Suman has a piece of land, which is in the shape of rhombus. She divides the land in two
equal parts by drawing a diagonal. Its perimeter is 400m and one of the diagonals is of length
120m.
a) Find the side of the land.
b) Find the perimeter of triangle ABD.
c) What is the area of triangle CBD?
d) Find the area of whole land.
e) Find the value of semi perimeter S.
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Answer:
perimeter of rhombus = 4 * A
where A is the side of Rhombus
so 4 * A = 400
A = 100 meter
Area of Rhombus= (P*Q)/2 ---------eq1
P=120 meter ----------------eq2
Q =2*{(AB)^2-(BO)^2}^(1/2)
=> Q= 2*{(100)^2-(60)^2}^(1/2)
=> Q= 2*{10000-3600}^(1/2)
=> Q= 2*{6400}^(1/2)
=> Q= 2*80=160 meter -----------eq3
where P and Q are two diagonals of Rhombus
a) side of land= 100 meter
b) Perimeter of triangle ABD = AB+BD+AD=100+120+100=320 meter
c) Area of Triangle CBD = (Area of Rhombus)/2={(P*Q)/2}/2
= (120 * 160)/4=4800 square meter
d) the area of land= Area of Rhombus= (P*Q)/2 =(120 *160)/2 = 9600 square meter
Step-by-step explanation:
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