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(1038 [C] )

9th (Maths)

Constructions

Construction of Angle and Angle Bisector

Bisecting an angle means drawing a ray in the interior of the angle, with its initial point at the vertex of the angle such that it divides the angle into two equal parts

In order to draw a ray AX bisecting a given angle $large angle BAC,$ we follow the following steps.
STEP I With centre A and any convenient radius draw an are cutting AB and AC at P and Q respectively.
STEP II With centre P and radius more than half of PQ draw an arc.
STEP III With centre Q and the same radius,as in step II, draw another arc intersecting the arc in step II at R.
STEP IV Join AR and produce it to any point X. The ray AX is the required bisector of $large angle BAC$ .

Verification: Measure $large angle BAX$ and $large angle CAX.$ You would find that $large angle BAX=angle CAX$

Justification: Now let us see how this method gives us the required angle bisector.

Join PR and QR.

In $Delta$ APR and $Delta$ AQR, we have

AP = AQ [ $large because$ AP and AQ are radii of the same arc]

PR = QR [ $large because$ PR and QR are arcs of equal radii.]

AR = AR [common] $large 72^{circ}$

So, $Delta$ APR $cong$ $Delta$ AQR [by SSS congruence criterion ]

$large Rightarrow$ $large angle PAR=angle QAR$ [C.P.C.T.]

Hence, AR is the bisector of $large angle BAC.$

ILLUSTRATION: Using a protractor, draw an angle of mesure $large 72^{circ}$ . With this angle as given, draw an angle of measure $large 36^{circ}$ .

SOLUTION We follow the following steps to draw an angle of $large 36^{circ}$ from an angle of .

Steps of Construction

STEP I Draw a ray OA as shown in fig. STEP II With the help of a protractor contruct an angle AOB of measure $large 72^{circ}$ .
STEP III WIth centre O and a convenient radius draw an arc cutting sides OA and OB at P and Q respectively.
STEP IV With centre P and radius more than half of PQ, draw an arc.
STEP V With centre Q and the same radius, as in the previous step, draw another arc intersecting the arc drawn in the previous step at R.
STEP VI Join OR and produce it to form ray OX.
The angle $large angle AOX$ so obtained the required angle of measure $large 36^{circ}$

Construction of 60 Degree Angle

CONSTRUCTION OF 60 DEGREE ANGLE:

In order to construct an angle of $large 60^{circ}$ with the help of ruler and compasses only, we follow the following steps:

Steps of Construction

STEP I Draw a ray OA. STEP II With centre O and any radius draw an arc PQ with the help of compasses, cutting the ray OA at P.
STEP III With centre P and the same radius draw an arc cutting the arc PQ at R.
STEP IV Join OR and produce it to obtain ray OB.
The angle $large angle AOB$ so obtained is the angle of measure $large 60^{circ}$ .

Justification : Now, let us see how this method gives us the required angle of $large 60^{circ}$ .

Join PR. Now In $large Delta OPR,$ we have

OP = OR = PR [PR is an arc equal to radius]

$large Rightarrow ;;Delta OPR$ is an equilateral triangle.

$large Rightarrow ;;angle OPR=60^{circ}$

$large Rightarrow ;;angle AOB=60^{circ}$

CONSTRUCTION OF AN ANGLE OF $large 120^{circ}$

In order to construct an angle of $large 120^{circ}$ by using ruler and compasses only, we follow the following steps:

Steps of Construction

STEP I Draw a ray OA. STEP II With O as centre and any convenient radius, draw an arc PQ cutting OA at P.
STEP III With P as centre and the same radius draw an arc, cutting the arc PQ at R. Then taking R as centre and the same radius draw an arc cutting the arc PQ at S.
STEP IV Join OS and produce it to any point B.
$large angle AOB$ so obtained is the angle of measure $large 120^{circ}$

Construction of 30 Degree Angle

Construction of 30 Degree Angle :

In order to contruct an angle of $large 30^{circ}$ with the help of ruler and compasses, we follow the following steps:

Steps of Construction

STEP I Draw $large angle AOB=60^{circ}$ by using the steps mentioned above. STEP II With centre O and any convenient radius daw an arc cutting OA and OB at P and R respectively.
STEP III With centre P and radius more than half of PQ, draw an arc in the interior of $large angle AOB.$
STEP IV WIth centre R and the same radius, as in step III, draw another arc intersecting the arc in step III at S.
STEP V Join OS and product it to any poiint C.
STEP VI The angle $large angle AOC$ is the angle of measure $large 30^{circ}$ .

