Enroll to 15 SSC CHSL 2022 Mocks At Just Rs. 149<\/a><\/p>\n Question 1:\u00a0<\/b>ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If the area of triangle ABC is 136 cm$^2$, a)\u00a0$36 cm^2$<\/p>\n b)\u00a0$38 cm^2$<\/p>\n c)\u00a0$24 cm^2$<\/p>\n d)\u00a0$34 cm^2$<\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b> Let the side of the equilateral triangle ABC = a<\/p>\n $\\Rightarrow$\u00a0 BC = a<\/p>\n D is the mid-point of BC<\/p>\n $\\Rightarrow$\u00a0 BD = $\\frac{a}{2}$<\/p>\n Side of the equilateral triangle BDE =\u00a0$\\frac{a}{2}$<\/p>\n Given, Area of the equilateral triangle ABC = 136 cm$^2$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{\\sqrt{3}}{4}a^2=136$<\/p>\n $\\therefore\\ $Area of the equilateral triangle BDE =\u00a0$\\frac{\\sqrt{3}}{4}\\left(\\frac{a}{2}\\right)^2$<\/p>\n $=\\frac{\\sqrt{3}}{4}\\times\\frac{a^2}{4}$<\/p>\n $=\\frac{1}{4}\\times\\frac{\\sqrt{3}a^2}{4}$<\/p>\n $=\\frac{1}{4}\\times136$<\/p>\n $=$ 34 cm$^2$<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 2:\u00a0<\/b>Two equilateral triangles of side $10\\sqrt{3}$ cm are joined to form a quadrilateral. What is the altitude of the quadrilateral?<\/p>\n a)\u00a012 cm<\/p>\n b)\u00a014 cm<\/p>\n c)\u00a016 cm<\/p>\n d)\u00a0i5 cm<\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b> Given that\u00a0\u00a0$10\\sqrt{3}$ cm<\/p>\n We know the area of equilateral triangle = $ \\frac{\\sqrt {3}} {4} a^2 $\u00a0 \u00a0…… Eq (1)<\/p>\n and on the $\\triangle DCB\u00a0 \u00a0is\u00a0 \u00a0also\u00a0 \u00a0given = \\frac{1}{2} \\times a \\times h $\u00a0 \u00a0…… Eq (2)<\/p>\n then Eq(1) = Eq (2)<\/p>\n $ \\frac{\\sqrt {3}} {4} a^2\u00a0 =\u00a0\u00a0 \\frac{1}{2} \\times a \\times h $<\/p>\n $\\Rightarrow h =\u00a0 \\frac{\\sqrt{3}} {2} a $<\/p>\n $\\Rightarrow h =\u00a0\\frac{\\sqrt{3}} {2} \\times\u00a010\\sqrt{3}$<\/p>\n $\\Rightarrow h = 15 cm Ans<\/p>\n Question 3:\u00a0<\/b>The perimeters of two similar triangles ABC and PQRare 78 cm and 46.8 cm, respectively. If PQ = 11.7, then the length of AB is:<\/p>\n a)\u00a019.5 cm<\/p>\n b)\u00a023.4 cm<\/p>\n c)\u00a024 cm<\/p>\n d)\u00a020 cm<\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Triangles ABC and PQR are similar. Question 4:\u00a0<\/b>The sides of an isosceles triangles are 10 cm, 10 cm and 12 cm. What is the area of the triangle?<\/p>\n a)\u00a060 $cm^2$<\/p>\n b)\u00a048 $cm^2$<\/p>\n c)\u00a040 $cm^2$<\/p>\n d)\u00a044 $cm^2$<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n To find the area of the isosceles triangle ABC draw a perpendicular line from A to the base of the triangle BC and name that point as D such that it will become a right angled triangle.<\/p>\n Question 5:\u00a0<\/b>Give that the ratio of altitudes of two triangles is 4 : 5, ratio of their areas is 3: 2. The ratio of their corresponding bases is<\/p>\n a)\u00a08:15<\/p>\n b)\u00a05:8<\/p>\n c)\u00a015:8<\/p>\n d)\u00a08:5<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given that ratio of altitudes of two triangles is 4:5<\/p>\n => $\\frac{h_{1}}{h_{2}}$=$\\frac{4}{5}$<\/p>\n Also, Given that, ratio of areas of two triangles is 3:2<\/p>\n => $\\frac{\\frac{1}{2} \\times b_{1} \\times h_{1}}{\\frac{1}{2} \\times b_{2} \\times h_{2}}$ = $\\frac{3}{2}$<\/p>\n =>\u00a0$\\frac{b_{1} \\times 4}{ b_{2} \\times 5}$ = $\\frac{3}{2}$<\/p>\n =>\u00a0$\\frac{b_{1}}{ b_{2}}$ =\u00a0$\\frac{15}{8}$<\/p>\n Therefore, ratios of the bases is 15:8<\/p>\n Take a free SSC CHSL Tier-1 mock test<\/a><\/p>\n Download SSC CGL Tier-1 Previous Papers PDF<\/a><\/p>\n Question 6:\u00a0<\/b>Equilateral triangles are drawn on the hypotenuse and one of the perpendicular sides of a right-angled isosceles triangles. Their areas are H and A respectively. $\\frac{A}{H}$ is equal to:<\/p>\n a)\u00a0$\\frac{1}{4}$<\/p>\n b)\u00a0$\\frac{1}{\\sqrt[2]{2}}$<\/p>\n c)\u00a0$\\frac{1}{\\sqrt{2}}$<\/p>\n d)\u00a0$\\frac{1}{2}$<\/p>\n 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the length of the two equal sides of\u00a0right-angled isosceles triangle be x cm then hypotenuse be $x\\sqrt{2} \u00a0cm$<\/p>\n Area of equilateral triangle drawn on hypotenuse (H) = $\\frac{\\sqrt{3}}{4}\\times a^{2}=\\frac{\\sqrt{3}}{4}\\times (x\\sqrt{2})^2$<\/p>\n Area of equilateral triangle drawn on side (A) = $\\frac{\\sqrt{3}}{4}\\times a^{2}=\\frac{\\sqrt{3}}{4}\\times x^2$<\/p>\n $\\frac{A}{H}=\\left(\\frac{\\sqrt{3}}{4}\\times x^{2}\\right)\\times\\frac{4}{\\sqrt{3}}\\times\\frac{1}{(x\\sqrt{2})^2} =\u00a0\\frac{1}{2}$<\/p>\n Question 7:\u00a0<\/b>There is a polygon of 11 sides. How many triangles can be drawn by only using the vertices of the polygon?<\/p>\n a)\u00a0180<\/p>\n b)\u00a0150<\/p>\n c)\u00a0165<\/p>\n d)\u00a0175<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,<\/p>\n Number of sides of the polygon = 11<\/p>\n $=$>\u00a0 Number of vertices of the polygon = 11<\/p>\n Number of vertices of the triangle = 3<\/p>\n Number of possibilities of selecting 3 from 11 =\u00a0$11_{C_3}$ =$\\frac{11\\times10\\times9}{1\\times2\\times3}$ =\u00a0$165$<\/p>\n $\\therefore\\ $Number of\u00a0triangles that can be drawn by only using the vertices of the polygon of 11 sides = 165<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 8:\u00a0<\/b>The ratio of the areas of two triangles ABC and PQR is 4: 5 and the ratio of their heights is 5 : 3. The ratio of the bases of triangle ABC to that of triangle PQR is:<\/p>\n a)\u00a025 : 12<\/p>\n b)\u00a012 : 25<\/p>\n c)\u00a011 : 15<\/p>\n d)\u00a015 : 11<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n We know that Question 9:\u00a0<\/b>The ratio of the areas of two triangles ABC and PQR is 3 : 5 and the ratio of their heights is 5 : 3. The ratio of the bases of triangle ABC to that of triangle PQR is:<\/p>\n a)\u00a025 : 9<\/p>\n b)\u00a09 : 25<\/p>\n c)\u00a02 : 1<\/p>\n d)\u00a01 : 1<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the Areas\u00a0 of $\\triangle ABC$ and $\\triangle PQR$\u00a0 are\u00a0 $3x:5x$<\/p>\n Let\u00a0the Heights\u00a0 of $\\triangle ABC$ and $\\triangle PQR$\u00a0 are\u00a0 $5y:3y$\u00a0 and Bases are $B_{1}$ and $B_{2}$ respectively.<\/p>\n Area of\u00a0$\\triangle ABC$ = $\\frac{1}{2} \\times 5y\\times B_{1}$ and\u00a0Area of\u00a0$\\triangle PQR$ = $\\frac{1}{2} \\times3y\\times B_{2}$<\/p>\n $\\therefore\u00a0\\frac{3x}{5x}\u00a0= \\frac{\\frac{1}{2} \\times 5y\\times B_{1}}{\\frac{1}{2} \\times3y\\times B_{2}}$<\/p>\n $\\Rightarrow \\frac{B_{1}}{B_{2}} = \\frac{9}{25}$<\/p>\n $\\therefore$\u00a0 \u00a0$B_{1}:B_{2} = 9:25$<\/p>\n Question 10:\u00a0<\/b>The ratio of the area of two triangles is 2 : 3 and ratio of their height is 3 : 2. The ratio of their bases is:<\/p>\n a)\u00a03:2<\/p>\n b)\u00a04:9<\/p>\n c)\u00a09:4<\/p>\n d)\u00a02:3<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let height of the two triangles be 3 and 2 units respectively.