3 UNDERSTANDING QUADRILATERALS · 2008. 12. 19. · 3 UNDERSTANDING QUADRILATERALS Exercise 3.1 Q.1. Given here are some figures. Classify each of them on the basis of the following.

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3 UNDERSTANDING QUADRILATERALS

Exercise 3.1

Q.1. Given here are some figures.

Classify each of them on the basis of the following. (a) Simple curve (b) Simple closed curve (c) Polygon (d) Convex polygon (e) Concave polygon Ans. (a) Simple curve : – (1), (2), (5), (6), (7) (b) Simple closed curve : – (1), (2), (5), (6), (7) (c) Polygon – (1), (2), (4) (d) Convex polygon – (2) (e) Concave polygon – (1), (4)

Q.2. How many diagonals does each of the following have? (a) A convex quadrilateral (b) A regular hexagon (c) A triangle

Ans. (a) A convex quadrilateral has 2 diagonals.



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(b) A regular hexagon has 9 diagonals. (c) A triangle has no diagonals.

Q.3. What is the sum of the measures of the angles of a convex quadrilateral? Will this property hold if the quadrilateral is not convex? (Make a non-convex quadrilateral and try!)

Ans. The sum of the measures of the angles of a convex quadrilateral is 360°.

Yes, this property holds if the quadrilateral is not convex.

e.g.

Here ABCD is a concave quadrilateral then AC is joined. Sum of Angles of ABCD = Sum of Angles of ΔABC and

ΔACD = 180° + 180° = 360°

Q.4. Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that.)

What can you say about the angle sum of convex

polygon with number of sides? (a) 7 (b) 8 (c) 10 (d) n

Ans. (a) We can say about the angle sum of a convex polygon with number of side 7 is = (n – 2) × 180 = (7 – 2) × 180° = 900° where n = number of sides



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(b) Angle sum of a convex polygon with number of sides 8 = (n – 2) × 180, where n = number of sides = (8 – 2) × 180 = 1080° (c) Angle sum of a convex polygon with number of sides

10 is = (n – 2) × 180°, where n = number of sides = (10 – 2) × 180 = 8 × 180° = 1440° (d) Angle sum of a convex polygon with number of sides

n is = (n – 2) × 180°, where n = number of sides.

Q.5. What is a regular polygon?

State the name of a regular polygon of (i) 3 sides (ii) 4 sides (iii) 6 sides

Ans. Regular Polygon : A regular polygon is a polygon which is both ‘equiangular’ and ‘equilateral’.

For Example : A square has sides of equal length and angles of equal measure. Hence, it is a regular polygon.

(i) The name of a regular polygon of 3 sides is equilateral triangle.

(ii) The name of a regular polygon of 4 sides is square. (iii) The name of a regular polygon of 6 sides is regular

hexagon.

Q.6. Find the angle measure x in the following figures.



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Ans. We know that the sum of angles of a quadrilateral is 360°. (a) Let ABCD is a quadrilateral in which ∠A = 130°, ∠B = 120°, ∠C = x, ∠D = 50° ⇒ ∠A + ∠B + ∠C + ∠D = 360° ⇒ 130° + 120° + x + 50° = 360° 300° + x = 360° x = 360° – 300° ⇒ x = 60° (b) Let ABCD be a quadrilateral ⇒ ∠DAX1 = 90° ⇒ ∠DAB = 90° ∴ ∠A + ∠B + ∠C + ∠D = 360° ⇒ 90° + 60° + 70° + x = 360° ⇒ 220° + x = 360° x = 360° – 220° ⇒ x = 140° (c) Let ABCDE be a pentagon ∴ ∠EAX1 + ∠EAX = 180° or 70° + ∠EAX = 180° ⇒ ∠EAX = 180° – 70° ∠EAX = 110 ------------ (i) similarly ∠CBX + CBX1 = 180° 60° + ∠CBX1 = 180° ∠CBX1 = 180° – 60°



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∠CBX1 = 120° ------------------- (ii) We know that the sum of angles of a pentagon is 540°. ⇒ ∠A + ∠B + ∠C + ∠D + ∠E = 540° ⇒ 110° + 120° + x + 30° + x = 540° ⇒ 260° + 2x = 540° ⇒ 2x = 280°

⇒ 22x =

o2802

⇒ x = 140°

(d) Let ABCDE be a pentagon. All sides of this pentagon are equal hence all angles of are also equal.

