Chapter 4 exercise answers: Quadrilaterals

Open NCERT p. 94Matches NCERT’s answer

(i)

1. In rectangle ABCD the diagonals AC and BD bisect each other at O, and OA = OB = OC = OD.
2. ∠DAB = 90° (a corner of the rectangle) and ∠BAC = 30° is given, so ∠CAD = 90° − 30° = 60°.
3. In triangle OAB, OA = OB, so it is isosceles: ∠OAB = ∠OBA = 30°. Since ∠OBA is the same as ∠ABD, ∠ABD = 30°.
4. AB ∥ DC, so with diagonal BD as transversal, alternate angles are equal: ∠BDC = ∠ABD = 30°.
5. AD ∥ BC, so with diagonal AC as transversal, alternate angles are equal: ∠ACB = ∠CAD = 60°.
6. Each corner is 90°, so ∠ADB = 90° − ∠BDC = 60°, and ∠ACD = 90° − ∠ACB = 30°.

Answer∠ABD = 30°, ∠CAD = 60°, ∠ADB = 60°, ∠BDC = 30°, ∠ACD = 30°, ∠ACB = 60°

(ii)

1. In rectangle QRSP the diagonals QS and RP bisect each other at O, and OQ = OR = OS = OP.
2. ∠QOR = 110° is given. Since P, O, R lie on the straight diagonal RP, ∠QOP = 180° − 110° = 70°. Since Q, O, S lie on the straight diagonal QS, ∠ROS = 180° − 110° = 70°.
3. In triangle OQR, OQ = OR, so its base angles are equal: ∠OQR = ∠ORQ = (180° − 110°) ÷ 2 = 35°.
4. In triangle OQP, OQ = OP and the angle between them is 70°, so ∠OQP = ∠OPQ = (180° − 70°) ÷ 2 = 55°.
5. In triangle ORS, OR = OS and the angle between them is 70°, so ∠ORS = ∠OSR = (180° − 70°) ÷ 2 = 55°.

Answer∠QOP = 70°, ∠ROS = 70°, ∠OQR = 35°, ∠ORQ = 35°, ∠OQP = 55°, ∠OPQ = 55°, ∠ORS = 55°, ∠OSR = 55°

Watch this explained “The two base angles”, 4:19 into The carpenter's problem: how to be sure a frame really is rectangular · हिंदी में देखें

Open NCERT p. 94One way to think about it

(i) 30°

1. Draw a line segment AC = 8 cm and mark its midpoint O, so that AO = OC = 4 cm.
2. At O, use a protractor to draw a line through O making 30° with AC.
3. On this line mark B and D, 4 cm from O on either side. Then BD = 8 cm and O is the midpoint of BD too.
4. Join A, B, C, D in order. The diagonals are equal and bisect each other, so ABCD is a rectangle (the Carpenter's Problem): every corner measures 90°.
5. Measuring, the sides come out about 2.1 cm and 7.7 cm.

In shortA rectangle with sides about 2.1 cm and 7.7 cm. It is not a square, because the diagonals do not cross at 90°.

(ii) 40°

1. Construct exactly as in (i), but draw the second diagonal at 40° to the first, through the midpoint O, with 4 cm on each side of O.
2. Join the four ends in order. The diagonals are equal and bisect each other, so the figure is a rectangle.
3. Measuring, the sides come out about 2.7 cm and 7.5 cm.

In shortA rectangle with sides about 2.7 cm and 7.5 cm (not a square).

(iii) 90°

1. Construct as in (i), but draw the second diagonal at 90° to the first, through O, with 4 cm on each side of O.
2. The diagonals are equal, bisect each other and meet at 90°, so the figure is a square, as the chapter showed.
3. Measuring, all four sides come out about 5.7 cm and every corner is 90°.

In shortA square with sides about 5.7 cm.

(iv) 140°

1. Construct as in (i), but draw the second diagonal at 140° to the first, through O, with 4 cm on each side of O.
2. Where two lines cross at 140°, the angle beside it is 180° − 140° = 40°. So these diagonals also cross at 40°, and the figure is the same rectangle as in (ii), only turned.
3. Measuring, the sides come out about 7.5 cm and 2.7 cm.

In shortA rectangle with sides about 7.5 cm and 2.7 cm, the same shape as in (ii).

