The triangles on Bhaskar's tent are similar triangles.
- <em>The length of IJ is 4.5 m
</em>
- <em>The applicable theorem is the mid-point theorem
</em>
- <em>The appropriate formula for area is the Heron's formula
</em>
- <em>The ratio of ABC to DEF is 1 : 9</em>
<u>(a) The length of IJ</u>
The given parameter are:
I and J are at the midpoint of DE and DF
The above highlight means that
--- midpoint theorem
Substitute 9 for EF
![\mathbf{IJ= \frac 12 \times 9m}](https://tex.z-dn.net/?f=%5Cmathbf%7BIJ%3D%20%5Cfrac%2012%20%5Ctimes%209m%7D)
<u>(b) The property used to find GH and IJ</u>
In (a), the midpoint theorem is used to calculate IJ
GH and IJ are corresponding sides of similar triangles,
So the midpoint theorem can also be used to calculate the length of GH
<u>(c) The area of the triangle</u>
For the given triangles, the lengths of the sides are known.
When side lengths are known, the formula to use for finding the triangle's area is the Heron's formula.
The Heron's formula is:
Where:
<u>(d): The ratio of the areas:</u>
For the small triangle, we have:
![\mathbf{a= 3.8,\ b = 4,\ c = 3}](https://tex.z-dn.net/?f=%5Cmathbf%7Ba%3D%203.8%2C%5C%20b%20%3D%204%2C%5C%20c%20%3D%203%7D)
So, we have:
So, the area is:
![\mathbf{Area = \sqrt{s \times (s -a) \times (s - b) \times (s - c)}}](https://tex.z-dn.net/?f=%5Cmathbf%7BArea%20%3D%20%5Csqrt%7Bs%20%5Ctimes%20%28s%20-a%29%20%5Ctimes%20%28s%20-%20b%29%20%5Ctimes%20%28s%20-%20c%29%7D%7D)
![\mathbf{A_{small} = \sqrt{10.8 \times (10.8 -3.8) \times (10.8 - 4) \times (10.8 - 3)}}](https://tex.z-dn.net/?f=%5Cmathbf%7BA_%7Bsmall%7D%20%3D%20%5Csqrt%7B10.8%20%5Ctimes%20%2810.8%20-3.8%29%20%5Ctimes%20%2810.8%20-%204%29%20%5Ctimes%20%2810.8%20-%203%29%7D%7D)
For the big triangle, we have:
![\mathbf{a= 11.4,\ b = 12,\ c = 9}](https://tex.z-dn.net/?f=%5Cmathbf%7Ba%3D%2011.4%2C%5C%20b%20%3D%2012%2C%5C%20c%20%3D%209%7D)
So, we have:
So, the area is:
![\mathbf{Area = \sqrt{s \times (s -a) \times (s - b) \times (s - c)}}](https://tex.z-dn.net/?f=%5Cmathbf%7BArea%20%3D%20%5Csqrt%7Bs%20%5Ctimes%20%28s%20-a%29%20%5Ctimes%20%28s%20-%20b%29%20%5Ctimes%20%28s%20-%20c%29%7D%7D)
![\mathbf{A_{big} = \sqrt{32.4 \times (32.4 -11.4) \times (32.4 - 12) \times (32.4 - 9)}}](https://tex.z-dn.net/?f=%5Cmathbf%7BA_%7Bbig%7D%20%3D%20%5Csqrt%7B32.4%20%5Ctimes%20%2832.4%20-11.4%29%20%5Ctimes%20%2832.4%20-%2012%29%20%5Ctimes%20%2832.4%20-%209%29%7D%7D)
The ratio of the small triangle to the big triangle is:
![\mathbf{Ratio = A_{small} : A_{big}}](https://tex.z-dn.net/?f=%5Cmathbf%7BRatio%20%3D%20A_%7Bsmall%7D%20%3A%20A_%7Bbig%7D%7D)
![\mathbf{Ratio = \sqrt{4009.824}:\sqrt{324795.744}}](https://tex.z-dn.net/?f=%5Cmathbf%7BRatio%20%3D%20%5Csqrt%7B4009.824%7D%3A%5Csqrt%7B324795.744%7D%7D)
Divide by 4009.824
![\mathbf{Ratio = \sqrt{1}:\sqrt{81}}](https://tex.z-dn.net/?f=%5Cmathbf%7BRatio%20%3D%20%5Csqrt%7B1%7D%3A%5Csqrt%7B81%7D%7D)
Take square roots
Hence, the ratio of ABC to DEF is 1 : 9
Read more about similar triangles at:
brainly.com/question/24874611