We know that the angles sum of a pentagon is 540°.

⇒ ∠A + ∠B + ∠C + ∠D + ∠E = 540° ⇒ x + x + x + x + x = 540° ⇒ 5x = 540°

⇒ x = o540

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∴ x = 108° Hence, each angle of the given pentagon is 108°.

Q.7.

(a) Find x + y + z (b) Find x + y + z + w

Ans. (a) Let ABC be a triangle.



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∴ ∠A + ∠B + ∠C = 180° (angle sum property) ⇒ 90 + 30 + ∠C = 180° ⇒ 120° + ∠C = 180° ∴ ∠C = 180° – 120° = 60° ∠CAB + ∠CAL = 180° (Linear pair) ⇒ x + 90 = 180° ⇒ x = 180° – 90° ∴ x = 90° --------------- (i) Similarly, ∠ACB + ∠QCB = 180° (Linear pair) ⇒ 60° + y = 180° ⇒ y° = 180° – 60° ∴ y = 120° --------------- (ii) Similarly, ∠ABN + ∠ABC = 180° (Linear pair) ⇒ z + 30° = 180° ⇒ z = 180° – 30° ∴ z = 150° -------------- (iii) So, x + y + z Adding eqn. (i), (ii) and (iii) we have x + y + z = 90° + 120° + 150° = 360° (b)



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Let ABCD be a quadrilateral. ∠A + ∠B + ∠C + ∠D = 360° ⇒ ∠A + 120° + 80° + 60° = 360° ⇒ ∠A + 260° = 360° ⇒ ∠A = 360° – 260° ∴ ∠A = 100° ∴ ∠BAN + ∠BAD = 180° (Linear pair) ⇒ 100° + w = 180° ⇒ w = 180° – 100° ∴ w = 80° --------------- (i) ∠LBP + ∠LBA = 180° (Linear pair) ⇒ x + 120° = 180° ∴ x = 60° ---------------- (ii) ∠LCD + ∠DCB = 180° ⇒ y + 80° = 180° (Linear pair) ∴ y = 100° ---------------- (iii) ∠MDA + ∠ADC = 180° ⇒ z + 60° = 180° ⇒ z = 180° – 60° ∴ z = 120° ----------------- (iv) Adding eqn. as (i), (ii), (iii) and (iv) we have x + y + z + w = 60° + 100° + 120° + 80° = 360°

Exercise 3.2



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Q.1. Find x in the following figures.

Ans. (a) Let ABC be a triangle. ∠A + ∠B + ∠C = 180° --------- (i)

(Angle sum property)

⇒ ∠ACB = 180° – x (Linear pair) ⇒ ∠C = 180° – x ---------- (ii) Similarly ∠B = 180° – 125° = 55° ----------- (iii) and, ∠A = 180 – 125° = 55° ------------- (iv) Putting the values of ∠A, ∠B and ∠C in eqn. (i) we have 55° + 55° + 180° – x = 180° ⇒ 110° + 180° – x = 180° ⇒ 290° – x = 180° ⇒ – x = – 110° ∴ x = 110° (b) Let ABCDE be a pentagon. ∴ ∠BAL + ∠BAE = 180° (Linear pair)



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⇒ 70° + ∠BAE = 180° ⇒ ∠BAE = 180 – 70° ∠BAE = 110° ⇒ A = 110° --------- (i) Similarly ∠EDY + ∠EDC = 180° (Linear pair) ⇒ 60° + ∠D = 180° ⇒ ∠D = 120° ------------ (ii) ∠C = 90° -------------- (iii) ∠MEA + ∠DEA = 180° ⇒ 90° + ∠E = 180° ⇒ ∠E = 90° -------------- (iv) ∠B = 180° – x ------------- (v) In the pentagon ABCDE ∠A + ∠B + ∠C + ∠D + ∠E = 540° -------------- (vi) Putting all values of A, B, C, D and E in eqn. (vi), we have 110° + 180° – x + 90° + 120° + 90° = 540° or 590° – x = 540° or – x = 540° – 590° ⇒ x = 50° Hence, the value of x is 50°.

