Figure it Out — Kites, Trapeziums and Relationships Between Quadrilaterals — Quadrilaterals

1. 13 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 107

   Find all the sides and the angles of the quadrilateral obtained by joining two equilateral triangles with sides 4 cm.

   Hint. Join the two triangles along a full side. Which angles get combined, and which stay as they are?

   Step 1 — Make the join. Place the two equilateral triangles so that one full 4 cm side of each is glued to the other. Call the result ABCD, with the join running along the diagonal.

   Step 2 — Find the sides. The joined side becomes an interior diagonal, not a side of the quadrilateral. The four remaining sides are the other two sides of each triangle, and every side of an equilateral triangle of side 4 cm measures 4 cm. So all four sides are 4 cm.

   Step 3 — Find the angles. Every angle of an equilateral triangle is 60°. • At the two vertices that were not on the join, the triangle's angle is the whole angle of the quadrilateral: 60° each. • At the two vertices on the join, one angle from each triangle meets, so they add: 60° + 60° = 120° each.

   The angles are therefore 60°, 120°, 60°, 120°. Check: 60 + 120 + 60 + 120 = 360° ✓

   Step 4 — Name the shape. All four sides are equal, so by definition this is a rhombus (and not a square, since its angles are not 90°).

   ✦ All four sides are 4 cm and the angles are 60°, 120°, 60°, 120° — the shape is a rhombus.

2. 23 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 107

   Construct a kite whose diagonals are of lengths 6 cm and 8 cm.

   Hint. In a kite one diagonal bisects the other at right angles — but unlike a rhombus, only one of them is bisected.

   Step 1 — Recall the kite's diagonal property. In a kite ABCD with AB = BC and CD = DA, the chapter shows that the diagonal BD bisects the other diagonal AC and is perpendicular to it. Note the asymmetry: BD cuts AC in half, but AC does not cut BD in half. That asymmetry is exactly what distinguishes a kite from a rhombus.

   Step 2 — Construction.

   1. Draw PQ = 6 cm — this is the diagonal that will be bisected.
   2. Construct the perpendicular bisector of PQ, meeting it at T (so PT = TQ = 3 cm).
   3. On that perpendicular, mark R on one side of PQ and S on the other, so that the total length RS = 8 cm. Do not make TR = TS — choose, say, TR = 3 cm and TS = 5 cm.
   4. Join PR, RQ, QS and SP.

   PRQS is the required kite.

   Step 3 — Check it is a kite. Triangles PTR and QTR are congruent by SAS (PT = TQ, right angle at T, TR common), so PR = RQ. The same argument below the line gives PS = QS. So the quadrilateral has two adjacent pairs of equal sides — a kite.

   Note there are many correct answers here, since only the total RS = 8 cm is fixed and the split either side of T is free. If you happened to choose TR = TS = 4 cm you would get a rhombus, which is a kite too but the special symmetric one — pick an unequal split to draw a typical kite.

   ✦ Draw the 6 cm diagonal, construct its perpendicular bisector, and mark the ends of the 8 cm diagonal on that perpendicular with an unequal split either side (e.g. 3 cm and 5 cm), then join the four endpoints.

3. 33 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 107

   Find the remaining angles in the following trapeziums. (i) Trapezium PQRS with PQ ∥ SR, ∠P = 135° and ∠Q = 105°. (ii) Trapezium ABCD with AB ∥ DC and AD = BC, given ∠D = 100°.

   Hint. A pair of parallel sides means the angles on each slanted side are co-interior, so they add to 180°.

   The governing fact: when two sides are parallel, the two angles at the ends of either non-parallel side are interior angles on the same side of a transversal, so they add up to 180°.

   (i) Trapezium PQRS, PQ ∥ SR The side PS is a transversal cutting the parallel pair, so ∠P and ∠S are co-interior: ∠S = 180° − ∠P = 180° − 135° = 45° Similarly QR is a transversal, so ∠Q and ∠R are co-interior: ∠R = 180° − ∠Q = 180° − 105° = 75° Check: 135 + 105 + 75 + 45 = 360° ✓

   (ii) Isosceles trapezium ABCD, AB ∥ DC with AD = BC AD is a transversal, so ∠A and ∠D are co-interior: ∠A = 180° − ∠D = 180° − 100° = 80°

   Now use the extra condition. Because the two non-parallel sides are equal, this is an isosceles trapezium, and the chapter proves that in one the angles opposite the equal sides are equal. Dropping perpendiculars XY and WZ from the shorter parallel side onto the longer one creates a rectangle in the middle and two congruent right triangles at the ends, which is what forces the base angles to match. Hence ∠B = ∠A = 80° and then ∠C = 180° − ∠B = 100° (or equally, ∠C = ∠D by the same symmetry). Check: 80 + 80 + 100 + 100 = 360° ✓

   ✦ (i) ∠R = 75° and ∠S = 45°. (ii) ∠A = 80°, ∠B = 80° and ∠C = 100°.

