CBSE Class 8 Mathematics Quadrilaterals Long Answer and Case Study Questions with Answers

Long Answer Questions

Question 1. A rectangle ABCD is drawn on a sheet of paper.
(i) Write the single-condition definition of a rectangle that the chapter arrives at, and list the four properties of a rectangle.
(ii) Using the fact that the sum of the interior angles on the same side of a transversal is 180°, justify that AD ∥ BC and AB ∥ DC.
(iii) If the diagonal AC of this rectangle is 13 cm long and the diagonals meet at O, write the length of BD and of OB, giving the property used.
Answer:
(i) A rectangle is a quadrilateral in which all the angles are 90°. Its properties:
(a) All the angles are 90°. (b) The opposite sides are equal. (c) The opposite sides are parallel. (d) The diagonals are equal and bisect each other.
(ii) Take AB as a transversal of AD and BC. The interior angles on the same side are ∠A and ∠B, and ∠A + ∠B = 90° + 90° = 180°. So AD ∥ BC.
Taking AD as a transversal of AB and DC, ∠A + ∠D = 180°, so AB ∥ DC.
(iii) The diagonals of a rectangle are equal, so BD = AC = 13 cm.
They bisect each other at O, so OB = \( \frac{13}{2} \) = 6.5 cm.
In simple words: Four right angles define a rectangle; co-interior right angles make the sides parallel; equal bisecting diagonals give BD = 13 cm and OB = 6.5 cm.

Teacher's Note:
1. Two right angles on one side always add to 180°.
2. OB is half of a diagonal, not half of a side.

Question 2. The angles of a quadrilateral always add up to the same total.
(i) Taking a quadrilateral SOME and drawing the diagonal SM, justify that the sum of its four angles is 360°.
(ii) The angles of a quadrilateral ABCD are 2x, 3x, 4x and 3x. Find the value of x and all four angles.
(iii) Explain why it is impossible to draw a quadrilateral having exactly three right angles.
Answer:
(i) The diagonal SM splits SOME into △SEM and △SOM. The angles of each triangle add up to 180°, so the six angles together make 180° + 180° = 360°.
These six angles are exactly the four angles of the quadrilateral, because the angles at S and M are each split into two parts. So the angle sum is 360°.
(ii) 2x + 3x + 4x + 3x = 360°, so 12x = 360° and x = 30°.
The angles are 60°, 90°, 120° and 90°. Check: 60 + 90 + 120 + 90 = 360.
(iii) Three right angles total 270°, so the fourth must be 360° − 270° = 90°. The fourth angle is forced to be a right angle too, so a quadrilateral with exactly three right angles cannot exist; it would be a rectangle.
In simple words: Two triangles give 360°; 12x = 360° gives x = 30°; three right angles leave exactly 90° for the fourth.

Teacher's Note:
1. Explain why the six triangle angles equal the four quadrilateral angles.
2. The word “exactly” is the key in part (iii).

Question 3. ABCD is a parallelogram with ∠A = 55°.
(i) Define a parallelogram and find ∠B, ∠C and ∠D.
(ii) Taking one angle of a general parallelogram as x, justify that the opposite angles of every parallelogram are equal.
(iii) State what happens to the two diagonals of a parallelogram at their point of intersection, and state whether they must be equal in length.
Answer:
(i) A parallelogram is a quadrilateral in which the opposite sides are parallel. Adjacent angles add to 180° and opposite angles are equal:
∠B = 180° − 55° = 125°, ∠C = ∠A = 55°, ∠D = ∠B = 125°. Check: 55 + 125 + 55 + 125 = 360.
(ii) Take parallelogram PEAR with ∠P = x. ∠P and ∠R are interior angles on the same side of a transversal between parallel sides, so ∠R = 180° − x.
Again ∠A + ∠R = 180°, so ∠A = 180° − (180° − x) = x = ∠P.
In the same way ∠E = 180° − x = ∠R. So opposite angles of every parallelogram are equal.
(iii) The diagonals bisect each other (they cross at the midpoint of each). They need not be equal; they are equal only when the parallelogram is a rectangle.
In simple words: Angles go 55°, 125°, 55°, 125°; using x shows opposite angles are equal; diagonals bisect each other but need not be equal.

