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JS script to test a quadrilateral property and output the determined shape?

I have just started my coding journey and was hoping to get some help. I have been asked to create a script that receives the length of each side and the angles of each corner from a user (these can be 'hard coded') and determines whether the shape is a square, rectangle, rhombus or a parrelellogram.

I am stuck with the logic while making use of conditional statements and I cannot get past the proof of a rhombus. I do not know if I can use a switch statement - I am not confident that I understand its use.

Please help :D

//Create a program that receives the length of each side and the angles of each corner (these can be hard-coded) and 
//determines whether the shape is a square, a rectangle, a rhombus or a parallelogram.

console.log('Good day! Time to see what shape you are dealing with today.')

sideLength_one = prompt("What is the length of the first side? ");
sideLength_two = prompt("What is the length of the second side?");
sideLength_three = prompt("What is the length of the third side?");
sideLength_four = prompt("What is the length of the fourth side?");

console.log("Awesome! Now let's move to the angles.");

angle_one = prompt("What is the measurement of angle one?");
angle_two = prompt("What is the measurement of angle two?");
angle_three = prompt("what is the measurement of angle three?");
angle_four = prompt("What is the measurement of angle four?");


//test the quadrilateral
if (sideLength_one && sideLength_two && sideLength_three == sideLength_four){
    console.log("Your shape is square!");
}

else if (sideLength_one ==sideLength_three || sideLength_two == sideLength_four){
    console.log("Your shape is a rectangle!");
}

else if ((angle_one < 90 || angle_two < 90) && (sideLength_one && sideLength_two == sideLength_three)){
    console.log ("Your shape is a rhombus!");
}

about 4 years ago · Juan Pablo Isaza
1 Respostas
Responde à pergunta

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It is not recommended to use switch with statements. I consider that to be abusing a side effect
switch (true) { case statement1 .... }

Instead you need to add more &&s

if (sideLength_one == sideLength_four && sideLength_two == sideLength_four && sideLength_three == sideLength_four)

instead of

if (sideLength_one && sideLength_two && sideLength_three == sideLength_four)

//Create a program that receives the length of each side and the angles of each corner (these can be hard-coded) and 
//determines whether the shape is a square, a rectangle, a rhombus or a parallelogram.

console.log('Good day! Time to see what shape you are dealing with today.')

sideLength_one = prompt("What is the length of the first side? ");
sideLength_two = prompt("What is the length of the second side?");
sideLength_three = prompt("What is the length of the third side?");
sideLength_four = prompt("What is the length of the fourth side?");

console.log("Awesome! Now let's move to the angles.");

angle_one = prompt("What is the measurement of angle one?");
angle_two = prompt("What is the measurement of angle two?");
angle_three = prompt("what is the measurement of angle three?");
angle_four = prompt("What is the measurement of angle four?");


//test the quadrilateral
if (sideLength_one == sideLength_four && sideLength_two == sideLength_four && sideLength_three == sideLength_four) {
  console.log("Your shape is square!");
} else if (sideLength_one == sideLength_three || sideLength_two == sideLength_four) {
  console.log("Your shape is a rectangle!");
} else if ((angle_one < 90 || angle_two < 90) && (sideLength_one == sideLength_three && sideLength_two == sideLength_three)) {
  console.log("Your shape is a rhombus!");
} else console.log("Your shape is weird!");

about 4 years ago · Juan Pablo Isaza Relatório
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