Diagonals of Square Are Equal and Perpendicular to Each Other.
Diagonals of Square Are Equal and Perpendicular to Each Other.
Theorem: In a square the diagonals are equal and perpendicular to each other.
Given: A square ABCD. To Prove: AC = BD and AC and BC intersect perpendicularly. Proof: In AD = BC [ AB = AB [ Common] Therfore As ABCD is a square Therefore it is a parallelogram. Also diagonals of a parallelogram bisect each other.
In BO = DO [ From Equation (1)] BC = CD [ Sides of a square] OC = OC [ Common ] Therfore
But
Similarly we can prove that Hence Proved that In a square the diagonals are equal and perpendicular to each other. |  |
| Theorem : If the diagonals of a parallelogram are equal and intersect at right angles, then the parallelogram is a square. | |
Given : ABCD is a parallelogram, such that AC = BD and AC To Prove: ABCD is a square Proof: In AO = OC [ Diagonals of parallelogram bisect each other] BO = OB [ Common] Therefore AB = BC [CPCT] As ABCD is a IIgm and the adjacent sides are equal AB = BC Therefore AB = BC = CD= DA. In AD = BC [ Proved above] BD = AC [Given] AB = AB [ Common] Therfore
As ABCD is a parallelogram, AD || BC Now AD || BC and AB is the transversal
Now ABCD is a IIgm in which all sides are equal and one angle is | |
Illustration: If the side of a square ABCD whose diagonals intersect at O is 6cm find the length of OA and OB;
Solution: As the diagonals of a square are equal and bisect each other perpendicularly We can say that OA = OB = x(say) ..............(1) and AB = 6cm ...................(2) [Given] Now
Substituting the values from Eq (1) and (2), we get
Hence OA = OB = | |
ABCD is a square. AC and BD intersect at O. State the measure of | |||
| Right Option : A | |||
| View Explanation | |||
If the side of a square ABCD whose diagonals intersect at O is | |||
| Right Option : A | |||
| View Explanation | |||
PQRS is a square. PR and SQ intersects at O. State the measure of | |||
| Right Option : B | |||
| View Explanation | |||
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