Teacher's Note:
1. Using a letter like x proves it for every parallelogram.
2. Equal diagonals happen only in rectangles.

Question 4. All four sides of a quadrilateral GAME are equal.
(i) Name this quadrilateral and explain, using alternate angles, why it must also be a parallelogram.
(ii) List the six properties of this quadrilateral given in the chapter.
(iii) If ∠G = 110°, find the remaining three angles and also the angle at which the diagonals cut each other.
Answer:
(i) It is a rhombus. Draw the diagonal AE. Since GE = GA and ME = MA, the two triangles formed are isosceles and congruent, so the four angles the diagonal makes with the sides are all equal.
The equal angles on opposite sides of the transversal AE are alternate angles, so EM ∥ GA; likewise GE ∥ AM. Both pairs of opposite sides are parallel, so it is a parallelogram.
(ii) (a) All sides are equal. (b) Opposite sides are parallel. (c) Adjacent angles add up to 180° and opposite angles are equal. (d) The diagonals bisect each other. (e) The diagonals bisect the angles. (f) The diagonals intersect at 90°.
(iii) ∠M = ∠G = 110° (opposite angles); ∠A = ∠E = 180° − 110° = 70° (adjacent angles). Sum: 110 + 70 + 110 + 70 = 360.
The diagonals cut each other at 90°.
In simple words: Equal sides make a rhombus, equal alternate angles make it a parallelogram, and its angles are 110°, 70°, 110°, 70° with diagonals at 90°.

Teacher's Note:
1. The diagonal AE creates the equal alternate angles.
2. The 90° between diagonals does not depend on ∠G.

Question 5. A carpenter wants to fix two flat strips of wood so that a thread passed through their four endpoints forms a square of diagonal 10 cm.
(i) State the length of each strip, the point at which they must be joined, and the angle between them.
(ii) Using the SSS condition of congruence on △BOA and △BOC, justify why the diagonals of a square must meet at 90°.
(iii) Find the angle that each diagonal of a square makes with a side of the square, and justify your answer.
Answer:
(i) Both strips must be 10 cm long (the diagonals of a square are equal), joined at the midpoint of each strip (the diagonals bisect each other), at an angle of 90°.
(ii) Let the square be ABCD with diagonals meeting at O. In △BOA and △BOC: BA = BC (sides of a square), OA = OC (diagonals bisect each other), BO is common.
By SSS, △BOA ≅ △BOC, so ∠BOA = ∠BOC. They form a straight angle: ∠BOA + ∠BOC = 180°, so each is 90°.
(iii) In △ADC, ∠ADC = 90° and AD = DC, so the angles opposite these sides are equal, say ∠1 each.
∠1 + ∠1 + 90° = 180°, so ∠1 = 45°. Each diagonal makes 45° with a side; that is, the diagonals bisect the angles of the square.
In simple words: Use two equal 10 cm strips crossed at their midpoints at 90°; congruent triangles give the right angle, and each diagonal makes 45° with a side.

Teacher's Note:
1. A square needs all three diagonal conditions.
2. The isosceles right triangle gives 45°.

Question 6. Rectangles, squares, rhombuses, kites, trapeziums and parallelograms form one family.
(i) Describe, in words or with a sketch, a Venn diagram that correctly places the sets of parallelograms, rectangles, rhombuses, squares and kites.
(ii) Which quadrilateral is both a kite and a parallelogram? Justify your answer.
(iii) Answer with justification: Is every kite a rhombus? Is every parallelogram a trapezium?
Answer:
(i) Draw a large closed curve for parallelograms. Inside it draw two overlapping curves for rectangles and rhombuses; their common region is the set of squares. Draw another large curve for kites that overlaps the parallelogram curve exactly in the rhombus region (which includes the squares).
So: square ⊂ rectangle ⊂ parallelogram, square ⊂ rhombus ⊂ parallelogram, and rhombus ⊂ kite.
(ii) A rhombus (and so also a square). A rhombus has all four sides equal, so it has two adjacent pairs of equal sides and is a kite; its opposite sides are parallel, so it is a parallelogram. No other kite is a parallelogram, since a kite with unequal pairs does not have equal opposite sides.
(iii) Not every kite is a rhombus: a kite needs only AB = BC and CD = DA, and the two pairs may differ, as in a kite with sides 5, 5, 9, 9 cm.
Every parallelogram is a trapezium, because a trapezium needs at least one pair of parallel opposite sides and a parallelogram has two.
In simple words: Squares sit inside rectangles and rhombuses, which sit inside parallelograms; rhombuses are where kites and parallelograms overlap.

