Enroll to CMAT crash course<\/a><\/p>\n Take Free CMAT mock tests here<\/a><\/p>\n Download CMAT previous papers PDF<\/a><\/p>\n Question 1:\u00a0<\/b>In two similar triangles ABC and MNP, if AB = 2.25 cm, MP = 4.5 cm and PN = 7.5 cm and m \u2220ACB = m \u2220MNP and m \u2220ABC = m \u2220MPN, then the length of side BC , in cm, is<\/p>\n a)\u00a04.5<\/p>\n b)\u00a03.75<\/p>\n c)\u00a04.75<\/p>\n d)\u00a03.5<\/p>\n Question 2:\u00a0<\/b>The heights of two similar right-angled triangles \u0394LMN and \u0394OPQ are 48 cm and 36 cm. If OP = 12 cm, then LM is<\/p>\n a)\u00a0$\\frac{10\\sqrt{6}}{3}$<\/p>\n b)\u00a016 cm<\/p>\n c)\u00a020 cm<\/p>\n d)\u00a012 cm<\/p>\n Question 3:\u00a0<\/b>Taking any three of the line segments out of segments of length 2 cm, 3 cm, 5 cm and 6 cm, the number of triangles that can be formed is :<\/p>\n a)\u00a03<\/p>\n b)\u00a02<\/p>\n c)\u00a01<\/p>\n d)\u00a04<\/p>\n Question 4:\u00a0<\/b>Two triangles ABC and PQR are congruent. If the area of A ABC is 60 sq. cm, then area of A PQR will be<\/p>\n a)\u00a060 sq.cm<\/p>\n b)\u00a030 sq.cm<\/p>\n c)\u00a015 sq.cm<\/p>\n d)\u00a0120 sq.cm<\/p>\n Question 5:\u00a0<\/b>The perimeters of two similar triangles \u0394ABC andCare 36cm and 24 cm respectively. If PQ = 10 cm, then AB is:<\/p>\n a)\u00a025 cm<\/p>\n b)\u00a010 cm<\/p>\n c)\u00a015 cm<\/p>\n d)\u00a020 cm<\/p>\n Get 5 CMAT mocks at just Rs.299<\/a><\/p>\n Question 6:\u00a0<\/b>\u0394 DEF and \u0394 GHI are similar triangles. Length of DE is 4 cm and length of the corresponding side GH is 9 cm. What is the ratio of areas of\u00a0\u0394 DEF and \u0394 GHI?<\/p>\n a)\u00a081:16<\/p>\n b)\u00a04:9<\/p>\n c)\u00a016:81<\/p>\n d)\u00a09:4<\/p>\n Question 7:\u00a0<\/b>\u0394 ABC and \u0394 DEF are similar triangles. Length of AB is 10 cm and length of the corresponding side DE is 6 cm. What is the ratio of Perimeter of \u0394ABC to \u0394DEF?<\/p>\n a)\u00a05:3<\/p>\n b)\u00a03:5<\/p>\n c)\u00a025:9<\/p>\n d)\u00a09:25<\/p>\n Question 8:\u00a0<\/b>The diagonals do not form at least two congruent triangles in a …………….<\/p>\n a)\u00a0Parallelogram<\/p>\n b)\u00a0Rhombus<\/p>\n c)\u00a0Trapezium<\/p>\n d)\u00a0Kite<\/p>\n Question 9:\u00a0<\/b>If the areas of two similar triangle are in the ratio 5 : 7, then what is the ratio of the corresponding sides of these two triangles?<\/p>\n a)\u00a0$5 : 7$<\/p>\n b)\u00a0$25 : 49$<\/p>\n c)\u00a0$\\sqrt5 : \\sqrt7$<\/p>\n d)\u00a0$125 : 343$<\/p>\n Question 10:\u00a0<\/b>\u0394PQR has sides PQ and PR measuring 983 and 893 units respectively. How many such triangles are possible with all integral sides?<\/p>\n a)\u00a01876<\/p>\n b)\u00a090<\/p>\n c)\u00a01785<\/p>\n d)\u00a01786<\/p>\n Question 11:\u00a0<\/b>The areas of two similar triangles \u0394ABC and \u0394PQR are 121 sq cms and 64 sq cms respectively. If PQ = 12 cm, what is the length (in cm) of AB?