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**Question:** Arshad has five measurements of a quadrilateral ABCD. These are AB = 5 cm, ∠A = 50°, AC = 4 cm, BD = 5 cm and AD = 6 cm. Can he construct a unique quadrilateral? Give reasons for your answer.**Solution:** No, the quadrilateral ABCD cannot be constructed with the given combination of measurements. Here, knowledge of BC or DC is must.**Question:** (i) We saw that 5 measurements of a quadrilateral can determine a quadrilateral uniquely. Do you think any five measurements of the quadrilateral can do this?

(ii) Can you draw a parallelogram BATS where BA = 5 cm, AT = 6 cm and AS = 6.5cm?

Why?

(iii) Can you draw a rhombus ZEAL where ZE = 3.5 cm, diagonal EL = 5 cm? Why?

(iv) A student attempted to draw a quadrilateral PLAY where PL = 3 cm, LA = 4 cm, AY = 4.5 cm, PY = 2 cm and LY = 6 cm, but could not draw it. What is the reason?**Hint: **Discuss it using a rough sketch.**Solution:**

(i) No, any 5 measurements (elements) cannot determine a quadrilateral. Actually to construct a quadrilateral, we need a specific combination of measurements such as:

(a) Four sides and one diagonal

or

(b) Three sides and two diagonals

or

(c) Two adjacent sides and three angles

or

(d) Three sides and two included angles.

or

(e) Some special properties are given.

(ii) Let us draw a rough sketch of BATS as given.

Here, we cannot locate the points T and B without knowing the measurements ST and SB respectively.

Thus, we cannot draw a parallelogram with the given measurements.

(iii) Let us draw a rough sketch of the (rhombus) quadrilateral ZEAL and mark the given measurements on it.

Yes, we can draw the required rhombus, because it's all sides are equal to 3.5 cm and a diagonal EL is known.

(iv) Let us draw the rough sketch of the quadrilateral PLAY and mark its measurements, No, this quadrilateral cannot be drawn

∵ Point P cannot be located. In D LPY, sum of the lengths of PL and PY is less than LY, i.e. (2 cm + 3 cm) < 6 cm.**EXERCISE 4.1****Question 1. **Construct the following quadrilaterals.**(i) Quadrilateral ABCD**

AB = 45 cm

BC = 5.5 cm

CD = 4 cm

AD = 6 cm

AC = 7 cm**(ii) Quadrilateral JUMP**

JU = 3.5 cm

UM = 4 cm

MP = 5 cm

MP = 5 cm

PJ = 4.5 cm

PU = 6.5 cm**(iii) Parallelogram MORE**

OR = 6 cm

RE = 4.5 cm

EO = 7.5 cm**(iv) Rhombus BEST**

BE = 4.5 cm

ET = 6 cm

**Solution:****Note:** Before making a fair construction for each of the above quadrilaterals, a rough sketch must be drawn and the given measurements be marked.**(i) Steps of construction:**

