Super Simultaneous Equations #3 (featuring shapes!)

Geometry Level 4

Shapes not to scale Shapes not to scale

Given that the area of S 1 S_1 is 10 cm 2 10 \text{ cm}^2 and the magnitude of the perimeter of S 2 S_2 is double the magnitude of the area of S 1 S_1 , find the perimeter of T T in meters

Side note: Sorry that my MS Paint skills are not too great. If anyone knows any software that makes images with the LaTeX \LaTeX and shapes, feel free to post that as a solution because it will solve one of my "problems"


The answer is 0.0735797331.

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2 solutions

Chew-Seong Cheong
Aug 12, 2020

Area of S 1 S_1 is 4 ( 2 x y ) = 10 8 x 4 y = 10 . . . ( 1 ) 4(2x-y) = 10 \implies 8x - 4y = 10 \ ...(1) also 2 x y = 10 4 = 5 2 \implies \blue{2x - y} = \dfrac {10}4 = \blue{\dfrac 52} .

Perimeter of S 2 S_2 is 2 ( 2 x y ) + 2 ( 3 x + 2 y ) = 20 5 + 6 x + 4 y = 20 6 x + 4 y = 15 . . . ( 2 ) 2\blue{(2x-y)} + 2(3x+2y) = 20 \implies \blue 5 + 6x + 4y = 20 \implies 6x + 4y = 15 \ ...(2) .

( 1 ) + ( 2 ) : 14 x = 25 x = 25 14 (1)+(2): \ 14x = 25 \implies x = \dfrac {25}{14} .

Perimeter of T T is given by:

x + 2 x y + x 2 + ( 2 x y ) 2 = 25 14 + 5 2 + ( 25 14 ) 2 + ( 5 3 ) 2 7.3579733 cm 0.0736 m \begin{aligned} x + \blue{2x-y} + \sqrt{x^2 + \blue{(2x-y)}^2} & = \frac {25}{14} + \blue{\frac 52} + \sqrt{\left(\frac {25}{14}\right)^2 + \left(\frac 53\right)^2} \\ & \approx 7.3579733 \text{ cm} \\ & \approx \boxed{0.0736} \text{ m} \end{aligned}

James Watson
Aug 11, 2020

First, I made both equations look nicer. Area of S 1 = 4 ( 2 x y ) = 8 x 4 y = set 10 cm 2 \text{Area of }S_1 = 4(2x-y) = 8x-4y \stackrel{\mathclap{\mbox{set}}}{=} 10 \text{ cm}^2 Perimeter of S 2 = 2 ( 3 x + 2 y + 2 x y ) = 2 ( 5 x + y ) = 10 x + 2 y = set 2 × Area of S 1 = 20 cm \text{Perimeter of }S_2 = 2(3x+2y + 2x-y) = 2(5x+y)=10x+2y \stackrel{\mathclap{\mbox{set}}}{=} 2 \times \text{Area of }S_1 = 20 \text{ cm}

I rearranged the top equation for y y , but there are other ways of solving these: 8 x 4 y = 10 add 4 y and subtract 10 8 x 10 = 4 y divide by 4 y = 2 x 5 2 cm 8x-4y = 10 \xRightarrow[]{\text{add }4y \text{ and subtract } 10} 8x-10=4y \xRightarrow[]{\text{divide by }4} y=2x-\frac{5}{2} \text{ cm}

Now we can substitute this into our second equation and solve for x x : 10 x + 2 y = 20 sub y 10 x + 2 ( 2 x 5 2 ) = 20 14 x 5 = 20 add 5 14 x = 25 divide by 14 x = 25 14 cm 10x+2y=20 \xRightarrow[]{\text{sub }y} 10x+2\left(2x-\frac{5}{2}\right)=20 \Longrightarrow 14x-5=20 \xRightarrow[]{\text{add }5} 14x=25 \xRightarrow[]{\text{divide by }14} x=\boxed{\frac{25}{14}\text{ cm}}

We can use this to find y y : y = 2 x 5 2 sub x y = 2 ( 25 14 ) 5 2 = 25 7 5 2 = 15 14 cm y=2x-\frac{5}{2} \xRightarrow[]{\text{sub }x} y=2\left(\frac{25}{14}\right)-\frac{5}{2} = \frac{25}{7}-\frac{5}{2} = \boxed{\frac{15}{14}\text{ cm}}

Now that we have both x x and y y , we can start to find the perimeter of T T . Firstly, we must know that the sides are of length x cm x \text{ cm} and 2 x y cm 2x-y \text{ cm} . Secondly, we need to find the hypotenuse of T T ; for this, we can use the Pythagorean theorem since it is a right angled triangle: Hypotenuse of T = x 2 + ( 2 x y ) 2 = ( 25 14 ) 2 + ( 5 2 ) 2 = 925 98 = 5 74 14 cm \text{Hypotenuse of }T = \sqrt{x^2+(2x-y)^2} = \sqrt{\left(\frac{25}{14}\right)^2 + \left(\frac{5}{2}\right)^2} = \sqrt{\frac{925}{98}} = \frac{5\sqrt{74}}{14}\text{ cm}

Finally, we can calculate the perimeter of T T : Perimeter of T = 5 74 14 + 25 14 + 2 ( 25 14 ) 15 14 = 60 + 5 74 14 cm \text{Perimeter of }T = \frac{5\sqrt{74}}{14}+\frac{25}{14}+2\left(\frac{25}{14}\right) - \frac{15}{14} = \boxed{\frac{60+5\sqrt{74}}{14} \text{ cm}}

However, the answer has to be in meters so the actual answer is 60 + 5 74 14 cm ÷ 100 = 12 + 74 280 meters = 0.0735797331 meters \large \frac{60+5\sqrt{74}}{14} \text{ cm } \div 100= \green{\boxed{\frac{12+\sqrt{74}}{280}\text{ meters}}} = \green{\boxed{0.0735797331\dots \text{ meters}}}

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