<\/p>\n Let bases of the two triangles be $b_1$ and $b_2$ units respectively.<\/p>\n => Ratio of area of triangles = $\\frac{\\frac{1}{2}\\times b_1\\times h_1}{\\frac{1}{2}\\times b_2\\times h_2}=\\frac{2}{3}$<\/p>\n => $\\frac{3b_1}{2b_2}=\\frac{2}{3}$<\/p>\n => $\\frac{b_1}{b_2}=\\frac{4}{9}$<\/p>\n $\\therefore$ Required ratio =\u00a04\u00a0: 9<\/strong><\/p>\n => Ans – (B)<\/p>\n Question 11:\u00a0<\/b>The ratio of the areas of two triangles is 1 : 2 and the ratio of their bases is 3 : 4. What will be the ratio of their height?<\/p>\n a)\u00a01 : 3<\/p>\n b)\u00a04 : 3<\/p>\n c)\u00a02 : 1<\/p>\n d)\u00a02 : 3<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the bases of the triangles be 3x and 4x units. Question 12:\u00a0<\/b>$PQRS$ is a square whose side is $20 cm$. By joining opposite vertices of $PQRS$ are get four triangles. What is the sum of the perimeters of the four triangles?<\/p>\n a)\u00a0$40\\surd2$<\/p>\n b)\u00a0$80\\surd2 + 80$<\/p>\n c)\u00a0$40\\surd2 + 40$<\/p>\n d)\u00a0$40\\surd2 + 80$<\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Side of square is 20 cm.<\/p>\n So, diagonal of square is 20\u221a2 cm.<\/p>\n So,to calculate perimeter of triangle,we get to count each diagonal 2 times and 4 sides only one time .<\/p>\n So, required perimeter=4\u00d720+4\u00d720\u221a2<\/p>\n =(80+80\u221a2) cm.<\/p>\n B is correct choice.<\/p>\n Question 13:\u00a0<\/b>In the given figure. $ABCDEF$ is a regular hexagon whose side is 6 cm. $APF, QAB, DCR$ and $DES$ are equilateral triangles. What is the area (in $cm^2$) of the shaded region? a)\u00a0$24\\surd3$<\/p>\n b)\u00a0$18\\surd3$<\/p>\n c)\u00a0$72\\surd3$<\/p>\n d)\u00a0$36\\surd3$<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n We have :<\/p>\n We here have 6 equilateral triangles of side 6cm so we can say a complete hexagon and half area of rectangle BCFE Question 14:\u00a0<\/b>Which of the following options is\/are CORRECT about the similarity of the two triangles?<\/p>\n a)\u00a0The corresponding sides are proportional to each other.<\/p>\n b)\u00a0The corresponding angles are equal.<\/p>\n c)\u00a0The corresponding sides may or may not be equal to each other.<\/p>\n d)\u00a0All option are correct.<\/p>\n 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n If two triangles are similar, then the corresponding sides are proportional to each other. Also, the corresponding angles are equal, but the corresponding sides may or may not be equal to each other. Thus, all are correct.<\/p>\n => Ans – (D)<\/p>\n Question 15:\u00a0<\/b>Consider the two equiangular triangles ABC and DEF having medians as AL and DM respectively as shown in figure below.<\/p>\n a)\u00a0$\\frac{BC}{EF}=\\frac{DM}{AL}$<\/p>\n b)\u00a0$\\frac{BC}{EF}=\\frac{AL}{DM}$<\/p>\n c)\u00a0$\\frac{EF}{BC}=\\frac{AL}{DM}$<\/p>\n d)\u00a0All options are Correct<\/p>\n 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n It is given that $\\triangle$ ABC and $\\triangle$ DEF are equiangular triangles, thus corresponding angles are equal.