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Q5:
The altitude of a right triangle is $7$ cm less than its base. If the hypotenuse is $13$ cm, find the other two sides.
Solution :
Given:
- A right-angled triangle.
- The length of the hypotenuse is $13$ cm.
- The altitude (height) is $7$ cm less than the base.
To find:
- The lengths of the base and the altitude of the triangle.
Step 1: Defining Variables
Let the base of the right triangle be $x$ cm.
According to the problem, the altitude is $7$ cm less than the base.
Therefore, the altitude = $(x - 7)$ cm.
Step 2: Applying the Pythagorean Theorem
In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides (Pythagorean Theorem):
$(\text{Hypotenuse})^2 = (\text{Base})^2 + (\text{Altitude})^2$
Substituting the given values:
$13^2 = x^2 + (x - 7)^2$
Step 3: Expanding and Simplifying the Equation
$169 = x^2 + (x^2 - 14x + 49)$ [Using the identity $(a - b)^2 = a^2 - 2ab + b^2$]
$169 = 2x^2 - 14x + 49$
Subtract $169$ from both sides to set the quadratic equation to zero:
$0 = 2x^2 - 14x + 49 - 169$
$2x^2 - 14x - 120 = 0$
Divide the entire equation by $2$ to simplify:
$x^2 - 7x - 60 = 0$
Step 4: Solving the Quadratic Equation by Factorization
We need to find two numbers that multiply to $-60$ and add to $-7$. These numbers are $-12$ and $5$.
$x^2 - 12x + 5x - 60 = 0$
$x(x - 12) + 5(x - 12) = 0$
$(x - 12)(x + 5) = 0$
Step 5: Determining the Possible Values for x
Setting each factor to zero:
1) $x - 12 = 0 \implies x = 12$
2) $x + 5 = 0 \implies x = -5$
Since the length of a side of a triangle cannot be negative, we discard $x = -5$.
Therefore, the base $x = 12$ cm.
Step 6: Calculating the Altitude
Altitude = $x - 7$
Altitude = $12 - 7 = 5$ cm.
Final Answer: The base of the triangle is 12 cm and the altitude is 5 cm.
More Questions from Class 10 Mathematics Quadratic Equations EXERCISE 4.2
- Q1(i): Find the roots of the following quadratic equations by factorisation: (i) $x^2 – 3x – 10 = 0$
- Q1(ii): Find the roots of the following quadratic equations by factorisation: (ii) $2x^2 + x – 6 = 0$
- Q1(iii): Find the roots of the following quadratic equations by factorisation: (iii) $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$
- Q1(iv): Find the roots of the following quadratic equations by factorisation: (iv) $2x^2 – x + \frac{1}{8} = 0$
- Q1(v): Find the roots of the following quadratic equations by factorisation: (v) $100x^2 – 20x + 1 = 0$
- Q2: Solve the problems given below. Represent the following situations mathematically: (i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with. (ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs.750. We would like to find out the number of toys produced on that day.
- Q3: Find two numbers whose sum is $27$ and product is $182$.
- Q4: Find two consecutive positive integers, sum of whose squares is $365$.
- Q6: A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was $3$ more than twice the number of articles produced on that day. If the total cost of production on that day was ₹ $90$, find the number of articles produced and the cost of each article.
CBSE Solutions for Class 10 Mathematics Quadratic Equations
Chapters in CBSE - Class 10 Mathematics
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