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**Question 1.** Can you construct the following quadrilateral MIST if we have 100Â° at M instead of 75Â°?

**Solution:** Yes, the quadrilateral MIST can be constructed with âˆ M = 100Â° instead of âˆ 75Â°.**Question 2.** Can you construct the quadrilateral PLAN if PL = 6 cm, LA = 9.5 cm, âˆ P = 75Â°, âˆ L = 150Â° and âˆ A = 140Â°?

(**Hint:** Recall angle-sum property)

**Solution**: Here, âˆ P + âˆ L + âˆ A + âˆ N = 75Â° + 150Â° + 140Â° + âˆ N

= 365Â° + âˆ N

But the sum of all the angles of a quadrilateral is 360Â°.

âˆ´ Construction of quadrilateral PLAN is not possible.**Question 3.** In a parallelogram, the lengths of adjacent sides are known. Do we still need measures of the angles to construct as in the Q-1 above?**Solution: **No, the measures of three angles are not necessary in case of a parallelogram as its opposite sides are parallel.**EXERCISE 4.1(Question 1.** Construct the following quadrilaterals:

**(i) Quadrilateral MORE **

MO = 6 cm

OR = 4.5 cm

âˆ M = 60Â°

âˆ O = 105Â°

âˆ R = 105Â° **(ii) Quadrilateral PLAN**

PL = 4 cm

LA = 6.5 cm

âˆ P = 90Â°

âˆ A = 110Â°

âˆ N = 85Â°**(iii) Parallelogram HEAR **

HE = 5 cm

EA = 6 cm

âˆ R = 85Â°**(iv) Rectangle OKAY**

OK = 7 cm

KA = 5 cm**Solution: **(i) Steps of construction:

I . Draw a line segment MO = 6 cm.

II. At M, draw , such that âˆ OMX = 60Â°.

III. At O, draw , such that âˆ MOY = 105Â°.

IV. From , cut off OR = 4.5 cm.

V. At R, draw , such that âˆ ORZ = 105Â°.

Let intersects at E.

Thus, MORE is the required quadrilateral.

**(ii) Steps of construction: **

I. Draw a line segment AL = 6.5 cm.

II. At A, draw such that âˆ LAX = 110Â°.

III. At L, draw such that âˆ ALY = 75Â°.**Note:** âˆ L = 75Â° is not given, but we can determine it using angle sum property

âˆµ Sum of the three given angles = 110Â° + 90Â° + 85Â° = 285Â°

âˆ´ The fourth angle âˆ L = 360Â° â€“ 285Â° = 75Â°.

IV. From, cut-off LP = 4 cm.

V. At P, draw such that âˆ LPZ = 90Â°

Let PZ and intersect at N.

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

I. Draw a line segment = 5 cm.

II. At E, draw such that âˆ HEA = 85Â°.

III. From , cut-off EA = 6 cm.

IV. With centre at A and radius = 5 cm, draw an arc towards H.

V. With centre at H and radius = 6 cm, draw an arc such that it intersects the previous arc at R.

VI. Join RA and RH.

Thus, HEAR is the required quadrilateral.

**(iv) Steps of constructions: **

I. Draw a line segment OK = 7 cm.

II. At O, draw such that âˆ KOP = 90Â°.

III. From , cut-off = 5 cm.

IV. At K, draw such that âˆ OKQ = 90Â°.

V. From cut-off = 5 cm.

VI. Join A and Y.

Thus, OKAY is the required quadrilateral.

**CONSTRUCTION OF A QUADRILATERAL WHEN THREE SIDES AND TWO INCLUDED ANGLES ARE GIVEN**

**Question**. We used some five measurements to draw quadrilateral so far. Can there be different sets of five measurements (other than seen so far) to draw a quadrilateral? The following problems may help you in answering the question.

(i) Quadrilateral ABCD with AB = 5 cm, BC = 5.5 cm, CD = 4 cm, AD = 6 cm and âˆ B = 80Â°.

(ii) Quadrilateral PQRS with PQ = 4.5 cm, âˆ P = 70Â°,âˆ Q = 100Â°, âˆ R = 80Â° and âˆ S = 110Â°.

Construct a few more examples of your own to find sufficiency/insufficiency of the data for construction of a quadrilateral**Solution:** (i) It is possible to construct a quadrilateral ABCD with the given measurements.

(ii) Not possible, because we cannot locate the points R and S with the help of given measurements.**EXERCISE 4.4 ****Question 1 **Construct the following quadrilaterals.

(i) Quadrilateral DEAR

DE = 4 cm

EA = 5 cm

AR = 4.5 cm

âˆ E = 60Â°

âˆ A = 90Â°**(ii) Quadrilateral TRUE**

TR = 3.5 cm

RU = 3 cm

UE = 4 cm

âˆ R = 75Â°

âˆ U = 120Â°**Solution:** **(i) Steps of construction:**

I. Draw a line segment DE = 4 cm.

II. At E, draw such that âˆ DEX = 60Â°.

III. From , cut-off EA = 5 cm.

IV. At A, draw ray such that âˆ EAY = 90Â°.

V. From , cut-off AR = 4.5 cm.

VI. Join R and D.

Thus, DEAR is the required quadrilateral.

**(ii) Steps of construction: **

I. Draw a line segment = 3.5 cm.

II. At R, draw a ray such that âˆ TRX = 75Â°.

III. From , cut-off RU = 3 cm.

IV. At U, draw such that âˆ RUY = 120Â°.

V. From UY, cut-off= 4 cm.

VI. Join E and T.

Thus, TRUE is the required quadrilateral.

**SOME SPECIAL CASES**

Using the properties of a square, rectangle, kite and rhombus, etc., we can construct some special quadrilaterals even with less than 5 measurements.

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