RS Aggarwal Class 9 Solutions Geometrical Constructions

RS Aggarwal Class 9 Solutions Chapter 12

Q.1: Draw a line segment AB = 5 cm and draw its perpendicular bisector.

Sol:

Steps of Construction:

(i) Draw a line segment AB= 5cm

(ii) With A as centre and radius equal to more than half of AB, draw two arcs, one above AB and the other below AB.

(iii) With B as a centre and the same radius draw two arcs which cut the previously drawn arcs at C and D.

(iv) Join CD, intersecting AB at point P.

Therefore, CD is the perpendicular bisector of AB at the point P.

 

https://lh6.googleusercontent.com/mVrsyzysKR_3Caks1MO7jy21nJEq8cLU9qZIgSH0TEoToiJAvUabvQ1_tjiOkSXohq71q9zgLMqPw-ZusJFfDeOK2Ps7PJ1YYoQPSIjxtKx70Ju99Ht51c4I098K5J99J82YGpMu

 

Q.2: Draw an angle of 450 using scale and compass only. Draw the bisector of this angle.

Sol:

Steps of Construction:

(i) Draw a line segment OA.

(ii) At A, draw AOE=90, using rulers and compass.

(iii) With B as centre and radius more than half of BD, draw an arc.

(iv) With D as centre and same radius draw another arc which cuts the previous arc at F.

(v) Join OF. AOF=45

(vi) Now with centre B and radius more than half of BC, draw an arc.

(vii) With centre C and same radius draw another arc which cuts the previously drawn arc at X.

(viii) Join OX.

Therefore, OX is the bisector of AOF.

 

https://lh5.googleusercontent.com/fNpwsshQDpRmH4EBoC-0CXWX7Sc4oAOwQvw9Ilu4XCUS6SnraWleZO80DPCoijj5Fx6wWOXaq3WMDR9DG5_ZpHJPRMI4xU7awA36sLPweDzInvaSFQdztT9EwnXcF5r7GxaiuFPT

 

Q.3: Draw an angle of 900 and draw its bisector.

Sol:

Steps of construction:

(i) Draw a line segment OA.

(ii) With O as the centre and any suitable radius draw an arc, cutting OA at B.

(iii) With B as the centre and the same radius cut the previously drawn arc at C.

(iv) With C as the centre and the same radius cut the arc at D.

(v) With C as the centre and the radius more than half CD draw an arc.

(vi) With D as the centre and the same radius draw another arc which cuts the previous arc at E.

(vii) Join E, So AOE=90

(viii) Now with B as centre and radius more than half of CB draw an arc.

(ix) With C as the centre and same radius draw an arc which cuts the previous at F.

(x) Join OF. Therefore, it is the bisector of the right AOE.

 

https://lh5.googleusercontent.com/KzX2QoFa03ujwAUgKGqaUnHh1AQwG7ou29FFjg4H8o9CjTvRQ6APZJUUkhyHZ9MSTK4P7bFuawo--_t5dl1v2eZddz9CLl2Ia4LSHp6JBlKSPAcotjpTl5xrSZ3KTUdTBQdndcQI

 

Q.4: Construct an equilateral triangle each of whose sides measures 5 cm.

Sol:

Steps of construction:

(i) Draw a line segment BC= 5cm

(ii) With B as centre and radius equal to BC draw an arc.

(iii) With C as the centre and the same radius draw another arc which cuts the previous arc at A.

(iv) Join AB and AC.

Then ABC is the required equilateral triangle.

https://lh6.googleusercontent.com/xfDrUytVZwS_jDUWwJ4kEa1Dyj_gUBcMBSzlbSYvxFV5-vxqWgi1FBQP1jqMvzZ6yM4gZKN01p2YwCMmMTsRUCLtVRou7dkz3e53Cox4lYY2BRFo4c24jGWSJ-p_BGH9Mw0H3zx_

 

Q.5: Construct an equilateral triangle each of whose altitudes measures 5.4 cm

Sol:

Steps of construction:

(i) Draw a line XY.

(ii) Mark any point P on it.

(iii) From P, draw PQXY.

(iv) From P, set off PA= 5.4 cm cutting PQ at A.

