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NCERT Solutions Ganita Prakash (Part 1) Chapter 4 –1074.6 Kite and Trapezium — In-text Questions

Book page 105 Updated on2026-09-05

Q1.
Property 1: In the kite, show that the diagonal BD (i) bisects ∠ABC and ∠ADC, (ii) bisects the diagonal AC, that is, AO = OC, and is perpendicular to it. Hint: Is ∆AOB ≅ ∆COB?
Answer

Kite ABCD has AB = BC and CD = DA. Let the diagonals meet at O.

Step 1 — the diagonal BD splits the kite into two congruent triangles.

AB = CB  (given)
AD = CD  (given)
BD = BD  (common)
⇒ ∆ABD ≅ ∆CBD by SSS
⇒ ∠ABD = ∠CBD and ∠ADB = ∠CDB

That is part (i): BD bisects both ∠ABC and ∠ADC.

Step 2 — now compare ∆AOB and ∆COB.

AB = CB  (given)
∠ABO = ∠CBO  (just proved)
BO = BO  (common)
⇒ ∆AOB ≅ ∆COB by SAS
⇒ AO = OC  and  ∠AOB = ∠COB

But ∠AOB + ∠COB = 180°  (linear pair along AC)
⇒ ∠AOB = ∠COB = 90°

So BD bisects AC and is perpendicular to it — part (ii).

Why only one diagonal has these powers: B and D are each equidistant from A and C, so both lie on the perpendicular bisector of AC — and BD is therefore that perpendicular bisector. Nothing similar holds for A and C, which is why AC does not bisect BD in general.
Tip: in a rhombus both pairs of adjacent sides are equal, so both diagonals get these properties. That is exactly why the rhombus's diagonals bisect each other, bisect the angles and meet at 90°.
Q2.
Construct a trapezium. Measure the base angles (marked in the figure). Can you find the remaining angles without measuring them?
Answer

Yes. In trapezium PQRS with PQ ∥ SR, each slanting side is a transversal cutting the two parallel sides, so the two angles at its ends are interior angles on the same side.

Property 1: ∠S + ∠P = 180°  and  ∠R + ∠Q = 180°

So measure the two base angles ∠P and ∠Q, and then simply subtract:

∠S = 180° – ∠P     ∠R = 180° – ∠Q

For example, if you measure ∠P = 70° and ∠Q = 55°, then ∠S = 110° and ∠R = 125°. Check: 70 + 55 + 110 + 125 = 360° ✓

Why only one pair of parallel sides is needed: the co-interior-angle rule applies to each leg separately, and each leg touches both parallel sides. The non-parallel pair contributes nothing extra — which is why a trapezium's opposite angles are not equal in general, unlike a parallelogram's.
Check it yourself: after computing ∠R and ∠S, measure them with a protractor. They should agree to within a degree — measurement error, not a flaw in the reasoning.
Q3.
How do we construct an isosceles trapezium? Construct an isosceles trapezium UVWX, with UV||XW. Measure ∠U. Can you find the remaining angles without measuring them?
Answer

Construction.

  1. Draw UV, the longer parallel side.
  2. Draw a line parallel to UV at the height you want.
  3. Mark X and W on that line so that UX = VW. The easiest way is to keep the figure symmetric about the perpendicular bisector of UV — set X and W the same distance in from each end.
  4. Join UX and VW.

Finding the rest without measuring. Measure ∠U once. Then:

∠V = ∠U  (angles opposite the equal sides — Property 2)
∠X = 180° – ∠U  (co-interior angles along the leg UX)
∠W = 180° – ∠V = ∠X

If ∠U = 72°, then ∠V = 72°, ∠X = ∠W = 108°, and 72 + 72 + 108 + 108 = 360° ✓

Why one measurement suffices: the trapezium supplies one relation (co-interior angles add to 180°) and the equal legs supply a second (∠U = ∠V). Two relations plus the 360° total leave only one unknown, so a single measured angle pins everything down.
Q4.
Does it appear that the angles opposite to the equal sides — ∠U and ∠V — are also equal? Can we find congruent triangles here? Consider line segments XY and WZ perpendicular to UV. What type of quadrilateral is XWZY?
Answer

XWZY is a rectangle, and that is what produces the congruent triangles.

Drop perpendiculars XY and WZ from X and W onto UV.

XW ∥ UV  (given), so YZ is a transversal
a = 180° – ∠XYZ = 180° – 90° = 90°
b = 180° – ∠WZY = 180° – 90° = 90°
So all four angles of XWZY are 90° ⇒ XWZY is a rectangle

Being a rectangle, its opposite sides are equal, so XY = WZ — the two perpendiculars have the same length.

In ∆UXY and ∆VWZ:
UX = VW  (equal legs of the isosceles trapezium)
XY = WZ  (opposite sides of rectangle XWZY)
∠UYX = ∠VZW = 90°
⇒ the triangles are congruent (RHS) ⇒ ∠U = ∠V
Why the perpendiculars are the key idea: ∠U and ∠V sit at opposite ends of the figure with nothing joining them. Dropping the two perpendiculars manufactures a rectangle in the middle, and the rectangle hands over the one equality (XY = WZ) needed to make the two end triangles congruent.
Q5.
Now, it can be shown that ∆UXY ≅ ∆VWZ. (How?)
Answer

By the RHS condition (right angle–hypotenuse–side), using the rectangle XWZY established above.

∠XYU = ∠WZV = 90°  (XY and WZ were drawn perpendicular to UV)
UX = VW  (hypotenuses — the equal non-parallel sides)
XY = WZ  (one leg each, opposite sides of rectangle XWZY)
⇒ ∆UXY ≅ ∆VWZ

Hence ∠U = ∠V  and also UY = VZ

This is Property 2: in an isosceles trapezium, the angles opposite to the equal sides are equal.

Why UY = VZ matters too: it says the two slanting ends stick out by the same amount at each side. That is the precise sense in which an isosceles trapezium is symmetric — it has a line of symmetry through the midpoints of UV and XW.
Tip: the same figure shows why the diagonals of an isosceles trapezium are equal: ∆UXV and ∆VWU are congruent by SAS, since UX = VW, ∠U = ∠V and UV is common. So XV = WU.
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