SPECIAL CASES QUADRILTERALS

Practical geometry of Class 8

CONSTRUCTING A QUADRILATERAL WHEN FOUR SIDES AND A DIAGONAL OF A QUADRILATERAL ARE GIVEN:

We consider the quadrilateral ABCD as a figure made of two triangles:

  • ΔABC and ΔADC when diagonal AC as the common side is given.
  • ΔABD and ΔBCD when diagonal BD as the common side is given.

In order to draw the quadrilateral when four sides and one diagonal are given, we first draw a rough sketch of the quadrilateral and write its dimensions along the sides and then we divide it into two triangles which can be drawn conveniently.

STEPS:

Step 1: Construct a triangle ABD (say).

Step 2: Find point C opposite to the vertex A as follows. With B as the centre and given radius, draw an arc on the other side of BD. Similarly with D as the centre and given radius, draw another arc intersecting the previous arc. The point of intersection of these arcs is marked as C.

Step 3: join points B and C, and D and C.

SPECIAL CASES QUADRILTERALS

question 1. Construct a quadrilateral ABCD, in which AB = 5.5 cm, AD = 4.4 cm, CD = 6.5 cm, and AC = 6.5 cm and BD = 7.1 cm.

Solution: First we draw a rough sketch of quadrilateral ABCD. It is evident from the rough sketch that we have sufficient data to draw triangles ADC and ABD.

  • Draw AC = 6.5 cm.
  •  With A as centre and radius AD = 4.4 cm, draw an arc.
  • With C as centre and radius CD = 6.5 cm, draw an arc to intersect the arc drawn in (ii) at D.
  • With A as centre and radius AB = 5.5 cm, draw an arc on the side of AC opposite to that of D.
  • With D as centre and radius BD = 7.1 cm, draw another arc to intersect the arc drawn in (iv) at B.

SPECIAL CASES QUADRILTERALS

  • Join AD, CD, AB and CB to obtain the required quadrilateral.

question 2. Construct a quadrilateral ABCD, when AB = 4 cm, BC = 6.0 cm, CD = DA = 5.2 cm and AC = 8 cm.

Sol. (i) Construct AC = 8 cm as shown in figure.

  • With A as centre and radius AB = 4 cm, draw an arc.
  • With C as centre and radius CB = 6.0 cm, draw another arc to cut the arc of step (ii) at B.
  • With A as centre and radius AD = 5.2 cm, draw an arc on the opposite side of AC.
  • With C as centre and radius CD = 5.2 cm, draw another arc to cut the arc of step (iv) at D.
  • Join AB, BC, CD and DA.

SPECIAL CASES QUADRILTERALS

Thus, ABCD is the required quadrilateral.

question 3. Construct the quadrilateral ABCD in which AB = 4.5 cm, BC = 5.5 cm, CD = 4, cm, AD = 6 cm and AC = 7 cm.

Solution: Steps of construction

  • Take a line segment AC = 7 cm.
  • Draw an arc of radius AB = 4.5 cm taking A as centre.
  • Draw another arc of radius 5.5 cm taking C as centre.
  • Let these two arc intersect at B.
  • Triangle ABC is obtained.

SPECIAL CASES QUADRILTERALS

  •  Repeat the construction to obtain ΔADC taking AD = 6 cm and DC = 4 cm.
  •  ABCD is the required quadrilateral 

SPECIAL CASES QUADRILTERALS


question 4. Construct the quadrilateral JUMP in which JU = 3.5 cm, UM = 4 cm, MP = 5 cm, PJ = 4.5 cm and PU = 4 cm.

Solution: Steps of construction

  • Draw a line segment PU = 4 cm.
  • Taking P as centre cut off an arc of radius 4.5 cm.
  • Taking U as centre cut off another arc of radius 3.5 cm.
  • Let these two arcs intersect at J.
  • Join PJ and UJ.
  • Similarly, obtain the point M on the opposite side of diagonal PU.
  • Hence, JUMP is the required quadrilateral.

SPECIAL CASES QUADRILTERALS

question 5. Construct a parallelogram ABCD in which, AB = 6 cm, BC = 4.5 cm and diagonal AC = 6.8 cm.

Solution: Steps of construction

  • Draw AB = 6 cm.
  • With A as centre and radius 6.8 cm draw an arc.
  • With B as centre and radius 4.5 cm draw another arc, so as it cuts the previous arc of step (ii) at C.

SPECIAL CASES QUADRILTERALS

  •  Join BC and AC.
  • With A as centre and radius 4.5 draw an arc.
  • With C as centre and radius 6 cm draw another arc so as it cuts the previously drawn arc of step (v) at D. 
  • Join DA and DC.

Then, ABCD is the required parallelogram.

CONSTRUCTING A QUADRILATERAL WHEN LENGTHS OF ITS THREE SIDES AND TWO INCLUDED ANGLES:

STEPS:
Step 1: Draw a line segment MN (say) of given length.

Step 2: Construct a given angle at M.

Step 3: Construct an angle 90° at N.

Step 4: Locate vertices L and O.

Step 5: Join L and O.

