{"id":212777,"date":"2022-07-14T17:00:37","date_gmt":"2022-07-14T11:30:37","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=212777"},"modified":"2024-03-29T14:24:09","modified_gmt":"2024-03-29T08:54:09","slug":"cat-triangles-and-circles-geometry-questions-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/cat-triangles-and-circles-geometry-questions-pdf\/","title":{"rendered":"CAT Triangles and Circles (Geometry) Questions PDF"},"content":{"rendered":"

Triangles and<\/b>\u00a0Circles<\/strong> questions are important concepts in the Geometry concept of the CAT<\/strong> Quant <\/strong>section. <\/span>These questions are not very tough; make sure you are aware of all the Important Formulas in CAT Geometry<\/span><\/a><\/strong>. S<\/span>olve more questions from CAT Triangles and Circles. <\/span>You can check out these CAT Geometry questions<\/span> from the CAT Previous year papers<\/strong><\/span><\/a>.<\/span> Practice a good number of questions in CAT Geometry – Triangles and<\/strong> Circles<\/strong> so that you can answer these questions with ease in the exam. In this post, we will look into some important CAT Geometry Questions. These are a good source of practice for CAT 2022 preparation; If you want to practice these questions, you can download these Important Triangles and Circles (Geometry) Questions for CAT<\/strong> (with detailed answers) PDF<\/strong> along with the video solutions below, which is completely Free.<\/p>\n

Download Triangles and Circles (Geometry) Questions for CAT<\/a><\/p>\n

Enroll for CAT 2022 Online Course<\/a><\/p>\n

Question 1:\u00a0<\/b>In a triangle ABC, the lengths of the sides AB and AC equal 17.5 cm and 9 cm respectively. Let D be a point on the line segment BC such that AD is perpendicular to BC. If AD = 3 cm, then what is the radius (in cm) of the circle circumscribing the triangle ABC?<\/p>\n

a)\u00a017.05<\/p>\n

b)\u00a027.85<\/p>\n

c)\u00a022.45<\/p>\n

d)\u00a032.25<\/p>\n

e)\u00a026.25<\/p>\n

1)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b>
\n<\/p>\n

Let x be the value of third side of the triangle. Now we know that Area = 17.5*9*x\/(4*R), where R is circumradius.<\/p>\n

Also Area = 0.5*x*3 .<\/p>\n

Equating both, we have 3 = 17.5*9 \/ (2*R)<\/p>\n

=> R = 26.25.<\/p>\n

Question 2:\u00a0<\/b>Consider obtuse-angled triangles with sides 8 cm, 15 cm and x cm. If x is an integer then how many such triangles exist?<\/p>\n

a)\u00a05<\/p>\n

b)\u00a021<\/p>\n

c)\u00a010<\/p>\n

d)\u00a015<\/p>\n

e)\u00a014<\/p>\n

2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

For obtuse-angles triangle, $c^2 > a^2 + b^2$ and c < a+b
\nIf 15 is the greatest side, 8+x > 15 => x > 7 and $225 > 64 + x^2$ => $x^2$ < 161 => x <= 12
\nSo, x = 8, 9, 10, 11, 12
\nIf x is the greatest side, then 8 + 15 > x => x < 23
\n$x^2 > 225 + 64 = 289$ => x > 17
\nSo, x = 18, 19, 20, 21, 22
\nSo, the number of possibilities is 10<\/p>\n

Question 3:\u00a0<\/b>Two circles, both of radii 1 cm, intersect such that the circumference of each one passes through the centre of the other. What is the area (in sq. cm.) of the intersecting region?<\/p>\n

a)\u00a0$\\frac{\\pi}{3}-\\frac{\\sqrt 3}{4}$<\/p>\n

b)\u00a0$\\frac{2\\pi}{3}+\\frac{\\sqrt 3}{2}$<\/p>\n

c)\u00a0$\\frac{4\\pi}{3}-\\frac{\\sqrt 3}{2}$<\/p>\n

d)\u00a0$\\frac{4\\pi}{3}+\\frac{\\sqrt 3}{2}$<\/p>\n

e)\u00a0$\\frac{2\\pi}{3}-\\frac{\\sqrt 3}{2}$<\/p>\n

3)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b>
\n<\/p>\n

The circumferences of the two circle pass through each other’s centers. Hence, O1A = O1B=O1O2 = 1cm<\/p>\n