Construction of 90 Degree Angle

Construction of 90 Degree Angle: In order to construct an angle of measure of $large 90^{circ}$ , we follow the following steps:

Steps of Construction

STEP I Draw a ray OA. STEP II With O as centre and any convenient radius, draw an arc, cutting OA at P. STEP III With P as centre and the same radius, draw an arc cutting the arc drawn in step II at R.
STEP IV With R as centre and the same radius as in steps II and III, draw an arc, cutting the arc drawn in step III at S.
STEP V With R as centre and the same radius, draw an arc.
STEP VI With S as centre and the same radius,draw an arc, cutting the arc drawn in step V at T.
STEP VII Draw OT and produce it to B.
$large angle AOB$ is the angle of measure $large 90^{circ}$ .

Construction of an angle of $large 45^{circ}$ : In order to construct an angle of $large 45^{circ}$ , we follow the following steps:

Steps of Construction

STEP I Draw $large angle AOB=90^{circ}$ by following the steps given above.
STEP II With P as centre and radius more than half PM , draw an arc.
STEP III With M as centre and radius same as step II draw an arc to intersect the arc at U.
Step IV: Join OU and extend it to point C
The angle $large angle AOC$ so obtained is the required angle of measure $large 45^{circ}$ .

Construction of a triangle given its base and base angle and sum of other two sides

In order to construct a triangle, when its base, sum of the other two sides and one of the base angles are given, we follow the following steps:

Steps of Construction: Obtain the base, base angle and the sum of other two sides. Let BC be the base, $dpi{100} large angle B$ be the base angle and x be the sum of the lengths of other two sides AB and AC of $large Delta ABC$ .

STEP I Draw the base BC. STEP II Draw $large angle B$ of measure given in the question. STEP III From ray through B making the specified angle, cut-off line segment AX equal to x (the sum of other two sides).
STEP IV Join XC.
STEP V: From point C taking radius more than half of CX Draw arc on both side of CX.
STEP VI : From point X taking same radius as in step V Draw arc on both side of CX to intersect the arcs drawn in step V at point P and Q.
STEP VII : Join PQ . This is the perpendicular bisector of CX meeting BX at A.
STEP VIII Join AC to obtain the required triangle ABC.

Justification: Let us now see how do we get the required triangle.

Since point A lies on the perpendicular bisector of CX.Therefore, AX = AC

Now, BA = BX - AX

$large Rightarrow$ BA = BX - AC [ $large because$ AX = AC ]

$large Rightarrow$ BX = BA + AC

PROOF

STEP I Obtain the base, base angle and the sum of other two sides. Let AB be the base, $large angle A$ be the base angle and l be the sum of the lengths of other two sides BC and CA of $large Delta ABC.$

STEP II Draw the base AB.

STEP III Draw $large angle BAX$ of measure to that of $large angle A$ .

STEP IV From ray AX, cut-off line segment AD equal to l ( the sum of other two sides).

STEP V Join BD.

STEP VI Construct an angle $large angle DBY$ equal to $large angle ADB$

STEP VII Suppose BY intersects AX at C.

Then, $large Delta ABC$ is the required triangle.

Justification: Let us now see how do we get the required triangle.

In $large Delta BCD$ , we have

$large angle CDB=angle DBC$

$large therefore$ BC = CD

Now, AC = AD - CD

$large Rightarrow$ AC = AD - BC

$large Rightarrow$ AD = AC + BC

Example : Construct a triangle ABC in which AB = 5.8 cm , BC + CA = 8.4 cm and $large angle B=60^{circ}$ .

SOLUTION In order to construct the $large Delta ABC$ we follow the following steps:

Steps of Construction

STEP I Draw AB = 5.8 cm

STEP II Draw $large angle ABX=60^{circ}$

STEP III From ray BX, cut off line segment BD = BC + CA = 8.4 cm.

STEP IV Join AD

STEP V Draw the perpendicular bisector of AD meeting BD at C.

STEP VI Join AC to obtain the required triangle ABC.

Justification: Clearly, C lies on the perpendicular bisector of AD.

$large therefore$ CA = CD

Now, BD = 8.4 cm

$large Rightarrow$ BD + CD = 8.4 cm

$large Rightarrow$ BD + CA = 8.4 cm

Hence, $large Delta ABC$ is the required triangle.

Construction of a triangle given a base and base angle and the difference of other two sides

Construction of a triangle given a base and base angle and the difference of other two sides : In order to contruct a triangle when its base, difference of the other two sides and one of the base angles are given, we follow the following steps:

Steps of Contruction: Obtain the base, base angle and the difference of two other sides. Let BC be the base, $dpi{100} large angle B$ be the base angle and x be the difference of the other two sides AB and AC of $large Delta ABC$ .. There can be two cases

Case I: AB > AC i.e., x =AB - AC

STEP I Draw the base BC of given length. STEP II Draw $large angle CBX$ of measure equal to that of $large angle B$ . STEP III AS AB > AC , then cut off segment BD = AB - AC from ray BX STEP IV . Join DC
STEP V: From point C taking radius more than half of CD Draw arc on both side of CD.
STEP VI : From point D taking same radius as in step V Draw arc on both side of CD to intersect the arcs drawn in step V at point P and Q.
STEP VII : Join PQ . This is the perpendicular bisector of CD meeting BX at A.
STEP VIII Join AC to obtain the required triangle ABC.

