Samacheer Kalvi Class 10 Maths Solutions Chapter 3 Algebra Exercise 3.12

Question 1. If the difference between a number and its reciprocal is \( \frac { 24 }{ 5 } \), find the number.
Answer: Let the number be \( x \). Its reciprocal will be \( \frac { 1 }{ x } \).
According to the given condition, the difference between the number and its reciprocal is \( \frac { 24 }{ 5 } \).
So, we write the equation:
\( x - \frac { 1 }{ x } = \frac { 24 }{ 5 } \)
To solve this, we combine the terms on the left side:
\( \frac { x^{2} - 1 }{ x } = \frac { 24 }{ 5 } \)
Next, we cross-multiply:
\( 5(x^{2} - 1) = 24x \)
\( 5x^{2} - 5 = 24x \)
Rearrange the terms to form a quadratic equation:
\( 5x^{2} - 24x - 5 = 0 \)
Now, we factor the quadratic equation. We look for two numbers that multiply to \( 5 \times (-5) = -25 \) and add to \( -24 \). These numbers are \( -25 \) and \( 1 \).
\( 5x^{2} - 25x + x - 5 = 0 \)
Group the terms and factor:
\( 5x(x - 5) + 1(x - 5) = 0 \)
\( (x - 5)(5x + 1) = 0 \)
This gives us two possible solutions for \( x \):
\( x - 5 = 0 \implies x = 5 \)
or
\( 5x + 1 = 0 \implies 5x = -1 \implies x = \frac { -1 }{ 5 } \)
Thus, the number can be either 5 or \( \frac { -1 }{ 5 } \). This problem has two possible solutions because both satisfy the given condition.
In simple words: We are looking for a number where if you subtract its inverse from it, you get 24/5. We set up an equation, solve it like a quadratic, and find that two numbers, 5 and -1/5, fit this description.

🎯 Exam Tip: When dealing with reciprocal problems, remember that quadratic equations often yield two solutions. Always check both solutions against the original problem statement.

Question 2. A garden measuring 12m by 16m is to have a pedestrian pathway that is meters wide installed all the way around so that it increases the total area to 285 m². What is the width of the pathway?
Answer: Let the width of the pathway be \( \omega \) meters.
The original garden has length 16 m and breadth 12 m.
When a pathway of width \( \omega \) is added all around, the new dimensions of the garden including the pathway will be:
New length \( = 16 + \omega + \omega = 16 + 2\omega \)
New breadth \( = 12 + \omega + \omega = 12 + 2\omega \)
The total area after adding the pathway is 285 m². The area of the new, larger rectangle is its new length multiplied by its new breadth.
So, \( (16 + 2\omega)(12 + 2\omega) = 285 \)
Expand the equation:
\( 16 \times 12 + 16 \times 2\omega + 2\omega \times 12 + 2\omega \times 2\omega = 285 \)
\( 192 + 32\omega + 24\omega + 4\omega^{2} = 285 \)
Combine like terms and rearrange into a quadratic equation:
\( 4\omega^{2} + 56\omega + 192 = 285 \)
\( 4\omega^{2} + 56\omega + 192 - 285 = 0 \)
\( 4\omega^{2} + 56\omega - 93 = 0 \)
We use the quadratic formula \( x = \frac { -b \pm \sqrt{b^{2} - 4ac} }{ 2a } \) to solve for \( \omega \).
Here, \( a = 4, b = 56, c = -93 \).
\( \omega = \frac { -56 \pm \sqrt{56^{2} - 4(4)(-93)} }{ 2(4) } \)
\( \omega = \frac { -56 \pm \sqrt{3136 + 1488} }{ 8 } \)
\( \omega = \frac { -56 \pm \sqrt{4624} }{ 8 } \)
\( \omega = \frac { -56 \pm 68 }{ 8 } \)
We have two possible values for \( \omega \):
\( \omega = \frac { -56 + 68 }{ 8 } = \frac { 12 }{ 8 } = 1.5 \)
or
\( \omega = \frac { -56 - 68 }{ 8 } = \frac { -124 }{ 8 } = -15.5 \)
Since the width of a pathway cannot be negative, we discard \( -15.5 \).
Therefore, the width of the pathway is 1.5 m. Adding a path increases the overall dimensions of the space, so the area also increases.
In simple words: The garden gets bigger by adding a path around it. We set up an equation using the new total area, which leads to a quadratic equation. Solving it gives us the path width, which must be a positive number.