(v) Construct PAC=30, meeting XY at B and C respectively.

Therefore, ABC is required equilateral triangle.

 

https://lh3.googleusercontent.com/uQzY-U68kb9IvESCaFsDibB0m9WN2EhjDA5xCvkK_hYGxddGye5whzZXfQVUtm8i5kWe7TxDZHbrbYQTa-QMSL31zL0yj7AShhWUrjSDnB0E38WrzZjPW7mshvJFLWtVUOW9gK7V

 

Q.6: Construct a triangle ABC in which BC = 5 cm, AB = 38 cm and AC = 26 cm.

Sol:

Steps of construction:

(i) Draw a line segment BC= 5 cm.

(ii) With centre B and radius equal to 3.8 cm draw an arc.

(iii) With centre B and radius equal to 3.8 cm draw another arc which cuts the previously drawn arc at A.

(iv) Join AB and AC

Therefore, ABC is the required equilateral triangle.

 

https://lh4.googleusercontent.com/95H2EV3us70gcvz3ABUW46t4Yuk9m_NR-w4tQlbzdAYw1XXsz0Kmau0mryYCU45dZvBE7_enXKLwktkN3c0GM2GlrjCO8Byn00LE1peYeXQ5HBHz4hBZWeUOUmYzJp1KJdCnw8kv

 

Q.7: Construct a ΔABCinwhichBC=4.7cm,B=60andC=30

Sol:

Steps of construction:

(i) Draw a line segment BC= 5cm.

(ii) At B draw XBC=60

(iii) At C draw YCB=30

Let XB and YC interact at A.

Therefore, ABC is the required triangle.

 

https://lh6.googleusercontent.com/WE--jUUA1O1sWV6Atc72-2JA-hBtCd1v9JQMo-W85_megxXD8xb2U9UGDVurTGZiUTd5eWXW_q_JBGkos8E2WAolS28ulH189iyxnWY9WWr_GQkXsiB9ostThsDcYV2qCMoHt2RA

 

Q.8: Construct an isosceles triangle PQR whose base measures 5 cm and each of equal sides measures 4.5 cm.

Sol:

Steps of construction:

(i) Draw a line segment QR= 5cm which is the base.

(ii) With centre Q and radius equal to 4.5 cm, draw an arc.

(iii) With centre P and the same radius is drawn another arc which cuts the previous arc at P.

(iv) Join PQ and PR.

Therefore, PQR is the required isosceles triangle.

 

https://lh6.googleusercontent.com/LlZJupne1AxZ87MWRRk7W34_moexKzPb386XulDCA-SVeHjzzCHob-aTgZnn3TKK9XXyDTtj0JjBNzoHDrNXX60c0g4PHfiIwY1rHIwqw6OrcX9SBaMNVbNR-z9cStOPXS92y6Wt

 

Q.9: Construct an isosceles triangle whose base is 4.8 cm and whose vertical angle is 800

Sol:

Steps of construction:

(i) Draw a line segment BC = 4.8 cm

(ii) Make CBX=80, below the line segment BC.

(iii) Make XBY=90

(iv) Draw the right bisector PQ of BC, intersecting BY at O.

(v) With O as centre and radius OB, draw a circle intersecting PQ at A.

(vi) Join AB and AC.

Therefore, PQR is the required isosceles triangle in which AB=AC.

 

https://lh6.googleusercontent.com/cAsdku2e0C2koFk6bCUeXPFl7pkzUDkb0KMRhrwnhtE-uEroMf6WACU5quD56wuTznrOZJPyhXXpGYvn7e8Z3DrgLhC4xcuTtMT8vIEJmfWqkGdcdgN2XyO4mwA08CdH_UtTU1CF

 

Q.10: Construct a right angled triangle whose hypotenuse measures 5.3 am and the length of one of whose sides containing the right angle measures 4.5 cm.

Sol:

Steps of construction:

(i) Draw a line segment BC= 5.3 cm.

(ii) Find the mid-point O of BC.

(iii) With O as a centre and radius OB, draw a semicircle on BC.

(iv) With B as centre and radius equal to 4.5 cm draw an arc cutting the semicircle at A.