4. 44 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 107

   Draw a Venn diagram showing the set of parallelograms, kites, rhombuses, rectangles and squares. Then answer: (i) What is the quadrilateral that is both a kite and a parallelogram? (ii) Can there be a quadrilateral that is both a kite and a rectangle? (iii) Is every kite a rhombus? If not, what is the correct relationship between these two types of quadrilaterals?

   Hint. Work from the definitions. A kite needs two adjacent pairs of equal sides; check which of the other shapes satisfy that.

   The Venn diagram. Draw a large oval for parallelograms and a separate, overlapping oval for kites. • Inside parallelograms, draw two overlapping ovals: rectangles and rhombuses. • Squares sit exactly in the overlap of rectangles and rhombuses. • The rhombus oval lies inside the kite oval as well — this is where the parallelogram and kite ovals overlap.

   So the nesting is: square ⊂ rhombus ⊂ parallelogram, square ⊂ rectangle ⊂ parallelogram, and rhombus ⊂ kite.

   (i) Both a kite and a parallelogram. A kite has two adjacent pairs of equal sides; a parallelogram has both pairs of opposite sides equal. Satisfying both forces all four sides to be equal — which is a rhombus (and the square, being a special rhombus). Answer: the rhombus, including its special case the square.

   (ii) Both a kite and a rectangle. A rectangle has all angles 90°. For it also to be a kite it needs two adjacent sides equal, which forces all four sides equal — that is a square. Answer: yes — the square is both a kite and a rectangle.

   A discrepancy in the printed answer key, worth knowing about. The book's key answers 'No' here. That answer is only defensible if 'kite' is defined to exclude shapes with all four sides equal, which some textbooks do. But Ganita Prakash does not define it that way — it says a kite is a quadrilateral labelled ABCD with AB = BC and CD = DA, which a square satisfies — and the key's own answer to part (i) counts the square as a kite, as does its own printed Venn diagram. So the key contradicts itself between (i) and (ii). Under the definition the book actually gives, the answer is yes, the square. If your school follows the printed key, write 'No' and add the one-line reason 'if kites are taken to exclude rhombuses'.

   (iii) Is every kite a rhombus? No. A kite only requires two adjacent pairs of sides to be equal, and those two pairs may have different lengths — for example a kite with sides 6 cm, 6 cm, 9 cm, 9 cm is not a rhombus. A rhombus requires all four sides equal. The correct relationship is the other way round: every rhombus is a kite, but not every kite is a rhombus. In the Venn diagram the rhombus oval lies wholly inside the kite oval.

   ✦ (i) The rhombus (and the square as its special case). (ii) Yes — the square, under the book's own definition of a kite; note the printed key says 'No', which conflicts with its own answer to (i). (iii) No — every rhombus is a kite, but a kite need not be a rhombus.

5. 52 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 108

   If PAIR and RODS are two rectangles, find ∠IOD.

   Hint. Both figures are rectangles, so every one of their own angles is 90°; the 30° travels through the shared vertex R.

   Answer from the book's key: ∠IOD = 30°.

   An honest note before the reasoning. This question depends entirely on how the two rectangles are placed relative to each other in the printed figure — which vertex is shared, which sides are marked 5 cm, and where the 30° sits. That placement cannot be recovered from the text alone, so the answer below is cited from the book's own answer key rather than re-derived here. Open your book to page 108 and read the configuration off the figure before writing the steps.

   The method the question is testing. The two facts you will need are the ones this chapter has been building:

   1. Every angle of a rectangle is 90°, so at the shared vertex the angles of the two rectangles partition a known total.
   2. The angles of any triangle sum to 180°, and the angles round a point sum to 360° — these let the given 30° be carried from one rectangle across to the other.

   The two 5 cm marks tell you a pair of sides are equal, which typically makes a triangle in the figure isosceles, so its base angles are equal — that is usually the step that transfers the 30°.

   Working it through on the printed figure gives ∠IOD = 30°, the same as the given angle.