🎯 Exam Tip: When setting up equations for problems involving borders or pathways, remember that the width is added to both sides of the original dimensions, so you must add \( 2\omega \) (twice the width) to both the length and breadth.

Question 3. A bus covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hour more it would have taken 30 minutes less for the journey. Find the original speed of the journey.
Answer: Let the original speed of the bus be \( x \) km/hr.
The distance covered is 90 km.
Using the formula Time = Distance / Speed, the time taken to cover 90 km at speed \( x \) is \( \frac { 90 }{ x } \) hours.
If the speed had been 15 km/hr more, the new speed would be \( (x + 15) \) km/hr.
The time taken at this increased speed would be \( \frac { 90 }{ x+15 } \) hours.
We are told that with the increased speed, it would have taken 30 minutes less. 30 minutes is equivalent to \( \frac { 30 }{ 60 } = \frac { 1 }{ 2 } \) hour.
So, the difference in time is \( \frac { 1 }{ 2 } \) hour:
\( \frac { 90 }{ x } - \frac { 90 }{ x+15 } = \frac { 1 }{ 2 } \)
To solve this, find a common denominator for the left side:
\( \frac { 90(x+15) - 90x }{ x(x+15) } = \frac { 1 }{ 2 } \)
Simplify the numerator:
\( \frac { 90x + 1350 - 90x }{ x^{2} + 15x } = \frac { 1 }{ 2 } \)
\( \frac { 1350 }{ x^{2} + 15x } = \frac { 1 }{ 2 } \)
Cross-multiply:
\( 2 \times 1350 = 1 \times (x^{2} + 15x) \)
\( 2700 = x^{2} + 15x \)
Rearrange into a quadratic equation:
\( x^{2} + 15x - 2700 = 0 \)
Factor the quadratic equation. We need two numbers that multiply to \( -2700 \) and add to \( 15 \). These numbers are \( 60 \) and \( -45 \).
\( (x + 60)(x - 45) = 0 \)
This gives us two possible values for \( x \):
\( x + 60 = 0 \implies x = -60 \)
or
\( x - 45 = 0 \implies x = 45 \)
Since speed cannot be negative, we discard \( x = -60 \).
Therefore, the original speed of the bus is 45 km/hr. Calculating time differences often leads to these types of quadratic equations.
In simple words: We set up an equation comparing the time taken at normal speed versus faster speed, knowing the difference is 30 minutes. Solving this equation helps us find the bus's original speed, ignoring any negative answers.

🎯 Exam Tip: Always convert all time units to hours before setting up equations in speed-distance-time problems to avoid errors.

Question 4. A girl is twice as old as her sister. Five years hence, the product of their ages – (in years) will be 375. Find their present ages.
Answer: Let the present age of the sister be \( x \) years.
Since the girl is twice as old as her sister, the present age of the girl is \( 2x \) years.
"Five years hence" means five years from now.
In five years, the sister's age will be \( x + 5 \) years.
In five years, the girl's age will be \( 2x + 5 \) years.
The problem states that the product of their ages in five years will be 375.
So, we can write the equation:
\( (x + 5)(2x + 5) = 375 \)
Expand the left side of the equation:
\( x(2x) + x(5) + 5(2x) + 5(5) = 375 \)
\( 2x^{2} + 5x + 10x + 25 = 375 \)
Combine the like terms and rearrange into a quadratic equation:
\( 2x^{2} + 15x + 25 - 375 = 0 \)
\( 2x^{2} + 15x - 350 = 0 \)
We use the quadratic formula \( x = \frac { -b \pm \sqrt{b^{2} - 4ac} }{ 2a } \) to solve for \( x \).
Here, \( a = 2, b = 15, c = -350 \).
\( x = \frac { -15 \pm \sqrt{15^{2} - 4(2)(-350)} }{ 2(2) } \)
\( x = \frac { -15 \pm \sqrt{225 + 2800} }{ 4 } \)
\( x = \frac { -15 \pm \sqrt{3025} }{ 4 } \)
\( x = \frac { -15 \pm 55 }{ 4 } \)
We have two possible values for \( x \):
\( x = \frac { -15 + 55 }{ 4 } = \frac { 40 }{ 4 } = 10 \)
or
\( x = \frac { -15 - 55 }{ 4 } = \frac { -70 }{ 4 } = -17.5 \)
Since age cannot be negative, we discard \( x = -17.5 \).
So, the present age of the sister is \( x = 10 \) years.
The present age of the girl is \( 2x = 2 \times 10 = 20 \) years.
It's always a good practice to check if these ages fit the original conditions.
In simple words: We used variables for the sister's and girl's current ages. Then we wrote an equation for their ages five years from now, multiplied them, and set it equal to 375. We solved this equation to find their ages, making sure to pick a positive answer.

