Q27 of 40 Page 1

Given that √5 is irrational, prove that (2√5−3) is an irrational number.

OR


If HCF of 144 and 180 is expressed in the form 13m-16. Find the value of m.


Proof :


Given - √5 is irrational


To prove - (2√5−3) is an irrational number


Property - Rational number is one which can be represented in the form p/q, where q is not equal to 0.


Answer –


As given √5 is irrational.


Now, let us assume that (2√5−3) is a rational number.


We know that rational numbers can be represented in the form of p/q where p & q are integers and q is not equal to 0.





In above equation p, q, 3, 2 are integers, hence must be rational.


But √5 is irrational number.


Therefore, above equation is contradiction.


Hence, our assumption is false.


Therefore, (2√5−3) is an irrational number.


OR


m = 4


Given – HCF of 144 and 180 is 13m – 16


To find – value of m


Property –


HCF of two integers m & n is


HCF = m mod n , where m>n


(m mod n = remainder after dividing m by n)


Answer –


HCF of 144 and 180 is


180 mod 144 = 36


(we get 36 as a remainder after dividing 180 by 144)


But as given HCF of 144 and 180 is 13m – 16


13m - 16 = 36


13m = 52


m = 4


Hence, the value of m is 4.


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OR


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26

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28

If the sum of first m terms of an AP is the same as the sum of its first n terms, show that the sum of its first (m+n) terms is zero.

29

In the figure, ABCDE is a pentagon with BE||CD and BC||DE. BC is perpendicular to CD. AB= 5cm, AE=5cm, BE= 7cm, BC= x-y and CD= x+y. If the perimeter of ABCDE is 27cm. find the value of x and y, given x, y ≠0.


OR


Solve the following system of equations: