Chapter 8: Quadrilaterals

Q1Written answer

One angle of a quadrilateral is of 108º and the remaining three angles are equal. Find each of the three equal angles.

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Worked solution

Let the remaining three equal angles be x.

We know,

Sum of all interior angles of a quadrilateral is = 360^o

108^o + x + x + x = 360^o

108^o + 3x = 360^o

3x = 360^o – 108^o

3x = 252^o

x = 252/3

x = 84^o

Each of three equal angles, x = 84^o.

Q2Written answer

ABCD is a trapezium in which AB || DC and ∠A = ∠B = 45º. Find angles C and D of the trapezium.

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Worked solution

According to the question,

ABCD is a trapezium

∠A = ∠B = 45º

NCERT Exemplar Solutions Class 9 Maths Chapter 8 Exercise 8.3-1

We know that,

Angles opposite to each other in quadrilateral are supplementary.

Then, we have,

∠A + ∠C = 180º

45^o + ∠C = 180º

∠C = 180^o – 45^o

∠C = 135^o

Similarly,

We have,

∠B + ∠D = 180º

45^o + ∠D = 180º

∠D = 180^o – 45^o

∠D = 135^o

Q3Written answer

The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60º. Find the angles of the parallelogram.

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Worked solution

According to the question,

ABCD is parallelogram,

DP ⊥ AB

DQ ⊥ BC.

∠PDQ = 60^o

In quad. DPBQ,

Using angle sum property of a quadrilateral,

We have,

∠PDQ + ∠Q + ∠P + ∠B = 360^o

60^o + 90^o + 90^o + ∠B = 360^o

240^o + ∠B = 360^o

∠B = 360^o – 240^o

∠B = 120^o

Since, opposite angles in parallelogram are equal,

We have,

∠B = ∠D = 120^o

Since, opposite sides are parallel in parallelogram,

We have,

AB||CD

Also, since sum of adjacent interior angles is 180^o,

We have,

∠B + ∠C = 180^o

120^o + ∠C = 180^o

∠C = 180^o – 120^o

∠C = 60^o

Since, opposite angles in parallelogram are equal,

We have,

∠C = ∠A = 60^o

Q4Written answer

ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the angles of the rhombus.

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Worked solution

NCERT Exemplar Solutions Class 9 Maths Chapter 8 Exercise 8.3-3

According to the question,

We have,

ABCD is a rhombus.

DE is the altitude on AB then AE = EB.

In ΔAED and ΔBED,

We have,

DE = DE (common line)

∠AED = ∠BED (right angle)

AE = EB (DE is an altitude)

∴ ΔAED ≅ ΔBED by SAS property.

∴ AD = BD (by C.P.C.T)

But AD = AB (sides of rhombus are equal)

⇒ AD = AB = BD

∴ ABD is an equilateral triangle.

∴ ∠A = 60^o

Since, opposite angles of rhombus are equal, we get,

⇒ ∠A = ∠C = 60^o

We also know that,

Sum of adjacent angles of a rhombus = supplementary.

So,

∠ABC + ∠BCD = 180^o

∠ABC + 60^o = 180^o

∠ABC = 180^o – 60^o = 120^o

Since, opposite angles of rhombus are equal, we get,

∠ABC = ∠ADC = 120^o

Hence, Angles of rhombus are:

∠A = 60^o, ∠C = 60^o, ∠B = 120^o, ∠D = 120^o

Q5Written answer

E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. Show that BFDE is a parallelogram.

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Worked solution

Construction:

Join BD, meeting AC at O.

According to the question,

Since diagonals of a parallelogram bisect each other,

We get,

OA = OC and OD = OB.

And,

OA = OC and AE = CF,

OA – AE = OC – CF

OE = OF

So, BFDE is a quadrilateral whose diagonals bisect each other.

Hence, BFDE is a parallelogram.

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Exercise 8.3