Watch this explained “Halving and equal”, 4:52 into What the diagonals alone tell you about a quadrilateral · हिंदी में देखें

Open NCERT p. 94Matches NCERT’s answer

1. O is the centre of the circle, so OP = OL = OA = OM, all being radii.
2. PL and AM are diameters through O, so O is the midpoint of both — the diagonals of quadrilateral APML bisect each other.
3. Since OP = OA = OL = OM, the diagonals PL and AM are also equal in length (each is twice the radius).
4. PL ⊥ AM is given, so the diagonals also cross at a right angle.
5. Diagonals that are equal, bisect each other, AND cross at a right angle belong only to a square.

AnswerAPML is a square.

Watch this explained “All three at once”, 6:19 into What the diagonals alone tell you about a quadrilateral · हिंदी में देखें

Open NCERT p. 94One way to think about it

1. Lay the thread along one stick and cut a piece exactly as long as the stick. Fold that piece in half: the fold marks half the length. Use it to mark the midpoint of each stick.
2. Cross the two sticks so that their midpoints meet, and tie them together at that point. Open them to any angle.
3. Pass the thread tightly round the four ends of the sticks and tie it off. The thread forms a quadrilateral whose diagonals are the two sticks.
4. The diagonals are equal (the sticks are equal) and bisect each other (they are joined at their midpoints), so the quadrilateral is a rectangle. Every corner of the thread is exactly 90°.
5. Why: the four half-sticks from the crossing point are equal, so the four triangles there are isosceles. If the sticks cross at x°, the base angles are 90° − x° ÷ 2 in one triangle and x° ÷ 2 in the next. Each corner of the thread holds one of each: 90° − x° ÷ 2 + x° ÷ 2 = 90°.
6. This is not the only way. You can also keep the sticks joined at their midpoints and adjust the angle until all four thread sides are equal. The figure is then a square, and the sticks themselves cross at exactly 90°.

In shortTie the two equal sticks together at their midpoints, open them to any angle and pull the thread tight round the four ends. Every corner of the thread is exactly 90°. (Another way: adjust the crossing until the four thread sides are equal; the sticks then cross at 90°.)

Watch this explained “Two strips and a pin”, 0:45 into The carpenter's problem: how to be sure a frame really is rectangular · हिंदी में देखें

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1. A quadrilateral with opposite sides parallel (and therefore equal) is, by definition, a parallelogram — that is all the condition guarantees.
2. Draw a parallelogram that is deliberately not right-angled — say with sides 5 cm and 3 cm meeting at 50°. Its opposite sides are still parallel and equal.
3. But no angle of this parallelogram is 90°, so it is not a rectangle.

AnswerNo — such a quadrilateral is a parallelogram in general, and a parallelogram need not have right angles, so it need not be a rectangle.

Watch this explained “Push a rectangle over”, 0:00 into The parallelogram: everything that follows from "opposite sides parallel" · हिंदी में देखें

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(i)

1. PRAE is a parallelogram: RA ∥ PE and PR ∥ EA, as the matching arrows show.
2. Opposite angles of a parallelogram are equal, so ∠A = ∠P = 40°.
3. Neighbouring angles add up to 180°, so ∠R = 180° − 40° = 140°. ∠E is opposite ∠R, so ∠E = 140°.

Answer∠R = 140°, ∠A = 40°, ∠E = 140°

(ii)

1. SRQP is a parallelogram: SR ∥ PQ and SP ∥ RQ, as the matching arrows show.
2. ∠R is opposite ∠P, so ∠R = ∠P = 110°.
3. ∠Q and ∠S are each next to ∠P, so each is 180° − 110° = 70°.

Answer∠Q = 70°, ∠R = 110°, ∠S = 70°

(iii)

1. All four sides of XWVU are marked equal, so it is a rhombus. A rhombus is a parallelogram, so opposite angles are equal and neighbouring angles add up to 180°.
2. The 30° is ∠UVX, between side VU and the diagonal VX. In triangle XUV, UX = UV, so the angles opposite these sides are equal: ∠UXV = ∠UVX = 30°.
3. So ∠U = 180° − 30° − 30° = 120°.
4. ∠V is next to ∠U, so ∠V = 180° − 120° = 60°. Opposite angles are equal, so ∠W = ∠U = 120° and ∠X = ∠V = 60°.