   ✦ ∠IOD = 30° (the book's answer key; the derivation depends on the printed figure on page 108, which is not reproducible from text).

6. 63 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 108

   Construct a square with diagonal 6 cm without using a protractor.

   Hint. The whole point is that the 90° must be built with compass and ruler, not measured.

   Step 1 — Recall what decides a square. The chapter proves that the diagonals of a square are equal, bisect each other, and meet at 90°. Conversely, drawing two such diagonals and joining the endpoints must give a square, since the endpoints of the diagonals fix the vertices.

   Step 2 — Construction, compass and ruler only.

   1. Draw AB = 6 cm.
   2. Construct the perpendicular bisector of AB using a compass: with centre A and radius more than half of AB, draw arcs above and below the line; repeat with centre B and the same radius; join the two crossing points. This line is perpendicular to AB and passes through its midpoint O.
   3. On this perpendicular, cut off OC = OD = 3 cm on either side of O.
   4. Join AC, CB, BD and DA.

   ACBD is the required square.

   Step 3 — Why no protractor is needed. The perpendicular bisector construction produces an exact right angle from the congruence of the arcs, not from a measurement, so the 90° between the diagonals is guaranteed. It also places O exactly at the midpoint of AB, giving the bisection for free, and cutting equal 3 cm lengths makes CD = 6 cm = AB.

   Step 4 — Check your drawing. Each side should measure √(3² + 3²) = √18 ≈ 4.2 cm, and all four must agree.

   ✦ Draw the 6 cm diagonal, construct its perpendicular bisector with a compass, cut off 3 cm on each side of the midpoint for the second diagonal, and join the endpoints; each side should come out to about 4.2 cm.

7. 74 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 108

   CASE is a square. The points U, V, W and X are the midpoints of the sides of the square. What type of quadrilateral is UVWX? Find this by using geometric reasoning, as well as by construction and measurement.

   Hint. Each corner of the big square is cut off by a right-angled triangle with two equal legs.

   Claim: UVWX is a square.

   Step 1 — Set up the lengths. Let each side of square CASE be 2x, so each half-side is x. Because U, V, W and X are midpoints, every one of the four corner triangles has its two legs equal to x.

   Step 2 — Show all four sides of UVWX are equal. In the right-angled triangle CUV, the right angle is at C (a corner of the square), so by the Pythagoras relation: UV² = CU² + CV² = x² + x² = 2x² UV = x√2. The same computation at each of the other three corners gives VW = WX = XU = x√2. So all four sides of UVWX are equal.

   Step 3 — Show all four angles are 90°. Triangle CUV has a right angle at C and CU = CV, so it is a right-angled isosceles triangle and its base angles are equal: ∠CUV = ∠CVU = (180° − 90°) ÷ 2 = 45°. The same holds at every corner, so the triangle on the other side of U also contributes 45°. Now look along the side of the big square at the point U. Three angles sit on that straight line: the 45° from one corner triangle, the angle of UVWX at U, and the 45° from the next corner triangle. Since angles on a straight line add to 180°: ∠XUV = 180° − 45° − 45° = 90°. The same argument at V, W and X gives 90° each.

   Step 4 — Conclude. All four sides equal and all four angles 90° — by definition, UVWX is a square. Its side is x√2, so its area is 2x², exactly half the area 4x² of the original square, a fact worth remembering.

   By construction: draw a square of side 6 cm, mark the midpoints, join them, and measure — every side should come out at 3√2 ≈ 4.24 cm and every angle at 90°.

   Other inner squares. The midpoints are not the only choice. Mark a point on each side at the same distance from the corner going round in the same rotational direction — say 2 cm clockwise from each corner. The four corner triangles are still congruent (by SAS), so the inner quadrilateral is still a square, just tilted differently. The midpoint case is the one where the inner square is smallest.

   ✦ UVWX is a square, with side x√2 where the outer square has side 2x — so it has exactly half the area of CASE.

8. 83 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 108

   If a quadrilateral has four equal sides and one angle of 90°, will it be a square? Find the answer using geometric reasoning as well as by construction and measurement.

   Hint. Four equal sides already names the shape. What do its angle properties then force?

   Yes, it will be a square.

   Step 1 — Name what four equal sides gives you. A quadrilateral in which all four sides have the same length is, by definition, a rhombus.

   Step 2 — Use the angle properties of a rhombus. The chapter establishes that a rhombus is a parallelogram, and therefore • its opposite angles are equal, and • its adjacent angles add up to 180°.

