10th cbse maths solution for exercise 4.2 part 3
This page 10th cbse maths solution for exercise 4.2 part 3 is going to provide you solution for every problems that you find in the exercise no 4.2
10th CBSE maths solution for Exercise 4.2 part 3
(5) The altitude of a right triangle is 7cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Solution:
Let 'x' be the base of the triangle
Altitude = x - 7
(Hypotenuse side)² = (Base)² + (Height)²
13² = x² + (x - 7)²
169 = x² + x² - 2 (x) (7) + 7²
169 = x² + x² - 14 x + 49
0 = 2 x²- 14 x + 49 - 169
2 x²- 14 x - 120 = 0
dividing the whole equation by 2 => x²- 7 x - 60 = 0
x²- 12 x + 5 x - 60 = 0
x (x - 12) + 5 (x - 12) = 0
(x + 5) (x - 12) = 0
x + 5 = 0 x - 12 = 0
x = -5 x = 12
Base = 12 cm
Height = 12 - 7 = 5 cm
(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 'x' be the number of articles produced on that day
cost of production of each article = 2 x + 3
Total cost of production = number of article x cost of one article
90 = x (2 x + 3)
90 = 2 x² + 3 x
2 x² + 3 x - 90 = 0
2 x² - 12 x + 15 x - 90 = 0
2 x (x - 6) + 15 (x - 6) = 0
(2 x + 15) (x - 6) = 0
2 x + 15 = 0 x - 6 = 0
2 x = -15 x = 6
x = -15/2
Therefore the number of articles produced on that particular day = 6
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