Math, asked by surajkumarsing4079, 1 month ago

find the length of the second diagonal of a rhombus whose side is 7cm and one of whose diagonal is 1 greater than its side

Answers

Answered by sjeevan11
0

Answer:

Since we know that the diagonals of a rhombus are perpendicular bisectors of each other, lets assume that ABCD is a rhombus and it's diagonals intersect at point O. Since all the sides of the rhombus are equal therefore AB=BC=CD=DA=5cm and let say diagonal AC=6cm. Since the diagonals intersect at point O this implies O is the mid-point of the diagonals AC & BD i.e. AO=3cm and OC=3cm. Also the diagonals of the rhombus bisect perpendicularly this implies that angle AOB = 90°. I.e. ΔAOB is a right angled triangle with AO=3cm,AB=5cm and OB=x cm, then by Pythagoras theorem we can write, AO^2 + OB^2 = AB^2.

I.e. 3^2+x^2=5^2

=> 9+x^2=25

=> x^2 = 25–9

=> x^2 = 16

=> x = +4 or -4

But since x is lenght it cannot be negative hence x=4cm.

I.e. OB = X = 4 cm

Therefore diagonal BD = 2×OB = 8cm.

Step-by-step explanation:

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