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
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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