Teacher's Note:
1. Inside a set means “is a special kind of”.
2. Use a counter-example for every “no”.

Question 7. Two identical cardboard triangles are joined along a pair of equal sides, edge to edge, to make a quadrilateral.
(i) Two isosceles triangles with sides 8 cm, 8 cm and 6 cm are joined along their 6 cm sides. Name the quadrilateral formed and justify your answer.
(ii) The same two triangles are instead joined along a pair of 8 cm sides, as mirror images. Name the quadrilateral formed and justify it.
(iii) Two scalene triangles with sides 6 cm, 9 cm and 12 cm are joined as mirror images along their 12 cm sides. Write the four sides of the quadrilateral formed and name it.
Answer:
(i) The 6 cm sides are hidden inside, so the four outer sides are all 8 cm. A quadrilateral with four equal sides is a rhombus. It is not a square, as its angles are not 90°.
(ii) Now an 8 cm side is hidden. Going round the figure, the sides are 8 cm, 6 cm, 6 cm, 8 cm. The two 8 cm sides are adjacent and so are the two 6 cm sides, so it is a kite. It is not a rhombus, since 8 ≠ 6.
(iii) The 12 cm sides are hidden. Going round, the sides are 6 cm, 9 cm, 9 cm, 6 cm. The equal sides come in adjacent pairs, so it is a kite.
In simple words: Hiding the 6 cm sides leaves four 8 cm sides (rhombus); mirror joins give adjacent equal pairs (kite).

Teacher's Note:
1. List the outer sides in order round the figure.
2. Adjacent equal pairs mean kite; all equal means rhombus.

Question 8. PQRS is a trapezium in which PQ ∥ SR, and UVWX is an isosceles trapezium in which UV ∥ XW and UX = VW.
(i) Define a trapezium and an isosceles trapezium.
(ii) In PQRS, ∠P = 125° and ∠Q = 95°. Find ∠S and ∠R, naming the property used, and check the angle sum.
(iii) By dropping perpendiculars XY and WZ from X and W to UV, justify that ∠U = ∠V in the isosceles trapezium UVWX.
Answer:
(i) A trapezium is a quadrilateral with at least one pair of parallel opposite sides. An isosceles trapezium is a trapezium whose two non-parallel sides are equal.
(ii) Property: interior angles on the same side of a transversal of parallel lines add up to 180°.
PS is a transversal of PQ ∥ SR: ∠S = 180° − 125° = 55°. QR is a transversal: ∠R = 180° − 95° = 85°.
Check: 125 + 95 + 85 + 55 = 360.
(iii) Drop XY ⟂ UV and WZ ⟂ UV. Since XW ∥ UV and ∠XYZ = ∠WZY = 90°, the angles at X and W in XWZY are also 90°, so XWZY is a rectangle and XY = WZ.
In △UXY and △VWZ: ∠UYX = ∠VZW = 90°, hypotenuse UX = VW (given) and XY = WZ.
By RHS, the triangles are congruent, so ∠U = ∠V.
In simple words: Co-interior angles give ∠S = 55° and ∠R = 85°; congruent right triangles at the ends give equal base angles.

Teacher's Note:
1. The rectangle in the middle gives equal heights.
2. RHS needs a right angle, hypotenuse and one side.

