Question for CAT Previous Year Questions: Geometry
Try yourself:Let C be the circle x^{2} + y^{2} + 4x - 6y - 3 = 0 and L be the locus of the point of intersection of a pair of tangents to C with the angle between the two tangents equal to 60°. Then, the point at which L touches the line x = 6 is
[2023]
Explanation
This is the equation of a circle with radius 4 units and centered at (-2, 3) From a point L we drop two tangents on the circle such that the angle between the tangents is 60^{o}.
Therefore the locus of the point L, is a circle centered at (-2, 3) and has a radius of (4 + x = 8) units.
The equation of this locus is thus, (x + 2)^{2} + (y - 3)^{2} = 82
When X = 6, we have, (8)^{2} + (y - 3)^{2} = 8^{2}, that is y = 3
The circle, (x + 2)^{2} + (y - 3)^{2} = 8^{2}, touches the line x = 6 at (6, 3).
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Question for CAT Previous Year Questions: Geometry
Try yourself:A quadrilateral ABCD is inscribed in a circle such that AB : CD = 2 : 1 and BC : AD = 5 : 4. If AC and BD intersect at the point E, then AE : CE equals
[2023]
Explanation
∠DAC=∠DBC (Angles subtended by the chord DC on the same side.) ∠ADB=∠ACB (Angles subtended by the chord AB on the same side.) ∠AED=∠BEC (Vertically Opposite angles.) Therefore, △AED∼△BEC
(Angles subtended by the chord AD on the same side.) ∠BAC=∠BDC (Angles subtended by the chord BC on the same side.) ∠AEB=∠DEC (Vertically Opposite angles.) Therefore, △AEB∼△DEC
Therefore, AE : EC = 8 : 5
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Question for CAT Previous Year Questions: Geometry
Try yourself:In a right-angled triangle ABC, the altitude AB is 5 cm, and the base BC is 12 cm. P and Q are two points on BC such that the areas of ΔABP, ΔABQ and ΔABC are in arithmetic progression. If the area of ΔABC is 1.5 times the area of ΔABP, the length of PQ, in cm, is
[2023]
Correct Answer : 2
Explanation
The construction of the triangle according to the question is as follows:
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Question for CAT Previous Year Questions: Geometry
Try yourself:In a rectangle ABCD, AB = 9 cm and BC = 6 cm. P and Q are two points on BC such that the areas of the figures ABP, APQ, and AQCD are in geometric progression. If the area of the figure AQCD is four times the area of triangle ABP, then BP : PQ : QC is
[2023]
Explanation
Let BP = x, PQ = y and QC = z.
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Question for CAT Previous Year Questions: Geometry
Try yourself:A triangle is drawn with its vertices on the circle C such that one of its sides is a diameter of C and the other two sides have their lengths in the ratio a : b. If the radius of the circle is r, then the area of the triangle is
[2023]
Explanation
Since the angle subtended by the diameter on the circle is a right-angle, such a triangle inscribed in a circle with the diameter as one of its sides will be right angled.
“…and the other two sides have their lengths in the ratio a: b.” Let the two sides be ax and bx.
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Question for CAT Previous Year Questions: Geometry
Try yourself:All the vertices of a rectangle lie on a circle of radius R . If the perimeter of the rectangle is P , then the area of the rectangle is
[2022]
Explanation
Let the dimensions of the rectangle inscribed in the circle be a and b. We are being asked to express the Area of the rectangle in terms of Perimeter and Radius. Area of the rectangle = a * b Perimeter, P = 2(a + b)
Since the vertices of the rectangle subtend right angle on the circle,
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Question for CAT Previous Year Questions: Geometry
Try yourself:A circle of diameter 8 inches is inscribed in a triangle ABC where∠ABC=90^{∘}. If BC = 10 inches then the area of the triangle in square inches is
[2021]
Correct Answer : 120
Explanation
We know that Inradius = (Perpendicular + Base − Hypotenuse) / 2
h - p = 2 or h = p + 2. Now, p^{2} + 100 = h^{2}
4p = 96 p = 24. Hence, Area = 1/2 x 24 x 10 = 120
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Question for CAT Previous Year Questions: Geometry
Try yourself:Suppose the length of each side of a regular hexagon ABCDEF is 2 cm.It T is the mid point of CD,then the length of AT, in cm, is
[2021]
Explanation
Since a regular hexagon can be considered to be made up of 6 equilateral triangles, a line joining the farthest vertices of a hexagon can be considered to be made up using the sides of two opposite equilateral triangle forming the hexagon. Hence, its length should be twice the side of the hexagon, in this case, 4 cm.
