Question

Prove that diagonal of a square makes an angle of 45 with the sides of the square

Answer 1

We know that
1)all 4 angles of a square are 90°.
2)the diagonals of a square bisect each other as well as the angle.
Let ABCD be a square..
\_ABD=\_BDC (diagonal of a square bisect the angles)
\_ABD+\_BDC= 90°
\_ABD+\_ABD=90°(\_ABD=\_BDC)
2\_ABD=90°
\_ABD=45°

Answer 2

Answer:

The diagonal of a square makes an angel of 45° with the sides of square.

Step-by-step explanation:

• Given: ABCD is a square in which diagonal intersect each other at O.

• To prove: Diagonal of a square makes an angle of 45° with the sides of a square.

Angle OAB=90°

• Proof: We know that diagonal of square intersect each other at 90°

• In ∆OAB,

angle AOB=90° and

OA=OB

Therefore, angle OAB=angle OBA

Let angle OAB=angle OBA= x

• We know that sum of all angles of a triangle is equal to 180°.

So, angle AOB+angle OAB+angle OBA=180°

=> 90°+x+x =180°

=> 2x=180°-90°

=> x = 90°/2

=> x = 45°

• Therefore, angle OAB= x= 45°