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NCERT Solutions Ganita Prakash (Part 1) Chapter 4 –1094.6 Kite and Trapezium — Figure it Out

Book page 107 Updated on2026-09-05

Q1.
Find all the sides and the angles of the quadrilateral obtained by joining two equilateral triangles with sides 4 cm.
Answer

All four sides are 4 cm; the angles are 60°, 120°, 60° and 120°. The figure is a rhombus.

Join the two cutouts along a full side. That side becomes a diagonal of length 4 cm, and the four remaining edges become the sides.

Sides: 4 cm, 4 cm, 4 cm, 4 cm ⇒ a rhombus

At the two ends of the join, two 60° angles come together:
60° + 60° = 120°  (twice)
At the two far vertices, a single 60° angle stands alone: 60°  (twice)

Check: 60 + 120 + 60 + 120 = 360° ✓
Why it is not a square: a square needs 90° angles. Here the angles are decided by the equilateral triangles, which only ever offer 60°. Four equal sides make it a rhombus, but the angles keep it from being a square.
Tip: the shorter diagonal is 4 cm (the join). The longer one measures 2 × (4 × √3 ∕ 2) = 4√3 ≈ 6.93 cm, since each equilateral triangle has height (√3∕2) × 4.
Q2.
Construct a kite whose diagonals are of lengths 6 cm and 8 cm.
Answer

Use the kite's diagonal property proved on page 105: one diagonal is the perpendicular bisector of the other.

  1. Draw PQ = 6 cm — this will be the diagonal that gets bisected.
  2. Construct the perpendicular bisector of PQ with a compass; let it meet PQ at T, so PT = TQ = 3 cm.
  3. On the perpendicular, mark R and S on opposite sides of PQ with RS = 8 cm — but not with T as the midpoint. For instance take TR = 2 cm and TS = 6 cm.
  4. Join PR, RQ, QS and SP.

PRQS is the required kite.

PR = QR = √(3² + 2²) = √13 ≈ 3.6 cm
PS = QS = √(3² + 6²) = √45 ≈ 6.7 cm
Two adjacent pairs of equal sides ⇒ a kite ✓
Why R and S must be placed unequally: if you took TR = TS = 4 cm, the second diagonal would also be bisected and you would get a rhombus — a special kite. Making TR ≠ TS keeps the two pairs of sides different, giving the ordinary kite shape. Many different kites have diagonals 6 cm and 8 cm; each choice of TR gives another one.
Q3.
Find the remaining angles in the following trapeziums —
Answer

Trapezium (i). The top and bottom sides carry arrowheads, so they are parallel. The two marked angles, 135° and 105°, are at the ends of the shorter (bottom) side. Each slanting side is a transversal, so the angles at its two ends add to 180°.

Above the 135° corner:  180 – 135 = 45°
Above the 105° corner:  180 – 105 = 75°
Check: 135 + 105 + 45 + 75 = 360° ✓

Trapezium (ii). Here the two slanting sides carry tick marks, so they are equal — this is an isosceles trapezium. The 100° angle lies between one equal side and the shorter parallel side.

Other end of that same leg:  180 – 100 = 80°
Property 2: the angles on each parallel side are equal, so
the other angle on the short parallel side = 100°
the other angle on the long parallel side = 80°
Check: 100 + 100 + 80 + 80 = 360° ✓
45° 75° 135° 105° (i) 100° 100° 80° 80° (ii)
Given angles in red/blue, computed angles in green. Both figures total 360°.
Why (ii) needs the tick marks: in an ordinary trapezium the two given angles could be anything, and 100° alone would only fix the angle at the other end of that leg. The equal legs supply the second relation — Property 2 — which is what lets a single measurement determine all four angles.
Q4.
Draw a Venn diagram showing the set of parallelograms, kites, rhombuses, rectangles, and squares. Then, answer the following questions — (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?
Answer
Parallelograms Kites Rectangles Rhombuses Squares
Rectangles and rhombuses both sit inside parallelograms; their overlap is the squares. The rhombuses also sit entirely inside the kites, which spill outside the parallelograms.

(i) The rhombus. A quadrilateral that is both a kite and a parallelogram must have two adjacent pairs of equal sides (kite) and equal opposite sides (parallelogram) — and that forces all four sides to be equal.

Kite: AB = BC and CD = DA
Parallelogram: AB = CD and BC = DA
Together: AB = BC = CD = DA ⇒ a rhombus  (the square is the special case)

(ii) Yes — the square. A square has AB = BC and CD = DA, so it meets the book's definition of a kite (page 105), and all its angles are 90°, so it is a rectangle.

A note on the answer key: the answers printed at the end of the chapter give “No” here. That would only be right if a kite were required to have its two pairs of different lengths. But part (i) of this very question expects a rhombus to count as a kite, so the same reading must let a square count as a kite — and a square is certainly a rectangle. The consistent answer is yes, the square.

(iii) No. Every rhombus is a kite, but not every kite is a rhombus.

