each side of rhombus is 13 cm and one diagonal is 10 cm then find the area of the rhombus
Answers
Answer:
Step-by-step explanation:
From the question,
Rhombus ABCD with centre O,
side (S) = 13 cm
shorter diagnol (P) = 10 cm
consider triangle OCB,
applying Pythagoras theorem,
OC = √ (BC^2 - OB^2 ) = √(169 - 25) = √ 144 = 12
OC = 12 cm
longer diagnol (AC) = AO + OC = 2OC = 2 x 12 = 24
Longer diagnol, AC = 24 cm.
Area of rhombus ABCD = [(AC x BD) / 2 ] = 10 x 24 / 2
Area of Rhombus ABCD = 120 sq.cm
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Oladipo Akinla, Engineer
Answered February 26, 2019
The 10cm diagonal divides the rhombus into two triangles, each with a base of 10cm, and the other two sides 13cm and 13cm.
The height of either triangle (h) is given by the line dropped to the base from the opposite vertex. This line bisects the base.
We thus have a right angled triangle, with hypotenuse of 13 cm, and the other sides of length h and 5cm.
Using Pythagoras, h is 12cm.
The diagonal of the rhombus is 2h, which is 24cm.
The area of the rhombus is 10 x h, which is 120 sq cm.
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Answered February 26, 2019
A rhombus is formed of 4-RATs.
Here, each RAT has one side as 5 cm and its hypotenuse as 13 cm.
So the other side of the RAT is [13^2–5^2]^0.5 = [169–25]^0.5 = 144^0.5 - 12 cm.
The other diagonal is 2x12 = 24 cm.
And the area of the rhombus = 10x24/2 = 120 sq cm.
The distance between the opposite sides = 120/13 = 9.23 cm.
The acute angle of the rhombus = 45.24 deg and the obtuse angle = 134.76 deg.
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