By symmetry, the line joining the two centres would be bisect AB and would be bisected by AB. As the line joining the center to the midpoint of a chord is perpendicular to the chord, O1O2 and AB are perpendicular bisectors of each other. Suppose they intersect at point P.<\/p>\n

O1P = Half of O1O2 = 1\/2 cm
\nSo, the angle AO1P = 60 degrees as cos 60 = 1\/2
\nBy symmetry, BO1P = 60 degrees.
\nSo, angle AO1B= 120 degrees<\/p>\n

<\/p>\n

In the above, the required area is 2 times A(segment ABO2)(blue region). And A(segment ABO2)(blue region) = A(sector O2AO1B)(blue + red) – A(triangleO1AB )(red)<\/p>\n

Area of sector = 120\u00b0\/360\u00b0 * $\\pi * 1^2 $ = $\\pi\/3$<\/p>\n

Area of triangle = 1\/2 * b * h = 1\/2 * (2* 1 cos 30\u00b0) * (1\/2) = \u221a3\/4<\/p>\n

Hence, required area = $\\frac{\\pi}{3}-\\frac{\\sqrt 3}{4}$ . Hence so the required area is 2 times the above value which is $\\frac{2\\pi}{3}-\\frac{\\sqrt 3}{2}$<\/p>\n

Question 4:\u00a0<\/b>In the figure below, AB is the chord of a circle with center O. AB is extended to C such that BC = OB. The straight line CO is produced to meet the circle at D. If $\\angle{ACD}$ = y degrees and $\\angle{AOD}$ = x degrees such that x = ky, then the value of k is
\n<\/p>\n

a)\u00a03<\/p>\n

b)\u00a02<\/p>\n

c)\u00a01<\/p>\n

d)\u00a0None of the above.<\/p>\n

4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b>
\n<\/p>\n

Since Angle BOC = Angle BCO = y.<\/p>\n

Angle OBC = 180-2y .<\/p>\n

Hence Angle ABO = \u00a0z = 2y = Angle OAB.<\/p>\n

Now since x is exterior angle of triangle AOC .<\/p>\n

We have x = z + y = 3y.<\/p>\n

Hence option A.<\/p>\n

Question 5:\u00a0<\/b>
\nIn the adjoining figure, chord ED is parallel to the diameter AC of the circle. If angle CBE = 65\u00b0, then what is the value of angle DEC?<\/p>\n

a)\u00a035<\/p>\n

b)\u00a055<\/p>\n

c)\u00a045<\/p>\n

d)\u00a025<\/p>\n

5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b>
\n<\/p>\n

If EBC = 65 then EOC = 130 then OEC = OCE = 25. NOw since OC and ED are parallel we have OCE = OED = 25. Hence option D.<\/p>\n

Checkout: CAT Free Practice Questions and Videos<\/strong><\/a><\/em><\/p>\n

Question 6:\u00a0<\/b>What is the distance in cm between two parallel chords of lengths 32 cm and 24 cm in a circle of radius 20 cm?<\/p>\n

a)\u00a01 or 7<\/p>\n

b)\u00a02 or 14<\/p>\n

c)\u00a03 or 21<\/p>\n

d)\u00a04 or 28<\/p>\n

6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b>
\n<\/p>\n

The distances of the chords from the center are 12 cm and 16 cm respectively.<\/p>\n

If the chords lie on the same side of the center, the distance between the chords is 4 cm, if they lie on opposite sides of the center, the distance between them is 28 cm.<\/p>\n

Question 7:\u00a0<\/b>Two identical circles intersect so that their centres, and the points at which they intersect, form a square of side 1 cm. The area in sq. cm of the portion that is common to the two circles is<\/p>\n

a)\u00a0$\\pi$\/4<\/p>\n

b)\u00a0$\\pi$\/2-1<\/p>\n

c)\u00a0$\\pi$\/5<\/p>\n

d)\u00a0$\\sqrt\\pi-1$<\/p>\n

7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b>
\n<\/p>\n

We know that quad ambn is a square of side 1.<\/p>\n

Area of\u00a0the sector a-mqn is $\\frac{90}{360}* \\pi *1*1$ = $\\frac{\\pi }{4}$.<\/p>\n

