 80°, ∠$\mathrm{\angle }$F = 60°
    Hence, ∠$\mathrm{\angle }$F = 60°.

18. Given : ΔABC∼ΔPQR$\mathrm{\Delta }ABC\sim \mathrm{\Delta }PQR$
    To Prove : ar(ΔABC)ar(ΔPQR)=(ABPQ)2=(BCQR)2=(ACPR)2$\frac{\mathrm{ar}(\mathrm{\Delta }ABC)}{\mathrm{ar}(\mathrm{\Delta }PQR)}={\left(\frac{AB}{PQ}\right)}^2={\left(\frac{BC}{QR}\right)}^2={\left(\frac{AC}{PR}\right)}^2$
    Construction: Draw AD ⊥$\mathrm{\perp }$BC and PE ⊥$\mathrm{\perp }$ QR
    Proof :
    
    ΔABC∼ΔPQR$\mathrm{\Delta }ABC\sim \mathrm{\Delta }PQR$
    ∴ABPQ=BCQR=ACPR$\therefore \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$ ( Ratio of corresponding sides of similar triangles are equal) ...(i)
    ∠B=∠Q$\mathrm{\angle }B=\mathrm{\angle }Q$ (Corresponding angles of similar triangles)......... (ii)
    In ΔADB and ΔPEQ$\mathrm{\Delta }ADB\text{ and }\mathrm{\Delta }PEQ$
    ∠B=∠Q$\mathrm{\angle }B=\mathrm{\angle }Q$ ( From (ii))
    ∠ADB=∠PEQ$\mathrm{\angle }ADB=\mathrm{\angle }PEQ$ [each90∘]$\left[\mathrm{each}{90}^{\circ }\right]$
    ∴$\therefore$ ΔADB∼ΔPEQ$\mathrm{\Delta }ADB\sim \mathrm{\Delta }PEQ$ [ By AA criteria]
    ⇒$\Rightarrow$ ADPE=ABPQ$\frac{AD}{PE}=\frac{AB}{PQ}$ (Corresponding sides of similar triangles) ...(iii)
    From equation (i) and equation (iii)
    ABPQ=BCQR=ACPR=ADPE$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}=\frac{AD}{PE}$ ...(iv)
    ar(ΔABC)ar(ΔPQR)=12×BC×AD12×QR×PE$\frac{\mathrm{ar}(\mathrm{\Delta }ABC)}{\mathrm{ar}(\mathrm{\Delta }PQR)}=\frac{\frac{1}{2}\times BC\times AD}{\frac{1}{2}\times QR\times PE}$
    =(BCQR)×(ADPE)$=\left(\frac{BC}{QR}\right)\times \left(\frac{AD}{PE}\right)$
    (ADPE=BCQR$\frac{AD}{PE}=\frac{BC}{QR}$)
    =BCQR×BCQR$=\frac{BC}{QR}\times \frac{BC}{QR}$
    ⇒$\Rightarrow$ ar(ΔABC)ar(ΔPQR)=BC2QR2$\frac{ar(\mathrm{\Delta }ABC)}{ar(\mathrm{\Delta }PQR)}=\frac{B{C}^2}{Q{R}^2}$ ....(v) [from eq. (iv)]
    From equation (iv) and equation (v),
    ∴ar(ΔABC)ar(ΔPQR)=(ABPQ)2=(BCQR)2=(ACPR)2$\therefore \frac{\mathrm{ar}(\mathrm{\Delta }ABC)}{\mathrm{ar}(\mathrm{\Delta }PQR)}={\left(\frac{AB}{PQ}\right)}^2={\left(\frac{BC}{QR}\right)}^2={\left(\frac{AC}{PR}\right)}^2$
    ∴$\therefore$ Ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