   Step 3 — Propagate the given right angle. Let the given angle be ∠A = 90°. • Its opposite angle: ∠C = ∠A = 90°. • Its adjacent angle: ∠B = 180° − 90° = 90°. • And ∠D = ∠B = 90°.

   So all four angles are forced to 90° — a single right angle is enough, because the parallelogram relations leave no freedom once one angle is fixed.

   Step 4 — Conclude. The quadrilateral now has all four sides equal and all four angles 90°, which is exactly the definition of a square.

   By construction: draw a 5 cm segment, erect a 90° angle at one end, mark 5 cm along it, then cut arcs of radius 5 cm from both free ends to locate the fourth vertex. Measuring the remaining three angles gives 90° each — you cannot force it to come out otherwise.

   Contrast worth noting: four equal sides alone does not give a square (any rhombus is a counter-example), and one right angle alone certainly does not. It is the combination that closes the case.

   ✦ Yes. Four equal sides make it a rhombus, and in a rhombus the opposite angles are equal while adjacent angles add to 180°, so one 90° angle forces all four to be 90° — giving a square.

9. 93 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 108

   What type of a quadrilateral is one in which the opposite sides are equal? Justify your answer. (Hint: Draw a diagonal and check for congruent triangles.)

   Hint. Equal sides alone give congruent triangles; alternate angles then convert that into parallelism.

   It is a parallelogram.

   Step 1 — Set up. Let ABCD be a quadrilateral with AB = CD and BC = AD. Draw the diagonal AC, splitting it into triangles ABC and CDA.

   Step 2 — Prove the triangles congruent. In △ABC and △CDA: • AB = CD (given) • BC = DA (given) • AC = CA (common side) So △ABC ≅ △CDA by the SSS condition.

   Step 3 — Read off the equal angles. Corresponding parts of congruent triangles are equal, so writing ∠1 = ∠BAC, ∠2 = ∠BCA, ∠3 = ∠DAC, ∠4 = ∠DCA: ∠1 = ∠4 and ∠2 = ∠3.

   Step 4 — Convert equal angles into parallel sides. ∠1 and ∠4 are alternate angles for the lines AB and CD with the transversal AC. Since alternate angles being equal is exactly the condition for two lines to be parallel, AB ∥ CD. Likewise ∠2 and ∠3 are alternate angles for BC and AD with the same transversal, so BC ∥ AD.

   Step 5 — Conclude. Both pairs of opposite sides are parallel, which is the definition of a parallelogram. Hence ABCD is a parallelogram.

   Why this result matters. It is the converse of the chapter's Deduction 7 (a parallelogram has equal opposite sides). Having both directions means 'opposite sides equal' and 'opposite sides parallel' pick out exactly the same class of quadrilaterals, so either may be used as the definition — which is what makes the constructions in questions 2 and 3 valid.

   ✦ It is a parallelogram: drawing a diagonal gives two triangles congruent by SSS, and the resulting equal alternate angles force both pairs of opposite sides to be parallel.

10. 103 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 108

    Will the sum of the angles in a quadrilateral such as the concave one shown (with a 'dent' at one vertex) also be 360°? Find the answer using geometric reasoning as well as by constructing this figure and measuring.

    Hint. The proof for ordinary quadrilaterals splits the shape into two triangles. Can that still be done here?

    Yes — the sum is still 360°.

    Step 1 — Recall why it is 360° for an ordinary quadrilateral. The chapter's proof draws one diagonal, splitting the quadrilateral into two triangles. Each triangle contributes 180°, and together their angles make up exactly the four angles of the quadrilateral, giving 180° + 180° = 360°.

    Step 2 — Check whether the same split works on a concave figure. In a concave quadrilateral ADCB, one interior angle is a reflex angle (greater than 180°) — the 'dent'. The care needed is in choosing which diagonal to draw. Draw the diagonal BD, the one that lies inside the figure. It splits the shape into triangles ADB and BDC.

    Step 3 — Add up. Sum of angles of △ADB = 180° Sum of angles of △BDC = 180° Adding: 360°.

    And these six angles reassemble exactly into the four interior angles of the quadrilateral — at the two vertices on the diagonal, the two triangle angles combine into one quadrilateral angle, and at the other two vertices the triangle angle is the quadrilateral angle unchanged. So the four interior angles total 360°.

    Step 4 — The point of the question. The result does not depend on the quadrilateral being convex, since the argument only needs the figure to be splittable into two triangles by an interior diagonal, and every quadrilateral is. What does change is that one of the four angles must be measured as a reflex angle — if you measure the 'outside' angle at the dent instead, your total will come out wrong.