Area of square = 1*1 = 1<\/p>\n

Area of common portion = 2 * Area of sector – Area of square<\/p>\n

= 2 * $\\frac{\\pi }{4}$ – 1 =\u00a0 $\\frac{\\pi }{2}$ – 1<\/p>\n

Question 8:\u00a0<\/b>In the following figure, the diameter of the circle is 3 cm. AB and MN are two diameters such that MN is perpendicular to AB. In addition, CG is perpendicular to AB such that AE:EB = 1:2, and DF is perpendicular to MN such that NL:LM = 1:2. The length of DH in cm is
\n<\/p>\n

a)\u00a0$2\\sqrt2 – 1$<\/p>\n

b)\u00a0$(2\\sqrt2 – 1)\/2$<\/p>\n

c)\u00a0$(3\\sqrt2 – 1)\/2$<\/p>\n

d)\u00a0$(2\\sqrt2 – 1)\/3$<\/p>\n

8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

Let EO = x, So, AE = 1.5 – x
\nAE : EB = 1:2 => x = 1\/2<\/p>\n

(1.5-x):(1.5+x) = 1:2.<\/p>\n

x=0.5.<\/p>\n

So, EO = 0.5
\nSimilarly, OL = 0.5
\nNow, EOLH is a parallelogram and EO = OL = 0.5
\nIn triangle DOL, DO = radius = 1.5 and OL = 0.5
\nSo, DL = $\\sqrt2$
\n=> DH =\u00a0$(2\\sqrt2 – 1)\/2$<\/p>\n

Question 9:\u00a0<\/b>In triangle DEF shown below, points A, B and C are taken on DE, DF and EF respectively such that EC = AC and CF = BC. If angle D equals 40 degress , then angle ACB is ?
\n<\/p>\n

a)\u00a0140<\/p>\n

b)\u00a070<\/p>\n

c)\u00a0100<\/p>\n

d)\u00a0None of these<\/p>\n

9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

Let angle EAC = x, so angle AEC = x and angle ACE = 180-2x
\nLet angle FBC = y, so angle BFC = y and angle BCF = 180-2y
\nSo, angle ACB = 180-(180-2x+180-2y) = 2(x+y) – 180
\nx+y = 180 – 40 = 140
\nSo, angle ACB = 280 – 180 = 100<\/p>\n

Question 10:\u00a0<\/b>Consider a circle with unit radius. There are 7 adjacent sectors, S1, S2, S3,….., S7 in the circle such that their total area is (1\/8)th of the area of the circle. Further, the area of the $j^{th}$ sector is twice that of the $(j-1)^{th}$ sector, for j=2, …… 7. What is the angle, in radians, subtended by the arc of S1\u00a0at the centre of the circle?<\/p>\n

a)\u00a0$\\pi\/508$<\/p>\n

b)\u00a0$\\pi\/2040$<\/p>\n

c)\u00a0$\\pi\/1016$<\/p>\n

d)\u00a0$\\pi\/1524$<\/p>\n

10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

Now area of 1st sector = $\\pi * r^2 * \\frac{x}{360}$ where x – angle subtended at center<\/p>\n

Now the next sector will have 2x as the angle, and similarly angles will be in GP with ratio\u00a0= 2.<\/p>\n

Sum of areas of all 7 sectors = $\\frac{127*x* \\pi * r^2}{360}$ which is equal to $\\frac{\\pi * r^2}{8}$<\/p>\n

We get x = $\\frac{360}{8*127}$<\/p>\n

Now if converted in radians we get \u00a0x =\u00a0$\\pi\/508$.<\/p>\n

Question 11:\u00a0<\/b>If a,b,c are the sides of a triangle, and $a^2 + b^2 +c^2 = bc + ca + ab$, then the triangle is:<\/p>\n

a)\u00a0equilateral<\/p>\n

b)\u00a0isosceles<\/p>\n

c)\u00a0right angled<\/p>\n

d)\u00a0obtuse angled<\/p>\n

11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) => 3(a^2 + b^2 + c^2)$<\/p>\n

