in a rhombus, one diagonal is 16 CM and one side is 17 CM. find its area
Answers
Answered by
2
Given:
Side of a rhombus, h = 17 cm
One Diagonal of Rhombus, D1 = 16 cm
Let another Diagonal of Rhombus be D2.
We know that in a Rhombus diagonals bisect each other at 90 degrees, there by forming 4 equal triangle in area with one side of triangle as D1/2 & another side as D2/2 and the side of Rhombus as hypotenuse.
By pythagorous theorem we can say that,
(D1/2)^2 + (D2/2)^2 = h^2
(16/2)^2 + (D2/2)^2 = 17^2
8^2 + (D2/2)^2 = 289
(D2/2)^2 = 289 - 64
(D2/2)^2 = 225
D2/2 = sqrt(225)
D2/2 = 15
D2 = 15*2 = 30 cm
Area of triangle = (1/2)*base*height = (1/2)*(D1/2)*(D2/2) = (1/2)*(16/2)*(30/2) = (1/2)*(8)*(15) = 4*15 = 60 sq cm
Area of Rhombus = 4* Area of triangle = 4*60 = 240 Sq cm ——> Answer
Side of a rhombus, h = 17 cm
One Diagonal of Rhombus, D1 = 16 cm
Let another Diagonal of Rhombus be D2.
We know that in a Rhombus diagonals bisect each other at 90 degrees, there by forming 4 equal triangle in area with one side of triangle as D1/2 & another side as D2/2 and the side of Rhombus as hypotenuse.
By pythagorous theorem we can say that,
(D1/2)^2 + (D2/2)^2 = h^2
(16/2)^2 + (D2/2)^2 = 17^2
8^2 + (D2/2)^2 = 289
(D2/2)^2 = 289 - 64
(D2/2)^2 = 225
D2/2 = sqrt(225)
D2/2 = 15
D2 = 15*2 = 30 cm
Area of triangle = (1/2)*base*height = (1/2)*(D1/2)*(D2/2) = (1/2)*(16/2)*(30/2) = (1/2)*(8)*(15) = 4*15 = 60 sq cm
Area of Rhombus = 4* Area of triangle = 4*60 = 240 Sq cm ——> Answer
Similar questions