Rhombus: AB = BC = CD = DA  ⇒ certainly AB = BC and CD = DA ⇒ a kite ✓
Kite with AB = BC = 6 cm and CD = DA = 9 cm ⇒ not a rhombus ✗

So the correct relationship is rhombuses ⊂ kites — the set of rhombuses lies wholly inside the set of kites.

Q5.
If PAIR and RODS are two rectangles, find ∠IOD.
Answer

∠IOD = 30°.

In the figure, PAIR is a rectangle with R at the bottom-left and I at the bottom-right, O lies on the side AI, and RODS is a second rectangle hanging off RO. The marked 30° is ∠ORI, the angle the segment RO makes with the side RI.

In ∆ORI:  ∠RIO = 90°  (angle of rectangle PAIR at I, since O lies on AI)
∠IRO = 30°  (given)
⇒ ∠ROI = 180 – 90 – 30 = 60°

RODS is a rectangle, so ∠ROD = 90°
I and D lie on the same side of RO, and ∠ROI + ∠IOD = ∠ROD
⇒ ∠IOD = 90 – 60 = 30°
30° P A O R I D S 5 cm 5 cm
Rectangle PAIR in grey, rectangle RODS in green, with RO shared between the two arguments.
Why the answer equals the given angle: both ∠IRO and ∠IOD are the “left-over” parts of right angles measured from the same line RO — one at R inside ∆ORI, the other at O inside the corner of RODS. Since ∠ROI = 90 – 30, the other left-over must also be 30. This is why the two 5 cm marks are not needed for the angle; they only fix the size of the drawing.
Q6.
Construct a square with diagonal 6 cm without using a protractor.
Answer

Use the diagonal property of a square: equal diagonals bisecting each other at 90°. A compass gives the right angle, so no protractor is needed.

  1. Draw AB = 6 cm.
  2. With centre A and any radius more than half of AB, draw arcs above and below AB. Repeat with centre B and the same radius. Join the two crossing points — this line is the perpendicular bisector of AB, and it meets AB at its midpoint O.
  3. On this perpendicular mark C and D with OC = OD = 3 cm, one on each side.
  4. Join AC, CB, BD and DA.

ACBD is the required square.

AB = CD = 6 cm  (equal diagonals)
OA = OB = OC = OD = 3 cm  (they bisect each other)
∠AOC = 90°  (by construction)
Each side = √(3² + 3²) = 3√2 ≈ 4.24 cm
Why the arc construction gives a true perpendicular: both crossing points are the same distance from A and from B, so both lie on the perpendicular bisector of AB. Two points determine that line — and a perpendicular bisector is by definition at 90° and through the midpoint, which is exactly the pair of conditions a square's diagonals need.
Q7.
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. Find other ways of constructing a square within a square such that the vertices of the inner square lie on the sides of the outer square, as shown in Figure (b).
Answer

UVWX is a square, with side 1⁄√2 times the side of CASE — so its area is exactly half.

Geometric reasoning. Let the side of CASE be x, so each half-side is x⁄2. Look at the four corner triangles cut off by UVWX. At corner C, the triangle has legs CU = CV = x⁄2 with a right angle between them, and the same is true at A, S and E.

UV² = CU² + CV² = x²⁄4 + x²⁄4 = x²⁄2
⇒ UV = x⁄√2
The four corner triangles are congruent (SAS), so
UV = VW = WX = XU = x⁄√2

Angles. Each corner triangle is right-angled and isosceles, so its two base angles are 45°. At the point U on side CA, three angles lie along a straight line:

45° + ∠XUV + 45° = 180°  ⇒ ∠XUV = 90°
The same at V, W and X ⇒ all four angles are 90°

Four equal sides and four right angles ⇒ a square.

C U A X S W E V 45° 45° 45° 45°
Joining the midpoints of a square gives a square of half the area, tilted through 45°.

By construction. Draw a square of side 6 cm, mark the four midpoints and join them. Measure: each inner side comes to about 4.2 cm (since 6 ∕ √2 ≈ 4.24), and each inner angle to 90°.

Other inner squares (Figure b). The midpoints are not the only choice. Take any distance d and, going round the outer square in the same rotational direction, mark one point on each side at distance d from the corner you are leaving. Join the four points.

Each corner triangle then has legs d and (x – d), all four congruent by SAS
⇒ the four inner sides are equal, each √(d² + (x – d)²)
⇒ and as before the angles come out 90° — a square for every value of d
Why turning the same way round is essential: it makes the four corner triangles congruent by a quarter-turn of the whole figure about the centre. A quarter-turn maps the inner quadrilateral onto itself, so its four sides and four angles must all be equal — which is what makes it a square. The midpoint case is simply d = x⁄2, the one that gives the smallest inner square.
Q8.
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.
Answer

Yes, it must be a square.

Four equal sides ⇒ the quadrilateral is a rhombus
Every rhombus is a parallelogram, so
  adjacent angles add to 180°  and  opposite angles are equal

Let ∠A = 90°. Then
∠B = 180 – 90 = 90°  (adjacent)
∠C = ∠A = 90°  (opposite)
∠D = ∠B = 90°

All angles 90° and all sides equal ⇒ a square

By construction. Draw AB = 5 cm and, at A, a perpendicular AD = 5 cm using compass arcs. With centre B and centre D draw arcs of radius 5 cm; they meet at C. Measure ∠B, ∠C and ∠D — each reads 90°, and BC = CD = 5 cm.