Case II: AC > AB i.e., x =AC - AB

STEP I Draw the base BC of given length. STEP II Draw $large angle CBX$ of measure equal to that of $large angle B$ . STEP III As AC > AB ,extend XB to D such that BD = AC - AB STEP IV . Join DC
STEP V: From point C taking radius more than half of CD Draw arc on both side of CD.
STEP VI : From point D taking same radius as in step V Draw arc on both side of CD to intersect the arcs drawn in step V at point P and Q.
STEP VII : Join PQ . This is the perpendicular bisector of CD meeting BX at A.
STEP VIII Join AC to obtain the required triangle ABC.

Justification: Let us now see how do we get the required triangle. Since A lies on the perpendicular bisector of DC.

$large therefore$ AD = AC

So, BD = AD - AB = AC - AB

Example : Construct a triangle ABC in which base AB = 5 cm, $large angle A=30^{circ}$ and AC - BC = 2.5 cm.

SOLUTION In order to construct the triangle ABC, we follow the following steps:

Steps of Construction:

STEP I Draw base AB = 5 cm

STEP II Draw $large angle BAX=30^{circ}$

STEP III From ray AX, cut off line segment AD = 2.5 cm(=AC-BC)

STEP IV Join BD.

STEP V Draw the perpendicular bisector of BD which cuts AX at C.

STEP VI Join BC to obtain the required triangle ABC.

Justification: Since C lies on the perpendicular bisector of DB. THerefore,

CD = CB

Now, AD = 2.5 cm

$large Rightarrow$ AC - CD = 2.5 cm

$large Rightarrow$ AC - BC = 2.5 cm

Hence, $large Delta ABC$ is the required triangle.

Construction of a triangle given its perimeter and its two base angles

Construction of Triangle Given Its Perimeter and Its Two Base Angles: In order to construct a triangle of given perimeter and two base angles, we follow the following steps:

Steps of Construction: Obtain the perimeter and the base angles of the triangle.Let ABC be a triangle of perimeter p cm and base BC.

STEP I Draw a line segment XY equal to the perimeter p of $large Delta ABC$ . STEP II Construct $dpi{100} large angle LXY=angle B$ and $large angle MYX=angle C.$
STEP III : Draw bisectors of angles $large angle LXY$ and $large angle ;MYX$ and mark their intersection point as A.
STEP IV: Draw the perpendicular bisectors PQ, RS of XA and YA meeting XY in B and C respectively.
STEP V Join AB and AC to obtain the required triangle ABC.

Justification: For the justification of the construction, we observe that B lies on the perpendicular bisector of AX.

$large therefore$ XB = AB $large Rightarrow ;;angle AXB=angle BAX$ [ Angles opposite to equal sides are equal]

Similarly, C lies on the perpendicular bisector of AY.

$large therefore$ YC = AC $large Rightarrow ;;angle AYC=angle YAC$

Now, XY = XB + BC + CY

$large Rightarrow$ XY = AB + BC + AC

In $large Delta AXB,$ we

$large angle ABC=angle AXB+angle BAX$ [Exterior angle is equal to the sum of interior opposite angle]

$large =2angle AXB$ [because the angles are equal Proved above]

$large =angle LXB$ [ AX is the bisector ]

= $large =angle B$ [By Step II of Construction]

In $large Delta AYC,$ we have

$large angle ACB=angle AYC+angle YAC$ [Exterior angle is equal to the sum of interior opposite angle]

$large =2angle AYC$ [because the angles are equal Proved above]

$large =angle MYC$ [ AY is the bisector ]

$large =angle C$ [By Step II of Construction]

Example: Construct a triangle PQR whose perimeter is equal to 14 cm, $large angle P=45^{circ}$ and $large angle Q=60^{circ}$ .

SOLUTION To draw $large Delta PQR,$ we follow the following steps:

Steps of Construction:

STEP I Draw a line segment XY = 14 cm STEP II Construct $large angle YXD=angle P= 45^{circ}$ and $large angle XYE=angle Q= 60^{circ}$
STEP III Draw the bisectors of angles $large angle YXD$ and $large angle XYE$ mark their point of intersection as R.
STEP IV Draw right bisector AB, CD of RX and RY meeting XY at P and Q respectively.
STEP V Join PR and QR to obtain the required triangle PQR.

Attention !!!!!

Construction of Angle and Angle Bisector

Construction of 60 Degree Angle

Construction of 30 Degree Angle

Construction of 90 Degree Angle

Construction of a triangle given its base and base angle and sum of other two sides

PROOF

Construction of a triangle given a base and base angle and the difference of other two sides

Construction of a triangle given its perimeter and its two base angles