Free Geometry Set II - 01 Practice Test - CAT
Question 1
A square has two of its vertices on a circle and the other two are on the tangent to the circle. Find the area of the square if the radius of the circle is 5 units.
SOLUTION
Solution : A
Area of the square: (x + 5)2
So the answer has to be a perfect square: 36 or 64
If area is 36 then x = 1; so EF = 6, OE = 1, OA = 5 and EA = 4
Angle OFB = ABF = BAE = 90, so angle OEA = 90
OE, EA and OA do not form Pythagoras triplet.
If area is 64 then x = 3, EF = 8, EA = 4, OA = 5
Shortcut 1
Reverse Gear approach-With a radius of 5, we will get a square of side 5%undefined2. area will be 50, but we know the side has to be slightly greater than 5%undefined2. hence, choose an option which is slightly greater than 50; i.e. 64
Shortcut 2
Graphical division:- The square gets divided into 8 parts. 4 parts are each right angled triangles of sides 3,4,5 and hence with area 6. 4 of them in total have an area of 24. The other 4 are triangles of the form 4, 5, √41 and whose areas are 10 each. Total area = (6+10)*4 = 64.
Question 2
What is the maximum number of times x circles of the same size can intersect each other?
SOLUTION
Solution : C
Deducing a Pattern:
Taking two circles. The number of points of intersection is 2 at most.
If u consider 3 circles, the number of points of intersection = 6 ( the third circles will have 4 points of intersection with the other 2 circles and those two circles will have 2 points together).
Shortcut:- Assumption & Reverse Gear:
Taking two circles. The number of points of intersection is 2 at most. From this itself, option a, b and d can be eliminated giving answer c.
Question 3
The interior angles of a polygon are in AP. If the smallest angle is 120∘ and the common difference is 5∘, then find the possible number of sides of that polygon?
SOLUTION
Solution : B
Sum of the interior angles of a polygon = (n-2) 180
n2 * (2a+ (n-1)d)= (n-2)180
n2 * [2*(120) + (n-1)5]=(n-2)180
n[48+(n-1)] = (n-2)72
n2 -25n + 144=0
n=9 and n=16
when n= 16, the greatest angle will be equal to a+15d = 120 + 15 x 5 = 195 and no interior angle of a
polygon can be equal to or greater than 180. hence answer =b
Question 4
Concentric circles are drawn with radii 1, 2, 3 … 100. The interior of the smallest circle is colored black and the annular regions are colored alternately red and black, so that no two adjacent regions are the same color. The total area of the red regions divided by the area of the largest circle is
SOLUTION
Solution : C
Areas of circles follow the sequence 1,4,9,16,…………1002
Areas of Black & Red annuals – 1,3,5,7,9……………199 – General term 2n – 1
Red annuals – 3, 7, 11, …………….. 199 Sum = (502)*(3 + 199)
Largest circle’s area = 1002
Ratio = 101200
Note : Ignore pi as it is a ratio of areas
Shortcut: Reverse gear approach –
We know that the denominator is 1002 and num is an integer, hence the denominator cannot be 101 as it is in no-way a factor of 10000. Hence option d) has been eliminated.
We need to find R(R+B). R + B = 10000. Each of the red annual is greater in area by 2 units than the subsequent black annual and there are 50 such cases. Hence R = 5050 and B = 4950. R(R+B). = 505010000 = 101200
Question 5
There is a circle of radius 10 units that circumscribes an equilateral triangle. A quadrilateral is drawn by joining mid-points of two adjacent sides of the triangle. Find the area of the quadrilateral
SOLUTION
Solution : B
In an equilateral triangle circum radius = 23 h,
10 = 23 * √32S
Side = 10\(\sqrt 3\) units.
Area of Triangle = √34(10√3)2
By graphical division, we need 34th of the area of the triangle =34×√34×(10√3)2 = 225√34
Question 6
Find the side of a piece of cloth in the shape of an equilateral triangle, whose area costs as much to paint at Rs 10/ square metre as it would cost to lay a border around the three sides at Rs 25 per metre?
SOLUTION
Solution : B
Option (b)
Perimeter = 3a
Area = √34a2
10x√34a2= 25x3a
a=30√3 = 17.32 m
Question 7
If sin α= P and tan α= Q, find the value of 1P2−1Q2 ?