I. Draw a line segments AB = 4.5 cm.

II. With centre at A and radius = 7 cm, draw an arc.

III. With centre at B and radius = 5.5 cm, draw another arc to intersect the previous arc at C.

IV. With centre at A and radius = 6 cm, draw an arc on the side opposite to that of B.

V. With centre at C and radius = 4 cm, draw another arc to intersect the previous arc at D.

VI. Join AC, BC, DA and DC.

Thus, ABCD is the required quadrilateral.**(ii) Steps of construction: **

I. Draw a line segments JU = 3.5 cm.

II. With centre at U and radius UP = 6.5 cm, draw an arc.

III. With centre at J and radius = 4.5 cm, draw another arc to intersect the previous arc at P.

IV. With centre at U and radius = 4 cm, draw an arc on the side opposite to that of J.

V. With centre at P and radius = 4 cm, draw another arc intersecting the previous arc at M.

VI. Join PJ, PU, PM and UM.

Thus, JUMP is the required quadrilateral. **(iii) Steps of construction: **

I. Draw a line segments ER = 4.5 cm.

II. With centre at E and radius = 7.5 cm, draw an arc.

III. With centre at R and radius = 6 cm, draw another arc to intersect the previous are at O.

IV. With centre at R and radius = 6 cm, draw an are on the side opposite to R.

V. With centre at O and radius = 4.5 cm, draw another arc to intersect the previous arc at M.

VI. Join RO, EM, EO and MO.

Thus, EROM or MORE is the required parallelogram.**(iv) Steps of construction:**

I. Draw a line segment BE = 4.5 cm.

II. With centre at E and radius = 6 cm, draw an arc.

III. With centre at B and radius = 4.5 cm, draw another arc to intersect the previous arc at T.**Note:** All the 4 sides of a rhombus are equal.

IV. Join TB and TE.

V. With centre at T and radius = 4.5 cm, draw an arc on the side opposite to that of B.

VI. With centre at E and radius = 4.5 cm, draw another arc to intersect the previous arc at S.

VII. Join ST and SE Thus, BEST is the required quadrilateral.**CONSTRUCTION OF A QUADRILATERAL WHEN ITS TWO DIAGONALS AND THREE SIDES ARE KNOWN**

In this case, we take one of the diagonal as the base of a triangle, then taking any two the given sides, we locate the vertex (third point of the quadrilateral) of the triangle. Again, using the second diagonal and third given side, we locate the fourth point of the quadrilateral.

**Question 1.** “Construct a quadrilateral ABCD, given that BC = 4.5 cm, AD = 5.5 cm, CD = 5 cm, the diagonal AC = 5.5 cm and diagonal BD = 7 cm.” Can we draw the quadrilateral by drawing D ABD first and then find the fourth point C?**Solution:** Since, the measurement of AB is not given.

∴ We cannot locate the point B. Thus, the quadrilateral cannot be drawn.**Question 2.** Can you construct a quadrilateral PQRS with PQ = 3 cm, RS = 3 cm, PS = 7.5 cm, PR = 8 cm and SQ = 4 cm? Justify your answer**Solution:** Yes a quadrilateral PQRS can be drawn using PQ = 3 cm, RS = 3 cm, PS = 7.5 cm, PR = 8 cm and SQ = 4 cm.**EXERCISE 4.2** **Question 1. **Construct the following quadrilaterals.**(i) Quadrilateral LIFT **

LI = 4 cm

I F = 3 cm

TL = 2.5 cm

LF = 4.5 cm

IT = 4 cm

**(ii) Quadrilateral GOLD**

OL = 7.5 cm

GL = 6 cm

GD = 6 cm

LD = 5 cm

OD = 10 cm**(iii) Rhombus BEND**

BN = 5.6 cm

DE = 6.5 cm**Solution: ****(i) Steps of constructions:**

I. Draw a line segment LI = 4 cm.

II. With centre at I and radius = 4 cm, draw an arc.

III. With centre at L and radius = 2.5 cm, draw another arc to intersect the previous arc at T.

IV. Join TI and TL.

V. With centre I and radius = 3 cm, draw an arc on the side opposite to that of L.

VI. With centre at L and radius = 4.5 cm, draw another arc to intersect the previous arc at F.

VII. Join FL and FI.

Thus, LIFT is the required quadrilateral.**(ii) Steps of construction:**

I. Draw a line segment LD = 5 cm.

II. With centre at L and radius = 6 cm, draw an arc.

III. With centre at D and radius = 6 cm, draw another arc to intersect the previous arc at G.

IV. Join GL and GD.

V. With centre at D and radius = 10 cm, draw an arc on the same side as that of G.

VI. With centre at L and radius = 7.5 cm, draw another arc to intersect the previous arc at O.

VII. Join OL and OD.

Thus, GOLD is the required quadrilateral.**(iii) Steps of construction:**

I. Draw a line segment DE = 6.5 cm.

II. Draw perpendicular bisector PQ of DE such that M being the mid-point of DE.

III. With centre M and radius = 2.8 cm, draw arcs to intersect PQ at B and N.

IV. Join ND, NE, BE and BD.

Thus, BEND is the required rhombus.

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