Question

A thin metal iron sheet is rhombus in shape with each side is 10m if one of its diagonals is 16m find the cost of painting its both sides at the rate of rupees 6 per m^2 Also find the distance between the opposite side of the rhombus

Answer 1

Solution :-

Each side of the rhombus = 10 m

Let the rhombus be ABCD and d1 = 16 m and d2 = ?

In Δ ADC 

⇒ a = AD = 10 m, b = AC = 16 m and c = DC = 10 m

Semi perimeter = s = (a + b + c)/2

⇒ (10 + 16 + 10)/2

⇒ 36/2

⇒ 18 m

Area of Δ ADC = √s(s - a)(s - b)(s -c)

⇒ √18(18 - 10)(18 - 16)(18 - 10)

⇒ √18*8*2*8

⇒ √2304

⇒ Area of Δ ADC = 48 m²

Area of Δ ADC = Area of Δ ABC

So, area of Δ ABC = 48 m²

Total area = 48 + 48 = 96 m²

Area of rhombus = (d1 × d2)/2

⇒ 96 = (16 × d2)/2

⇒16d2 = 96*2

⇒ d2 = 192/16

⇒ d2 = 12 m

Cost of painting both sides of the rhombus = Rs. 6 m²

Area of both the sides of the rhombus = 96*2 = 192 m²

Total cost of painting both the sides of rhombus = 192*6

Total cost of painting both the sides of rhombus = Rs. 1152

Answer.

Answer 2

Answer:

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