Question 9. CASE is a square and U, V, W, X are the midpoints of its four sides, as shown.
(i) Using congruent triangles, justify that UV = VW = WX = XU.
(ii) Using the angles of the corner triangles, justify that each angle of UVWX is 90°.
(iii) Hence name the quadrilateral UVWX, and state whether the same conclusion would follow if CASE were only a rectangle and not a square. Give a reason.
Answer:
(i) Let each side of the square be 2a. Since U, V, W, X are midpoints, each corner is cut off by a triangle with two legs a and a: for example △AUV has AU = AV = a and ∠A = 90°, and △CUX has CU = CX = a and ∠C = 90°.
All four corner triangles are congruent by SAS, so their third sides are equal: UV = VW = WX = XU.
(ii) In △AUV, AU = AV, so ∠AUV = ∠AVU; with ∠A = 90°, each is 45°. Similarly ∠CUX = 45°.
C, U, A lie on one side, so ∠CUX + ∠XUV + ∠AUV = 180°, giving ∠XUV = 180° − 45° − 45° = 90°. The same reasoning at V, W, X gives 90° each.
(iii) UVWX has four equal sides and four right angles, so it is a square.
If CASE were only a rectangle, each corner triangle would have legs l and b with l ≠ b. The four triangles would still be congruent, so the sides of UVWX would still be equal, but the base angles would not be 45°, so its angles would not be 90°. Then UVWX would be a rhombus, not a square.
In simple words: Four congruent corner triangles give equal sides, and 45° + 45° leaves 90° at each midpoint, so UVWX is a square (only a rhombus for a rectangle).

Teacher's Note:
1. Name the corner triangles by their vertices.
2. Equal legs are what make the angles 45°.

Question 10. Decide whether each statement below is true or false and justify your answer, giving a counter-example wherever the statement is false.
(i) A quadrilateral whose diagonals bisect each other must be a parallelogram.
(ii) A quadrilateral whose diagonals are perpendicular to each other must be a rhombus.
(iii) A quadrilateral in which all the angles are equal must be a rectangle.
(iv) Every isosceles trapezium is a parallelogram.
Answer:
(i) True. Let the diagonals of ABCD meet at O with OA = OC and OB = OD. In △AOB and △COD these pairs are equal and ∠AOB = ∠COD (vertically opposite), so the triangles are congruent by SAS.
Hence ∠OAB = ∠OCD; these are alternate angles, so AB ∥ DC. The other pair of triangles gives BC ∥ AD. So it is a parallelogram.
(ii) False. A kite also has perpendicular diagonals but its sides are not all equal. For example, the kite with AB = BC = 5 cm and CD = DA = 9 cm has BD ⟂ AC but is not a rhombus.
(iii) True. If each angle is x, then 4x = 360°, so x = 90°. A quadrilateral with every angle 90° is, by definition, a rectangle.
(iv) False. An isosceles trapezium has only one pair of parallel sides; its equal sides are the non-parallel ones. For example, a trapezium with angles 72°, 72°, 108°, 108° has equal slant sides that would meet if extended, so it is not a parallelogram.
In simple words: Bisecting diagonals and equal angles do force a parallelogram and a rectangle; perpendicular diagonals and an isosceles trapezium do not.

Teacher's Note:
1. A “true” needs a reason for every case.
2. A “false” needs only one counter-example.

Case Study Based Questions

Question 11. A carpenter’s way of making a rectangular photo frame. A carpenter in a small workshop makes rectangular photo frames. Before cutting the wood, she checks the shape in a clever way. She takes two flat strips, places one across the other, drives a single nail through the crossing point, and then stretches a thread through the four endpoints of the strips. If the thread forms a rectangle, she goes ahead and cuts the frame. She has already cut one strip of length 12 cm and now has to decide the length of the second strip and the place where the two strips must be nailed together. Carpenters in Europe and house builders in Mozambique are known to use exactly this method.
(i) What must be the length of the second strip?
(ii) At which point of the two strips should the nail be driven?
(iii) Must the angle between the two strips have one particular value for the thread to form a rectangle? Give a reason.
(iv) What one extra condition should the strips satisfy so that the thread forms a square?
Answer:
(i) 12 cm. The strips are the diagonals of the frame, and the diagonals of a rectangle are equal.
(ii) Through the midpoint of each strip, because the diagonals of a rectangle bisect each other.
(iii) No. Whatever the angle, if the strips are equal and bisect each other, every angle of the figure works out to 90°, so the thread always forms a rectangle.
(iv) The strips must also be at right angles to each other, since the diagonals of a square bisect each other at 90°.
In simple words: Use an equal 12 cm strip nailed at both midpoints at any angle for a rectangle, and at 90° for a square.

