Question

Show that the diagonals of a square are equal and bisect each other at right angle​

Answer 1

Answer:

Step-by-step explanation:

Let the common ratio between the angles be x. Therefore, the angles will be 3x, 5x, 9x, and

13x respectively.

As the sum of all interior angles of a quadrilateral is 3600,

=> 3x + 5x + 9x + 13x = 3600

=> 30x = 3600

=> x = 3600/30

=> x = 120

Hence, the angles are

3x = 3 * 120 = 360

5x = 5 * 120 = 600

9x = 9 * 120 = 1080

13x = 13 * 120 = 1560

Question 2:

If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Answer:

              Class_9_Maths_Quadrilaterals_Rectangle1                                              

Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that one of

its interior angles is 900.

In ΔABC and ΔDCB,

AB = DC (Opposite sides of a parallelogram are equal)

BC = BC (Common)

AC = DB (Given)

By SSS congruence rule,

ΔABC  ≅ ΔDCB

So, ∠ABC = ∠DCB

It is known that the sum of measures of angles on the same side of traversal is 1800

     ∠ABC + ∠DCB = 1800                  [AB || CD]

=> ∠ABC + ∠ABC = 1800

=> 2∠ABC = 1800

=> ∠ABC = 900

Since ABCD is a parallelogram and one of its interior angles is 900, ABCD is a rectangle.

Question 3:

Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Answer:

Class_9_Maths_Quadrilaterals_Quadrilateral3

Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right angle i.e.,

OA = OC, OB = OD, and ∠AOB = ∠BOC = ∠COD = ∠AOD = 90º. To prove

 

ABCD a rhombus, we have to prove ABCD is a parallelogram and all the sides of ABCD are

equal.

In ΔAOD and ΔCOD,

OA = OC                      (Diagonals bisect each other)

∠AOD = ∠COD          (Given)

OD = OD                    (Common)

So, ΔAOD ≅ ΔCOD (By SAS congruence rule)

Hence, AD = CD …………..1

Similarly, it can be proved that

AD = AB and CD = BC ………..2

From equation 1 and 2, we get

AB = BC = CD = AD

Since opposite sides of quadrilateral ABCD are equal, it can be said that ABCD is a

parallelogram. Since all sides of a parallelogram ABCD are equal, it can be said that ABCD is a

rhombus.

Answer 2

Step-by-step explanation:

Given that ABCD is a square.

To prove : AC=BD and AC and BD bisect each other at right angles.

Proof:

(i) In a ΔABC and ΔBAD,

AB=AB ( common line)

BC=AD ( opppsite sides of a square)

∠ABC=∠BAD ( = 90° )

ΔABC≅ΔBAD( By SAS property)

AC=BD ( by CPCT).

(ii) In a ΔOAD and ΔOCB,

AD=CB ( opposite sides of a square)

∠OAD=∠OCB ( transversal AC )

∠ODA=∠OBC ( transversal BD )

ΔOAD≅ΔOCB (ASA property)

OA=OC ---------(i)

Similarly OB=OD ----------(ii)

From (i) and (ii) AC and BD bisect each other.

Now in a ΔOBA and ΔODA,

OB=OD ( from (ii) )

BA=DA

OA=OA ( common line )

ΔAOB=ΔAOD----(iii) ( by CPCT

∠AOB+∠AOD=180° (linear pair)

2∠AOB=180°

∠AOB=∠AOD=90°

∴AC and BD bisect each other at right angles.