Problem 1 :
Simplify √20 - √225 + √80
Solution :
Decompose 20, 225 and 80 into prime factors.
√20 = √2 ⋅ 2 ⋅ 5 = 2√5
√225 = √5 ⋅ 5 ⋅ 3 ⋅ 3 = 5 ⋅ 3 = 15
√225 = √2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 5 = (2 ⋅ 2)√5 = 4√5
Then, we have
√20 - √225 + √80 = 2√5 - 15 + 4√5
√20 - √225 + √80 = 6√5 - 15
√20 - √225 + √80 = 6√5 - 15
√20 - √225 + √80 = 3(2√5 - 5)
Problem 2 :
The perimeter of the rhombus is 20 cm. One of the diagonals is of length 8 cm. Find the length of the other diagonal.
Solution :
In a rhombus, the diagonals are intersecting at right angles.
Perimeter of rhombus = 20 cm
4a = 20
a = 5 cm
Length of diagonal = 8 cm
Half of diagonal = 4
Let x be the half of the other diagonal.
5^{2 } = 4^{2} + x^{2}
25 = 16 + x^{2}
x^{2 }= 25 - 16
x^{2 }= 9
x = 3
Length of other diagonal = 2(3)
= 6 cm
Problem 3 :
Jerry and Silvia wanted to go from the southwest corner of a square field to the northeast corner. Jerry walked due east and then due north to reach the goal, but Silvia headed northeast and reached the goal walking in a straight line. How much shorter Silvia's trip was, compared to Jerry's trip?
Solution :
Let AB = 1, BC = 1
Then AC = √1^{2} + 1^{2}
AC = √2
Distance covered by Jerry = 1 + 1 ==> 2
Distance covered by Silvia = √2
Difference of distance covered = 2 - √2
= 2 - 1.414
= 0.586
= (0.586/2) ⋅ 100%
= 30%
So Silvia's trip is 30% shorter than the Jerry's trip.
Problem 4 :
Solve log_{10} (2x + 50) = 3
Solution :
log_{10} (2x + 50) = 3
2x + 50 = 10^{3}
2x = 1000-50
2x = 950
x = 425
So, the value of x is 425.
Problem 5 :
A set has only one element is called ____________ set
Solution :
A set has only one element is called as singleton set.
Problem 6 :
Which of the following statement represents this Venn diagram?
Solution :
In the given venn diagram, the common region for both A and B is not shaded, then the remaining part of B is shaded. So the correct statement is B-A.
Problem 7:
If (x+p) (x+q) = x^{2}-5x-300, find the value of p² + q²
Solution :
(x+p) (x+q) = x^{2}-5x-300
x^{2}+(p+q)x+pq = x^{2}-5x-300
p+q = -5 and pq = -300
p^{2}+q^{2} = (p+q)^{2} - 2pq
= (-5)^{2}-2(-300)
= 25 + 600
p^{2}+q^{2} = 625
Problem 8 :
The point of concurrence of the angle bisector of a triangle is called the _____________ of the triangle.
Solution :
The point of concurrence of the angle bisector of a triangle is called the incenter of the triangle. 9
Problem 9 :
The angles are supplementary and larger angle is twice the smaller angle.Find the angles.
Solution :
Let x be the smaller angle
2x - larger angle
2x+x = 180
3x 180
x = 60
So, the required angles are 60 and 120.
Problem 10 :
Find the distance between the points
A (-15,-3) and B (7, 1)
Solution :
Distance between two points = √(7+15)^{2} + (1+3)^{2}
= √(-22)^{2} + 4^{2}
= √(484 + 16)
= √500
= 10√5
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