Plz help me solve these sums i will be great full
1.How many non congruent right triangles with positive integer leg lengths have areas that are numerically equal to 3 times their perimeters
2.A triangle with side lengths in the ratio 3:4:5 is inscribed in a circle of radius 3.What is the area the triangle
3.Two points A and B .Find the set of feet of perpendiculars dropped from the point A onto all possible straight lines passing through the point B.
Sure, I'd be happy to help you solve these problems!
1. Let's solve the first problem together. We are looking for non-congruent right triangles with positive integer leg lengths whose areas are equal to 3 times their perimeters.
Let's call the lengths of the two legs a and b, and the hypotenuse c.
The perimeter of a right triangle is given by the formula P = a + b + c.
The area of a right triangle is given by the formula A = (1/2) * a * b.
We are given that A = 3P. Using the formulas above, we can substitute these values:
(1/2) * a * b = 3 * (a + b + c)
Now, we need to find the distinct pairs of positive integer values (a, b) that satisfy this equation.
To make things easier, we can rewrite the equation as:
2ab = 6(a + b + c)
ab = 3(a + b + c)
Now, we can start by testing different values for a and b. Since we are looking for positive integer values, let's start with small integers. We can also consider that a > b, since the triangles are non-congruent.
By substituting some values, we can find that a = 4, b = 3 is a valid solution. Let's check it:
ab = 4 * 3 = 12
a + b + c = 4 + 3 + c = 7 + c
3(a + b + c) = 3(7 + c) = 21 + 3c
We can see that ab = 12 is equal to 3(a + b + c) = 21 + 3c.
So, a = 4 and b = 3 is a valid solution to the problem. This means there is one non-congruent right triangle with positive integer leg lengths that has an area numerically equal to 3 times its perimeter.
You can continue to test other pairs of positive integer values for (a, b) to see if any other solutions exist.