Question

find the area of parallelogram whose diagonal are 24cm and 10 cm where as one side of the parallelogram is 13cm ​

Answer 1

Since we have both the diagonals, we can start by thinking about the properties of diagonals in parallelograms – they bisect each other. So if we have diagonals of 24 and 10 units, they bisect each other into segments of 12 units and 5 units, respectively:

So now we have a triangle with sides 5, 12 and 13 – a Pythagorean Triple, which means the triangle is a right triangle, and we can easily compute its area as leg x leg /2, or 5×12/2=30.

But if m∠AOB=90°, then m∠AOD=90°, as ∠AOB and ∠AOD are a linear pair which add up to 180°. Triangles ΔAOB and ΔAOD are then congruent (by Side-Angle-Side), as are triangles ΔAOB and ΔCOB (also Side-Angle-Side). So we have 4 triangles of area 30 making up the area of the parallelogram, whose area is thus 4×30=120.

Another way to think about the problem is to remember that if the parallelogram is a rhombus, then its area is the product of the diagonals divided by two.

So the areas of the parallelogram is (diagonal x diagonal /2 ), or 24×10/2=120, as above.

Solution:

(1) AC=24 //Given

(2) BD=10 //Given

(3) AO=OC=12 //Diagonals of a parallelogram bisect each other

(4) BO=OD=5 //Diagonals of a parallelogram bisect each other

(5) AB=13 //Given

(6) m∠AOB=90° //(3), (4), (5), Converse of the Pythagorean theorem (5,12,13 are a Pythagorean Triple)

(7) AC⊥DB  //(6), definition of perpendicular lines