19. 
    In a trapezium ABCD, AB|| DC ,. EF || AB and CD=2AB
    and also BEEC=43$\frac{BE}{EC}=\frac{4}{3}$ -------------(1)
    AB || CD and  AB || EF
    ∴$\therefore$ AFFD=BEEC=43$\frac{AF}{FD}=\frac{BE}{EC}=\frac{4}{3}$
    In ΔBGE and ΔBDC$\mathrm{\Delta }BGE\text{ and }\mathrm{\Delta }BDC$
    ∠BEG=∠BCD$\mathrm{\angle }BEG=\mathrm{\angle }BCD$ (∵ corresponding angles)
    ∠GBE=∠DBC$\mathrm{\angle }GBE=\mathrm{\angle }DBC$ (Common)
    ∴ΔBGE∼ΔBDC$\therefore \mathrm{\Delta }BGE\sim \mathrm{\Delta }BDC$ [ By AA similarity]
    ⇒$\Rightarrow$EGCD=BEBC$\frac{EG}{CD}=\frac{BE}{BC}$ ...........(2)
    Now,$Now,$ from (1) BEEC=43$\frac{BE}{EC}=\frac{4}{3}$
    ⇒$\Rightarrow$ ECBE=34$\frac{EC}{BE}=\frac{3}{4}$
    ⇒$\Rightarrow$ ECBE+1=34+1$\frac{EC}{BE}+1=\frac{3}{4}+1$
    ⇒$\Rightarrow$ EC+BEBE=74$\frac{EC+BE}{BE}=\frac{7}{4}$
    ⇒$\Rightarrow$ BCBE=74 or BEBC=47$\frac{BC}{BE}=\frac{7}{4}\text{ or }\frac{BE}{BC}=\frac{4}{7}$
    from equation (2), EGCD=47$\frac{EG}{CD}=\frac{4}{7}$
    So EG=47CD$EG=\frac{4}{7}CD$ ......(3)
    Similarly, ΔDGF∼ΔDBA$\mathrm{\Delta }DGF\sim \mathrm{\Delta }DBA$ (by AA similarity)
    ⇒$\Rightarrow$ DFDA=FGAB$\frac{DF}{DA}=\frac{FG}{AB}$
    ⇒$\Rightarrow$ FGAB=37$\frac{FG}{AB}=\frac{3}{7}$
    ⇒$\Rightarrow$ FG=37AB$FG=\frac{3}{7}AB$  ...(4)
    [∵AFAD=47=BEBC⇒ECBC=37=DEDA]$\left[\begin{array}{c}\because \frac{AF}{AD}=\frac{4}{7}=\frac{BE}{BC}\\ \Rightarrow \frac{EC}{BC}=\frac{3}{7}=\frac{DE}{DA}\end{array}\right]$
    Adding equations (3) and (4),we get,
    EG+FG=47CD+37AB$EG+FG=\frac{4}{7}CD+\frac{3}{7}AB$
    ⇒$\Rightarrow$EF=47×(2AB)+37AB$EF=\frac{4}{7}\times (2AB)+\frac{3}{7}AB$
    =87AB+37AB=117AB$=\frac{8}{7}AB+\frac{3}{7}AB=\frac{11}{7}AB$
    ∴7EF=11AB$\therefore 7EF=11AB$

20. 

    1. Given: In △$\mathrm{△}$ABC such that
       AC2 = AB2 + BC2
       To prove: Triangle ABC is right angled at B
       Construction: Construct a triangle DEF such that
       DE = AB, EF = BC and ∠E=90∘$\mathrm{\angle }E={90}^{\circ }$
       Proof: ∵$\because$ △$\mathrm{△}$ DEF is a right angled triangle right angled at E [construction]
       By Pythagoras theorem, we have
       DF2 = DE2 + EF2
       ⇒$\Rightarrow$ DF2 = AB2 + BC2 [ ∵$\because$ DE = AB and EF = BC]
       ⇒$\Rightarrow$ DF2 = AC2[ ∵$\because$ AB2 + BC2 = AC2]
       ⇒$\Rightarrow$ DF = AC
       Thus,in △$\mathrm{△}$ ABC and △$\mathrm{△}$DEF, we have
       AB = DE
       BC = EF
       and AC = DF [By Construction and (i)]
       ∴$\therefore$ △$\mathrm{△}$ ABC ≅$\cong$ △$\mathrm{△}$DEF (SSS)
       ⇒$\Rightarrow$ ∠$\mathrm{\angle }$ B = ∠$\mathrm{\angle }$ E = 90o
       Hence, △$\mathrm{△}$ ABC is a right triangle.
    2. In △$\mathrm{△}$QPR , ∠$\mathrm{\angle }$ QPR = 90o
       ⇒$\Rightarrow$ 242 + x2 = 262
       ⇒$\Rightarrow$ x = 10
       ⇒$\Rightarrow$ PR = 10 cm
       Now in △$\mathrm{△}$PKR, PR2 = PK2 + KR2[as 102 = 82 + 62]
       ∴$\therefore$ PKR is right angled at K
       ⇒$\Rightarrow$ ∠$\mathrm{\angle }$ PKR =90o