Teacher's Note:
1. The thread is the quadrilateral; the strips are its diagonals.
2. Only a square needs a fixed angle between diagonals.

Question 12. Measuring a glass pane in a window grill. A glazier is replacing a four-sided glass pane fitted into an old wooden window grill. The pane has four straight edges, so it is a quadrilateral. Because the frame is old, the glazier does not assume that the pane is a rectangle. He places his protractor at three of the corners, one after the other, and notes the readings 100°, 80° and 100°. He cannot reach the fourth corner, as it is hidden behind the hinge of the grill, so he decides to calculate that angle instead of measuring it.
(i) What is the sum of the four angles of the pane?
(ii) Find the measure of the hidden fourth angle.
(iii) Are the opposite angles of the pane equal? What kind of quadrilateral does this suggest?
(iv) Can the pane be a rectangle? Justify your answer.
Answer:
(i) 360°, as for every quadrilateral.
(ii) Hidden angle = 360° − (100° + 80° + 100°) = 360° − 280° = 80°.
(iii) Yes. The two 100° angles are opposite each other, and so are the two 80° angles. Equal opposite angles suggest a parallelogram.
(iv) No. A rectangle needs all four angles to be 90°, but this pane has angles of 100° and 80°.
In simple words: The hidden angle is 80°; equal opposite angles point to a parallelogram, which is not a rectangle.

Teacher's Note:
1. The corners were measured in order, so 100° and 100° are opposite.
2. Angles of 100° rule out a rectangle at once.

Question 13. A floor tiled with equal-sided tiles. The floor of a temple hall is covered with identical tiles. Every tile has four straight edges and a measurement shows that all four edges of a tile are 15 cm long. The tiles are not laid upright: each tile has two sharp corners and two blunt corners, and the sharp corner of a tile measures 60°. A child notices that the pattern of light and dark lines on the floor is formed by the two cross lines of each tile, that is, by its diagonals.
(i) Name the shape of one tile.
(ii) Find the other three angles of a tile.
(iii) Find the perimeter of one tile.
(iv) State the angle at which the two diagonals of a tile cut each other, and say whether a tile is a square.
Answer:
(i) A rhombus, since all four sides are equal (and the angles are not right angles).
(ii) Adjacent angles add to 180°, so the blunt angle = 180° − 60° = 120°. Opposite angles are equal, so the angles are 60°, 120°, 60°, 120° (sum 360°).
(iii) Perimeter = 4 × 15 = 60 cm.
(iv) The diagonals cut each other at 90°. The tile is not a square, because its angles are 60° and 120°, not 90°.
In simple words: Each tile is a rhombus with angles 60°, 120°, 60°, 120°, perimeter 60 cm and diagonals at 90°.

Teacher's Note:
1. Perpendicular diagonals do not make a square by themselves.
2. Perimeter of a rhombus = 4 × side.

Question 14. Making a paper kite with two sticks. For a kite festival, Meera makes a kite from two thin bamboo sticks and a sheet of coloured paper. One stick is 60 cm long and the other is 40 cm long. She lays the short stick across the long one so that the long stick passes through the midpoint of the short stick, and the two sticks are at right angles to each other. The crossing point is not the midpoint of the long stick. She then pastes the paper and cuts it along the four straight lines that join the endpoints of the sticks.
(i) Name the quadrilateral formed by the cut paper.
(ii) Which pairs of sides of this quadrilateral are equal in length?
(iii) Which of the two sticks is bisected by the other?
(iv) Is every kite a rhombus? Give a reason.
Answer:
(i) A kite.
(ii) Call the long stick BD and the short stick AC. The two sides from B are equal (AB = BC) and the two sides from D are equal (CD = DA): two adjacent pairs of equal sides.
(iii) The short stick (40 cm) is bisected by the long one, which passes through its midpoint at right angles. The long stick is not bisected.
(iv) No. A rhombus needs all four sides equal, but here the pair AB = BC differs from the pair CD = DA. Every rhombus is a kite, but a kite need not be a rhombus.
In simple words: The paper is a kite: the long stick cuts the short one in half at 90°, and its two pairs of equal sides are different.