Q.2. Find the measure of each exterior angle of a regular polygon of

(i) 9 sides (ii) 15 sides

Ans. ∴ We know that measure of exterior angle of a

regular polygon = o360

number of sides



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(i) Sides = 9

∴ Measure of each exterior angle = o360

9 = 40°

(ii) Sides = 15

∴ Measure of exterior angle = o360

15 = 24°

Q.3. How many sides does a regular polygon have if the measure of an exterior angle is 24°?

Ans. Given measure of an exterior angle = 24° We know that the sum of the measures of the external

angles of a polygon is 360°.

∴ Number of sides of regular polygon = o

o

36024

= 15

Hence, if the measure of an exterior angle is 24°, then number of sides of regular polygon is 15.

Q.4. How many sides does a regular polygon have if each of its interior angles is 165°?

Ans. Given interior angles of a polygon is 165°, so exterior angle will be 180° – 165° = 15°.

Hence, the number of sides of the polygon = 36015

°°

= 24.

Q.5. (a) Is it possible to have a regular polygon with measure of each exterior angle as 22°?

(b) Can it be an interior angle of a regular polygon? Why?

Ans. (a) No, it is not possible to have a regular polygon with measure of each exterior angle as 22°. Because, 22° is not a divisor of 360°.



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(b) No, it can not be an interior angle of a regular polygon. Because each exterior angle is 180° – 22° = 158°, which is not a divisor of 360°.

Q.6. (a) What is the minimum interior angle possible for a regular polygon? Why?

(b) What is the maximum exterior angle possible for a regular polygon?

Ans. (a) The minimum interior angle possible for regular polygon is 60°. Because an equilateral triangle being a regular polygon of 3 sides has the least measure of an interior angle = 60°.

(b) The maximum exterior angle possible for a regular polygon is 120°. As exterior angle and interior angles are supplementary.

Maximum exterior angle = 180° – least interior angle = 180° – 60° = 120°.

Exercise 3.3

Q.1. Given a parallelogram ABCD. Complete each statement along with the definition or property used.

(i) AD = …… (ii) ∠DCB = …… (iii) OC = …… (iv) m ∠DAB + m ∠CDA = ……

Ans. (i) AD = BC Because, in a parallelogram opposite sides are equal

(ii) ∠DCB = ∠DAB Because, in a parallelogram opposite angles are equal

(iii) OC = OA Because, diagonal of parallelogram is bisect each other



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(iv) m ∠DAB + m ∠CDA = 180°, since AB || DC and AD is transversal ∠DAB and ∠CDA are opposite angles, interior opposite angles,

Q.2. Consider the following parallelograms. Find the values of unknowns x, y, z.

Ans. (i) ABCD is a parallelogram (Given) ∴ AD || BC ∴ AB || DC ⇒ ∠x = ∠z (opposite angles of parallelogram ABCD) or ∠y = 100° (Opposite angles of parallelogram ABCD) ⇒ ∠y + ∠x = 180° (Co-interior angles as AD || BC) or 100° + ∠x = 180° ⇒ x = 80° ⇒ z = 80° Hence, x, y and z are 80°, 100° and 80° (ii) Let ABCD be a || gm. ∠B = ∠D (opposite angles property) ∴ ∠D = 50° ∴ ∠B = 50° ∴ ∠CBA + ∠CBL = 180° (Linear pair)



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⇒ 50° + z = 180° ⇒ z = 180° – 50° = 130° ∴ z = 130° (opposite angles are equal) ∴ ∠z = ∠x ⇒ x = 130° ∴ x = y (opposite angles property) ⇒ y = 130° (iii) Let ABCD be a parallelogram. ∴ x = 90° (AC ⊥ BD)

⇒ ∠x + ∠y + 30° = 180° ⇒ 90° + ∠y + 30° = 180° ⇒ ∠y + 120° = 180° ⇒ ∠y = 180° – 120° ⇒ ∠y = 60° ⇒ ∠z = ∠y = 60° Hence, ∠z = 60° (iv) Let ABCD be a parallelogram ∠D = ∠B = 80° (opposite angles property) Also, ∠C + ∠B = 180° ⇒ ∠C + 80° = 180° ⇒ ∠C = 180° – 80°