Answer∠X = 60°, ∠W = 120°, ∠V = 60°, ∠U = 120°. The remaining angles are 60° and 120°.

(iv)

1. All four sides of OIEA are marked equal, so it is a rhombus, and so a parallelogram.
2. The 20° is ∠AEO, between side EA and the diagonal EO. In triangle OAE, AO = AE, so ∠AOE = ∠AEO = 20°.
3. So ∠A = 180° − 20° − 20° = 140°.
4. ∠E is next to ∠A, so ∠E = 180° − 140° = 40°. Opposite angles are equal, so ∠I = ∠A = 140° and ∠O = ∠E = 40°.

Answer∠O = 40°, ∠I = 140°, ∠E = 40°, ∠A = 140°. The remaining angles are 40° and 140°.

Watch this explained “The angles, two ways”, 5:46 into The rhombus, and what its diagonals do · हिंदी में देखें

Open NCERT p. 102One way to think about it

1. Draw a line segment of length 7 cm — this is one diagonal.
2. Mark its midpoint O (3.5 cm from each end).
3. At O, use a protractor to draw a second line through O making a 140° angle with the first diagonal.
4. On this second line, mark points 2.5 cm on each side of O — this gives a 5 cm diagonal, also bisected at O.
5. Join the four end points in order; because the diagonals bisect each other, this is guaranteed to be a parallelogram.

In shortA parallelogram with diagonals 7 cm and 5 cm crossing at 140° (it is not a rectangle, since the two diagonals are not equal).

Watch this explained “Halving alone”, 4:10 into What the diagonals alone tell you about a quadrilateral · हिंदी में देखें

Open NCERT p. 102One way to think about it

1. Draw a line segment of length 5 cm (the longer diagonal) and mark its midpoint O.
2. At O, construct a line through O perpendicular to the first diagonal (using a set square, or by paper folding).
3. On this perpendicular line, mark points 2 cm on each side of O — this gives a 4 cm diagonal, also bisected at O.
4. Join the four end points in order; because the diagonals bisect each other AND cross at 90°, the result is a rhombus.

In shortA rhombus whose diagonals are 4 cm and 5 cm, crossing at right angles at their common midpoint.

Watch this explained “Halving and square”, 5:36 into What the diagonals alone tell you about a quadrilateral · हिंदी में देखें

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1. Joining two equilateral triangles of side 4 cm along a common side puts that shared side as a diagonal, with the two triangles' apex corners as the other two corners of the quadrilateral.
2. Every side of the resulting quadrilateral is a side of one of the equilateral triangles, so all four sides are 4 cm.
3. Each equilateral triangle contributes its own 60° angle whole to one corner (its apex), so two opposite corners of the quadrilateral are 60°.
4. At the two ends of the shared side, the 60° angles of both triangles combine (60° + 60°), giving 120° at each of the other two corners.

AnswerAll four sides are 4 cm; the angles are 60°, 120°, 60°, 120°, so the quadrilateral is a rhombus.

Watch this explained “Two equilateral cutouts”, 3:06 into Which quadrilaterals you can build by joining two triangles · हिंदी में देखें

Open NCERT p. 107One way to think about it

1. In a kite, one diagonal cuts the other in half at right angles (Property 1 of the kite). Here we let the 8 cm diagonal cut the 6 cm diagonal in half.
2. Draw BD = 8 cm. Mark a point O on BD. It can be anywhere between B and D; take BO = 2 cm, so OD = 6 cm.
3. At O, draw a line perpendicular to BD, using a set square or by folding BD onto itself at O.
4. On this line mark A and C, 3 cm from O on either side. Then AC = 6 cm and O is its midpoint.
5. Join AB, BC, CD and DA. Triangles AOB and COB are congruent (SAS: AO = CO, right angles at O, OB common), so AB = CB. In the same way AD = CD. So ABCD is a kite. Here AB = CB ≈ 3.6 cm and AD = CD ≈ 6.7 cm.
6. O could have been put anywhere between B and D, so many different kites have these two diagonals. If O is the midpoint of BD, the kite is a rhombus. You could also let the 6 cm diagonal cut the 8 cm one in half instead.