    By measurement: construct the figure and measure carefully, remembering to take the reflex angle at the dented vertex; the four readings will total 360°.

    ✦ Yes, 360°. Drawing the interior diagonal BD splits even a concave quadrilateral into two triangles contributing 180° each — provided the reflex angle at the dent is the one measured.

11. 117 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 108-109

    State whether the following statements are true or false, and justify your answers. (i) A quadrilateral whose diagonals are equal and bisect each other must be a square. (ii) A quadrilateral having three right angles must be a rectangle. (iii) A quadrilateral whose diagonals bisect each other must be a parallelogram. (iv) A quadrilateral whose diagonals are perpendicular to each other must be a rhombus. (v) A quadrilateral in which the opposite angles are equal must be a parallelogram. (vi) A quadrilateral in which all the angles are equal is a rectangle. (vii) Isosceles trapeziums are parallelograms.

    Hint. For each false statement, find one shape that satisfies the condition but is not the named shape.

    (i) False. Equal diagonals that bisect each other force all four angles to 90°, so the shape is certainly a rectangle — but not necessarily a square. To get a square the diagonals must also be perpendicular to each other, and nothing here says they are. Counter-example: a 6 cm by 3 cm rectangle has equal diagonals bisecting each other and is not a square.

    (ii) True. The angles of any quadrilateral sum to 360°. If three of them are 90°, the fourth is 360° − (90° + 90° + 90°) = 90°. So all four are right angles, and a quadrilateral with all angles 90° is by definition a rectangle. This is why you cannot draw a quadrilateral with exactly three right angles.

    (iii) True. Let the diagonals of ABCD meet at O with AO = OC and BO = OD. In triangles AOB and COD: AO = CO, BO = DO, and ∠AOB = ∠COD as vertically opposite angles. So the triangles are congruent by SAS, giving AB = CD. The same argument on the other pair gives AD = CB. A quadrilateral with both pairs of opposite sides equal is a parallelogram (question 9), so the statement holds.

    (iv) False. Perpendicular diagonals are not enough — a kite has perpendicular diagonals but its four sides are not all equal. Counter-example: a kite with sides 6 cm, 6 cm, 9 cm, 9 cm. What a rhombus additionally requires is that the diagonals bisect each other; in a kite only one diagonal bisects the other.

    (v) True. Let ∠A = ∠C = x and ∠B = ∠D = y. The angle sum gives 2x + 2y = 360°, so x + y = 180°. That means each pair of adjacent angles is supplementary, which is precisely the co-interior-angle condition for the opposite sides to be parallel. So both pairs of opposite sides are parallel and the shape is a parallelogram.

    (vi) True. If all four angles are equal, each must be 360° ÷ 4 = 90°. A quadrilateral with all angles 90° is a rectangle by definition. (Note the contrast with sides: a quadrilateral with all sides equal is a rhombus, not necessarily a square.)

    (vii) False. An isosceles trapezium has only one pair of parallel sides; the other pair is equal in length but not parallel. A parallelogram needs both pairs parallel. Indeed if the second pair were also parallel the figure would be a parallelogram and the base angles would no longer be equal unless it were a rectangle.

    ✦ (i) False — it is a rectangle; a square also needs perpendicular diagonals. (ii) True — the fourth angle is forced to 90°. (iii) True — SAS congruence makes opposite sides equal. (iv) False — a kite also has perpendicular diagonals. (v) True — equal opposite angles force adjacent angles to be supplementary. (vi) True — each angle must be 90°. (vii) False — only one pair of sides is parallel.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 1, Reprint 2026-27 (hegp104.pdf). The chapter builds every property by deduction (congruence and transversal arguments) rather than assertion, and its three 'Figure it Out' blocks sit on pages 94, 102 and 107-109. Every angle answer here was independently recomputed from the stated configuration and then checked against the book's own printed answer key. TWO DEVIATIONS ARE FLAGGED IN PLACE: (1) the key's answer to the Venn-diagram question 4(ii) contradicts its own answer to 4(i) and its own printed diagram — the mathematically correct answer under the book's stated definition of a kite is given, with the discrepancy explained; (2) question 5 (∠IOD in rectangles PAIR and RODS) has an answer in the key but its derivation depends on the exact printed figure, which cannot be recovered from the text, so the answer is cited rather than derived and this is stated openly in the solution.. Questions are referenced from the NCERT textbook for identification.