## Exercise 4.2 Part 3

Question 4: Find two consecutive positive integers, sum of whose squares is 365.

Solution: Let us assume, first integer = x

Then, second integer = x + 1

As per question; x^2 + (x + 1)^2 = 365

Or, x^2 + x^2 + 2x + 1 = 365

Or, 2x^2 + 2x + 1 – 365 = 0

Or, 2x^2 + 2x – 364 = 0

Or, x^2 + x – 182 = 0

Or, x^2 + 14x – 13x – 182 = 0

Or, x(x + 14) – 13(x + 14) = 0

Or, (x – 13)(x + 14) = 0

Hence, x = 13 and x = - 14

Since integers are positive, hence they are 13 and 14

Question 5: The altitude of a right triangle is 7cm less than its base. If the hypotenuse is 13cm, find the other two sides.

Solution: Let us assume, base = x

Therefore; altitude = x – 7

As per question; using Pythagoras Theorem:

13^2 = x^2 + (x – 7)^2

Or, 169 = x^2 + x^2 – 14x + 49

Or, 2x^2 – 14x + 49 – 169 = 0

Or, 2x^2 – 14x – 120 = 0

Or, x^2 – 7x – 60 = 0

Or, x^2 – 12x + 5x – 60 = 0

Or, x(x – 12) + 5(x – 12) = 0

Or, (x + 5)(x – 12) = 0

Hence, x = - 5 and x = 12

Ruling out the negative value; we have x = 12

So, altitude = 12 – 5 = 7

Thus, two sides are 12 cm and 5 cm

Question 6: 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 Rs 90, find the number of articles produced and the cost of each article.

Solution: Let us assume, number of pottery in a day = x

So, cost of production of each article = 2x + 3

As per question; x(2x + 3) = 90

Or, 2x^2 + 3x = 90

Or, 2x^2 + 3x – 90 = 0

Or, 2x^2 – 12x + 15x – 90 = 0

Or, 2x(x – 6) + 15(x – 6) = 0

Or, (2x + 15)(x – 6) = 0

Hence, x = - (15)/(2) and x = 6

Ruling out the negative value; x = 6`

Cost of article = Rs. 15