In shortKite ABCD with BD = 8 cm and AC = 6 cm, where BD cuts AC in half at right angles at O. With BO = 2 cm, AB = CB ≈ 3.6 cm and AD = CD ≈ 6.7 cm. Many kites are possible, because O may be any point between B and D.

Watch this explained “Two diagonals do not pin a kite down”, 8:08 into The kite: why one diagonal bisects the other at right angles, and the angles too · हिंदी में देखें

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(i)

1. The top and bottom sides of the trapezium are parallel (shown by the matching arrows).
2. Each slanting side is a transversal cutting the two parallel sides, so the two angles on the same slanting side add up to 180° (co-interior angles).
3. On the left leg: top-left angle = 180° − 135° = 45°.
4. On the right leg: top-right angle = 180° − 105° = 75°.

AnswerTop-left angle = 45°, top-right angle = 75°.

(ii)

1. The tick marks show the two slanting sides (legs) are equal in length, so this is an isosceles trapezium.
2. In an isosceles trapezium, the two angles at the ends of the same parallel side are equal to each other, so the other angle on the same (top) side is also 100°.
3. Each leg is a transversal between the two parallel sides, so the angle at the bottom of a leg is 180° − 100° = 80° (co-interior with the 100° angle at the top of that same leg).
4. By the same isosceles-trapezium rule, the angle at the bottom of the other leg is also 80°.

AnswerThe other angle on the same parallel side = 100°; the angle at the bottom of the same leg = 80°; the remaining angle at the fourth vertex = 80°.

Watch this explained “Filling in the other two”, 4:09 into The trapezium: what a single pair of parallel sides forces, and what "isosceles" adds · हिंदी में देखें

Open NCERT p. 107Checked by computer

Venn diagram

1. Draw a large oval for parallelograms.
2. Inside it draw two overlapping ovals: rectangles (all angles 90°) and rhombuses (all sides equal). The region where they overlap is the squares, since a square is both.
3. Draw an oval for kites that overlaps the parallelogram oval, so that the part they share is exactly the rhombus oval. Every rhombus (and so every square) is a kite, and no other parallelogram is.

AnswerRectangles and rhombuses lie inside parallelograms, and they overlap in the squares. The rhombuses also lie inside the kites, and the only parallelograms that are kites are the rhombuses.

(i) What is the quadrilateral that is both a kite and a parallelogram?

1. Let ABCD be a kite with AB = BC and CD = DA (the chapter's definition).
2. If it is also a parallelogram, its opposite sides are equal: AB = CD and BC = DA.
3. Then AB = BC = CD = DA, so all four sides are equal and it is a rhombus. Every rhombus is also a parallelogram and has AB = BC and CD = DA, so it is both.

AnswerA rhombus (a square is a special rhombus).

(ii) Can there be a quadrilateral that is both a kite and …

1. A rectangle has opposite sides equal. If it is also a kite, then AB = BC as well, so all four sides are equal.
2. A rectangle with all four sides equal is a square, and a square has AB = BC and CD = DA, so it is a kite.
3. The answer key at the back of the book prints 'No' for (ii), but a square has AB = BC and CD = DA, so it is a kite by the book's own definition, and it is also a rectangle, so the answer is yes — a square.

AnswerYes. A square is both a kite and a rectangle, and it is the only such quadrilateral.

(iii) Is every kite a rhombus?

1. A kite with sides 2 cm, 2 cm, 5 cm and 5 cm (two equal pairs of different lengths) is a kite, but its sides are not all equal, so it is not a rhombus.
2. A rhombus has all four sides equal, so it certainly has AB = BC and CD = DA. Every rhombus is a kite.

AnswerNo. Every rhombus is a kite, but not every kite is a rhombus.

Watch this explained “Where the kite sits”, 9:02 into The kite: why one diagonal bisects the other at right angles, and the angles too · हिंदी में देखें

Open NCERT p. 107Matches NCERT’s answer

1. In rectangle PAIR, I is a corner, so ∠RIA = 90°. Since O lies on side AI, ∠RIO = 90° too.
2. In triangle ROI, ∠ORI = 30° (given) and ∠RIO = 90°, so ∠ROI = 180° − 90° − 30° = 60°.
3. In rectangle RODS, O is a corner, so ∠ROD = 90°.
4. Ray OI lies inside angle ROD, so ∠IOD = ∠ROD − ∠ROI = 90° − 60° = 30°.

