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