Enroll to 15 SSC MTS 2022 Mocks At Just Rs. 149<\/a><\/p>\n Question 1:\u00a0<\/b>$\\triangle$ABC $\\sim$ $\\triangle$DEF and the area of $\\triangle$ABC is 13.5 cm$^2$ and the area of $\\triangle$DEF is 24 cm$^2$ . If BC = 3.15 cm, then the length (in cm) of EF is:<\/p>\n a)\u00a04.8<\/p>\n b)\u00a03.9<\/p>\n c)\u00a05.1<\/p>\n d)\u00a04.2<\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b> $\\triangle$ABC $\\sim$ $\\triangle$DEF<\/p>\n $\\Rightarrow$\u00a0 $\\frac{\\text{Area of triangle ABC}}{\\text{Area of triangle DEF}}$ = $\\left(\\frac{BC}{EF}\\right)^2$<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\frac{13.5}{24}=\\left(\\frac{3.15}{EF}\\right)^2$<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\frac{135}{240}=\\left(\\frac{3.15}{EF}\\right)^2$<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\frac{9}{16}=\\left(\\frac{3.15}{EF}\\right)^2$<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\frac{3.15}{EF}=\\frac{3}{4}$<\/p>\n $\\Rightarrow$\u00a0 EF = 4.2 cm<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 2:\u00a0<\/b>In $\\triangle$ABC, $\\angle$C = 90$^\\circ$ and Q is the midpoint of BC. If AB = 10 cm and AC = $2\\sqrt{10}$ cm, then the length of AQ is:<\/p>\n a)\u00a0$\\sqrt{55}$ cm<\/p>\n b)\u00a0$5\\sqrt{3}$ cm<\/p>\n c)\u00a0$5\\sqrt{2}$ cm<\/p>\n d)\u00a0$3\\sqrt{5}$ cm<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b> From right angled triangle ABC,<\/p>\n AC$^2$ +\u00a0BC$^2$ =\u00a0AB$^2$<\/p>\n $\\left(2\\sqrt{10}\\right)^2$ +\u00a0BC$^2$ =\u00a0$\\left(10\\right)^2$<\/p>\n 40 +\u00a0BC$^2$ = 100<\/p>\n BC$^2$ = 60<\/p>\n BC =\u00a0$2\\sqrt{15}$ cm<\/p>\n Q is the midpoint of BC.<\/p>\n CQ = $\\frac{BC}{2}$ =\u00a0$\\sqrt{15}$ cm<\/p>\n From right angled triangle ACQ,<\/p>\n AC$^2$ + CQ$^2$ = AQ$^2$<\/p>\n $\\left(2\\sqrt{10}\\right)^2$ +\u00a0$\\left(\\sqrt{15}\\right)^2$ =\u00a0AQ$^2$<\/p>\n 40 + 15 =\u00a0AQ$^2$<\/p>\n AQ$^2$ = 55<\/p>\n AQ = $\\sqrt{55}$ cm<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 3:\u00a0<\/b>In $\\triangle ABC$ and $\\triangle DEF$ we have $\\frac{AB}{DF} = \\frac{BC}{DE} = \\frac{AC}{EF}$, then which of the following is true?<\/p>\n a)\u00a0$\\triangle BCA \\sim \\triangle DEF$<\/p>\n b)\u00a0$\\triangle DEF \\sim \\triangle ABC$<\/p>\n c)\u00a0$\\triangle DEF \\sim \\triangle BAC$<\/p>\n d)\u00a0$\\triangle CAB \\sim \\triangle DEF$<\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b> Hence, the correct answer is Option A<\/p>\n Question 4:\u00a0<\/b>In a triangle ABC, length of the side AC is 4 cm more than 2 times the length of the side AB. Length of the side BC is 4 cm less than the three times the length of the side AB. If the perimeter of $\\triangle$ABC is 60 cm, then its area (in cm$^2$) is:<\/p>\n a)\u00a0150<\/p>\n b)\u00a0144<\/p>\n c)\u00a0100<\/p>\n d)\u00a0120<\/p>\n 4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Length of the side AC is 4 cm more than 2 times the length of the side AB.