Answer∠IOD = 30°.

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Open NCERT p. 108One way to think about it

1. Draw AC = 6 cm. This is one diagonal.
2. Open the compass to more than 3 cm. With centre A, and then centre C, draw arcs above and below AC. Join the two points where the arcs cross: this line is the perpendicular bisector of AC. It meets AC at its midpoint O, at 90°, without a protractor. (Folding the paper so that A falls on C gives the same line.)
3. With the compass set to OA = 3 cm and centre O, cut the perpendicular bisector at B and D, one on each side of AC. Then BD = 6 cm and O is its midpoint.
4. Join AB, BC, CD and DA. The diagonals are equal, bisect each other and meet at 90°, so ABCD is a square, as the chapter showed.
5. Check by measuring: each side is about 4.2 cm and each corner is 90°.

In shortSquare ABCD whose diagonals AC = BD = 6 cm bisect each other at right angles at O; each side is about 4.2 cm.

Watch this explained “All three at once”, 6:19 into What the diagonals alone tell you about a quadrilateral · हिंदी में देखें

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1. Let each side of square CASE be a. U, V, W and X are midpoints, so CU = CV = a ÷ 2, and the same is true at every corner.
2. Joining the midpoints cuts off a right-angled triangle at each corner, with both shorter sides a ÷ 2 (for example triangle UCV, with ∠C = 90°). The four corner triangles are congruent (SAS), so UV = VW = WX = XU: all four sides are equal.
3. Each corner triangle is right-angled and isosceles, so its other two angles are 45° each; for example ∠CUV = ∠CVU = 45°.
4. At U, the angles along the straight side CA add up to 180°: ∠CUV + ∠VUX + ∠XUA = 180°. So ∠VUX = 180° − 45° − 45° = 90°, and the same holds at V, W and X.
5. Four equal sides and four right angles: UVWX is a square. Measuring a drawing with CA = 6 cm gives sides of about 4.2 cm and corners of 90°.
6. Other squares inside a square (Figure (b)): going round CASE in one direction, mark a point on each side at the same distance from the corner, for example 2 cm from C along CA, 2 cm from A along AS, 2 cm from S along SE and 2 cm from E along EC. Join the four points. The four corner triangles are again congruent (SAS: a right angle between sides of 2 cm and a − 2 cm), so the four sides are equal. At each point, the two angles of the neighbouring triangles are the two acute angles of the same right triangle, so they add up to 90°, and the corner left between them is 90°. Any distance between 0 and a works, and the midpoints are the case a ÷ 2.

AnswerUVWX is a square. More generally, points taken at the same distance from each corner of CASE, going round in one direction, always form a square.

Open NCERT p. 108Matches NCERT’s answer

1. Four equal sides make the quadrilateral a rhombus, and every rhombus is a parallelogram.
2. In a parallelogram, angles next to each other add up to 180°, so if one angle is 90°, the angle next to it is 180° − 90° = 90° too.
3. Opposite angles of a parallelogram are equal, so the remaining two angles are also 90° — all four angles come out 90°.
4. Four equal sides together with four right angles is exactly the definition of a square.

AnswerYes — a rhombus with one right angle is forced to have all four angles equal to 90°, so it must be a square.

Watch this explained “So it is a parallelogram”, 4:55 into The rhombus, and what its diagonals do · हिंदी में देखें

Open NCERT p. 108Matches NCERT’s answer

1. In quadrilateral ABCD, draw diagonal BD. We are given AB = CD and AD = BC (opposite sides equal), and BD is common to both triangles ABD and CDB.
2. By the SSS test, triangle ABD ≅ triangle CDB, so ∠ABD = ∠CDB and ∠ADB = ∠CBD (matching angles of congruent triangles).
3. ∠ABD and ∠CDB are alternate angles for lines AB and DC with BD as transversal, so their being equal means AB ∥ DC.
4. Likewise, ∠ADB and ∠CBD being equal means AD ∥ BC.
5. Both pairs of opposite sides are parallel, so the quadrilateral is a parallelogram.

AnswerIt must be a parallelogram.