SOLUTION
Solution : B
Option (b)
Assume a 45-45-90 triangle
sin α= P = 1√2 . (Sin= opposite side/ hypotenuse)
tan α= Q = 1 (Tan = opposite side/ adjacent side)
1P2−1Q2 = 2-1 =1
Shortcut
cosec2α – cot2α = 1 is a trigonometric identity
If you are not aware of this identity, substitute α = 45.
Question 8
In Δ ABC, D,E and F are taken on AB, AC and BC respectively, so that EF=CF and DF=BF. If ∠A=40∘, then find ∠DFE?
SOLUTION
Solution : A
Option (a)
Take DBF = a∘ and ECF = b∘.
40 + a∘ + b∘ = 180 ⇒ a∘ + b∘= 140
DFB + EFC + DFE = 180 ⇒ 180 – 2a∘ + 180 – 2b∘ + DFE = 180 ⇒ 180∘ – 2(a + b) = - DFE
DFE = 280 – 180 = 100∘
Question 9
What is the circumradius of a triangle whose sides are 7, 24 and 25 respectively?
SOLUTION
Solution : B
Option b
It’s important that we note that it is a Pythagoras triplet72 = 252+242
Now,
In a right-angled triangle, the median to the hypotenuse is half the hypotenuse and is also the circum radius of the triangle.
As the hypotenuse is 25, the circum radius is 12.5
Question 10
If O is the centre of a circle and the radius of the circle is 4 units, find the value of length of AC.
SOLUTION
Solution : D
Option (d)
AC has to less than 8 as the diameter will be of 8 units. So the correct option is “D” as all are greater than 8
Question 11
A trapezium DEFG is circumscribed about a circle that has centre C and radius 2 cm. If DE = 3 cm and the measure of∠DEF=∠EFG=90∘, then find the area of trapezium DEFG
SOLUTION
Solution : A
Draw perpendicular from D to GF meeting at X.
Also, let GN = GP = k
Since, DE = XF = 3 cm
NX = 3 – 2 = 1 = DM = DP
Therefore, GX = k – 1
ΔDXG is right angled triangle.
Hence, (k + 1)2 – (k – 1)2 = 42
Hence, k = 4 and so GF = 6 cm
Area of trapezium DEFG = (3+6)2 * 4 = 18 cm2
Question 12
P,Q,R,S,T are points on a circle with centre O so that ∠PTQ= 2 x ∠RTS and PQ %undefined RS. Find ∠SOQ, if P,O,Q are in a straight line
SOLUTION
Solution : A
∠PTQ= 90 (angle in a semi circle)
, hence ∠RTS = 45.
∠ROS = 90 ( Inscribed angle= 2 *Angle at circumference)
OR=OS=OR (radius)
∠RSO = ∠SOQ= 45∘ (interior alternate angles)
Option (a)
Question 13
In an equilateral triangle ΔABC, D divides BC in the ratio 2 : 1 while E divides AC in the ratio 1 : 2. Find BE : ED ?
SOLUTION
Solution : C
Let the side of the equilateral triangle ΔABC be 3 units. Clearly, BD = 2, CD = 1 and CE = 2.
CD/CE = 12 So ∠DEC=30∘
Hence, ΔCDE is a right angled triangle.
Similarly, ΔBDE is also a right angled triangle.
BE = √7 and ED = √3 hence BE:ED = √7 : √3
Question 14
In a triangle ABC, M is the mid-point of BC. If ∠AMB = 45∘, and ∠ACM = 30∘, then ∠BAM is (∠B is not obtuse)
SOLUTION
Solution : D
Consider any triangle ABC, AM is a line connecting A to mid point of BC. Drop a perpendicular from A to BC touching BC at D. The angle ADM is a right angled triangle with 45 – 45 – 90 angles. Hence DAM = 45 and BAM > 45. BAM= 45∘ If ∠ABC =90∘. hence, answer is option (d)
Question 15
What is the minimum area of quadrilateral ABCD, if the area of a pair of diagonally opposite triangles (Formed by joining the diagonals of the quadrilateral) is 12 and 27 respectively?
SOLUTION
Solution : C
Option c
In a quadrilateral, product of area of diagonally opposite triangles is same.
So, the product of other set of opposite triangles will be 12*27
Now, we have to find minimum a and b such that a*b=12*27 and a+b is minimum.
So a = b = 18 (as 12*27 = 324)
Hence the area of the quadrilateral = 18 + 18 + 12 + 27 = 75