Answer:
the new height of the triangle = 2.449
Step-by-step explanation:
Given that:
the sides of the triangle are 4m in length i.e they are equal. It shows that the triangle is known to be an equilateral triangle.
Let say the triangle is a triangle IJK
Let the length of the side to be i = 4
Definitely
IJ = JK = IK = i = 4
If a midline is drawn and cuts the equilateral triangle in two equal halves of a right-angle triangle. Then, suppose the midline is L
Then ;
[tex]JL = \dfrac{1}{2} JK[/tex]
[tex]JL = \dfrac{1}{2} i[/tex]
Let consider triangle IJL
(IL)² = (IJ)² - (JL)²
[tex](IL)^2 = i^2 - \dfrac{i^2}{4}[/tex]
[tex](IL)^2 = \dfrac{3i^2}{4}[/tex]
[tex](IL) =\sqrt{ \dfrac{3i^2}{4}}[/tex]
[tex]IL =\sqrt{3} \dfrac{i}{2}[/tex]
Area of triangle IJK can be expressed as:
[tex]Area = \dfrac{1}{2}\times Base \times Height[/tex]
[tex]Area = \dfrac{1}{2}\times i \times \sqrt{3}\dfrac{i}{2}[/tex]
[tex]Area = \sqrt{3}\dfrac{i^2}{4}[/tex]
where, i = 4
Then:
[tex]Area = \sqrt{3}\dfrac{4^2}{4}[/tex]
[tex]Area = \sqrt{3} \dfrac{16}{4}[/tex]
[tex]Area =4 \sqrt{3}[/tex]
when the area is exactly half of the original triangle's area, the new height is :
[tex]A = \dfrac{1}{2} \times Area[/tex]
[tex]\sqrt{3} \dfrac{i^2}{4} = \dfrac{1}{2} \times 4\sqrt{3}[/tex]
[tex]\dfrac{i^2}{4} = \dfrac{1}{2} \times 4[/tex]
[tex]\dfrac{i^2}{4} =2[/tex]
[tex]i^2 = 8[/tex]
[tex]i= \sqrt{8}[/tex]
[tex]i= \sqrt{4 \times 2}[/tex]
[tex]i= 2 \sqrt{2}[/tex]
Finally, the new height of the new triangle is:
[tex]IL =\sqrt{3} \dfrac{i}{2}[/tex]
[tex]IL =\sqrt{3} \dfrac{2 \sqrt{2}}{2}[/tex]
[tex]IL =\sqrt{3 \times 2}[/tex]
[tex]IL =\sqrt{6}[/tex]
IL = 2.449 m