Now, AD divided the hexagon into two symmetrical halves. Hence, AD bisects angle D, and hence, angle ADC is 60^{∘}. We can find out the value of AT using cosine formula:
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Question for CAT Previous Year Questions: Geometry
Try yourself:If the area of a regular hexagon is equal to the area of an equilateral triangle of side 12 cm, then the length, in cm, of each side of the hexagon is
[2021]
Explanation
Area of a regular hexagon = Area of an equilateral triangle = ; where a = side of the triangle Since the area of the two figures are equal, we can equate them as folllows:
On simplifying: x^{2} = 24 ∴ x = 2√6
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Question for CAT Previous Year Questions: Geometry
Try yourself:AB is a diameter of a circle of radius 5 cm. Let P and Q be two points on the circle so that the length of PB is 6 cm, and the length of AP is twice that of AQ. Then the length, in cm, of QB is nearest to
[2019]
Explanation
We know that angle APB = AQB = 90^{o} We know PB = 6, AB = 10 and AP = 2AQ From using Pythagorean triplet, we find that AP = 8 cms So, AQ = 4cms AQ^{2} + QB^{2} = AB^{2} 16 + QB^{2 }= 100 QB^{2 }= 84 QB = 9.3
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Question for CAT Previous Year Questions: Geometry
Try yourself:In a triangle ABC, medians AD and BE are perpendicular to each other, and have lengths 12 cm and 9 cm, respectively. Then, the area of triangle ABC, in sq cm, is
[2019]
Explanation
We know that intersection of two medians is called a Centroid We know, a Centroid divides the median in the ratio 2:1 AD = 12cms, AG = 8cms, GD = 4cms And similarly, BG = 6cms, GE = 3cms Now, area (ΔABE) can be easily found out area (ΔABE) = 1 / 2 x BE X AG area (ΔABE) = 1 / 2 x 9 x 8 = 36 cm^{2} area (ΔABE) = area (ΔBEC) area (ΔABC) = 2 x 36 = 72 cm^{2}
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Question for CAT Previous Year Questions: Geometry
Try yourself:Two circles, each of radius 4 cm, touch externally. Each of these two circles is touched externally by a third circle. If these three circles have a common tangent, then the radius of the third circle, in cm, is
[2019]
Explanation
For three circles to have a common tangent, the third circle must lie in between the two circles. Let the radius of the third circle be r cms Now, Connect the centres of the circles with one another and with the intersection points. Let them be A,B,C,D,E respectively. Now, from observing the constructed diagram, we find AB = 8cms AC = (4 + r) cms BC = (4 + r) cms OC = (4 - r) cms Now, consider the right triangle OBC, Applying Pythagoras theorem, (4 + r)^{2} = 4^{2} + (4 - r)^{2} 16 + r^{2} + 8r = 16 + 16 + r^{2} - 8r 16r = 16 r = 1 cms
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Question for CAT Previous Year Questions: Geometry
Try yourself:Let A and B be two regular polygons having a and b sides, respectively. If b = 2a and each interior angle of B is 3 / 2 times each interior angle of A, then each interior angle, in degrees, of a regular polygon with a + b sides is
[2019 TITA]
Correct Answer : 150
Explanation
Sum of the interior angles of a n-sided Polygon = (n - 2) x 180 Measure of each interior angle = So, Interior angle of the polygon with side a = Interior angle of the polygon with side b = It is given that interior angle of side b is 3 / 2 times to that of the polygon with side A = 3 / 2 x (2a-2) x 180 = (3a - 6) x 180 2a - 2 = 3a - 6 a = 4 So, b = 8 (a + b) sides = 8 + 4 = 12 sides Interior angle = = 150^{°}
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Question for CAT Previous Year Questions: Geometry
Try yourself:Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm. If AP is perpendicular on BC, then the maximum possible length of AP, in cm, is
[2019]
Explanation
Now, the length of AP would be maximum, when AB = AC (ABC is an Isosceles Right Triangle) So, AB = AC = 10√2cms Applying Pythagoras theorem to ΔAPB, we get AB^{2} = AP^{2} + PB^{2} (10√2)^{2} = 10^{2} + x^{2} x^{2} = 100 x = 10 cms
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Question for CAT Previous Year Questions: Geometry
Try yourself:In a circle with centre O and radius 1 cm, an arc AB makes an angle 60 degrees at O. Let R be the region bounded by the radii OA, OB and the arc AB. If C and D are two points on OA and OB, respectively, such that OC = OD and the area of triangle OCD is half that of R, then the length of OC, in cm, is
[2018]
Explanation
Given ∠AOB = 60° Area of Sector AOB = Given OC = OD => ∠OCD = ∠ODC = 60° △OCD is an Equilateral Triangle with side = a Area(△OCD) = Its given that Area(OCD) = 1/2 × Area(OAB)
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Question for CAT Previous Year Questions: Geometry
Try yourself:Given an equilateral triangle T1 with side 24 cm, a second triangle T2 is formed by joining the midpoints of the sides of T1. Then a third triangle T3 is formed by joining the midpoints of the sides of T2. If this process of forming triangles is continued, the sum of the areas, in sq cm, of infinitely many such triangles T1, T2, T3,... will be
[2018]
Explanation
As the triangle progresses infinitely and the side length decreases, it follows an infinite GP series As the sides decrease by half, their areas decrease by 1 / 4 We know, Area of an Equilateral Triangle = Area of T1 =
Sum of an Infinite GP = a / 1 − r where a = 144 √3 , r = 1 / 4 Sum of areas ( T1, T2, T3,..) = Therefore, Sum of areas= 192 √3 sq cms.
