#### Data Interpretation

• ( 1 ) The distance between the points A(5, -7) and B(2, 3) is:

• 1) 109
• 2) 5√ 7
• 3) √ 109
• 4) 7√ 5
• Discussion in forum
Answer : 3) √ 109
Solution : AB2 = (2 - 5)2 + (3 + 7)2
=> (-3)2 + (10)2
=> 9 + 100 => √ 109

discussion

Answer : 3) √ 109

• ( 2 ) Find the value of k for which the points A(-2, 5), B(3, k) and C(6, 1) are collinear.

• 1) 5
• 2) 7
• 3) 12
• 4) 4
• Discussion in forum
Answer : 1) 5
Solution : x1 = -2, x2 = 3, x3 = 6 and y1 = 5, y2 = k, y3 = -1
Now Δ = 0 <=> -2(k + 1) + 3(-1 + 5) + 6(5 - k) = 0
=> -2 (k+1) + 3(4) + 6(5-k) = 0
=> -2k-2 + 12+30 -6k = 0
=> 40 - 8k = 0
=> -8k = -40
=> k = 5.

discussion

Answer : 1) 5

• ( 3 ) The distance between the points A(b, 0) and B(0, a) is.

• 1) √(a2-b2)
• 2) √(a2+b2)
• 3) √(a+b)
• 4) a-b
• Discussion in forum
Answer : 2) √(a2+b2)
Solution : AB = √((b-0)2 - (0 - a)2)
= √(b2 + a2)
= √(a2 + b2).

discussion

Answer : 2) √(a2+b2)

• ( 4 ) The points A(0, 6), B(-5, 3) and C(3, 1) are the vertices of a triangle which is ?

• 1) equilateral
• 2) right angled
• 3) isosceles
• 4) scalene
• Discussion in forum
Answer : 3) isosceles
Solution : AB2 = (-5 - 0)2 + (-3 - 0)2 = 25 + 9 = 34
BC2 = (3 + 5)2 + (1-3)2 = 82 + (-2)2 = 64 + 4 = 68
AC2 = (3 - 0)2 + (1 - 6)2 = 9 + 25 = 34.
AB = AC. => Δ ABC is isosceles.

discussion

Answer : 3) isosceles

• ( 5 ) The vertices of a quadrilateral ABCD are A(0, 0), B(3,3), C(3, 6) and D(0, 3). Then , ABCD is a

• 1) square
• 2) parallelogram
• 3) rectangle
• 4) rhombus
• Discussion in forum
Answer : 2) parallelogram
Solution : AB2 = (3-0)2 + (3-0)2 = 18
BC2 = (3-3)2 + (6-3)2 = 9 CD2 = (0-3)2 + (3-6)2 =18
AD2 = (0-0)2 + (3-0)2 = 9
AB = CD = √ 18 => 3√ 2,
BC = AD = √ 9
AC2 = (3-0)2 + (6-0)2 = 9 + 36 = 45
BD2 = (0-3)2 + (3-3)2 = 9 + 0 = 9
AC ≠ BD ABCD is a parallelogram.

discussion

Answer : 2) parallelogram

• ( 6 ) Find the distance between the points A(-4, 7) and B(2, -5).

• 1) 8√ 5 Units
• 2) 6√ 4 Units
• 3) 6√ 5 Units
• 4) 7√ 5 Units
• Discussion in forum
Answer : 3) 6√ 5 Units
Solution : AB = √((2+4)2 + (-5-7)2)
= √(62 + (-12)2)
= √(36 + 144) = √ 180
=√ (36*5) = 6√ 5 units.

discussion

Answer : 3) 6√ 5 Units

• ( 7 ) Find the distance of the point A(4, -4) from the origin.

• 1) 3√ 2
• 2) 2√ 8
• 3) 6√ 2
• 4) 8√ 2
• Discussion in forum
Answer : 4) 8√ 2
Solution : OA = √(42 + (-4)2) = √(16 + 16) = √ 32 = 8√ 2

discussion

Answer : 4) 8√ 2

• ( 8 ) A is a point on y-axis at a distance of 5 units from x-axis lying below x-axis. The co-ordinates of A are:

• 1) (5, 0)
• 2) (-5, 0)
• 3) (0, 5)
• 4) (0, -5)
• Discussion in forum
Answer : 4) (0, -5)
Solution : The co-ordinates of A are A(0, -5)

discussion

Answer : 4) (0, -5)

• ( 9 ) Find the area of ΔABC whose vertices are A(9, -5), B(3, 7) and (-2, 4).

• 1) 29 units
• 2) 35.9 sq.units
• 3) 39 sq.units
• 4) 39.5 sq.units
• Discussion in forum
Answer : 3) 39 sq.units
Solution : Here, x1 = 9, x2 = 3, x3 = -2 and y1 = -5, y2 = 7, y3 = 4
= 1/2 [9(7 - 4) + 3(4 + 5) + (-2)(-5 - 7)]
= 1/2 [9(3) + 3(9) - 2(-12)]
= 1/2 [27 + 27 + 24]
= 1/2 
= 39 sq.units

discussion

Answer : 3) 39 sq.units

• ( 10 ) If the sum of the dimensions of rectangular parallel piped is 24 cm and the length of the diagonal is 15 cm, then the total surface are of its is

• 1) 420 cm2
• 2) 275 cm2
• 3) 351 cm2
• 4) 378 cm2
• Discussion in forum
Answer : 3) 351 cm2
Solution : Let the dimensions of the rectangular parallelepiped be x, y and z
Then x + y + z = 24
Length of the diagonal √(x2 + y2 + z2) = 15
x2 + y2 + z2 = 225
Now , (x+y+z)2 = x2 + y2 + z2 + 2(xy+yz+zx)
(24)2 = 225+2(xy+yz+zx)
2(xy+yz+zx) = 576-225 =351
Total surface area = 2(xy+yz+zx) = 351 cm2

discussion

Answer : 3) 351 cm2