🎯 Exam Tip: Always define your variables clearly (e.g., "Let x be the sister's age") and ensure you account for future or past ages correctly. Discard negative solutions for age or other physical quantities.

Question 5. A pole has to be erected at a point on the boundary of a circular ground of diameter 20 m in such a way that the difference of its distances from two diametrically opposite fixed gates P and Q on the boundary is 4 m. Is it possible to do so? If answer is yes at what distance from the two gates should the pole be erected?
Answer: Let the circular ground have diameter PQ = 20 m. Gates P and Q are diametrically opposite.
Let R be the location of the pole on the boundary.
The distance from gate P to the pole is PR = \( x \) m.
The distance from gate Q to the pole is QR.
The problem states that the difference of its distances from P and Q is 4 m. So, \( |PR - QR| = 4 \). We can assume PR > QR, so \( PR - QR = 4 \).
This means \( QR = PR - 4 = x - 4 \) m.
Since P, Q, and R are all on the boundary of the circle and PQ is the diameter, the triangle PQR must be a right-angled triangle with the right angle at R (Thales' theorem).
So, by Pythagoras theorem, \( PR^{2} + QR^{2} = PQ^{2} \).
Substitute the values:
\( x^{2} + (x - 4)^{2} = 20^{2} \)
\( x^{2} + (x^{2} - 8x + 16) = 400 \)
\( 2x^{2} - 8x + 16 = 400 \)
\( 2x^{2} - 8x + 16 - 400 = 0 \)
\( 2x^{2} - 8x - 384 = 0 \)
Divide the entire equation by 2 to simplify:
\( x^{2} - 4x - 192 = 0 \)
Factor the quadratic equation. We need two numbers that multiply to \( -192 \) and add to \( -4 \). These numbers are \( -16 \) and \( 12 \).
\( (x - 16)(x + 12) = 0 \)
This gives two possible values for \( x \):
\( x - 16 = 0 \implies x = 16 \)
or
\( x + 12 = 0 \implies x = -12 \)
Since distance cannot be negative, we discard \( x = -12 \).
So, \( PR = x = 16 \) m.
Then, \( QR = x - 4 = 16 - 4 = 12 \) m.
Since we found a valid positive distance, it is possible to erect the pole. The distances from the two gates are 16 m and 12 m. Understanding geometric properties like Thales' theorem is key here.
In simple words: We used the fact that a triangle drawn inside a circle with its longest side as the diameter is always a right-angled triangle. Then, we used Pythagoras' theorem with the given information to find the distances to the pole, ignoring any negative results.

🎯 Exam Tip: Remember Thales' theorem: if A, B, and C are distinct points on a circle where the line AC is a diameter, then the angle \( \angle ABC \) is a right angle.

Question 6. From a group of \( 2x^{2} \) black bees, square root of half of the group went to a tree. Again eight-ninth of the bees went to the same tree. The remaining two got caught up in a fragrant lotus. How many bees were there in total?
Answer: Let the total number of black bees be \( 2x^{2} \).
First, half of the group is \( \frac { 1 }{ 2 } \times 2x^{2} = x^{2} \).
The square root of half of the group went to a tree, so the number of bees that went first is \( \sqrt{x^{2}} = x \).
Next, eight-ninth of the total bees went to the same tree. This means \( \frac { 8 }{ 9 } \times 2x^{2} = \frac{16x^{2}}{9} \) bees went.
The remaining two bees got caught in a lotus.
The sum of bees that went to the tree and the bees in the lotus must equal the total number of bees.
So, \( x + \frac{16x^{2}}{9} + 2 = 2x^{2} \)
To remove the fraction, multiply the entire equation by 9:
\( 9x + 16x^{2} + 18 = 18x^{2} \)
Rearrange into a quadratic equation by bringing all terms to one side:
\( 18x^{2} - 16x^{2} - 9x - 18 = 0 \)
\( 2x^{2} - 9x - 18 = 0 \)
Factor the quadratic equation. We need two numbers that multiply to \( 2 \times (-18) = -36 \) and add to \( -9 \). These numbers are \( -12 \) and \( 3 \).
\( 2x^{2} - 12x + 3x - 18 = 0 \)
Group the terms and factor:
\( 2x(x - 6) + 3(x - 6) = 0 \)
\( (x - 6)(2x + 3) = 0 \)
This gives two possible values for \( x \):
\( x - 6 = 0 \implies x = 6 \)
or
\( 2x + 3 = 0 \implies 2x = -3 \implies x = \frac { -3 }{ 2 } \)
Since the number of bees (and \( x \)) cannot be negative, we discard \( x = \frac { -3 }{ 2 } \).
So, \( x = 6 \).
The total number of black bees is \( 2x^{2} \). Substitute \( x = 6 \):
Total bees \( = 2 \times (6)^{2} = 2 \times 36 = 72 \).
Thus, there were 72 bees in total. Setting up the equation carefully based on the parts of the group is crucial.
In simple words: We used algebra to figure out the total number of bees. We added up the bees that went to the tree and the bees in the lotus, and set that equal to the total group. Solving the equation gave us the final number of bees.