Watch this explained “Reading them as alternate pairs”, 4:04 into The rhombus, and what its diagonals do · हिंदी में देखें

Open NCERT p. 108Matches NCERT’s answer

1. ABCD has a dent at D: the angle inside the figure at D is a reflex angle, more than 180°.
2. The diagonal AC lies outside this figure, so it cannot split it into two triangles. Use the other diagonal, BD, which lies inside the figure and splits it into triangles ABD and CBD.
3. In triangle ABD: ∠BAD + ∠ABD + ∠ADB = 180°. In triangle CBD: ∠BCD + ∠CBD + ∠CDB = 180°.
4. Adding the two: ∠A + (∠ABD + ∠CBD) + ∠C + (∠ADB + ∠CDB) = 360°. Here ∠ABD + ∠CBD = ∠B, and ∠ADB + ∠CDB is the reflex angle at D, the angle inside the figure.
5. So ∠A + ∠B + ∠C + ∠D = 360°, where ∠D is the reflex angle.
6. Measuring a drawing, for example one with ∠A ≈ 37°, ∠B ≈ 53° and ∠C ≈ 37°, the reflex angle at D is about 233°, and 37° + 53° + 37° + 233° = 360°. Measuring the smaller angle at D (about 127°) instead would wrongly give 254°.

AnswerYes. The angles still add up to 360°, as long as the angle at the dent D is taken as the reflex angle inside the figure.

Watch this explained “The figure that caves in”, 6:19 into Why the angles of any quadrilateral add to 360° · हिंदी में देखें

Open NCERT p. 108Matches NCERT’s answer

(i) diagonals are equal and bisect each other must be a square

1. Every rectangle has equal diagonals that bisect each other. Take one that is not a square, say 6 cm by 3 cm.
2. Its sides are not all equal, so it is not a square.

AnswerFalse. Equal diagonals that bisect each other give a rectangle, which need not be a square.

(ii) having three right angles must be a rectangle

1. The four angles of any quadrilateral add up to 360°.
2. Three right angles use up 90° + 90° + 90° = 270°, leaving exactly 90° for the fourth angle.
3. All four angles are 90°, which is the chapter's definition of a rectangle.

AnswerTrue.

(iii) diagonals bisect each other must be a parallelogram

1. Let the diagonals AC and BD bisect each other at O, so OA = OC and OB = OD.
2. In triangles AOB and COD, OA = OC, OB = OD and ∠AOB = ∠COD (vertically opposite angles). So the triangles are congruent (SAS), and ∠OAB = ∠OCD.
3. These are alternate angles made by the line AC with AB and DC, so AB ∥ DC.
4. In the same way, triangles AOD and COB are congruent, so ∠OAD = ∠OCB and AD ∥ BC.
5. Both pairs of opposite sides are parallel, so ABCD is a parallelogram.

AnswerTrue.

(iv) diagonals are perpendicular to each other must be a rhombus

1. Take a kite whose diagonals are 8 cm and 6 cm, with the 8 cm diagonal cutting the 6 cm one in half at right angles, 2 cm from one end.
2. Its diagonals are perpendicular, but its sides are about 3.6 cm, 3.6 cm, 6.7 cm and 6.7 cm, which are not all equal. So it is not a rhombus.

AnswerFalse. A kite has perpendicular diagonals but need not be a rhombus.

(v) opposite angles are equal must be a parallelogram

1. Let ∠A = ∠C and ∠B = ∠D. The four angles add up to 360°, so 2∠A + 2∠B = 360° and ∠A + ∠B = 180°.
2. With AB as a transversal, ∠A and ∠B are interior angles on the same side adding up to 180°, so AD ∥ BC.
3. Since ∠B = ∠D, also ∠A + ∠D = 180°. With AD as a transversal, this gives AB ∥ DC.
4. Both pairs of opposite sides are parallel, so it is a parallelogram.

AnswerTrue.

(vi) all the angles are equal is a rectangle

1. Four equal angles adding up to 360° must each be 360° ÷ 4 = 90°.
2. All angles 90° is the chapter's definition of a rectangle.

AnswerTrue.

(vii) Isosceles trapeziums are parallelograms

1. In an isosceles trapezium, the two equal sides are the non-parallel sides (that is how the chapter defines it). So one pair of its opposite sides is not parallel.
2. A parallelogram needs both pairs of opposite sides parallel, so an isosceles trapezium is not a parallelogram.

AnswerFalse.

Watch the lesson What the diagonals alone tell you about a quadrilateral · हिंदी में देखें