Teacher's Note:
1. The long stick is the axis of symmetry.
2. If the long stick were also bisected, it would be a rhombus.

Question 15. A farmer’s field with two parallel boundaries. A farmer owns a four-sided field ABCD. The boundary AB along the village road is 40 m long and the opposite boundary DC, along a canal, is 25 m long. The surveyor’s sketch shows that these two boundaries never meet however far they are extended, that is, AB ∥ DC. The other two boundaries AD and BC slope inwards. Standing at the corner A, the surveyor measures the angle between the road and the boundary AD as 75°, and at the corner B the angle between the road and the boundary BC as 65°.
(i) Which pair of sides of the field is parallel, and what is the field’s shape called?
(ii) Find ∠D, stating the property of parallel lines you used.
(iii) Find ∠C and verify that the four angles add up to 360°.
(iv) Is this field an isosceles trapezium? Justify your answer.
Answer:
(i) AB ∥ DC. A quadrilateral with at least one pair of parallel opposite sides is a trapezium.
(ii) AD is a transversal of the parallel lines AB and DC, and interior angles on the same side add up to 180°. So ∠D = 180° − 75° = 105°.
(iii) Similarly ∠B + ∠C = 180°, so ∠C = 180° − 65° = 115°. Check: 75 + 65 + 115 + 105 = 360.
(iv) No. In an isosceles trapezium the angles at the two ends of a parallel side are equal, so we would need ∠A = ∠B. Here 75° ≠ 65°, so AD ≠ BC.
In simple words: The field is a trapezium with ∠D = 105° and ∠C = 115°; unequal base angles show it is not isosceles.

Teacher's Note:
1. Each slanting boundary is a transversal.
2. Equal base angles are the test for an isosceles trapezium.

Question 16. Braces behind a cricket sightscreen. A cricket ground has a large sightscreen ABCD whose front face is a rectangle. To stop the frame from twisting in the wind, two straight metal braces are bolted behind it along the two diagonals AC and BD. The braces are bolted to each other at the point O where they cross. The ground staff find that each brace is 7.2 m long, and that one of the angles between the two braces at O measures 70°.
(i) What is the measure of each angle of the rectangular face?
(ii) Explain why the two braces must be of equal length.
(iii) Find the distance from O to each corner of the screen.
(iv) If ∠AOB = 70°, find ∠OAB and ∠OBA.
Answer:
(i) Each angle is 90°.
(ii) The braces lie along the diagonals, and the diagonals of a rectangle are always equal (from △ADC ≅ △DAB, which gives AC = BD).
(iii) The diagonals bisect each other, so O is the midpoint of each brace. Distance = \( \frac{7.2}{2} \) = 3.6 m.
(iv) In △AOB, OA = OB = 3.6 m, so the base angles are equal: ∠OAB = ∠OBA = \( \frac{180^\circ - 70^\circ}{2} \) = 55°.
In simple words: The braces are equal diagonals meeting at their midpoints, 3.6 m from each corner, and the base angles are 55° each.

Teacher's Note:
1. Equal half-diagonals make △AOB isosceles.
2. Subtract the top angle, then halve.

Question 17. A folding safety gate made of equal rods. The entrance of a shop is closed by a folding safety gate. The gate is made of flat metal rods, all of the same length 30 cm, which are pinned together so that they form a long row of identical four-sided openings. When the gate is pulled, the openings become tall and narrow; when it is pushed back, they become short and wide. Through all this, the rods neither bend nor change length. At one stage of the pulling, the shopkeeper notices that the sharp angle of an opening measures 50°.
(i) Name the shape of one opening, giving a reason.
(ii) Find the other three angles of that opening when the sharp angle is 50°.
(iii) As the gate is pushed and pulled, which facts about an opening do not change?
(iv) At one particular position every opening becomes a square. What must the sharp angle be then? Justify.
Answer:
(i) A rhombus, because its four sides are rods of the same length, 30 cm.
(ii) Adjacent angles add to 180° and opposite angles are equal, so the angles are 50°, 130°, 50°, 130° (sum 360°).
(iii) Unchanged: all sides stay 30 cm; it stays a rhombus; opposite angles stay equal; adjacent angles add to 180°; the angle sum stays 360°; the diagonals always bisect each other at 90°. Only the angle sizes and the diagonal lengths change.
(iv) The sharp angle must be 90°. A quadrilateral with four equal sides and one angle of 90° has all angles 90°, so it is a square.
In simple words: Each opening is a rhombus (50°, 130°, 50°, 130°) that keeps its side length and becomes a square at 90°.