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Question for CAT Previous Year Questions: Geometry
Try yourself:Let ABCD be a rectangle inscribed in a circle of radius 13 cm. Which one of the following pairs can represent, in cm, the possible length and breadth of ABCD?
[2018]
Explanation
Since, ABCD is a rectangle inscribed inside a circle ABC must be a Right triangle Given that radius of Circle = 13 cms 5, 12, 13 forms a Pythagorean Triplet 10, 24, 26 is also a Pythagorean triplet So, 10 and 24 are possible length and breadth of ABCD
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Question for CAT Previous Year Questions: Geometry
Try yourself:Points E, F, G, H lie on the sides AB, BC, CD, and DA, respectively, of a square ABCD. If EFGH is also a square whose area is 62.5% of that of ABCD and CG is longer than EB, then the ratio of length of EB to that of CG is:
[2018]
Explanation
Given, Area(EFGH) = 62.5% Area(ABCD) Area(EFGH) = 5 / 8 Area(ABCD) Let EB = 1 and CG = r Similarly, the rest other dimensions are also of lengths 1 or r units Applying pythagoras theorem on △DHG we get GH = √(1 + r^{2}) Area(ABCD) = (1 + r)^{2} and Area(EFGH) = (√(1 + r^{2}))^{2} => (1 + r^{2}) = 5 / 8(1 + r)^{2} 1 + r^{2} = 5 / 8 (1 + r^{2 }+ 2r) 8 + 8r^{2} = 5 + 5r^{2} + 10r 3r^{2} – 10r + 3 = 0 (3r - 1) × (r - 3) = 0 r = 1 / 3 or r = 3 As CG > EB r = 3 and ratio (EB : CG)= (1 : r) = (1 : 3)
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Question for CAT Previous Year Questions: Geometry
Try yourself:In a circle, two parallel chords on the same side of a diameter have lengths 4 cm and 6 cm. If the distance between these chords is 1 cm, then the radius of the circle, in cm, is
[2018]
Explanation
Given that Chords lie on the same side of diameter with lengths 4 cms and 6 cms Draw a perpendicular from the origin to both the chords and mark the points of intersection as P and Q respectively Consider radius of circle as ‘r’ and distance OP as ‘x’ Draw lines from origin to the end of the chord and mark the points as D and B respectively Thus, △OQB and △OPD form a right Triangle Applying Pythagoras Theorem on both triangles, (x+1)^{2} +2^{2} = r^{2} ---(1) x^{2} +3^{2} = r^{2} ---(2) We find that there is an increase and decrease by 1 in both equations So, x^{2} = 2^{2} , x = 2 r^{2 }= 2^{2 }+ 3^{2} = 4 + 9 = 13 r = √13 cms
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Question for CAT Previous Year Questions: Geometry
Try yourself:On a triangle ABC, a circle with diameter BC is drawn, intersecting AB and AC at points P and Q, respectively. If the lengths of AB, AC, and CP are 30 cm, 25 cm, and 20 cm respectively, then the length of BQ, in cm, is
[2018 TITA]
Correct Answer : 24
Explanation
Let ABC be the triangle on which a circle of diameter BC is drawn, intersecting AB and AC at points P and Q respectively.The lengths of AB, AC and CP are 30cm, 25cm and 20 cm respectively we have to find the length of BQ in cm. Key thing here is ,this is semicircle so this angle P and Q should be 90° and now we are looking to do Pythagoras theorem
So think about triangle PAC, by Pythagoras theorem AC^{2} = AP^{2} + PC^{2} we can find that AP = 15
Since AP = 15 we can find BP by AB = AP + PB 30 = 15 + PB PB = 15 Now we can look at the triangle BPC, by Pythagoras theorem we can find that BC = 25