∴ ∠C = 100°



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∠z + ∠C = 180° (Linear pair) ⇒ ∠z + 100° = 180°; ⇒ ∠z = 180° – 100° = 80° ⇒ ∠y = 80° (opposite angles) ∠C = ∠x = 100° (opposite angles property)

(v) Let ABCD be a parallelogram. ∴ ∠D = y = ∠B = 112°

(opposite angles property) In ΔDAC ∠D + ∠DAC + ∠DCA = 180° (angle sum property) ⇒ 112° + 40° + x = 180° ⇒ 152° + x = 180° ∴ x = 180° – 152° = 28° ⇒ x = z (Alternate angles) ∴ z = 28° In Δ BAC ∠B + ∠BAC + ∠ACB = 180° ⇒ 112° + 28° + ∠C = 180° ⇒ ∠C = 180° – 140° = 40° Hence, y = 112° z = 28° x = 28°

Q.3. Can a quadrilateral ABCD be a parallelogram if (i) ∠D + ∠B = 180°? (ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?



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(iii) ∠A = 70° and ∠C = 65°?

Ans. (i) ∠D = ∠B If, then ABCD can be a parallelogram. If ∠D = ∠B then ABCD is not a parallelogram.

(ii) AB = DC = 8 cm, AD = 4 cm, BC = 4.4 cm. We know that in a parallelogram, opposite sides are

always equal, but here AD ≠ BC Hence, ABCD is not a parallelogram. (iii) ∠A = 70° and ∠C = 65°. We know that in a parallelogram, opposite angles are

equal, but here opposite angles ∠A ≠ ∠C Hence, ABCD is not a parallelogram.

Q.4. Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.

Ans. Example – Kite In the quadrilateral ABCD AB = AD

and BC = CD Also, ∠A = ∠C and ∠B = ∠D But it is not a parallelogram as opposite

sides are not equal. Hence, ABCD is a kite.

Q.5. The measures of two adjacent angles of a parallelogram are in the ratio 3:2. Find the measure of each of the angles of the parallelogram.

Ans. Let ABCD be a parallelogram. A and B are two adjacent angles.



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∴ ∠A + ∠B = 180° (Co interior angles) ⇒ 3x + 2x = 180° ⇒ 5x = 180° ⇒ x = 36° ∴ ∠A = 3 × 36° = 108° ∠B = 2 × 36° = 72° ∠B = ∠D (opposite angle of a parallelogram) ∠B = ∠D = 72° ∠A = ∠C = 108° (opposite angle of a parallelogram)

Q.6. Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.

Ans. Let ABCD be a parallelogram. ∠B = ∠C (adjacent angles) ∠B = ∠D (opposite angles of a parallelogram) ⇒ ∠C = ∠A (opposite angles of a parallelogram) We know that sum of angles of a quadrilateral is 360°. So, ∠A + ∠B + ∠C + ∠D = 360° ⇒ ∠A + ∠A + ∠A + ∠A = 360° ⇒ 4∠A = 360° or ∠A = 90° Hence, each angle of the parallelogram ABCD is 90°.

Q.7. The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.



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Ans. In a given a parallelogram HOPE ∠POH + ∠POX = 180° (Linear pair) ⇒ ∠POH + 70° = 180° ⇒ ∠POH = 180° – 70° ⇒ ∠POH = 110° ⇒ ∠OPH = ∠EHP (Alternate angles) ∴ y = 40° In ΔPOH ∠y + ∠z + 110° = 180° (By angle sum property of triangle) 40° + z + 110° = 180° 150° + z = 180° ⇒ z = 30° ∠E = ∠O (Opposite angles of a parallelogram are equal) or x = 110° In Δ PHE ∠EPH + ∠PHE + ∠E = 180° (By angle sum property) ⇒ ∠EPH + 40 + 110° = 180° ⇒ ∠EPH = 40° Hence, x = 110° y = 40° z = 30°

Q.8. The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm)



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Ans. (i) Given GUNS is a parallelogram. We know that opposite sides of a parallelogram are equal ∴ GU = NS ⇒ 3y – 1 = 26 ⇒ 3y = 27

∴ y = 273

= 9 cm.