This is possible only if a = b = c.<\/p>\n

So, the triangle is an equilateral triangle.<\/p>\n

Question 12:\u00a0<\/b>In the figure given below, ABCD is a rectangle. The area of the isosceles right triangle ABE = 7 $cm^2$ ; EC = 3(BE). The area of ABCD (in $cm^2$) is
\n<\/p>\n

a)\u00a021 $cm^2$<\/p>\n

b)\u00a028 $cm^2$<\/p>\n

c)\u00a042 $cm^2$<\/p>\n

d)\u00a056 $cm^2$<\/p>\n

12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b>
\n<\/p>\n

Let AB = BE = x<\/p>\n

Area of triangle ABE = $x^2\/2$ = 14; we get x = $\\sqrt{14}$<\/p>\n

So we have side BC = 4*$\\sqrt{14}$<\/p>\n

Now area is AB*BC = 14 *4 = 56 $cm^2$<\/p>\n

Question 13:\u00a0<\/b>The area of the triangle whose vertices are (a,a), (a + 1, a + 1) and (a + 2, a) is
\n[CAT 2002]<\/p>\n

a)\u00a0$a^3$<\/p>\n

b)\u00a0$1$<\/p>\n

c)\u00a0$2a$<\/p>\n

d)\u00a0$2^{1\/2}$<\/p>\n

13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

The triangle we have is :<\/p>\n

<\/p>\n

The length of three sides is $\\sqrt 2, \\sqrt 2$ and $2$.
\nThis is a right-angled triangle.
\nHence, it’s area equals $1\/2 * \\sqrt 2 * \\sqrt 2 = 1$
\nSo, the correct answer is b)
\nAlternate Approach :
\nArea of triangle =\u00a0$\\frac{1}{2}\\times\\ base\\times\\ height$
\nSo we get\u00a0$\\frac{1}{2}\\times2\\times\\ 1$=1 square units<\/p>\n

Question 14:\u00a0<\/b>In the following figure, ACB is a right-angled triangle. AD is the altitude. Circles are inscribed within the triangle ACD and triangle BCD. P and Q are the centers of the circles. The distance PQ is
\n
\nThe length of AB is 15 m and AC is 20 m<\/p>\n

a)\u00a07 m<\/p>\n

b)\u00a04.5 m<\/p>\n

c)\u00a010.5 m<\/p>\n

d)\u00a06 m<\/p>\n

14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b>
\n<\/p>\n

By Pythagoras theorem we get BC = 25 . Let BD = x;Triangle ABD is similar to triangle CBA => AD\/15 = x\/20 and also triangle ADC is similar to triangle ACB=> AD\/20 = (25-x)\/15. From the 2 equations, we get x = 9 and DC = 16<\/p>\n

We know that AREA = (semi perimeter ) * inradius<\/p>\n

For triangle ABD, Area = 1\/2 x BD X AD = 1\/2 x 12 x 9 = 54 and semi perimeter = (15 + 9 + 12)\/2 = 18. On using the above equation we get, inradius, r = 3.<\/p>\n

Similarly for triangle ADC we get inradius R = 4 .<\/p>\n

PQ = R + r = 7 cm<\/p>\n

Question 15:\u00a0<\/b>A circle is inscribed in a given square and another circle is circumscribed about the square. What is the ratio of the area of the inscribed circle to that of the circumscribed circle?<\/p>\n

a)\u00a02 : 3<\/p>\n

b)\u00a03 : 4<\/p>\n

c)\u00a01 : 4<\/p>\n

d)\u00a01 : 2<\/p>\n

15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

As we know that area of the circle is directly proportional to the square of its radius.
\nHence $\\frac{A_{ic}}{A_{cc}} = \\frac{\\frac{x^2}{4}}{\\frac{x^2}{2}}$
\nWhere $x$ is side of square (say), ic is inscribed circle with radius $\\frac{x}{2}$, cc is circumscribed circle with radius $\\frac{x}{\\sqrt{2}}$
\nSo ratio will be 1:2<\/p>\n