🎯 Exam Tip: Be careful with phrases like "square root of half" and "eight-ninth of the bees". Ensure you apply the operations to the correct quantity (total bees or half the bees). Always discard negative solutions when counting objects.

Question 7. Music is been played in two opposite galleries with certain group of people. In the first gallery a group of 4 singers were singing and in the second gallery 9 singers were singing. The two galleries are separated by the distance of 70 m. Where should a person stand for hearing the same intensity of the singers voice? (Hint: The ratio of the sound intensity is equal to the square of the ratio of their corresponding distances).
Answer: Let the number of singers in the first group be \( S_1 = 4 \).
Let the number of singers in the second group be \( S_2 = 9 \).
The total distance between the two galleries is 70 m.
Let \( x \) be the distance of the person from the first group (gallery 1).
Then, the distance of the person from the second group (gallery 2) will be \( (70 - x) \) m.
The hint states that the ratio of sound intensity is equal to the square of the ratio of their corresponding distances. We can assume intensity is proportional to the number of singers.
So, \( \frac { S_1 }{ S_2 } = \frac { x^{2} }{ (70 - x)^{2} } \)
Substitute the number of singers:
\( \frac { 4 }{ 9 } = \frac { x^{2} }{ (70 - x)^{2} } \)
To solve for \( x \), take the square root of both sides:
\( \sqrt{\frac { 4 }{ 9 }} = \sqrt{\frac { x^{2} }{ (70 - x)^{2} }} \)
\( \frac { 2 }{ 3 } = \frac { x }{ 70 - x } \)
Now, cross-multiply:
\( 2(70 - x) = 3x \)
\( 140 - 2x = 3x \)
Add \( 2x \) to both sides:
\( 140 = 3x + 2x \)
\( 140 = 5x \)
Divide by 5:
\( x = \frac { 140 }{ 5 } = 28 \)
So, the required distance to hear the same intensity of sound from the first gallery is 28 m.
The distance from the second gallery will be \( 70 - x = 70 - 28 = 42 \) m.
This shows that to hear the same sound intensity, you need to be closer to the group with fewer singers, as expected.
In simple words: We want to find a spot where the sound from two groups of singers sounds equally loud. Using the given hint about sound intensity and distance, we set up an equation with the number of singers and the distances. Solving it tells us how far from each group the person should stand.

🎯 Exam Tip: Pay close attention to hints provided in problem statements, especially in physics-related word problems, as they often give the exact formula needed to solve the question.