<\/p>\n a)\u00a07.25<\/p>\n b)\u00a06.25<\/p>\n c)\u00a016.5<\/p>\n d)\u00a05.25<\/p>\n Question 12:\u00a0<\/b>The areas of two similar triangles \u0394ABC and \u0394PQR are 36 sq cms and 9 sq cms respectively. If PQ = 4 cm then what is the length of AB (in cm)?<\/p>\n a)\u00a016<\/p>\n b)\u00a012<\/p>\n c)\u00a08<\/p>\n d)\u00a06<\/p>\n Question 13:\u00a0<\/b>The perimeter of two similar triangles ABC and PQR are 36 cms and 24 cms respectively. If PQ = 10 cm then the length of AB is<\/p>\n a)\u00a018 cm<\/p>\n b)\u00a012 cm<\/p>\n c)\u00a015 cm<\/p>\n d)\u00a030 cm<\/p>\n Download CMAT Previous Papers PDF<\/a><\/p>\n Question 14:\u00a0<\/b>\u2206ABC and \u2206DEF are two similar triangles and the perimeter of \u2206ABC and \u2206DEF are 30 cm and 18 cm respectively. If length of DE = 36 cm, then length of AB is<\/p>\n a)\u00a060 cm<\/p>\n b)\u00a040 cm<\/p>\n c)\u00a045 cm<\/p>\n d)\u00a050 cm<\/p>\n Question 15:\u00a0<\/b>If the triangle\u00a0 ABC and DEF follows the given equation, then these two triangles are similar by which of the following criterion ? a)\u00a0SAS similarity<\/p>\n b)\u00a0SSS similarity<\/p>\n c)\u00a0AAA similarity<\/p>\n d)\u00a0None of the these<\/p>\n Question 16:\u00a0<\/b>Which of the following is the CORRECT option for the triangles having sides in the ratio of 3:4:6?<\/p>\n a)\u00a0Acute angled<\/p>\n b)\u00a0Obtuse angled<\/p>\n c)\u00a0Right angled<\/p>\n d)\u00a0Either acute or right angled<\/p>\n Question 17:\u00a0<\/b>If the ratio of the angle bisector segments of the two equiangular triangles are in the ratio of 3:2 then what is the ratio of the corresponding sides of the two triangles?<\/p>\n a)\u00a02:3<\/p>\n b)\u00a03:2<\/p>\n c)\u00a06:4<\/p>\n d)\u00a04:6<\/p>\n Question 18:\u00a0<\/b>If the $\\angle A=\\angle D$ and $\\frac{AB}{DE}=\\frac{AC}{DF}$ then both triangles ABC and DEF is similar by which of the following criteria?<\/p>\n a)\u00a0SAS similarity<\/p>\n b)\u00a0ASA similarity<\/p>\n c)\u00a0AAA similarity<\/p>\n d)\u00a0None of these<\/p>\n Question 19:\u00a0<\/b>In a triangle ABC, a line is drawn from C which bisects AB at point D. Find the ratio of area of the triangles DBC and ABC.<\/p>\n a)\u00a01:1<\/p>\n b)\u00a02:1<\/p>\n c)\u00a01:2<\/p>\n d)\u00a01:3<\/p>\n Question 20:\u00a0<\/b>Consider the following two triangles as shown in the figure below<\/p>\n a)\u00a0$\\triangle BAC \\sim\\triangle NMP$<\/p>\n b)\u00a0$\\triangle BAC \\sim\\triangle MNP$<\/p>\n c)\u00a0$\\triangle CAB \\sim\\triangle NMP$<\/p>\n d)\u00a0$\\triangle BAC \\sim\\triangle PMN$<\/p>\n Take CMAT Mock Tests<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n As we can see, the 2 triangles are similar. Their sides are in the ratio 2:1 (MPN: ABC). Hence, BC will be 7.5\/2 = 3.75 cm. Option B is the right answer.