Constructing Irregular quadrilaterals


question 1. Construct a quadrilateral ADBC, in which AB = 4.4 cm, BC = 4 cm,  CD = 6.4 cm, DA = 2.8 cm, and BD = 6.6 cm. Measure the length of AC.

Solution: Steps of construction

  • Draw a line segment AB = 4.4 cm.
  • With centre A and radius 2.8 cm, draw an arc. Then, with centre B and radius 6.6 cm, draw another arc which cuts the earlier arc of this step at D.
  •  Join AD and BD.

SPECIAL CASES QUADRILTERALS

  • With centre B and radius 4 cm, draw an arc. Then, with centre D and radius 6.4 cm, draw another arc which cuts the earlier arc of this step at the point C.
  • Join BC and DC.

Thus, ABCD is the required quadrilateral.

Join AC. Length of AC = 6 cm. 

question 2. Construct a quadrilateral ABCD in which BC = 4 cm, CA = 5.6 cm, AD = 4.5 cm, CD = 5 cm, and BD = 6.5 cm.

Solution:Steps of construction

  • Draw a line segment CD = 5 cm.
  • With centre D and radius 4.5 cm, drawn an arc. Then, with centre C and radius 5.6 cm, draw another arc. Which cuts the earlier arc of this step at A.

SPECIAL CASES QUADRILTERALS

  • Join DA and CA.
  • With centre D and radius 6.5 cm, draw an arc. Then with centre C, and radius 4 cm, draw another arc which cuts the earlier arc of this step at point B.
  • Join CB and B.

Thus, ABCD is the required quadrilateral.

question 3. Construct the quadrilateral LIFT in which LI = 4 cm, IF = 3 cm, TL = 2.5 cm, LF = 4.5 cm and IT = 4 cm.

Solution:Steps of construction

  • Take a line segment LF = 4.5 cm
  • With centre as L and F cut off arcs of radii 4 cm and 3 cm respectively.
  • Triangle LFI is obtained.
  • Now, with L and I as centre cut off arcs of length TL = 2.5 cm and TI = 4 cm.
  • Join TI, TL and TF.
  • LIFT is the required quadrilateral.

SPECIAL CASES QUADRILTERALS

question 4. Construct the quadrilateral GOLD in which OL = 7.5 cm, GL = 6 cm, GD = 6 cm, LD = 5 cm and OD = 10 cm.

Solution: Here, GL = 6 cm

  GD = 6 cm

and   LD = 5 cm.

Steps of construction

  • Take a line segment GL = 6 cm.
  • Cut off GD = 6 cm and LD = 5 cm.
  • Join GD and LD.
  • Triangle GLD is obtained.

SPECIAL CASES QUADRILTERALS

  • Now, with L and D as centre cut off two arcs of length 7.5 cm and 10 cm respectively to obtain a point O.
  • Join OL, OD and OG.
  •  Required quadrilateral GOLD is obtained.

SPECIAL CASES QUADRILTERALS

CONSTRUCTING A QUADRILATERAL WHEN LENGTHS OF ITS ADJACENT SIDES AND THREE ANGLES ARE GIVEN:

Step 1: Draw a line segment of EF (say) of given length.

Step 2: Construct a given angle at E.

Step 3: Construct a given angle at F.

Step 4: Locate point G.

Step 5: Locate point H.

Constructing Irregular quadrilaterals

question 1. Construct the quadrilateral MORE in which MO = 6 cm, OR = 4.5 cm, ∠M = 60º, ∠O = 105º, ∠R = 105º.

Solution:

Steps of construction

  • Take a line segment MO = 6 cm.
  • Construct ∠MOR = 105º and cut off OR = 4.5 cm.
  • At R construct ∠ORX = 105º.
  • Now, construct ∠OMY = 60º.
  • Let MY intersect RX at E.
  • Thus, MORE is the required quadrilateral.

question 2. Construct a rectangle OKAY in which OK = 7 cm and KA = 5 cm.

Solution:

SPECIAL CASES QUADRILTERALS

Steps of construction

  • Draw a line segment OK = 7 cm.
  • Construct ∠OKX = 90º and cut off KA = 5 cm.
  • Also, construct ∠KOZ = 90º and cut off OY = 5 cm.
  • Join OY and AY.
  • Thus, OKAY is the required rectangle.

SPECIAL CASES QUADRILTERALS

question 3. Construct a quadrilateral ABCD in which, AB = 4.3 cm, BC = 5 cm, ∠A = 60°, ∠B = 100° and ∠C = 125°.

Solution: Steps of construction

  • Draw AB = 4.3 cm.
  • Make ∠ABX = 100°.
  • With B as centre and radius 5 cm, draw an arc, so as it cuts BX at C.
  • Make ∠BCY = 125° and ∠BAZ = 60°.
  • Extend, so as CY and AZ cuts at D.

SPECIAL CASES QUADRILTERALS

Then, ABCD is the required quadrilateral.

question 4. (i) Can a quadrilateral PLAN be constructed if PL = 6 cm, LA = 9.5 cm,      ∠P = 75º, ∠L = 150º and ∠A = 140º.

(ii) In a parallelogram, the lengths of adjacent sides are known. Do we still need measures of the angles to construct the parallelogram?