<\/p>\n =>\u00a0$\\triangle$ ABC $\\sim\\triangle$ DEF<\/p>\n => Ratio of perimeter = $\\frac{AB}{DE}=$\u00a0$\\frac{BC}{EF}=$\u00a0$\\frac{AC}{DF}$\u00a0————-(i)<\/p>\n Since, AL and DM are medians, => Ratio of perimeter of $\\triangle$ ABC to $\\triangle$ DEF = $\\frac{AL}{DM}$ ———–(ii)<\/p>\n From, equations (i) and (ii), we get\u00a0:<\/p>\n =>\u00a0$\\frac{BC}{EF}=\\frac{AL}{DM}$<\/p>\n => Ans – (B)<\/p>\n Question 16:\u00a0<\/b>Consider the two similar triangles ABC and DEF. Which of the following is correct about the ratio of the area of the triangle ABC and DEF?<\/p>\n a)\u00a0$(\\frac{AB}{DE})=(\\frac{EF}{BC})=(\\frac{AC}{DF})$<\/p>\n b)\u00a0$(\\frac{AB}{DE})^2=(\\frac{BC}{EF})^2=(\\frac{AC}{DF})^2$<\/p>\n c)\u00a0$(\\frac{AB}{DE})=(\\frac{BC}{EF})=(\\frac{DF}{AC})$<\/p>\n d)\u00a0None of these<\/p>\n 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n It is given that $\\triangle$ ABC $\\sim\\triangle$ DEF<\/p>\n => $\\frac{AB}{DE}=$\u00a0$\\frac{BC}{EF}=$\u00a0$\\frac{AC}{DF}$<\/p>\n Also, ratio of areas of two similar triangles is equal to the ratio of the square of the corresponding sides.<\/p>\n => $\\frac{ar(\\triangle ABC)}{ar(\\triangle DEF)}=$\u00a0$(\\frac{AB}{DE})^2=(\\frac{BC}{EF})^2=(\\frac{AC}{DF})^2$<\/p>\n => Ans – (B)<\/p>\n Question 17:\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 17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/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 Question 18:\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 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/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 Question 19:\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 19)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/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 Question 20:\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 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/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 Question 21:\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 21)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/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 Question 22:\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 22)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/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 Question 23:\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 23)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/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 Question 24:\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 24)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/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 Question 25:\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 25)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/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 shortlink to=”https:\/\/cracku.in\/pay\/c8f0w” color=”red” css_class=””]Enroll to 15 SSC MTS 2022 Mocks At Just Rs. 149[\/shortlink]<\/p>\n
\nthen the area of triangle BDE is equal to:<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/354607.png\")
<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/geogebra-export_zWtG8q9.png\")
<\/p>\n
\nSo,
\n$\\frac{perimeter of ABC}{perimeter of PQR} =\u00a0\\frac{side of\u00a0ABC}{side of PQR}$
\n$\\frac{78}{46.8} = \\frac{AB}{11.7}$
\n$\\frac{78}{46.8} \\times 11.7 = AB$
\nAB = 19.5 cm<\/p>\n
\n
\nNow,BD = CD =6cm.The area of the right angled triangle is given as = $\\frac{1}{2}\\times{b}\\times{h}$,where b and h are base and height.
