Tarush is a landscape architect. For his first public project he is asked to create a small scale drawing of a garden to be plac
ed in the corner of a city park. The garden is a right triangle with base 25\text{ m}25 m25, start text, space, m, end text and height 30\text{ m}30 m30, start text, space, m, end text. Draw the garden such that 111 unit on the grid below represents 5\text{ m}5 m5, start text, space, m, end text.
1 answer:
Answer: Check the explanation for the solution.
Step-by-step explanation:
Given that the garden is a right triangle with base 25m start and height 30 m
The length of the hypothenus can be achieved by using pythagorean theorem.
Please find the attached file for the diagram.
I made a large unit for the sake of clarity.
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Answer:
<h2><DEF = 40</h2><h2><EBF = <EDF = 56</h2><h2><DCF = <DEF =40</h2><h2><CAB = 84</h2>Step-by-step explanation:
In triangle DEF, we have:
<u>Given</u>:
<EDF=56
<EFD=84
So, <DEF =180 - 56 - 84 =40 (sum of triangle angles is 180)
____________
DE is a midsegment of triangle ACB
( since CD=DA(given)=>D is midpoint of [CD]
and BE = EA => E midpoint of [BA] )
According to midsegment Theorem,
(DE) // (CB) "//"means parallel
and DE = CB/2 = FB =CF
___________
DEBF is a parm /parallelogram.
<u>Proof</u>: (DE) // (FB) ( (DE) // (CB))
AND DE = FB
Then, <EBF = <EDF = 56
___________
DEFC is parm.
<u>Proof</u>: (DE) // (CF) ((DE) // (CB))
And DE = CF
Therefore, <DCF = <DEF =40
___________
In triangle ACB, we have:
<CAB =180 - <ACB - <ABC =180 - 40 - 56 =84(sum of triangle angles is 180)