Question 16:\u00a0<\/b>If ABCD is a square and BCE is an equilateral triangle, what is the measure of \u2220DEC?
\n<\/p>\n

a)\u00a015\u00b0<\/p>\n

b)\u00a030\u00b0<\/p>\n

c)\u00a020\u00b0<\/p>\n

d)\u00a045\u00b0<\/p>\n

16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

According to given diagram, as triangle BCE is equilateral CE=BE=DC
\nHence angle CDE = angle CED<\/p>\n

So angle DEC = $\\frac{180-150}{2} = 15$
\nQuestion 17:\u00a0<\/b>In triangleABC, \u2220B is a right angle, AC = 6 cm, and D is the mid-point of AC. The length of BD is<\/p>\n

a)\u00a04cm<\/p>\n

b)\u00a06 cm<\/p>\n

c)\u00a03cm<\/p>\n

d)\u00a03.5cm<\/p>\n

17)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

A circle is circumscribed to triangle ABC.<\/p>\n

For this circle D will be centre of circle, and AD , DC , BD will radius of this circle.<\/p>\n

Hence AD=BD=DC=3 cm<\/p>\n

Question 18:\u00a0<\/b>The sum of the areas of two circles, which touch each other externally, is $153\\pi$. If the sum of their radii is 15, find the ratio of the larger to the smaller radius.<\/p>\n

a)\u00a04<\/p>\n

b)\u00a02<\/p>\n

c)\u00a03<\/p>\n

d)\u00a0None of these<\/p>\n

18)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

Given:
\n$\\pi((r_1)^2 + (r_2)^2) = 153\\pi$
\nSo
\n$(r_1)^2 + (r_2)^2 = 153$
\nOr $((r_1) + (r_2))^2 – 2(r_1)(r_2) = 153$
\nOr\u00a0\u00a0$(r_1)(r_2) = 36$ and $(r_1) + (r_2) = 15$
\n$r_1 = 12$
\n$r_2 = 3$
\nRatio = 4<\/p>\n

Question 19:\u00a0<\/b>
\nAB is the diameter of the given circle, while points C and D lie on the circumference as shown. If AB is 15 cm, AC is 12 cm and BD is 9 cm, find the area of the quadrilateral ACBD.<\/p>\n

a)\u00a054sq. cm<\/p>\n

b)\u00a0216sq. cm<\/p>\n

c)\u00a0162sq. cm<\/p>\n

d)\u00a0None of these<\/p>\n

19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

Since ACBD is cyclic quadrilateral with diagonals as AB = 15 ad CD.
\nSo area = $\\frac{1}{2} \\times (AB) \\times (CD)$ \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0eq.(1)
\nTriangle ACB and ABD is right angled triangle
\nLet’s say angle CBA = $\\theta$
\nso $CD = 2 \\times (9 sin\\theta)$ \u00a0(Where\u00a0\u00a0$sin\\theta$ = $\\frac{12}{15}$)
\nCD = $\\frac{72}{5}$
\nPutting values in eq. (1) , we will get area = 108 sq.cm.
\nHence answer will be D)<\/p>\n

Question 20:\u00a0<\/b>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<\/p>\n

a)\u00a0$225\\sqrt{3}$<\/p>\n

b)\u00a0$\\frac{500}{\\sqrt{3}}$<\/p>\n

c)\u00a0$\\frac{275}{\\sqrt{3}}$<\/p>\n

d)\u00a0$\\frac{250}{\\sqrt{3}}$<\/p>\n

20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

The lengths are given as 40, 25 and 35.<\/p>\n

The perimeter = 100<\/p>\n

Semi-perimeter, s = 50<\/p>\n

Area = $ \\sqrt{50 * 10 * 25 * 15}$ = $250\\sqrt{3}$<\/p>\n

<\/figure>\n

The triangle formed by the centroid and two vertices is removed.<\/p>\n

Since the cenroid divides the median in the ratio 2 : 1<\/p>\n

The remaining area will be two-thirds the area of the original triangle.<\/p>\n

Remaining area = $\\frac{2}{3} * 250\\sqrt{3}$ = $\\frac{500}{\\sqrt{3}}$<\/p>\n

\n

CAT Geometry Preparation Videos | CAT 2022 Preparation Videos<\/span><\/h2>\n