Similarly, GS = NU (opposite sides of a parallelogram) ⇒ 3x = 18

⇒ x = 183

= 6 cm

Hence, y = 9 cm x = 6 cm (ii) Given RUNS is a parallelogram. Let O be the intersecting point. ⇒ SO = OU (Diagonals bisect each other at point

O) ⇒ 20 = y + 7 y + 7 = 20 ------------------ (i) ⇒ RO = ON (Diagonals bisect each other at point O) ⇒ x + y = 16 ---------------- (ii) From eqn. (i) y = 20 – 7



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∴ y = 13 Putting y = 13 in eqn. (ii) we have x + 13 = 16 ⇒ x = 16 – 13 x = 3 cm Hence, y = 13 cm x = 3 cm.

Q.9.

In the above figure both RISK and CLUE are parallelograms. Find the value of x.

Ans.

Given RISK and CLUE are parallelograms. ∴ ∠R + ∠K = 180° (Adjacent angles) ⇒ ∠R + 120° = 180° ⇒ ∠R = 60° ∴ ∠R = ∠ISK = 60° (Opposite angle of a || gm are equal) In parallelogram CLUE ∠L = ∠CEU = 70° ⇒ ∠L + ∠U = 180° (Adjacent angle of parallelogram CLUE) ∴ 70° + ∠U = 180°



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∠U = 110° In Δ OSE ∠O + ∠OSE + ∠OES = 180° (By angle sum property of a triangle) ⇒ x + 60° + 70° = 180° ⇒ x + 130° = 180° ⇒ x = 180° – 130° ∴ x = 50° Hence, the value of x is 50°.

Q.10. Explain how this figure is a trapezium. Which of its two sides are parallel?

Ans. We know that trapezium is a quadrilateral with a pair of parallel sides. ∠M and ∠L are adjacent angles and ∠M + ∠L ⇒ 100° + 80° = 180° ∴ MN || KL Hence, KLMN is a trapezium

Q.11. Find m∠C in figure if AB || DC.

Ans. ABCD is a trapezium. We know that in a trapezium sum of interior opposite angles (Co interior angles

as AB || DC) is 180°. ⇒ ∠B + ∠C = 180°



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⇒ 120° + ∠C = 180° ⇒ ∠C = 60° Hence, m∠C = 60°

Q.12. Find the measure of ∠P and ∠S if SP || RQ in figure (If you find m∠R, is there more than one method to find m∠P?)

Ans. Let PQRS be a trapezium with SP || RQ.

∠S + ∠R = 180° (Co interior angles as SP || RQ) ∠P + ∠Q = 180° (Co-interior angles as SP || RQ) ∠S + 90° = 180°

∠S = 90° ⇒ ∠P + 130° = 180° ⇒ ∠P = 180° – 130° ∴ ∠P = 50° Hence, m∠P = 50° m∠S = 90°

Exercise 3.4

Q.1. State whether True or False. (a) All rectangles are squares (b) All rhombuses are parallelograms (c) All squares are rhombuses and also rectangles (d) All squares are not parallelograms. (e) All kites are rhombuses. (f) All rhombuses are kites. (g) All parallelograms are trapeziums. (h) All squares are trapeziums.



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Ans. (a) All rectangle are squares (False) (b) All rhombuses are parallelograms (True) (c) All squares are rhombuses and also rectangles (True) (d) All squares are not parallelograms. (False) (e) All kites are rhombuses. (False) (f) All rhombuses are kites. (True) (g) All parallelograms are trapeziums. (True)

(h) All squares are trapeziums. (True)

Q.2. Identify all the quadrilaterals that have. (a) four sides of equal length (b) four right angles

Ans. (a) Rhombus – A rhombus is a quadrilateral in which all sides are equal length.

Square – A square is a quadrilateral in which all sides are equal length and angles are 90°.

Hence, rhombus and square are the quadrilaterals that have four sides of equal length.