Teacher's Note:
1. A rhombus can change shape without changing its sides.
2. At 90° the “sharp” and “blunt” angles become equal.

Question 18. Slanting tiles on a wall. An interior designer covers a wall with identical tiles. Each tile is a four-sided slab whose opposite edges are parallel, with edges of lengths 12 cm and 7 cm. One of the angles of a tile measures 65°. The tiles are laid in slanting rows so that the wall looks like a woven mat. While ordering the tiles, the shop assistant writes “rhombus tiles” on the bill, and the designer asks him to correct it.
(i) Name the shape of a tile and find all four of its angles.
(ii) Find the perimeter of one tile.
(iii) The diagonals of a tile meet at O. What can be said about the two parts into which each diagonal is divided? Must the two diagonals be equal?
(iv) Was the shop assistant right in calling the tiles rhombuses? Justify.
Answer:
(i) A parallelogram, since its opposite edges are parallel. Adjacent angles add to 180° and opposite angles are equal, so the angles are 65°, 115°, 65°, 115°.
(ii) Perimeter = 2 × (12 + 7) = 38 cm.
(iii) The diagonals bisect each other, so each diagonal is cut into two equal parts at O. They need not be equal; that happens only in a rectangle, which this tile is not.
(iv) No. A rhombus must have all four sides equal, but this tile has sides of 12 cm and 7 cm. It is a parallelogram that is neither a rhombus nor a rectangle.
In simple words: The tile is a parallelogram (65°, 115°, 65°, 115°, perimeter 38 cm), not a rhombus, since its sides are unequal.

Teacher's Note:
1. Parallel opposite sides define a parallelogram.
2. A rhombus needs four equal sides.

Teacher's Note:
1. Match each card with a definition first.
2. A kite is a parallelogram only when it is a rhombus.

Question 20. A surveyor names a plot from its diagonals. A surveyor has to name the shape of a four-sided plot KLMN without measuring its sides. She walks along the two straight paths KM and LN that cut across the plot, that is, along its diagonals, and marks the point O where they cross. Her field book records KO = OM = 21 m, LO = ON = 14 m, and the angle between the two paths at O is a right angle.
(i) What do the recorded lengths tell you about the diagonals of the plot?
(ii) Are the two diagonals equal in length? Show the working.
(iii) Name the shape of the plot and justify your answer from the diagonal properties alone.
(iv) Keeping KM unchanged, what should the length of LN be made so that the plot becomes a square? Justify.
Answer:
(i) Since KO = OM and LO = ON, O is the midpoint of both diagonals, so the diagonals bisect each other; they also cross at 90°, so they are perpendicular bisectors of each other.
(ii) KM = 21 + 21 = 42 m and LN = 14 + 14 = 28 m. Since 42 ≠ 28, the diagonals are not equal.
(iii) The plot is a rhombus. The diagonals bisect each other at right angles, so the four triangles KOL, LOM, MON, NOK are congruent (SAS) and KL = LM = MN = NK. It is not a square, because the diagonals are unequal.
(iv) LN should be 42 m (LO = ON = 21 m). Then the diagonals are equal and bisect each other, making a rectangle, and since they also cross at 90°, all sides are equal: a square.
In simple words: Perpendicular bisecting diagonals of 42 m and 28 m give a rhombus; making LN 42 m too gives a square.

Teacher's Note:
1. Diagonals alone can identify the shape.
2. Equal + perpendicular + bisecting diagonals = square.