5 Mark Question

Question 15 Marks

Draw a triangle ABC with AB = 3cm, BC = 4cm and $\angle\text{B}=60^\circ.$ Also, draw the bisector of angles C and A of the triangle, meeting in a point O. Measure $\angle\text{COA}.$

Answer

Steps of construction:
Step I: Draw a line segment BC = 4cm.
Step II: Draw $\angle\text{CBX}=60^\circ.$
Step III: Draw an arc on BX at a radius of 3cm cutting BX at A.
Step IV: Join AC to get the required triangle.
Angle bisector for angle A:
Step I: With A as centre, cut arcs of the same radius cutting AB and AC at P and Q, respectively.
Step II: From P and Q cut arcs of same radius intersecting at R.
Step II: Join AR to get the angle bisector of angle A.
Angle bisector for angle C:
Step I: With A as centre, cut arcs of the same radius cutting CB and CA at M and N, respectively.
Step II: From M and N, cut arcs of the same radius intersecting at T
Step III: Join CT to get the angle bisector of angle C.
Step IV: Mark the point of intersection of CT and AR as 0.
Step V: Angle $\angle\text{COA}=120^\circ.$

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Question 25 Marks

Draw a $\triangle\text{ABC}$ with AB = 6cm, BC = 7cm and CA = 8cm. Using ruler and compass alone, draw (i) the bisector AD of $\angle\text{A}$ and (ii) perpendicular AL from A on BC. Measure LAD.

Answer

Steps of construction:
Step I: Draw a line segment BC of length 7cm.
Step II: With centre B, draw an arc of radius 6cm.
Step III: With centre C, draw an arc of radius 8cm intersecting the previously drawn arc at A.
Step IV: Join AC and BC to get the required triangle.
Angle bisector steps:
Step I: From A, cut arcs of equal radius intersecting AB and AC at E and F respectively.
Step II: From E and F, cut arcs of equal radius intersecting at point H.
Step III: Join AH and extend to produce the angle bisector of angle A, meeting line BC at D.

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Question 35 Marks

Take any three non-collinear points A, B, C and draw $\angle\text{ABC}.$ Through each vertex of the triangle, draw a line parallel to the opposite side.

Answer

Steps of construction:
Step I: Mark three non collinear points A, B and C such that none of them lie on the same line.
Step II: Join AB, BC and CA to form triangle ABC.
Step III: Parallel line to AC
Step IV: With A as centre, draw an arc cutting AC and AB at T and U, respectively.
Step V: With centre B and the same radius as in the previous step, draw an arc on the opposite side of AB to cut AB at X.
Step VI: With centre X and radius equal to TU, draw an arc cutting the arc drawn in the previous step at Y.
Step VII: Join BY and produce in both directions to obtain the line parallel to AC.
Parallel line to AB:
Step I: With B as centre, draw an arc cutting BC and BA at W and V, respectively.
Step II: With centre C and the same radius as in the previous step, draw an arc on the opposite side of BC to cut BC at P.
Step III: With centre P and radius equal to WV, draw an arc cutting the arc drawn in the previous step at Q.
Step IV: Join CQ and produce in both directions to obtain the line parallel to AB.

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Question 45 Marks

Draw any $\triangle\text{ABC}.$ Bisect side AB at D. Through D, draw a line parallel to BC, meeting AC in E. Measure AE and EC.

Answer

Steps of construction:
Step I: We first draw a triangle ABC with each side = 6cm.
Steps to bisect line AB:
Step I: Draw an arc from A on either side of line AB.
Step II: With the same radius as in the previous step, draw an arc from B on either side of AB intersecting the arcs drawn in the previous step at P and Q.
Step III: Join PQ cutting AB at D. PQ is the perpendicular bisector of AB.
Parallel line to BC:
Step I: With B as centre, draw an arc cutting BC and BA at M and N, respectively.
Step II: With centre D and the same radius as in the previous step, draw an arc on the opposite side of AB to cut AB at Y.
Step II: With centre Y and radius equal to MN, draw an arc cutting the arc drawn in the previous step at X.
Step III: Join XD and extend it to intersect AC at E.
Step IV: DE is the required parallel line.

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Question 55 Marks

Draw a line PQ. Draw another line parallel to PQ at a distance of 3cm from it.

Answer

Steps of construction:
Step I: Draw a line PQ.
Step II: Take any two points A and B on the line.
Step III: Construct $\angle\text{PBF}=90^\circ$ and $\angle\text{QAE}=90^\circ$
Step IV: With A as centre and radius 3cm cut AE at C.
Step V: With B as centre and radius 3cm cut BF at D.
Step VI: Join CD and produce it on either side to get the required line parallel to AB and at a distance of 5cm from it.

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Question 65 Marks

Draw an $\angle\text{BAC}$ of measure 50° such that AB = 5cm and AC = 7cm. Through C draw a line parallel to AB and through B draw a line parallel to AC, intersecting each other at D. Measure BD and CD.

