Question

The perimeter of a triangle is 2004. One side of the triangle is 21 times the other. The shortest side is of integral length. If length of one side of the triangle in every possible case, is \[x\], then \[x = \]
A.47 or 48
B.46 or 47
C.45 or 46
D.44 or 48

Answer 1

Hint: Here, we will first assume that a triangle \[\Delta ABC\] with side lengths, AB is the largest side and BC is the shortest side and then assume that the length of one side AB of the triangle in every possible case is \[x\]. Since AB is the largest side, which is 21 times that of the shortest side, BC and add the sides of a triangle and take it equal to 2004. Then we will use the sum of two sides of a triangle is always greater than the third side and the difference of two sides of a triangle is always less than the third side to find the required value.

Complete step-by-step answer:
We are given the perimeter of a triangle is 2004 and one side of the triangle is 21 times the other.
Let us assume that a triangle \[\Delta ABC\] with side lengths, AB is the largest side and BC is the shortest side.

Let us assume that the length of one side AB of the triangle in every possible case, is \[x\].
\[ \Rightarrow AB = x\]

Since AB is the largest side, which is 21 times that of the shortest side, BC, so we have
\[
   \Rightarrow BC = 21AB \\
   \Rightarrow BC = 21x{\text{ ......eq.(1)}} \\
 \]
We know that the perimeter of a triangle is the sum of the all sides of the triangles.
Adding the sides of a triangle and taking it equal to 2004, we get
\[
   \Rightarrow x + AC + 21x = 2004 \\
   \Rightarrow AC + 22x = 2004 \\
 \]
Subtracting the above equation on both sides by \[22x\], we get
\[
   \Rightarrow AC + 22x - 22x = 2004 - 22x \\
   \Rightarrow AC = 2004 - 22x{\text{ ......eq.(2)}} \\
 \]
We know that the sum of two sides of a triangle is always greater than the third sides, so we have
\[
   \Rightarrow x + 21x > 2004 - 22x \\
   \Rightarrow 22x > 2004 - 22x \\
 \]
Adding the above equation on both sides by \[22x\], we get
\[
   \Rightarrow 22x + 22x > 2004 - 22x + 22x \\
   \Rightarrow 44x > 2004 \\
 \]
Dividing the above equation by 44 on both sides, we get
\[
   \Rightarrow \dfrac{{44x}}{{44}} > \dfrac{{2004}}{{44}} \\
   \Rightarrow x > 45.54{\text{ ......eq.(3)}} \\
 \]
We know that the difference of two sides of a triangle is always less than the third sides, so we have
\[
   \Rightarrow 21x - x < 2004 - 22x \\
   \Rightarrow 20x < 2004 - 22x \\
 \]
Adding the above equation on both sides by \[22x\], we get
\[
   \Rightarrow 20x + 22x < 2004 - 22x + 22x \\
   \Rightarrow 42x < 2004 \\
 \]
Dividing the above equation by 42 on both sides, we get
\[
   \Rightarrow \dfrac{{42x}}{{42}} < \dfrac{{2004}}{{42}} \\
   \Rightarrow x < 47.71{\text{ ......eq.(4)}} \\
 \]
From equation (3) and (4), we get
\[ \Rightarrow 45.54 < x < 47.71\]
Thus, the value of AB cannot be 45 or 48.
Let us take \[x = 47\] for AB.
Substituting the value of \[x\] in the equation (1), we get
\[
   \Rightarrow BC = 21\left( {47} \right) \\
   \Rightarrow BC = 987 \\
 \]
Substituting the value of \[x\] in the equation (2), we get
\[
   \Rightarrow AC = 2004 - 22\left( {47} \right) \\
   \Rightarrow AC = 970 \\
 \]
Adding the value of AB, BC and AC to find the perimeter of the triangle, we get
\[
   \Rightarrow 47 + 970 + 987 \\
   \Rightarrow 2004 \\
 \]
Let us now take \[x = 46\] for AB.
Substituting the value of \[x\] in the equation (1), we get
\[
   \Rightarrow BC = 21\left( {46} \right) \\
   \Rightarrow BC = 966 \\
 \]
Substituting the value of \[x\] in the equation (2), we get
\[
   \Rightarrow AC = 2004 - 22\left( {46} \right) \\
   \Rightarrow AC = 992 \\
 \]
Adding the value of AB, BC and AC to find the perimeter of the triangle, we get
\[
   \Rightarrow 46 + 992 + 966 \\
   \Rightarrow 2004 \\
 \]
Hence, option B is correct.

Note: In solving these types of questions, we should be used to compute the area of a triangle where sides of the triangle are given. Students find the perimeter of triangle of using the formula, \[{\text{Area}} = \dfrac{1}{2} \times {\text{base}} \times {\text{height}}\], but this is wrong. We know that integral side lengths do not have anything related to integration and calculus. It simply means that the length must not be a decimal or fraction, so It must be an integer.