<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n ratio of heights of two similar triangles is always equal to the ratio of the corresponding sides of the triangles<\/p>\n i.e, $\\frac{\\text{height of } \\triangle LMN}{\\text{height of } \\triangle OPQ}$ = $\\frac{\\text{length of LM}}{\\text{length of OP}}$<\/p>\n put in values from the question<\/p>\n $\\frac{48}{36}$ = $\\frac{\\text{length of LM}}{12}$<\/p>\n length of LM = 16<\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n We have to select three values out of the four length values given so that sum of any two values in the chosen set is larger than the third value.<\/p>\n => Three out of four combination problem and we can choose in $C_3^4$ = 4 ways.<\/p>\n But because of the triangle formation constraint, sides 2cm, 3cm can’t be taken together in any choice.<\/p>\n This reduces number of combinations by 2, [2, 3, 5] and [2, 3, 6]<\/p>\n => The only two possibilities, [2, 5, 6] and [3, 5, 6].<\/p>\n Ans – (B)<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Two congruent triangles are always equal in area.<\/p>\n $\\because$ $\\triangle$ABC $\\cong$ $\\triangle$PQR<\/span><\/p>\n => area($\\triangle$ABC) = area($\\triangle$PQR) = 60 $cm^2$<\/span><\/span><\/span><\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n In Similar triangles , corresponding sides are of same proportion. 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n It is given that \u0394DEF $\\sim$ \u0394GHI<\/p>\n Also, length of DE = 4 cm and length of the corresponding side GH = 9 cm<\/p>\n => Ratio of Area of \u0394DEF : Area of \u0394GHI = Ratio of square of corresponding sides = $(DE)^2$ : $(GH)^2$<\/p>\n =\u00a0$\\frac{(4)^2}{(9)^2} = \\frac{16}{81}$<\/p>\n $\\therefore$ The required ratio is 16 : 81<\/p>\n => Ans – (C)<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n It is given that \u0394AB $\\sim$ \u0394DEF<\/p>\n Also, length of AB = 10 cm and length of the corresponding side DE = 6 cm<\/p>\n => Ratio of Perimeter of \u0394ABC : Perimeter of \u0394DEF = Ratio of corresponding sides = AB : DE<\/p>\n =\u00a0$\\frac{10}{6} = \\frac{5}{3}$<\/p>\n $\\therefore$ The required ratio is 5 : 3<\/p>\n => Ans – (A)<\/p>\n 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n In a parallelogram, rhombus or kite, both pairs of opposite sides are parallel, and thus there are at least two congruent triangles which is not the case in a trapezium which has only one pair of parallel sides.<\/p>\n => Ans – (C)<\/p>\n 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Ratio of areas of two similar triangles is equal to the ratio of square of the corresponding sides.<\/p>\n Let the ratio of corresponding sides = $\\frac{x}{y}$<\/p>\n => $\\frac{x^2}{y^2}=\\frac{5}{7}$<\/p>\n => $(\\frac{x}{y})^2=\\frac{5}{7}$<\/p>\n => $\\frac{x}{y}=\\frac{\\sqrt5}{\\sqrt7}$<\/p>\n => Ans – (C)<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let us assume that the length of the third side is x units.