(b) Rectangle: – A rectangle is a parallelogram of equal angles. Thus each angle of a rectangle is a right angle.

Square: – A square is a quadrilateral with all sides and angles are equal. Thus each angle of a square is right angle.

Hence, rectangle and square are the quadrilaterals that have four right angles.

Q.3. Explain how a square is. (i) a quadrilateral (ii) a parallelogram (iii) a rhombus (iv) a rectangle



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Ans. (i) Square is a quadrilateral – A square is a closed figure formed by 4 sides of equal length, so it is a quadrilateral.

(ii) Square is a parallelogram – In a square opposite sides are parallel and equal so it is a parallelogram.

(iii) Square is a rhombus – A square is a parallelogram with all the 4 sides are of equal length, so it is a rhombus.

(iv) Square is a rectangle – A square is a parallelogram with each angle is a right angle, so it is a rectangle.

Q.4. Name the quadrilaterals whose diagonals. (i) bisect each other (ii) are perpendicular bisectors of each other (iii) are equal

Ans. (i) Parallelogram, rhombus, rectangle, and square are the quadrilaterals whose diagonals bisect each other.

(ii) Square and rhombus are the quadrilaterals whose diagonals are perpendicular bisectors of each other.

(iii) Rectangle and square are the quadrilaterals whose diagonals are equal.

Q.5. Explain why a rectangle is a convex quadrilateral.

Ans. A rectangle is a convex quadrilateral because both of its diagonals lie in its interior.

Q.6. ABC is a right-angled triangle and O is the mid point of the side opposite to the right angle. Explain why O is equidistant from A, B and C. (The dotted lines are drawn additionally to help you).

Ans. Given O is the mid point of AC, i.e. AO = OC



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Draw, AD || BC and DC || AB shown in dotted line)

Now, ABCD is a parallelogram and, ∠B = ∠D (Opposite angles of a parallelogram) ∴ ∠D = 90° (∴ ∠B = 90°) Then, ABCD is a rectangle. So, diagonals, AC and BD are

equal and bisects each other at 0. ∴ AO = OC = OB = OD Hence, O is equidistant from A, B and C.



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4 PRACTICAL GEOMETRY

Exercise 4.1

Q.1. Construct the following quadrilaterals. (i) Quadrilateral ABCD. (ii) Quadrilateral JUMP AB = 4.5 cm JU = 3.5 cm BC = 5.5 cm UM = 4 cm CD = 4 cm MP = 5 cm AD = 6 cm PJ = 4.5 cm AC = 7 cm PU = 6.5 cm (iii) Parallelogram MORE (iv) Rhombus BEST OR = 6 cm BE = 4.5 cm RE = 4.5 cm ET = 6 cm EO = 7.5 cm

Ans. (i) Step I ΔABC can be constructed using SSS construction condition. Draw ΔABC

Step II Locate the fourth point D. This ‘D’ would be on the side opposite to B with reference to AC. D is 6.0 cm away from A. So, with A as centre draw an arc of radius 6.0 cm.



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Step III D is 4.0 cm away from C. So with C as centre, draw an arc of radius 4.0 cm cutting the arc of Step II

Step IV D should lie on both the arcs drawn. So it is the point of intersection of the two arcs, mark it as D and join AD and CD. ABCD is the required quadrilateral.

(ii) Step I ΔPJU can be constructed using SSS

construction condition. Draw ΔPJU (Figure) Step II Locate the fourth point M. This ‘M’ would

be on the side opposite to J with reference to



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PU. M is 5.0 cm away from P. So, with P as centre draw an arc of radius 5.0 cm (Figure)

Step III M is 4.0 cm away from U. So with U as

centre, draw an arc of radius 4.0 cm cutting the arc of Step II.

Step IV M should lie on both the arcs drawn. So it is

the point of intersection of the two arcs. Mark M and join PM and UM. JUMP is the required quadrilateral.

(iii) Step I ΔERO can be constructed using SSS

construction condition. Draw ΔERO (Figure)



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Step II Locate the fourth point M. This ‘M’ would be on the side opposite to R with reference to EO. Since, MORE is a parallelogram EM = RO. M is 6.0 cm away from E. So, with E as centre draw an arc with radius = 6.0 cm.

