Quadrilaterals: class 9 math: 9 Math

NCERT Exercise 8.1: 9th math

Quadrilaterals Class 9 math NCERT Exercise 8.1 Question (1) The angles of quadrilateral are in the ratio 3:5:9:13. Find all the angles of the quadrilateral.

Solution

Given, ratio of angles of quadrilateral = 3 : 5 : 9 : 13

Then all angles = ?

Let angles of the given quadrilateral = 3x, 5x, 9x and 13x

We know that sum of all angles of a quadrilateral = 360^o

Thus, 3x + 5x + 9x + 13x = 360^o

⇒ 30 x = 360^o

`=>x = 360^o/30 = 12^o`

Thus, 3x = 3 × 12^o = 36^o

And, 5x = 5 × 12^o = 60^o

And, 9x = 9 × 12^o = 108^o

And, 13x = = 13 × 12^o = 156^o

Thus, angles of given quadrilateral are 36^o, 60^o, 108^o and 156^o respectively Answer

Quadrilaterals Class 9 math NCERT Exercise 8.1 Question (2) If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Solution

Let, ABCD is the given parallelogram.

And as per question diagonals of this parallelogram are equal.

That is AC = BD - - - - - (i)

Thus, to prove that ABCD is a rectangle.

Now, In triangle ABC and triangle BCD

AB = DC

[Because Opposite sides of given parallelogram]

AC = BD

[As given in question, diagonals are equal {From equation (i)}]

Side BC is common in both of the triangles.

Thus, SSS (Side Side Side) congruency

&Detla; ABC ≅ Δ BCD

Now, by CPCT, we know that corresponding parts of congruent triangles are equal.

Thus, ∠ABC = ∠BCD - - - - (ii)

Now, since ABCD is a parallelogram

Thus, AB||DC

Now, we know that, the angles on the same side of a transversal are supplementary.

Here a transversal BC is intersecting two parallel lines AB and DC

Therefore, ∠ABC + ∠BCD = 1800^o

⇒ ∠ABC + ∠ABC = 180^o

[From equatin (ii)]

⇒ 2∠ABC = 180^o

⇒ ∠ABC = 180^o/2

⇒ ∠ABC = 90^o

Now, we know that if one of the angle of a parallelogram is a righ angle, then it is a rectangle.

Here since ∠ABC = 90^o

Therefore, ABCD is a rectangle. Proved

Quadrilaterals Class 9 math NCERT Exercise 8.1 Question (3) Show that if diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Solution

As given, Let PQRS is a quadrilateral.

And its diagonals PR and QS bisects each other at right angle.

i.e. ∠POS = ∠QOP = ∠ROQ = ∠SOR = 90^o

Then to prove that PQRS is a Rhombus

Proof

We know that in a Rhombus all sides are equal and opposite sides are parallel.

In ΔPOS and ΔQOP

Side OS = OQ

[Because as per question, diagonals bisect each other]

And, ∠POS = ∠QOP = 90^o

And side OP is common in both of the triangles

Thus, by SAS (Side Angle Side) congruency

ΔPOS ≅ ΔQOP

Now, by CPCT, we know that corresponding parts of congruent triangles are equal.

Thus, PS = QP - - - - (i)

Similarly, in ΔQOP and ΔROQ,

OP = OR

[Because diagonals bisect each other as per question]

And ∠QOP = ∠ROQ = 90^o

And OQ is common side in both of the triangles

Therefore, by SAS (Side Angle Side) congruency

Δ QOP ≅ Δ ROQ

Now, by CPCT, we know that corresponding parts of congruent triangles are equal

Thus, PQ = QR - - - - (ii)

Similarly, in ΔROQ and Δ ROS,

OP = OQ

[Because as per question diagonals bisect each other]

And ∠ROQ = ∠SOR = 90^o

And OR is the common side in both of the triangles

Thus by SAS (Side Angle Side) congrueny

ΔROQ ≅ ΔROS

Now, by CPCT, we know that corresponding parts of congruent triangles are equal

Thus, QR = RS - - - - (iii)

Now, by equation (i), equation (ii) and equation (iii), we get

PQ = QR = RS = PS

Now, since all sides of the given quadrilateral are equal, thus its opposite sides are parallel.

Thus, PQRS is a Rhombus. Proved

Quadrilaterals Class 9 math NCERT Exercise 8.1 Question (4) Show that diagonals of a square are equal and bisects each other at right angle.

Solution

As given,

Let PQRS is a square.

And PQ and QS are its diagonals.

Then to prove PR = QR

And, OP = OR and OQ = OS

And, ∠ROS = 90^o

Since, PQRS is a square thus, its all sides are equal.

i.e. PQ = QR = SR = PS

Now, in ΔQRS and ΔPQR

Sides RS = PQ

[Since all sides of a square are equal]

∠QRS = ∠PQR = 90^o

[Because each interior angles of a square is equal to 90^o]

And side QR is common in both of the triangle

Therefore, By SAS (Side Angle Side) congruency

ΔQRS ≅ ΔPQR

Now, by CPCT we know that corresponding sides of congruent triangles are equal

Therefore, PR = QS

Hence diagonals of given square are equal.

Now in ΔQOP and ΔROS

∠ROS = ∠QOP

[Because vertically opposite angles are equal]

And side RS = side PQ

[Since these are sides of a square and hence are equal]

∠ORS = ∠QPO

[Because these are alternate interior angles formed by parallel lines RS and PQ and a transversal PR, and hence are equal]

Thus by AAS (Angle Angle Side) congruency

ΔQOP ≅ ΔROS

Now, by CPCT we know that corresponding sides of congruent triangles are equal

Therefore, OS = OQ and OP = OR - - - - (i)

Thus, diagonals bisect each other

Now, in ΔQOR and ΔROS

OS = OQ [From equation (i)]

And SR = QR [Sides of given square and hence are equal]

And OR is common in both of the triangles

Thus by SSS (Side Side Side) congruency

ΔQOP ≅ Δ ROS

Now, by CPCT we know that corresponding sides of congruent triangles are equal

Therefore, ∠ROS = ∠ROQ - - - - (ii)

Now, ∠ROS + ∠ROQ = 180^o

[Because these angles together form a linear pair of angles.]

⇒ ∠ROS + ∠ROS = 180^o

[From equation (ii)]

⇒ 2∠ROS = 180^o

⇒ ∠ROS = 180^o/2

⇒ ∠ROS = 90^o

Hence diagonals of a square bisect each other at right angle Proved

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