Now we have to find BQ .So we can take the area of the triangle formula which is equal to 1/2 × base × height area of the triangle formula = 1/2 × base × height 1/2 × AB × PC = 1/2 × AC × BQ 1/2 × 30 × 20 = 1/2 × 25 × BQ BQ = 24 cm
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Question for CAT Previous Year Questions: Geometry
Try yourself:A chord of length 5 cm subtends an angle of 60° at the centre of a circle. The length, in cm, of a chord that subtends an angle of 120° at the centre of the same circle is
[2018]
Explanation
Given that a chord of length 5 cm subtends an angle of 60° at the centre of a circle. We have to find the length of a chord that subtends an angle of 120° at the centre of the same circle sin 60° = BC / OB ⟹ √3 / 2 = BC / 5 ⟹ BC = 5√3 / 2 and AC = 5√3 / 2 So AB = 5√3 / 2 + 5√3 / 2 = 5√3 The length of a chord that subtends an angle of 120° at the centre of the same circle is 5√3
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Question for CAT Previous Year Questions: Geometry
Try yourself:From a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC. The area, in sq ft, of the remaining portion of triangle ABC is:
[2017]
Explanation
Given that from a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC. Here GBC is the one third of the area of the triangle. We can join AG and GD which is the median. Each of the shaded triangle has the same area and therefore the remaining area is two-thirds of ABC The ratio of the sides are 8 : 5 : 7 Area of triangle = √(s(s − a)(s − b)(s − c)) where, semi perimeter (s) = 8 + 5 + 7 / 2 = 10 Area = √(10(10 − 8)(10 − 5)(10 − 7)) Area = √(10(2)(5)(3)) Area = 10√3 Area of ABC = 25 × 10 √3 (As 8 : 5 : 7 multiplied by 5 gives the sides of triangle ABC) Therefore area of the remaining traingle = 2 / 3 × 250√3 = 500 / √3
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Question for CAT Previous Year Questions: Geometry
Try yourself:Let ABC be a right-angled triangle with BC as the hypotenuse. Lengths of AB and AC are 15 km and 20 km, respectively. The minimum possible time, in minutes, required to reach the hypotenuse from A at a speed of 30 km per hour is:
[2017 TITA]
Correct Answer : 24
Explanation
Given that ABC be a right-angled triangle with BC as the hypotenuse. Lengths of AB and AC are 15 km and 20 km, respectively. We have to find the minimum possible time, in minutes, required to reach the hypotenuse from A at a speed of 30 km per hour. We should first find the minimum distance in order to find the minimum possible time, in minutes, required to reach the hypotenuse from A at a speed of 30 km per hour. Therefore minimum distance AD has to be found and then it should be divided by the 30 km per hour. Using the idea of similar triangles Area of the triangle ABC ⟹ 1 / 2 × BA × AC = 1 / 2 × BC × AD ⟹ 1 / 2 × 15 × 20 = 1 / 2 × 25 × AD ⟹ AD = 12 units Hence 12 kms is travelled at 30km per hour ⟹ 12 / 30 = 2 / 5 The minimum possible time, in minutes, required to reach the hypotenuse from A at a speed of 30 km per hour is 2 / 5 × 60 = 24 minutes Key thing to be noted here is using Pythagoras theorem to find the altitude AD and then using Speed, Time and Distance formula to find the time.