Step III O as centre draw an arc with radius = 4.5 cm

cutting arc of Step II. Step IV M should lie on both the arcs drawn. So it is

the point of intersection of the two arcs. Mark M and complete MORE by joining EM and MO. MORE is the required parallelogram.

(iv) Step I ΔBET can be constructed using SAS

construction condition. Draw ΔBET



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Step II Locate the fourth point S, this ‘S’ would be on the side opposite to B with reference to ET. Since, BEST is a rhombus, all sides are equal. So, S is 4.5 cm away from E. E as centre draw an arc with radius 4.5 cm (Figure)

Step III T as centre draw an arc with radius = 4.5 cm Step IV S should lie on both the arcs drawn. So, it

will be at the point of intersection of two arcs. Mark S then rhombus BEST is completed by joining ES and ST BEST is the required rhombus.



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Exercise 4.2

Q.1. Construct the following quadrilaterals. (i) quadrilateral LIFT. (ii) quadrilateral GOLD LI = 4 cm OL = 7.5 cm IF = 3 cm GL = 6 cm TL = 2.5 cm GD = 6 cm LF = 4.5 cm LD = 5 cm IT = 4 cm OD = 10 cm (iii) Rhombus BEND BN = 5.6 cm DE = 6.5 cm

Ans. (i) Step I ΔLIF can be constructed using SSS construction condition. Draw ΔLIF (Figure)

Step II With I as centre, draw an arc of radius

4.0 cm (Figure) Step III With L as centre, draw an arc of radius

2.5 cm (Figure) cutting the arc of Step II



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Step IV Since T lies on both the arcs, T is the point of intersection of the two arcs. Mark T and complete LIFT by joining LT and TF. LIFT is the required quadrilateral (Figure)

(ii) Step I ΔODL can be constructed using SSS

construction condition. Draw ΔODL (Figure)

Step II L as centre, draw an arc of radius 6.0 cm

(Figure) Step III With D as centre draw an arc of radius

6.0 cm (Figure) cutting the arcs of Step II



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Step IV Since G lies on both the arcs. G is the point of intersection of the two arcs. Mark G and complete GOLD by joining OG and DG. GOLD is the required quadrilateral (Figure)

(iii) In the rhombus the diagonals bisects each other at 90°. Step I Draw a DE = 6.5 cm and construct its

perpendicular bisector XY Step II With O as centre draw an arc of radius

5.62

= 2.8 cm on either side marked as B and N.

Join BDBE, EN and DN This, BEND is the required rhombus.



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Exercise 4.3

Q.1. Construct the following quadrilaterals. (i) Quadrilateral MORE. (ii) Quadrilateral PLAN MO = 6 cm PL = 4 cm OR = 4.5 cm LA = 6.5 cm ∠M = 60° ∠P = 90° ∠O = 105° ∠A = 110° ∠R = 105° ∠N = 85° (iii) Parallelogram HEAR (iv) Rectangle OKAY HE = 5 cm OK = 7 cm EA = 6 cm KA = 5 cm ∠R = 85°

Ans. (i) Step I Draw a line segment MO which measures 6.0 cm.

Step II Take M as centre, make an angle of 60° (Figure)

Step III Take O as centre, make an angle of 105°. Such that ∠MOR = 105°. Join OR.



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Step IV Take R as centre, make an angle of 105° (Figure)

Step V Take R as centre, draw an arc of radius 4.5 cm. Join RE and ME. We get required quadrilateral MORE. (Figure)

(ii) Step I Draw a line segment AL which measure 6.5 cm. (Figure)

Step II Take A as centre, make an angle of 110°

(Figure).



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Step III Since, ∠P = 90°, ∠A = 110°, ∠N = 85° ∠L = 360° – (90° + 110° + 85°) = 75° Take L as centre, draw an ∠A ∠P = 75° Taking L as centre and radius 4.0 cm draw

an arc cutting LY at P.

Step IV Take P as centre, make an angle of 90°. Join PN, N make an angle of 85°. We get required quadrilateral PLAN (Figure)

(iii) Step I Draw a line segment AR = AE = 5 cm, as HEAR is a parallelogram.