Answer

Steps of construction:
Step I: Draw angle BAC = 50° such that AB = 5cm and AC = 7cm.
Step II: Cut an arc through C at an angle of 50°
Step III: Draw a straight line passing through C and the arc. This line will be parallel to AB since $\angle\text{CAB}=\angle\text{RCA}=50^\circ$
Step IV: Alternate angles are equal; therefore the line is parallel to AB.
Step V: Again through B, cut an arc at an angle of 50° and draw a line passing through B and this arc and say this intersects the line drawn parallel to AB at D.
Step VI: $\angle\text{SBA}=\angle\text{BAC}=50^\circ,$ since they are alternate angles. Therefore BD parallel to AC.
Step VII: Also we can measure BD = 7cm and CD = 5cm.

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Question 75 Marks

Draw a triangle ABC in which BC = 4cm, AB = 3cm and $\angle\text{B}=45^\circ.$ Also, draw a perpendicular from A on BC.

Answer

Steps of construction:
Step I: Draw a line segment AB of length 3cm.
Step II: Draw an angle of 45° and cut an arc at this angle at a radius of 4cm at C.
Step III: Join AC to get the required triangle.
Step IV: With A as centre, draw intersecting arcs at M and N.
Step V: With centre M and radius more than half of MN, cut an arc on the opposite side of A.
Step VI: With N as centre and radius the same as in the previous step, cut an arc intersecting the previous arc at E.
Step VII: Join AE, it meets BC at D, then AE is the required perpendicular.

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Question 85 Marks

Draw $\triangle\text{PQR}$ in which PQ = 3cm, QR. 4cm and RP = 5cm. Also, draw the bisector of $\angle\text{Q}.$

Answer

Steps of construction:
Step I: Draw a line segment PQ of length 3cm.
Step II: With Q as centre and radius 4cm, draw an arc.
Step III: With P as centre and radius 5cm, draw an arc intersecting the previously drawn arc at R.
Step IV: Join PR and OR to obtain the required triangle.
Step V: From Q, cut arcs of equal radius intersecting PQ and QR at M and N, respectively.
Step VI: From M and N, cut arcs of equal radius intersecting at point S.
Step VII: Join QS and extend to produce the angle bisector of angle PQR.
Step VIII: Verify that $\angle\text{PQS}$ and $\angle\text{SQR}$ are equal to 45° each.

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Question 95 Marks

Draw a triangle whose sides are of lengths 4cm, 5cm and 7cm. Draw the perpendicular bisector of the largest side.

Answer

Steps of construction:
Step I: Draw a line segment PR of length 7cm.
Step II: With centre P, draw an arc of radius 5cm.
Step III: With centre R, draw an arc of radius 4cm intersecting the previously drawn arc at Q.
Step IV: Join PQ and QR to obtain the required triangle.
Step V: From P, draw arcs with radius more than half of PR on either side.
Step VI: With the same radius as in the previous step, draw arcs from R on either sides of PR intersecting the arcs drawn in the previous step at M and N.
Step VII: MN is the required perpendicular bisector of the largest side.

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Question 105 Marks

Draw $\triangle\text{DEF}$ such that DE= DF= 4cm and EF = 6cm. Measure $\angle\text{E}$ and $\angle\text{F}.$

Answer

Steps of construction:
Step I: Draw a line segment EF of length 6cm.
Step II: With E as centre, draw an arc of radius 4cm.
Step III: With F as centre, draw an arc of radius 4cm intersecting the previous arc at D.
Step IV: Join DE and DF to get the desired triangle DEF.
Step V: By measuring we get, $\angle\text{E}=\angle\text{F}=40^\circ.$

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Question 115 Marks

Draw two parallel lines at a distance of 5cm apart.

Answer

Steps of construction:
Step I: Draw a line PQ.
Step II: Take any two points A and B on the line.
Step III: Construct $\angle\text{PBF}=90^\circ$ and $\angle\text{QAE}=90^\circ$
Step IV: With A as centre and radius 5cm cut AE at C.
Step V: With B as centre and radius 5cm cut BF at D.
Step VI: Join CD and produce it on either side to get the required line parallel to AB and at a distance of 5cm from it.

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Question 125 Marks

Draw $\triangle\text{ABC}$ in which AB = 5.5cm. BC = 6cm and CA = 7cm. Also, draw perpendicular bisector of side BC.

Answer

Steps of construction:
Step I: Draw a line segment AB of length 5.5cm.
Step II: From B, cut an arc of radius 6cm.
Step III: With centre A, draw an arc of radius 7cm intersecting the previously drawn arc at C.
Step IV: Join AC and BC to obtain the desired triangle.
Step V: With centre B and radius more than half of BC, draw two arcs on both sides of BC.
Step VI: With centre C and the same radius as in the previous step, draw two arcs intersecting the arcs drawn in the previous step at X and Y.
Step VII: Join XY to get the perpendicular bisector of BC.

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Maths 5 Mark Question

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