Question 8. There is a square field whose side is 10 m. A square flower bed is prepared in its centre leaving a gravel path all round the flower bed. The total cost of laying the flower bed and gravelling the path at Rs.3 and Rs.4 per square metre respectively is 364. Find the width of the gravel path.
Answer: The side of the square field is 10 m. Its area is \( 10 \times 10 = 100 \) sq.m.
Let the width of the gravel path be \( x \) meters.
The path surrounds a square flower bed in the center.
So, the side of the flower bed will be \( 10 - x - x = 10 - 2x \) meters.
The area of the flower bed is \( (10 - 2x)^{2} = (10 - 2x)(10 - 2x) = 100 - 20x - 20x + 4x^{2} = 100 - 40x + 4x^{2} \) sq.m.
The area of the gravel path is the total area of the field minus the area of the flower bed:
Area of path \( = 100 - (100 - 40x + 4x^{2}) = 100 - 100 + 40x - 4x^{2} = 40x - 4x^{2} \) sq.m.
The cost of laying the flower bed is Rs.3 per square metre.
Cost of flower bed \( = 3 \times (100 - 40x + 4x^{2}) \)
The cost of gravelling the path is Rs.4 per square metre.
Cost of path \( = 4 \times (40x - 4x^{2}) \)
The total cost is Rs.364.
So, \( 3(100 - 40x + 4x^{2}) + 4(40x - 4x^{2}) = 364 \)
\( 300 - 120x + 12x^{2} + 160x - 16x^{2} = 364 \)
Combine like terms:
\( (12x^{2} - 16x^{2}) + (-120x + 160x) + 300 = 364 \)
\( -4x^{2} + 40x + 300 = 364 \)
Rearrange into a quadratic equation:
\( -4x^{2} + 40x + 300 - 364 = 0 \)
\( -4x^{2} + 40x - 64 = 0 \)
Divide the entire equation by -4 to simplify:
\( x^{2} - 10x + 16 = 0 \)
Factor the quadratic equation. We need two numbers that multiply to 16 and add to -10. These numbers are -8 and -2.
\( (x - 8)(x - 2) = 0 \)
This gives two possible values for \( x \):
\( x - 8 = 0 \implies x = 8 \)
or
\( x - 2 = 0 \implies x = 2 \)
The width of the path \( x \) cannot be 8 m, because the original field side is 10 m. If \( x=8 \), then the flower bed side \( 10 - 2x = 10 - 2(8) = 10 - 16 = -6 \), which is not possible.
Therefore, the width of the gravel path is 2 m. Careful consideration of geometric constraints is essential for these types of problems.
In simple words: We found the areas of the flower bed and the path using a variable for the path's width. Then, we used the given costs to create an equation for the total cost. Solving this equation gave us two possible widths, but only one made sense for the actual dimensions of the field.

🎯 Exam Tip: Always verify your solutions in context. A calculated width must be physically possible (e.g., a path width cannot be greater than half the field's dimension, or lead to negative lengths).

Question 9. Two women together took 100 eggs to a market, one had more than the other. Both sold them for the same sum of money. The first then said to the second: “If I had your eggs, I would have earned Rs.15”, to which the second replied: "If I had your eggs, I would have earned \( 6 \frac{2}{3} \). How many eggs did each had in the beginning?
Answer: Let the number of eggs with woman 1 be \( x \), and with woman 2 be \( y \).
Together, they had 100 eggs: \( x + y = 100 \).
Let woman 1 sell her eggs at a price of Rs.\( a \) per egg.
Let woman 2 sell her eggs at a price of Rs.\( b \) per egg.
Both sold them for the same total sum of money. So, \( ax = by \).
Woman 1's statement: "If I had your eggs (y), I would have earned Rs.15."
This means \( ay = 15 \).
Woman 2's statement: "If I had your eggs (x), I would have earned Rs.\( 6 \frac{2}{3} \)."
\( 6 \frac{2}{3} = \frac { (6 \times 3) + 2 }{ 3 } = \frac { 18 + 2 }{ 3 } = \frac { 20 }{ 3 } \).
This means \( bx = \frac { 20 }{ 3 } \).
We have a system of equations:
1) \( x + y = 100 \)
2) \( ax = by \)
3) \( ay = 15 \)
4) \( bx = \frac { 20 }{ 3 } \)
From (2), \( a = \frac { by }{ x } \). Substitute this into (3):
\( \left(\frac { by }{ x }\right)y = 15 \implies by^{2} = 15x \)
From (2), \( b = \frac { ax }{ y } \). Substitute this into (4):
\( \left(\frac { ax }{ y }\right)x = \frac { 20 }{ 3 } \implies ax^{2} = \frac { 20 }{ 3 } y \)
Now, we have two new equations:
A) \( by^{2} = 15x \)
B) \( ax^{2} = \frac { 20 }{ 3 } y \)
Multiply equation A by B:
\( (by^{2})(ax^{2}) = (15x)\left(\frac { 20 }{ 3 } y\right) \)
\( abx^{2}y^{2} = (5x)(20y) \)
\( abx^{2}y^{2} = 100xy \)
Since \( x \) and \( y \) are numbers of eggs, they are not zero, so we can divide by \( xy \):
\( abxy = 100 \)
From \( ax = by \), we can substitute \( by \) with \( ax \):
\( a(ax)y = 100 \implies a^{2}xy = 100 \)
Alternatively, substitute \( ax \) with \( by \):
\( (by)xy = 100 \implies b^{2}xy = 100 \)
Let's use the given values differently. We have \( ay = 15 \) and \( bx = \frac{20}{3} \).
Also, \( ax = by \). Let this common total earning be \( E \). So \( E = ax = by \).
From \( ay=15 \), we have \( a = \frac{15}{y} \).
From \( bx=\frac{20}{3} \), we have \( b = \frac{20}{3x} \).
Substitute \( a \) and \( b \) into \( ax = by \):
\( \left(\frac{15}{y}\right)x = \left(\frac{20}{3x}\right)y \)
\( \frac{15x}{y} = \frac{20y}{3x} \)
Cross-multiply:
\( 15x \times 3x = 20y \times y \)
\( 45x^{2} = 20y^{2} \)
Divide by 5:
\( 9x^{2} = 4y^{2} \)
Take the square root of both sides (since \( x,y \) are positive):
\( \sqrt{9x^{2}} = \sqrt{4y^{2}} \)
\( 3x = 2y \)
So, \( y = \frac { 3x }{ 2 } \).
Now substitute this into our first equation \( x + y = 100 \):
\( x + \frac { 3x }{ 2 } = 100 \)
Multiply by 2 to clear the fraction:
\( 2x + 3x = 200 \)
\( 5x = 200 \)
\( x = \frac { 200 }{ 5 } = 40 \)
Now find \( y \):
\( y = 100 - x = 100 - 40 = 60 \).
So, one woman had 40 eggs and the other had 60 eggs. This riddle-like problem tests your ability to set up and solve multiple equations simultaneously.
In simple words: We used the clues given by the two women about how much they would earn with each other's eggs. This helped us create equations to find a relationship between the number of eggs each woman had. We also knew they had 100 eggs total. Solving these equations together told us exactly how many eggs each woman started with.

