Free Geometry Set I - 03 Practice Test - CAT
Question 1
The coordinates of circumcentre of the triangle with vertices (2,3), (4, - 1) and (4,3) are
SOLUTION
Solution : C
1st method: - In a right angled triangle
Circumradius = hypotenuse2
The coordinates of circumcentre = (3,1)
2nd method: - If O(x1, y1) is the circumcentre, then OA2=OB2.
(x1−2)2+(Y1−3)2=(X1−4)2+(Y1+1)2
−4X1−6Y1+8X1−2Y1=17−13=4
i.e., 4X1−8Y1=4 or X1−2Y1=1 ..................(I)
OB2=OC2
X1−4)2+(Y1+1)2=(X1−4)2+(Y1−3)2
2Y1+6Y1=8
8Y1=8 ....................(II)
Y1=1
From (i) X1=3
X1,Y1)=(3,1)
Question 2
Find the area of a triangle whose vertices are (0,0), (12, 9) and (14, 6).
SOLUTION
Solution : A
The simplest way to find this is to put a rectangle around the given triangle and then subtract off the areas of the right triangles that surround it. (See Figure) The area of the entire rectangle is 126, but after subtracting the areas of the 3 right triangles, whose areas are 54, 3, and 42, we are left with an area of 27.
Question 3
Two opposite sides of a square are given by the lines 5x + 12y + 18 = 0 and 5x + 12y – 21 = 0. Find the length of diagonal of the square.
SOLUTION
Solution : A
Given lines are 5x + 12y + 18 = 0……………….(i) and 5x + 12y – 21 = 0………………….(ii)
Length of the side = distance between lines (i) and (ii) is given by
D = ∣∣∣C2−C1√a2+b2∣∣∣=∣∣∣−21−18√25+144∣∣∣=∣∣∣−3913∣∣∣=3
Length of the diagonal = 3√2; Option(a).
Question 4
Find the area of quadrilateral formed by joining the points (- 4, 2), (- 1, 0), (4, 1) and (2, 5):
20
40
36
21 .5
SOLUTION
Solution : D
1st method: - The simplest way to find this is to put a rectangle around the given quadrilateral and then subtract off the areas of the right triangles that surround it. (See Figure) The area of the entire rectangle is 40, but after subtracting the areas of the 4 right triangles, whose areas are 52, 4, 9 and 3. we are left with an area of 21.5.
2nd method: - Area of quadrilateral = 2(x1y2−x2y1)+(x2y3−x3y2)+(x3y4−x4y3)+(x4y1−x1y4)
=12(−4×0−(−1)×2)+(−1×1−4×0)+(4×5−2×1)+(2×2−(−4)×5)=21.5
Question 5
The lines 2x + 2y + 1 = 0 and 8x – 2y + 9 = 0 intersect in the
SOLUTION
Solution : B
2x + 2y + 1 = 0 …………….(i)
8x – 2y + 9 = 0 …………………………(ii)
After solving equation (i) and (ii)
x = - 1 and y = 12
(−1,12); which is in 2nd quadrant.
Question 6
Find the slope of the line joining the points (7, 5) and (9, 7):
SOLUTION
Solution : A
Option(a)
Slope of the line = y2−y1x2−x1=7−59−7=1
Here (x1,y1) = (7, 5) and (x2,y2) = (9, 7)
Question 7
Which of the following points is not on the line y = 5x + 3?
SOLUTION
Solution : D
Even without calculations it is clear that (√2,313) cannot lie on the line y = 5x + 3. If x is irrational, y must also be irrational. Option(d)
Question 8
Three points A, B and C are collinear such that AB = 2BC. If the coordinates of the points A and B are (1, 7) and (6, -3) respectively, then the coordinates of the point C can be
SOLUTION
Solution : B
Given, AB = 2BC
AB : BC = 2 : 1
Let C BE (h, k), then
(2h±12±1;2k±72±1) = (6, - 3) [B may divide AC internally or externally]
2h + 1 = 18 and 2k + 7 = - 9
(or) 2h - 1 = 6 , 2k - 7 = - 3
2h = 17 and 2k = - 16 (or) 2h = 7, 2k = 4
h = 172 and k = - 8 (or) h = 72; k = 2
C = (172;−8) or (172;2)
Option(b)
Question 9
The circumradius of a triangle with its vertices at (1, -2), (-3, 0) and (5, 6) is
SOLUTION
Solution : A
A(1, - 2) ,B(- 3, 0) and C(5, 6)
AB = √20, BC = √100 , AC = √80
BC2=AB2+AC2
ABC is a right angled triangle with the right angle at A.
Circumradius is half the length of the hypotenuse.
R = 12× BC, BC = 5. Option(a).
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Question 10
Find the area of pentagon formed by joining the points (4, 4), (2, 2), (6, 2), (2, - 3) and (6, -3):
SOLUTION
Solution : A
Area of pentagon = Area of triangle ABE + Area of rectangle BCDE12×4×2+5×4=24
Question 11
Determine the radius of the circle, two of whose tangents are the lines 2x+3y-9 = 0 and 4x+6y+19 =0
SOLUTION
Solution : D
The given tangents are
2x+3y-9 = 0
4x+6y+19 =0
Or 2x + 3y + 192 = 0
Which are parallel, than distance between parallel tangents must be diameter of the circle than diameter.
∣∣∣192−(−9)∣∣∣√22+32=372√13Radius = 12 diameter = 374√13
Question 12
What is the equation of a line which has an x-intercept of 4 units and passes through the point (-2,3)?
SOLUTION
Solution : B
option (b)
To have an x-intercept 4, the value at y=0 should give x=4.
Put y=0 in all 3 answer options.
Option a and b give x=4
Question 13
Find the equation of the straight line passing through the point (2, 1) and through the point of intersection of the lines x + 2y = 3 and 2x – 3y= 4.
SOLUTION
Solution : A
1st method: - equation of any straight line passing through the intersection of the lines x + 2y = 3 and 2x – 3y= 4 is
λ(x + 2y – 3) + (2x – 3y – 4) = 0
Since it passes through the point (2, 1)
λ(2 + 2 – 3) + (4 – 3 – 4) = 0
λ - 3 = 0
λ = 3
Now substituting this value of λ in (i), we get
3(x + 2y – 3) + (2x – 3y – 4) = 0
5x + 3y – 13 = 0
2nd method: - The straight line passing through the point (2, 1), put x = 2 and y = 1only option (a) and (b) will satisfy. Now, intersection point of the lines x + 2y = 3 and 2x – 3y= 4 is (177,27), Put x = 177 and y = 27 only option(a) will satisfy. Option(a).
Question 14
Find the equation of circle whose radius is 5 and which touches the circle x2+y2−2x−4y−20=0 at the point (5, 5).
SOLUTION
Solution : A
The given circle is x2+y2−2x−4y−20=0 with centre (1, 2) and radius = √1+4+20=5.
And required circle has radius 5 hence circles touch each other externally.
Since point of contact is P(5, 5).
P is the mid point of C1 and C2, let co – ordinate of centre C2 is (h, k) then
1+h2=5;h=9
2+k2=5; k = 8
And, Equation of required circle is (x−9)2+(y−8)2=52
x2+y2−18x−16y+120=0; Option(a)
Question 15
Find the co-ordinates of the foot of the perpendicular from (2, 4) upon the line x + y = 4
SOLUTION
Solution : A
From the given figure, we can easily determine that the co-ordinates of the foot of the perpendicular from (2, 4) upon the line x + y = 4 are (1, 3)