Mathematics, 20.10.2019 12:30 Angel1107

The areas of two similar triangles are 72dm2 and 50dm2. the sum of their perimeters is 226dm. what is the perimeter of each of these triangles?

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Answer from: lilspike420

123 3/11 dm102 8/11 dm

Step-by-step explanation:

The ratio of the linear dimensions (perimeter) is the square root of the ratio of area dimensions, so the larger : smaller ratio is ...

√(72/50) = √1.44 = 1.2

The sum of ratio units is 1.2 + 1 = 2.2, so each ratio unit stands for ...

(226 dm)/2.2 = 102 8/11 dm . . . . . the perimeter of the smaller triangle

Then the perimeter of the larger triangle is ...

1.2 × 102 8/11 dm = 123 3/11 dm . . . . the perimeter of the larger triangle

Answer from: Weser17

$\large\boxed{\dfrac{1130}{11}\ dm=102\dfrac{8}{11}\ dm\ and\ \dfrac{1356}{11}\ dm=123\dfrac{3}{11}\ dm}$

Step-by-step explanation:

We know: The ratio of the areas of similar triangles is equal to the square of the similarity scale.

Threfore:

$\text{If}\ \triangle_1\sim\triangle_2,\ \text{then}\ \dfrac{A_{\triangle_1}}{A_{\triangle_2}}=k^2$

We have

$A_{\triangle_1}=72\ dm^2\\\\A_{\triangle_2}=50\ dm^2$

Susbtitute:

$k^2=\dfrac{72}{50}\\\\k^2=\dfrac{72:2}{50:2}\\\\k^2=\dfrac{36}{25}\to k=\sqrt{\dfrac{36}{25}}\\\\k=\dfrac{\sqrt{36}}{\sqrt{25}}\\\\\boxed{k=\dfrac{6}{5}}$

We have the similarity scale.

We know: The ratio of the perimeters of similar triangles is equal to the similarity scale.

Therefore:

$\text{If}\ \triangle_1\sim\triangle_2,\ \text{then}\ \dfrac{P_1}{P_2}=k\to P_1=kP_2$

We have:

$P_1+P_2=226\ dm^2$

Substitute:

$k=\dfrac{6}{5},\ P_1=kP_2\to P_1=\dfrac{6}{5}P_2$

$\dfrac{6}{5}P_2+P_2=226\\\\\dfrac{6}{5}P_2+\dfrac{5}{5}P_2=226\\\\\dfrac{11}{5}P_2=226\qquad\text{multiply both sides by 5}\\\\11P_2=1130\qquad\text{divide both sides by 11}\\\\P_2=\dfrac{1130}{11}\\\\\boxed{P_2=102\dfrac{8}{11}\ dm}\\\\\\P_1=kP_2\to P_1=\dfrac{6}{5}\cdot\dfrac{1130}{11}\\\\P_1=\dfrac{6}{1}\cdot\dfrac{226}{11}\\\\P_1=\dfrac{1356}{11}\\\\\boxed{P_1=123\dfrac{3}{11}\ dm}$

Answer from: Quest

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

Answer from: Quest

step-by-step explanation: it is d

Answer this multiple choice question correctly for 30 points and

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The areas of two similar triangles are 72dm2 and 50dm2. the sum of their perimeters is 226dm. what i...