Question

good afternoon guys i need ur help hope u will help me here are ur questions 8,12,13,and 14 . hope for the best have a good day​

Answer 1

Answer:

(8) option c

(12)Given as :

The dimension of the room

Length of room = L = 20 meters

Breadth of room = B = 20 meters

Height of the room = H = 10 meters

Let The length of the longest rod that can place in room = L'

According to question

The length of the longest rod that can place in room = Diagonal of the room

∵ Diagonal of room = \sqrt{Lenght^{2} + breadth^{2} +height^{2} }

Lenght

2

+breadth

2

+height

2

So, The length of the longest rod that can place in room = \sqrt{L^{2}+B^{2}+H^{2} }

L

2

+B

2

+H

2

Or, L' = \sqrt{20^{2}+20^{2}+10^{2} }

20

2

+20

2

+10

2

Or, L' = \sqrt{400+400+100}

400+400+100

Or, L' = \sqrt{900}

900

Or, L' = 30 meters

∴ The length of the longest rod that can place in room = L' = 30 meters

(13) The formula for the total surface area of a right cone is T. S. A=πrl+πr2

u can use it and solve this problem