Why one right angle is enough: a rhombus has only one degree of freedom left after its sides are fixed — the angle. Pinning that single angle at 90° removes the last freedom, and the shape has no choice but to be a square. Contrast this with a general quadrilateral, where four equal sides and one 90° angle would not be nearly enough information.
Q9.
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.
Answer

It is a parallelogram.

Let ABCD have AB = CD and BC = AD. Draw the diagonal AC.

In ∆ABC and ∆CDA:
AB = CD  (given)
BC = DA  (given)
AC = CA  (common)
⇒ ∆ABC ≅ ∆CDA by SSS

So ∠BAC = ∠DCA  and  ∠BCA = ∠DAC  (CPCT)

Now read those equal angles as alternate angles across the transversal AC:

∠BAC = ∠DCA  ⇒ AB ∥ DC
∠BCA = ∠DAC  ⇒ BC ∥ AD

Both pairs of opposite sides are parallel, so ABCD is a parallelogram by definition.

Why this is the converse of Deduction 7: Deduction 7 started from “opposite sides parallel” and concluded “opposite sides equal”. Here we run the implication the other way. Because both directions hold, “opposite sides equal” is an equally valid test for a parallelogram — which is why joining two congruent triangles by a half turn (page 104) reliably produces one.
Q10.
Will the sum of the angles in a quadrilateral such as the following one also be 360°? Find the answer using geometric reasoning as well as by constructing this figure and measuring.
Answer

Yes — 360°, provided the angle at the caved-in vertex D is measured as the reflex angle, the one on the inside of the figure.

ABCD here is a concave quadrilateral: D has been pushed inside triangle ABC, so the interior angle at D is more than 180°.

The diagonal AC now lies outside the figure, so use the other diagonal, BD, which does lie inside.

BD splits ABCD into ∆ABD and ∆BCD
Angle sum of ∆ABD = 180°
Angle sum of ∆BCD = 180°
Total = 360°

Regrouping: ∠A + (∠ABD + ∠DBC) + ∠C + (∠ADB + ∠BDC) = 360°
that is, ∠A + ∠B + ∠C + reflex ∠D = 360°
A B C D diagonal BD reflex angle at D is the interior one
In a concave quadrilateral one diagonal falls outside; the other still cuts it into two triangles.

By construction. Draw the figure, then measure ∠A, ∠B, ∠C and the reflex angle at D. To measure a reflex angle, measure the ordinary angle ∠ADC and subtract it from 360°. The four values add to 360°.

Why the result survives: the proof on page 95 needed only one diagonal that lies inside the figure. Every quadrilateral, convex or concave, has at least one such diagonal — a concave one has exactly one. So the “two triangles” argument never breaks down, and the 360° total holds for every quadrilateral.
Q11.
State whether the following statements are true or false. 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.
Answer

(i) False. Equal diagonals that bisect each other give a rectangle (Deduction 3), not necessarily a square. To force a square you need one more condition — the diagonals must also meet at 90° (Deduction 5). A 6 cm by 3 cm rectangle is a counter-example.

(ii) True. The angle sum of a quadrilateral is 360°, so the fourth angle is 360 – 90 × 3 = 90°. All four angles are then right angles, which is the definition of a rectangle.

(iii) True. Let the diagonals of ABCD meet at O with OA = OC and OB = OD.

∆AOB ≅ ∆COD by SAS  (∠AOB = ∠COD, vertically opposite)
⇒ AB = CD and ∠BAO = ∠DCO ⇒ AB ∥ CD
Similarly ∆AOD ≅ ∆COB ⇒ AD = CB and AD ∥ CB
⇒ a parallelogram

(iv) False. A kite also has perpendicular diagonals, and a kite with sides 6, 6, 9, 9 is not a rhombus. Perpendicularity alone is not enough — the diagonals must also bisect each other.

(v) True. Let ∠A = ∠C = x and ∠B = ∠D = y.

x + y + x + y = 360  ⇒ 2(x + y) = 360  ⇒ x + y = 180°
So each pair of adjacent angles is supplementary
⇒ AB ∥ DC and AD ∥ BC  (co-interior angles) ⇒ a parallelogram

(vi) True. Four equal angles adding to 360° means each is 360 ∕ 4 = 90°, which is exactly the book's definition of a rectangle.

(vii) False. An isosceles trapezium is only guaranteed one pair of parallel sides. Its two equal sides slant towards each other and are not parallel, so it is not a parallelogram. Indeed, its base angles are equal (∠U = ∠V), whereas in a parallelogram adjacent angles must add to 180° — the two conditions can only hold together when the angles are all 90°.

The pattern in these seven: the false ones (i), (iv) and (vii) each drop one condition from a correct characterisation — equal diagonals without perpendicularity, perpendicularity without bisection, one pair of parallel sides instead of two. A property can be necessary without being sufficient, and telling the two apart is the whole skill this exercise is testing.
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