Question

All three sides of a triangle are initially 4 m in length. One of the triangle's sides is oriented horizontally. The triangle is scaled down in size without changing any of the angles. What is the new height of the triangle when the area is exactly half of the original triangle's area

Answer 1

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 ;

$JL = \dfrac{1}{2} JK$

$JL = \dfrac{1}{2} i$

Let consider triangle IJL

(IL)² = (IJ)² - (JL)²

$(IL)^2 = i^2 - \dfrac{i^2}{4}$

$(IL)^2 = \dfrac{3i^2}{4}$

$(IL) =\sqrt{ \dfrac{3i^2}{4}}$

$IL =\sqrt{3} \dfrac{i}{2}$

Area of triangle IJK can be expressed as:

$Area = \dfrac{1}{2}\times Base \times Height$

$Area = \dfrac{1}{2}\times i \times \sqrt{3}\dfrac{i}{2}$

$Area = \sqrt{3}\dfrac{i^2}{4}$

where, i = 4

Then:

$Area = \sqrt{3}\dfrac{4^2}{4}$

$Area = \sqrt{3} \dfrac{16}{4}$

$Area =4 \sqrt{3}$

when the area is exactly half of the original triangle's area, the new height is :

$A = \dfrac{1}{2} \times Area$

$\sqrt{3} \dfrac{i^2}{4} = \dfrac{1}{2} \times 4\sqrt{3}$

$\dfrac{i^2}{4} = \dfrac{1}{2} \times 4$

$\dfrac{i^2}{4} =2$

$i^2 = 8$

$i= \sqrt{8}$

$i= \sqrt{4 \times 2}$

$i= 2 \sqrt{2}$

Finally, the new height of the new triangle is:

$IL =\sqrt{3} \dfrac{i}{2}$

$IL =\sqrt{3} \dfrac{2 \sqrt{2}}{2}$

$IL =\sqrt{3 \times 2}$

$IL =\sqrt{6}$

IL = 2.449 m