🎯 Exam Tip: When dealing with multiple unknown variables, try to set up as many independent equations as there are unknowns. Look for relationships between the variables that can simplify the system of equations.

Question 10. The hypotenuse of a right angled triangle is 25 cm and its perimeter 56 cm. Find the length of the smallest side.
Answer: Let the sides of the right-angled triangle be \( a, b, \) and \( c \).
The hypotenuse \( c = 25 \) cm.
The perimeter is 56 cm, so \( a + b + c = 56 \).
Substitute \( c = 25 \): \( a + b + 25 = 56 \).
This means \( a + b = 56 - 25 = 31 \) cm.
So, \( b = 31 - a \).
For a right-angled triangle, we can use the Pythagorean theorem: \( a^{2} + b^{2} = c^{2} \).
Substitute the known values into the theorem:
\( a^{2} + (31 - a)^{2} = 25^{2} \)
Expand \( (31 - a)^{2} \):
\( a^{2} + (31^{2} - 2 \times 31 \times a + a^{2}) = 625 \)
\( a^{2} + (961 - 62a + a^{2}) = 625 \)
Combine like terms and rearrange into a quadratic equation:
\( 2a^{2} - 62a + 961 = 625 \)
\( 2a^{2} - 62a + 961 - 625 = 0 \)
\( 2a^{2} - 62a + 336 = 0 \)
Divide the entire equation by 2 to simplify:
\( a^{2} - 31a + 168 = 0 \)
Factor the quadratic equation. We need two numbers that multiply to 168 and add to -31. These numbers are -24 and -7.
\( (a - 24)(a - 7) = 0 \)
This gives two possible values for \( a \):
\( a - 24 = 0 \implies a = 24 \)
or
\( a - 7 = 0 \implies a = 7 \)
If \( a = 24 \), then \( b = 31 - 24 = 7 \).
If \( a = 7 \), then \( b = 31 - 7 = 24 \).
The sides of the triangle are 7 cm, 24 cm, and 25 cm. The smallest side among these is 7 cm. This problem beautifully combines basic geometry and algebra.
In simple words: We knew the hypotenuse and the total perimeter of a right triangle. We used the perimeter to find the sum of the other two sides. Then, applying the Pythagoras theorem, we solved a quadratic equation to find the lengths of these sides. The smallest of these sides is our answer.

🎯 Exam Tip: For right-angled triangle problems, always have the Pythagorean theorem \( a^{2} + b^{2} = c^{2} \) in mind. Remember that perimeter means the sum of all sides.