<\/p>\n Case 1:\u00a0When the side PQ is the largest among three sides. Case 2: When the side QR is the largest among the three sides. Hence, the number of with integral value =\u00a0 1875 – 91 +1 = 1785.<\/p>\n 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n It is given that \u0394ABC $\\sim$ \u0394PQR<\/p>\n Let length of AB = $x$ cm and length of the corresponding side PQ = 12 cm<\/p>\n => Ratio of Area of \u0394ABC : Area of \u0394PQR = Ratio of square of corresponding sides = $(AB)^2$ : $(PQ)^2$<\/p>\n => $(\\frac{x}{12})^2 = \\frac{121}{64}$<\/p>\n => $\\frac{x}{12}=\\sqrt{\\frac{121}{64}}=\\frac{11}{8}$<\/p>\n => $x=\\frac{11}{8}\\times12=16.5$ cm<\/p>\n => Ans – (C)<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n For similar triangles<\/p>\n Ratio of sides = $\\sqrt{ \\text(ratio of areas)}=\\sqrt{36:9}=2:1$<\/p>\n AB\/PQ = 2\/1<\/p>\n AB\/4 = 2\/1<\/p>\n AB = 8<\/p>\n So the answer is option C.<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n It is given that \u0394ABC $\\sim$ \u0394PQR<\/p>\n Also, perimeter of \u2206ABC and \u2206PQR are 36 cm and 24 cm<\/p>\n => Ratio of Perimeter of \u0394ABC : Perimeter of \u0394PQR = Ratio of corresponding sides = AB : PQ<\/p>\n =\u00a0$\\frac{36}{24} = \\frac{AB}{10}$<\/p>\n => AB = $\\frac{3}{2} \\times 10=15$ cm<\/p>\n => Ans – (C)<\/p>\n 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n It is given that \u0394ABC $\\sim$ \u0394DEF<\/p>\n Also, perimeter of \u2206ABC and \u2206DEF are 30 cm and 18 cm<\/p>\n => Ratio of Perimeter of \u0394ABC : Perimeter of \u0394DEF = Ratio of corresponding sides = AB : DE<\/p>\n =\u00a0$\\frac{30}{18} = \\frac{AB}{36}$<\/p>\n => AB = $\\frac{5}{3} \\times 36=60$ cm<\/p>\n => Ans – (A)<\/p>\n 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given :\u00a0$\\frac{AB}{DE}=\\frac{BC}{EF}=\\frac{AC}{DF}$<\/p>\n => AB = DE , BC = EF , AC = DF<\/p>\n Thus, all corresponding sides of the two triangles are equal.<\/p>\n $\\therefore$ The\u00a0two triangles are similar by\u00a0SSS similarity.<\/p>\n => Ans – (B)<\/p>\n 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the sides of $\\triangle$ ABC be $a,b,c$, where the largest side = $’c’$<\/p>\n If $c^2=a^2+b^2$, then the angle at $C$ is right angle.<\/p>\n If $c^2<a^2+b^2$, then the angle at $C$ is acute angle.<\/p>\n If $c^2>a^2+b^2$, then the angle at $C$ is obtuse angle.<\/p>\n Now, according to ques, => $6^2=36$<\/p>\n and $3^2+4^2=9+16=25$<\/p>\n $\\therefore c^2>a^2+b^2$, hence it is an obtuse angled triangle.<\/p>\n => Ans – (B)<\/p>\n 17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given\u00a0: $AD:PS=3:2$<\/p>\n To find\u00a0: $AB : PQ=?$<\/p>\n Solution\u00a0: The given triangles are equiangular, i.e.