\nHeight we can calculate from pythagoras theorem
\n$AD^{2}$ = $(AB^{2} – BD^{2})$
\n$\\Rightarrow$ $AD^{2}$ = $(10^{2} – 6^{2})$
\n$\\Rightarrow$ AD = 8cm.
\nThe area of right angled now will be = $\\frac{1}{2}\\times{6}\\times{8}$ = 24$cm^{2}$
\nThe area of triangle ABC will be two times the area of triangle ADB = 24+24=48$cm^{2}$.
\nHence option B is correct.<\/p>\n
\nArea =\u00a0$\\frac{1}{2}\\times\\ b\\times\\ h$
\nNow we can say
\nThe ratio of areas of two triangles with bases b1 and b2 and heights h1 and h2 will be\u00a0$\\frac{\\left(b1\\ h1\\right)}{b2\\ h2}$
\nso we get
\n$\\frac{4}{5}=\\frac{\\left(5b1\\ \\right)}{3b2}$
\nso b1:b2 =12:25<\/p>\n
\nHeights of the triangles be a and b units respectively.
\nThen, Ratio of their areas = $\\dfrac{1}{2}\\times3x\\times a : \\dfrac{1}{2}\\times4x\\times b = 3a : 4b$
\nGiven, $3a :\u00a04b = 1 :\u00a02$
\n=> $3a\\times2 = 4b\\times1 => 6a = 4b$
\n=> $\\dfrac{a}{b} = \\dfrac{4}{6} = \\dfrac{2}{3}$
\nTherefore, Their heights will be in the ratio 2 :\u00a03.<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/Q36(2).png\")
<\/p>\n![](\"https:\/\/media-cdn.cracku.in\/uploads\/image_sE5I4s1.png\")
<\/p>\n
\nNow In triangle ABF
\nusing cosine rule
\nwe get cos A = (6^2+6^2-BF^2)\/2AF AB
\nwe get BF $6\\sqrt{\\ 3}$
\nArea of shaded region = Area of hexagon + 0.5(Area of rectangle BFCE)
\nwe get area = $6\\times\\ \\frac{\\sqrt{\\ 3}}{4}\\times\\ 6\\times\\ 6\\ +\\ 6\\sqrt{\\ 3}\\times\\ 6\\times\\ 0.5$
\n= $54\\sqrt{\\ 3}+18\\sqrt{\\ 3}\\ =72\\sqrt{\\ 3}$<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/137697.png\")
Choose the correct option for the above situation.<\/figure>\n![](\"https:\/\/cracku.in\/media\/uploads\/136890_HKlbu4O.png\")
<\/figure>\n<\/figure>\n![](\"https:\/\/cracku.in\/media\/uploads\/136379.png\")
<\/figure>\n<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/1503_mpiLESm.PNG\")
In $\\triangle$ ABC and\u00a0$\\triangle$\u00a0DEF,<\/p>\n<\/figure>\n
\n$\\frac{AB}{DE}=\\frac{BC}{EF}=\\frac{AC}{DF}$<\/p>\n<\/figure>\n