Download Geometry Triangles<\/strong><\/span> Questions for CAT<\/a><\/p>\n Enroll for CAT 2022 Crash 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> 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 Question 3:\u00a0<\/b>If the lengths of diagonals DF, AG and CE of the cube shown in the adjoining figure are equal to the three sides of a triangle, then the radius of the circle circumscribing that triangle will be? a)\u00a0equal to the side of the cube<\/p>\n b)\u00a0$\\sqrt 3$ times the side of the cube<\/p>\n c)\u00a01\/$\\sqrt 3$ times the side of the cube<\/p>\n d)\u00a0impossible to find from the given information<\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Consider side of the cube as x.<\/p>\n So diagonal will be of length $\\sqrt{3}$ * x.<\/p>\n Now if diagonals are side of equilateral triangle we get area = 3*$\\sqrt{3}*x^2$ \/4 .<\/p>\n Also in a triangle<\/p>\n 4 * Area * R = Product of sides<\/p>\n 4*\u00a03*$\\sqrt{3}*x^2$ \/4 * R = .3*$\\sqrt{3}*x^3$<\/p>\n R = x<\/p>\n Question 4:\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 ? a)\u00a0140<\/p>\n b)\u00a070<\/p>\n c)\u00a0100<\/p>\n d)\u00a0None of these<\/p>\n 4)\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 Question 5:\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 5)\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 6:\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 a)\u00a021 $cm^2$<\/p>\n b)\u00a028 $cm^2$<\/p>\n c)\u00a042 $cm^2$<\/p>\n d)\u00a056 $cm^2$<\/p>\n 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> 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 Checkout: CAT Free Practice Questions and Videos<\/strong><\/a><\/em><\/p>\n Question 7:\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 a)\u00a021 $cm^2$<\/p>\n b)\u00a028 $cm^2$<\/p>\n c)\u00a042 $cm^2$<\/p>\n d)\u00a056 $cm^2$<\/p>\n 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> 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 8:\u00a0<\/b>The area of the triangle whose vertices are (a,a), (a + 1, a + 1) and (a + 2, a) is a)\u00a0$a^3$<\/p>\n b)\u00a0$1$<\/p>\n c)\u00a0$2a$<\/p>\n d)\u00a0$2^{1\/2}$<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The triangle we have is :<\/p>\n The length of three sides is $\\sqrt 2, \\sqrt 2$ and $2$. Question 9:\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 a)\u00a07 m<\/p>\n b)\u00a04.5 m<\/p>\n c)\u00a010.5 m<\/p>\n d)\u00a06 m<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> 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 10:\u00a0<\/b>If ABCD is a square and BCE is an equilateral triangle, what is the measure of \u2220DEC? a)\u00a015\u00b0<\/p>\n b)\u00a030\u00b0<\/p>\n c)\u00a020\u00b0<\/p>\n d)\u00a045\u00b0<\/p>\n 10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n
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\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
\n![](\"https:\/\/cracku.in\/media\/uploads\/2015\/07\/08\/geom12.PNG\")
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\n![](\"https:\/\/cracku.in\/media\/question\/4795.png\")
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\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
\n![](\"https:\/\/cracku.in\/media\/question\/5100_1.png\")
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\n[CAT 2002]<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/image_N5kgqHD.png\")
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\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
\n![](\"https:\/\/cracku.in\/media\/uploads\/2015\/07\/01\/pic-1.png\")
\nThe length of AB is 15 m and AC is 20 m<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/2015\/07\/08\/geom-11.png\")
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\n![](\"https:\/\/cracku.in\/media\/uploads\/2015\/08\/11\/untitled_3_wsoA32i.png\")
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