\u00a0$\\angle$\u00a0A =\u00a0$\\angle$\u00a0P ,\u00a0$\\angle$\u00a0B =\u00a0$\\angle$\u00a0Q ,\u00a0$\\angle$\u00a0C =\u00a0$\\angle$\u00a0R<\/p>\n Now, in\u00a0$\\triangle$\u00a0ABD and $\\triangle$ PQS,<\/p>\n $\\angle$\u00a0B =\u00a0$\\angle$\u00a0Q<\/p>\n $\\angle$\u00a0BAD =\u00a0$\\angle$\u00a0QPS \u00a0 \u00a0 [$\\because$ $\\angle$\u00a0A = $\\angle$\u00a0P => $\\frac{1}{2}$ $\\angle$ A = $\\frac{1}{2}$ $\\angle$ P =>\u00a0$\\angle$\u00a0BAD =\u00a0$\\angle$\u00a0QPS]<\/p>\n So, by A-A criterion of similarity, we have :<\/p>\n $\\triangle$ ABD $\\sim$ $\\triangle$ PQS<\/p>\n => $\\frac{AB}{PQ}=\\frac{AD}{PS}=\\frac{3}{2}$<\/p>\n => Ans – (B)<\/p>\n 18)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $\\angle A=\\angle D$ \u00a0 \u00a0 (given)<\/p>\n $\\frac{AB}{DE}=\\frac{AC}{DF}$ \u00a0 \u00a0(given)<\/p>\n $\\therefore$\u00a0$\\triangle$\u00a0ABC $\\sim$\u00a0$\\triangle$\u00a0DEF by\u00a0SAS<\/span><\/em> similarity.<\/p>\n => Ans – (A)<\/p>\n 19)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given\u00a0: CD bisects AB, => AD = DB = $\\frac{8}{2}=4$ cm<\/p>\n To find\u00a0: $\\frac{ar(\\triangle DBC)}{ar(\\triangle ABC)}=?$<\/p>\n Solution\u00a0: Clearly $\\triangle$ ABC is a right angled triangle, $\\because (10)^2=(8)^2+(6)^2$<\/p>\n Thus, AC is the hypotenuse and $\\triangle$ ABC is right angled at B.<\/p>\n => AB = 8 cm is the height of triangle<\/p>\n $\\therefore$\u00a0$\\frac{ar(\\triangle DBC)}{ar(\\triangle ABC)}=\\frac{\\frac{1}{2}\\times(DB)\\times(BC)}{\\frac{1}{2}\\times(AB)\\times(BC)}$<\/p>\n = $\\frac{4\\times6}{8\\times6}=\\frac{1}{2}$<\/p>\n => Ans – (C)<\/p>\n 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n According to angle sum property\u00a0: $\\angle$ C = $50^\\circ$ and\u00a0$\\angle$ N = $80^\\circ$<\/p>\n Now, in $\\triangle$ ABC and\u00a0$\\triangle$\u00a0\u00a0MNP,<\/p>\n $\\angle$ A =\u00a0$\\angle$\u00a0N = $80^\\circ$<\/p>\n The remaining 2 angles are equal to $50^\\circ$, thus either $\\angle$\u00a0B =\u00a0$\\angle$\u00a0P or\u00a0$\\angle$\u00a0B =\u00a0$\\angle$\u00a0M<\/p>\n But\u00a0$\\angle$\u00a0A is corresponding to\u00a0$\\angle$\u00a0N.<\/p>\n Thus,\u00a0$\\triangle BAC \\sim\\triangle MNP$<\/p>\n => Ans – (B)<\/p>\n
\n$\\frac{AB}{DE}=\\frac{BC}{EF}=\\frac{AC}{DF}$<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/136379.png\")
<\/figure>\n![](\"https:\/\/cracku.in\/media\/uploads\/136890_HKlbu4O.png\")
<\/figure>\n
\n$\\frac{ Perimeter \u0394ABC }{ Perimeter \u0394PQR }=\\frac{AB}{PR}$
\n$\\frac{ 36 }{ 24 }=\\frac{AB}{10}$
\nAb = 15
\n<\/span><\/p>\n
\n893 + x > 983
\nx > 90<\/p>\n
\n893 + 983 > x
\nx < 1876
\nHence we can say that x $\\epsilon$ (90, 1876)<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/1503_mpiLESm.PNG\")
<\/figure>\n<\/figure>\n
In $\\triangle$ ABC and\u00a0$\\triangle$\u00a0DEF,<\/p>\n<\/p>\n<\/figure>\n