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Category: Homework
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Experience:  BA (Hons) in Mathematics from Oxford University
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A building lot has the dimensions shown in the figure. Find

Resolved Question:

A building lot has the dimensions shown in the figure. Find the length of road frontage for this lot, and the area of the lot.
Submitted: 5 years ago.
Category: Homework
Expert:  M Hasan replied 5 years ago.
Hi there!

Can you please post the figure?

Thanks!
Customer: replied 5 years ago.

I'm only able to paste image now. I'm not able to paste in the first box where the question is typyed.
Expert:  M Hasan replied 5 years ago.
The site is experiencing technical difficulties and that is why I am unable to see the picture. Can you please upload it to www.mediafire.com and post the sharing link here. Thanks! :)
Customer: replied 5 years ago.
Expert:  M Hasan replied 5 years ago.
Customer: replied 5 years ago.
can you please explain the steps needed to do that, thanks
Expert:  M Hasan replied 5 years ago.
Customer: replied 5 years ago.

let me know if this is right
Expert:  M Hasan replied 5 years ago.

We first find the length of CA:

CA = √(61^2 + 68.6^2)
CA = 91.8

Angle BCA:

tan BCA = 68.6/61
BCA = 48.36º

ACD = 105 - 48.36º = 56.64º

We will find the length of road frontage (AD) using the cosine rule.

AD = √(CD^2 + CA^2 - 2(CD)(CA) cos ACD)
AD = √(109.5^2 + 91.8^2 - 2(109.5)(91.8) cos 56.64º)

Therefore, the road frontage is 96.76 m.

Area = (1/2)(61)(68.6) + (1/2)(109.5)(91.8) sin 56.64º
Area = 6290.22

Therefore, area is 6290.22 m^2.

Hope this helps. If so, please click 'Accept'. Thanks! :)
Customer: replied 5 years ago.
can you show formula for the area
Expert:  M Hasan replied 5 years ago.
Yes, sure.

Area = Area of right triangle ABC + Area of triangle ADC
Area = (1/2)(BC)(BA) + (1/2)(CD)(CA) sin ACD
Area = (1/2)(61)(68.6) + (1/2)(109.5)(91.8) sin 56.64º
Area = 6290.22

Hope this helps. If so, please click 'Accept'. Thanks! :)
Customer: replied 5 years ago.
can you explain where/ how in ABC the A = (1/2)
Expert:  M Hasan replied 5 years ago.