Question

12. Prove that. (i) diagonals of a rectangle are equal. (ii) diagonals of a square are equal.​

Answer 1

Answer:

(i)ABCD is a rectangle with AC and BD as its diagonals.

ABCD is a rectangle

A = 90o, AD = BC

ADBC and AB is a transversal

A + B = 180o

B = 90o

In ABD and BAC

AB = BA

A = B

AD = BC

ABD  BAC (SAS)

BD = AC  (c.p.c.t)

(ii)Let the diagonals AC and BD intersect each other at a point O.

To prove that the diagonals of a square are equal and bisect each other at right angles,

we have to prove AC = BD,

OA = OC,

OB = OD,

and AOB = 90º

the diagonals of a square are equal in length.

Answer 2

i) Given ABCD is a rectangle,

then AC and BD are diagonals, then in triangle ABC and BCD, angle b is common angle

BC is common side,

AB = CD

so by SAS congruency,

triangle ABC is congruent to BCD.

so by cpct, AC = BD

so diagonals are equal

hence proved

ii) Given that ABCD is a square.

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

Proof: (i) In a AABC and ABAD, AB=AB (common line)

BC=AD (opposite sides of a square)

ABC=BAD (= 90°)

AABC=ABAD( By SAS property) AC-BD (by CPCT).

(ii) In triangle OAD and OCB,

AD=CB (opposite sides of a square)

OAD=OCB (transversal AC )

ODA =OBC (transversal BD)

OAD= OCB (ASA property)

OA=OC--- -(i)

Similarly OB=OD---(ii)

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

Now in a AOBA and AODA,

OB=OD (from (ii))

BA=DA

OA OA (common line)

AOB=AOD----(iii) ( by CPCT

AOB+AOD=180° (linear pair)

2AOB=180°

AOB=AOD=90°

AC and BD bisect each other at right

angles.