Question

(152 QUESTIUNS)
1. The sum of three positive numbers is 26. The second number is 3 times as large as the first. If
the sum of the squares of these numbers is least, find the numbers.
(ISC 2000)
BC is a right angled triangle of given area S. Find the sides of the triangle for which the area​

Answer 1

Answer:

260

Step-by-step explanation:

first number is a then second number is 3a let third number be b. sum is 26 implies 4a+b=26

sum of squares(S)=a^2+(3a)^2+b^2=10a^2+b^2

=8*(5a^2/4)+5*(b^2/5)

=8*( sqrt(5)*a/2 )^2+ 5* (b/sqrt(5))^2

using root mean square>=mean

we have sqrt(S/13)>= (4*sqrt(5)*a+sqrt(5)*b)/13

sqrt(S/13)>=26*sqrt(5)/13

S>=260