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Question for CAT Previous Year Questions: Geometry
Try yourself:Let ABCDEF be a regular hexagon with each side of length 1 cm. The area (in sq cm) of a square with AC as one side is
[2017]
Explanation
Given that ABCDEF be a regular hexagon with each side of length of 1 cm. We have to find the area of a square with AC as one side. ABO is the equilateral triangle with each side having length of 1 cm and ABCO is the rhombus. Altitude of an equilateral triangle = √3/2 a So AP = √3/2 × 1 Where AP = PC ⟹ AC = AP + PC ⟹ AC = √3 / 2 × 2 ⟹ AC = √3 Area of a square = a^{2} sq.units Area of the square with AC as one side = √3 × √3 = 3 sq.units
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Question for CAT Previous Year Questions: Geometry
Try yourself:The base of a vertical pillar with uniform cross section is a trapezium whose parallel sides are of lengths 10 cm and 20 cm while the other two sides are of equal length. The perpendicular distance between the parallel sides of the trapezium is 12 cm. If the height of the pillar is 20 cm, then the total area, in sq cm, of all six surfaces of the pillar is
[2017]
Explanation
We have to find the total area of all six surfaces of the pillar which is 4 rectangles + 2 trapeziums one at the top and one at the bottom Area of the rectangle = b × h ⟹ Area of rectangle with l = 20 b = 10 ⟹ Area = 200 sq.cm ⟹ Area of rectangle with l = 20 b = 20 Area = 400 sq.cm ⟹ Area of 2 rectangles with l = 20 b = 13 is 2 × 13 × 20 = 520 sq. cm Sum of areas of 4 rectangles = 520 + 200 + 400 = 1120 sq.cm Area of the trapezium = 1/2 × 12 (10 + 20) Area of the trapezium = 6 × 30 = 180 sq.cm Area of both the trapezium = 2 × 180 = 360 sq.cm Sum of areas of 4 rectangles = 1120 sq.cm. Area of both the trapezium = 360 sq.cm Total surface area of the pillar = 4 rectangles + 2 trapeziums Total surface area of the pillar = 1120 + 360 Total surface area of the pillar = 1480 sq.cm
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Question for CAT Previous Year Questions: Geometry
Try yourself:ABCD is a quadrilateral inscribed in a circle with centre O. If ∠COD = 120 degrees and ∠BAC = 30 degrees, then the value of ∠BCD (in degrees) is
[2017 TITA]
Correct Answer : 90
Explanation
ABCD is a quadrilateral inscribed in a circle with centre O. ∠COD = 120° ∠BAC = 30° ∠BOC = 60° (twice that of ∠BAC) BOD is a straight line (60 + 120) = 180° or BD is the diameter of the circle ∠BOD = 180° We have to find the value of ∠BCD ∠BCD is an angle in a semicircle = 90°
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Question for CAT Previous Year Questions: Geometry
Try yourself:If three sides of a rectangular park have a total length 400 ft., then the area of the park is maximum when the length (in ft.) of its longer side is
[2017 TITA]
Correct Answer : 200
Explanation
Given that the three sides of a rectangular park have a total length 400ft i.e. 2x + y = 400 ft xy should be maximum y = 400 – 2x x × (400 – 2x) = max x × (200 – x) = max Let us check for different values
The area of the park is maximum when x = 100 and y = 200. The length of the longer side y is equal to 200 ft.
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Question for CAT Previous Year Questions: Geometry
Try yourself:Let P be an interior point of a right-angled isosceles triangle ABC with hypotenuse AB. If the perpendicular distance of P from each of AB, BC, and CA is 4(√2 - 1) cm, then the area, in sq. cm, of the triangle ABC is
[2017 TITA]
Correct Answer : 16
Explanation
Let ABC be the right angled isosceles triangle with hypotenuse AB and P be an interior point r (inradius) = 4(√2 - 1) In any right triangle with a , b , h being its sides, r = a+b−h / 2 Here we have the right angled isosceles triangle with a , a , a√2 as its sides
⟹ a / √2 = 4 ⟹ a = 4√2 ⟹ The area of the triangle ABC = 1 / 2 × 4√2 × 4√2 The area of the triangle ABC = 16 sq.cm
FAQs on Geometry CAT Previous Year Questions with Answer PDF
1. What are the basic principles of geometry?
Ans. Geometry is based on fundamental principles such as points, lines, planes, angles, and shapes. These elements form the basis for all geometric concepts and calculations.
2. How is geometry used in real life?
Ans. Geometry is used in various real-life applications such as architecture, engineering, design, and navigation. It helps in creating structures, solving spatial problems, and understanding the relationships between shapes and objects.
3. What is the importance of geometry in mathematics?
Ans. Geometry plays a crucial role in mathematics as it helps in developing spatial reasoning, problem-solving skills, and logical thinking. It provides a foundation for understanding complex mathematical concepts and theories.
4. What are the different types of angles in geometry?
Ans. In geometry, angles are classified into different types based on their measures. Some common types of angles include acute angles (less than 90 degrees), obtuse angles (greater than 90 degrees), right angles (exactly 90 degrees), and straight angles (exactly 180 degrees).
5. How can I improve my geometry skills?
Ans. To improve your geometry skills, practice solving geometric problems regularly, use visualization techniques to understand spatial relationships, and study the properties of different shapes and angles. Additionally, seeking help from a tutor or using online resources can also be beneficial in enhancing your geometry skills.