Step II Take R as centre, make an angle of 85° (Figure). Taking R as centre and radius 6 cm, draw an arc on RX and marked it as 4.



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Step III Take H as centre, draw an arc of radius 5.0 cm.

Step IV Take A as centre, draw an arc of radius 6.0 cm both arcs intersect each other at the point E. Join HE and EA we get required parallelogram HEAR. (Figure)

(iv) Step I Draw a line segment OK which measure 7.0 cm. (Figure)

Step II Take K as centre make, ∠K = 90° and with O as centre draw ∠O = 90°.

Step III With O as centre and radius 5 cm draw an arc cutting OP at Y. With K as centre and radius 5 cm draw an arc cutting KX at A.

Join AY. This, OKAY is the required rectangle.



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Exercise 4.4

Q.1. Construct the following quadrilaterals. (i) Quadrilateral DEAR. (ii) Quadrilateral TRUE DE = 4 cm TR = 3.5 cm EA = 5 cm RU = 3 cm AR = 4.5 cm UE = 4 cm ∠E = 60° ∠R = 75° ∠A = 90° ∠U = 120°

Ans. (i) Step I Draw a line segment EA which measure 5.0 cm. (Figure)

Step II Take E as centre, make an ∠XEA = 60°. (Figure)

Step III Take E as centre, draw an arc of radius 4.0 cm. (Figure) cutting EX at D.

Step IV Take A as centre, make an angle of 90°. (Figure)



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Step V Take A as centre. Draw an arc of radius 4.5 cm. Join cutting AR at R. DR. We get required quadrilateral DEAR. (Figure)

(ii) Step I Draw a line segment RU which measure 3.0 cm. (Figure)

Step II Take R as centre make an angle of 75° i.e. ∠XRU = 75° (Figure)

Step III Take R as centre, draw an arc of radius 3.5 cm cutting RX at T.

Step IV Take U as centre, make an angle of 120°, ∠YUR = 120° (Figure)



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Step V Take U as centre, draw an arc of radius 4.0 cm cutting UY at E. Join TE, then we get the required quadrilateral TRUE. (Figure)

Exercise 4.5

Q.1. Draw the following. (1) The square READ with RE = 5.1 cm.

Ans. We know that all sides of a square are equal and each angle is 90°

Step I Draw a line segment RE which measure 5.1 cm. (Figure)

Step II Take R and E as centre, make an angle of 90°. (Figure) ∠XRE = 90° and ∠YER = 90°

Step III Take R and E as centre, draw an arc of radius 5.1 cm cutting RX at D and EY at A.



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Step IV Join DA then we get the required square READ. (Figure)

(2) A rhombus whose diagonals are 5.2 cm and 6.4 cm long.

Ans. Step I Draw AC = 6.4 cm, construct its perpendicular bisector, XY intersecting AC at O.

Step II Take O as centre draw two arcs on either side cutting XY at B and D.

Step III Join AB, BC, CD and DA. We get required rhombus ABCD.



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(3) A rectangle with adjacent sides of lengths 5 cm and 4 cm.

Ans. Step I Draw a line segment AB which measure 5.0 cm. (Figure)

Step II Take A and B as centre, make an angle 90°. (Figure)

Step III Take A and B as centre, draw arcs of radius 4.0 cm, arcs intersect AX at the point C and BY at D. Join DC. We get required rectangle ABCD.

(4) A parallelogram OKAY where OK = 5.5 cm and KA = 4.2 cm.

Ans. Step I Draw a line segment OK = 5.5 cm.

Step II With O and K as centres and radius equal to 4.2 cm, draw two arcs.



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Take any point y on the arc draw above (with centre O).

Step III With y as centre and radius equal to 5.5 cm, draw an arc which cuts the arc draw in step 2 at A. Join KA and AY, to get the required parallelogram.

Note : Many answer are possible to this question.

3 UNDERSTANDING QUADRILATERALS · 2008. 12. 19. · 3 UNDERSTANDING QUADRILATERALS Exercise 3.1 Q.1. Given here are some figures. Classify each of them on the basis of the following.

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