Solution: (i) ∠P + ∠L + ∠A = 365º

75º + 150º + 140º = 365º

Sum of four angles in a quadrilateral is 360º.

∴ Quadrilateral is not possible to construct.

(ii) No, it is not required to have the measure of angles because opposite sides of a parallelogram are equal.

If AB and BC are given 

then, AB = DC, BC = AD.         

CONSTRUCTING A QUADRILATERAL WHEN LENGTHS OF ITS THREE SIDES AND TWO DIAGONALS ARE GIVEN:

STEPS:

Step1: Draw a rough sketch of the quadrilateral PQRS.

Step 2: Draw a line segment of given length.

Step 3: Construct the two given angles at two ends of the line.

Step 4: With centre O and radius of the given length, draw an arc.

Step 5: With centre R and radius of thje second side, draw another arc.

Step 6:  Join the two arcs

Constructing Irregular Quadrilaterals

question 1. Construct a quadrilateral PQRS in which PQ = 3.5 cm, QR = 2.5 cm, RS = 4.1 cm, ∠Q = 75° and ∠R = 120°. Measure side PS.

Solution:Steps of construction

  •  Draw a rough sketch of the quadrilateral PQRS.
  •  
  •  
  •  Draw a line segment QR = 2.5 cm.
  • Draw ∠XQR = 75° and ∠YRQ = 120°.
  • With centre Q and radius 3.5 cm, draw an arc to cut QX at P.
  • With centre R and radius 4.1 cm, draw an arc to cut RY at S.

  • Join PS.

Thus, PQRS is the required quadrilateral.

Measuring PS we find that PS = 3.7 cm.

question 2. Construct a quadrilateral ABCD in which, AB = 4.2 cm, BC = 5 cm, CD = 5.3 cm, ∠B = 120° and ∠C = 75°.

Solution: Steps of construction

  • Draw BC = 5 cm.
  • Make ∠CBX = 120°.
  • With B as centre and radius 4.2 cm draw an arc BA = 4.2 cm, so as it cuts BX at A.
  • Make ∠BCY = 75°.
  • With C as centre and radius 5.3 cm draw an arc so as it cuts CY at D.

SPECIAL CASES QUADRILTERALS

  •  Join DA.

Thus, ABCD is the required quadrilateral.

question 3.  Construct a quadrilateral ABCD given AB = 5.1 cm, AD = 4 cm, BC = 2.5 cm, ∠A = 60° and ∠B = 85°.

Sol. Steps of construction

  • Draw AB = 5.1 cm.
  • Construct ∠XAB = 60° at A.
  • With A as centre and radius AD = 4 cm, cut off AD = 4 cm along AX.
  • Construct ∠ABY = 85° at B.
  • With B as centre and radius BC = 2.5 cm, cut off BC = 2.5 cm along BY.

SPECIAL CASES QUADRILTERALS

  • Join CD.

The quadrilateral ABCD so obtained is the required quadrilateral.

question 4. (i) Can a quadrilateral ABCD be constructed with AB = 5 cm, BC = 5.5 cm, CD = 4 cm, AD = 6 cm and ∠B = 80º?

(ii) Can a quadrilateral PQRS be constructed with PQ = 4.5 cm, ∠P = 70º,   ∠Q = 100º, ∠R = 80º and ∠S = 110º?

Solution:(i)

SPECIAL CASES QUADRILTERALS

We can construct a quadrilateral ABCD with four sides and one angle.

(ii)

Here, at least one side RQ or PS is required. So, data is in-sufficient. Construction is possible only if 

(a) two angles and two sides, e.g. if ∠A = 90º, ∠B = 90º, AD = 6 cm, BC = 5 cm. are given.

(b) four angles and two sides e.g. PQ = 4.5 cm, ∠P = 65º, ∠Q = 110º,              ∠QR = 4.7 cm ∠R = 115º are given.

SPECIAL CASES QUADRILTERALS

A quadrilateral can be constructed using properties of specified quadrilateral.

PARALLELOGRAM

  • Opposite sides are equal.
  • Opposite angles are equal.
  • Sum of adjacent angles is 180º.
  • Diagonals bisect each other.

SQUARE

  • All sides are equal.
  • Each angle 90º.
  • Diagonals are equal.

RECTANGLE

SPECIAL CASES QUADRILTERALS

  • Opposite sides are equal.
  • Diagonals are equal.
  • Each angle 90º.

RHOMBUS

  • Diagonals bisect each other at 90º.
  • All sides are equal.
  • Diagonals are unequal.

STEPS:

Step 1: Draw a side of given leng

Step 3: Draw side CE (say) of length th (say) CL

Step 2: Draw side LU (say) of given length perpendicular to CL at L. equal to LU and perpendicular to CL at C.

Step 4: Draw side UE.

Steps to Construct a Rectangle, constructing a rectangle

STEPS:

Step 1: Draw diagonal (say) AY and its perpendicular bisector.

Step 2:  Draw sides say AL and AZ of given length.

Step 3: Draw sides LY and YZ.

Steps to Construct a Kite, constructing a Kite

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