(v) Join AB and AC.

Therefore, ABC is the required triangle.

 

https://lh3.googleusercontent.com/J34yYfhy0HaVLMlVrtc86HCH8zN_rYMVTYBQv5LSEftb7Fnkg35AVAWDAQU6aMsrrJN_652mmdwaBeJKi9VlmUbVUoGszW0a9wMkE1UvjkHWZrVSWo4xhfHEiR-WudavCB_WmPMt

 

Q.11: Construct a ΔABC in which B=30,B=60 and the length of the perpendicular form the vertex A to the base BC is 4.8 cm.

Sol:

Steps of construction:

(i) Draw a line segment XY.

(ii) Take any point P on XY and draw PQXY.

(iii) Along PQ, set off PA= 4.8 cm.

(iv) Through A, draw LMXY.

(v) Construct LAB=30 and MAC=60 meeting XY at B and C respectively.

Therefore, ABC is the required triangle.

 

https://lh6.googleusercontent.com/wWhFeIXWBKYTmfKnOIfQ1VfMeWfX-sgxiq2R3BVerLOjDP-avggTfKOrDMw5BokGMmLpWjGmcWB_03XANKkWrW4YLMq_Rmpn-XMgeG13-5DkeWmUgYKfniaLRArCVtwujeLx_85V

 

Q12: Construct a triangle PQR whose perimeter is 12 am and lengths of whose sides are in the ratio 3:2:4.

Sol:

Steps of construction:

(i) Draw a line segment AB = 12 cm.

(ii) Draw a ray AX, making an acute angle with AB and draw in the downward direction.

(iii) From A set of (3+2+4) = 9 equal distances along AX.

(iv) Mark points L,M,N on AX such as that AL = 3 units, Lm = 2 units and MN = 4 units.

(v) Join NB.

(vi) Through L and M, draw LQNBandMRNB cutting AB at Q and R respectively.

(vii) With Q as centre and radius AQ, draw an arc.

(viii) With R as centre and radius RB, draw another arc, cutting the previous arc at P.

(ix) Join PQ and PR.

Therefore, PQR is the required triangle.

 

https://lh4.googleusercontent.com/rx-NctlZKIdKBd8amj1c99IKi88rSaelMEG1uj5l4C9-ddXPb7KWXkItAt_x94XuqbQTVvPR0Z40IM7wIqJT0E5cDYiyCP60FGM1mzMrqngtw6tjDDqAYd3xie9dHeY2Dpz2BRlv

 

Q13: Construct a ΔABC in which BC = 4.5 cm , B=60 and the sum of the other two sides is 8 cm.

Sol:

Steps of Construction:

(i) Draw BC = 4.5 cm

(ii) Construct CBX=60

(iii) Along BX set off BP = 8 cm

(iv) Join CP.

(v) Draw the perpendicular bisector of CP to intersecting BP at A.

(vi) Join AC.

Therefore, ABC is the required triangle.

 

https://lh5.googleusercontent.com/aqcVx3xCM9n5azkfWMCGAMvpD6INB3KRA2YK1I2ckFaRye7524r9AaY8NhVdiHe8_2xMz1i6a930rbOFhH4NZWZTKTUAIp41ySOMwNFi4W9gQC_yDJzakUFsVxpIYmnpQVM5aI2K

 

Q14: Construct a ΔABC in which BC = 5.2 cm, B=30 and the difference of the other two sides is 3.5 cm.

Sol:

Steps of Construction:

(i) Draw BC= 5.2 cm

(ii) Construct CBX=30

(iii) Along BX set off BP = 3.5 cm

(iv) Join PC.

(v) Draw the perpendicular bisector of PC meeting BP produced at A.

(vi) Join AC.

Therefore, ABC is the required triangle.

 

https://lh4.googleusercontent.com/8Fr4_jZ3NhA9DVj_vMgvrPKSn15j6Edlj7--vCq7tJoFDQHu0gFkOw4kfzVfPPk-uOBUkRlQ-OaGQhxVZWG5JvplYdSw3lvOrT09ffWK9zfE4YVp4TK_2-YdxYVaTdNuh1XYbmqJ

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