<\/p>\n b = 2c + 4……(1)<\/p>\n Length of the side BC is 4 cm less than the three times the length of the side AB.<\/p>\n a = 3c – 4…….(2)<\/p>\n Perimeter of $\\triangle$ABC is 60 cm.<\/p>\n a + b + c = 60<\/p>\n 3c – 4 + 2c + 4 + c = 60<\/p>\n 6c = 60<\/p>\n c = 10 cm<\/p>\n b = 2c + 4 = 24 cm<\/p>\n a = 3c – 4 = 26 cm<\/p>\n Half of the perimeter(s) = 30 cm<\/p>\n Area of the triangle =\u00a0$\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}$<\/p>\n =\u00a0$\\sqrt{30\\left(30-26\\right)\\left(30-24\\right)\\left(30-10\\right)}$<\/p>\n =\u00a0$\\sqrt{30\\left(4\\right)\\left(6\\right)\\left(20\\right)}$<\/p>\n =\u00a0$\\sqrt{120\\times120}$<\/p>\n = 120 cm$^2$<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 5:\u00a0<\/b>The area of a table top in the shape of an equilateral triangle is $9\\sqrt{3}$ cm$^2$. What is the length (in cm) of each side of the table?<\/p>\n a)\u00a06<\/p>\n b)\u00a02<\/p>\n c)\u00a04<\/p>\n d)\u00a03<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the length of each side of the table = a<\/p>\n The area of a table top in the shape of an equilateral triangle is $9\\sqrt{3}$ cm$^2$.<\/p>\n $\\frac{\\sqrt{3}}{4}$a$^2$ =\u00a0$9\\sqrt{3}$<\/p>\n a$^2$ = 36<\/p>\n a = 6 cm<\/p>\n Length of each side of the table = 6 cm<\/p>\n Hence, the correct answer is Option A<\/p>\n Take a free SSC MTS Tier-1 mock test<\/a><\/p>\n Download SSC CGL Tier-1 Previous Papers PDF<\/a><\/p>\n Question 6:\u00a0<\/b>Let $\\triangle$ABC $\\sim$ $\\triangle$RPQ and $\\frac{ar(\\triangle \\text{ABC})}{ar(\\triangle \\text{PQR})}=\\frac{16}{25}$. If PQ = 4 cm, QR = 6 cm and PR = 7 cm, then AC (in cm) is equal to:<\/p>\n a)\u00a06<\/p>\n b)\u00a04.8<\/p>\n c)\u00a03.6<\/p>\n d)\u00a07.2<\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b> $\\triangle$ABC $\\sim$ $\\triangle$RPQ<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\left(\\frac{AC}{QR}\\right)^2=\\frac{ar(\\triangle\\text{ABC})}{ar(\\triangle\\text{PQR})}$<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\left(\\frac{AC}{6}\\right)^2=\\frac{16}{25}$<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\frac{AC}{6}=\\frac{4}{5}$<\/p>\n $\\Rightarrow$\u00a0 AC = 4.8 cm<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 7:\u00a0<\/b>In $\\triangle$ABC, D is a point on BC such that $\\angle$BAD = $\\frac{1}{2} \\angle$ADC and $\\angle$BAC = 77$^\\circ$ and $\\angle$C = 45$^\\circ$. What is the measure of $\\angle$ADB?<\/p>\n a)\u00a0$77^\\circ$<\/p>\n b)\u00a0$64^\\circ$<\/p>\n c)\u00a0$58^\\circ$<\/p>\n d)\u00a0$45^\\circ$<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b> $\\angle$ADC = x<\/p>\n Given, $\\angle$BAD = $\\frac{1}{2} \\angle$ADC<\/p>\n $\\angle$BAD = $\\frac{\\text{x}}{2}$<\/p>\n $\\angle$DAC = 77 –\u00a0$\\frac{\\text{x}}{2}$<\/p>\n From triangle ADC,<\/p>\n $\\angle$ADC +\u00a0$\\angle$ACD +\u00a0$\\angle$DAC = 180$^\\circ$<\/p>\n x +\u00a045$^\\circ$ + 77 –\u00a0$\\frac{\\text{x}}{2}$ =\u00a0180$^\\circ$<\/p>\n $\\frac{\\text{x}}{2}$ = 58$^\\circ$<\/p>\n x =\u00a0116$^\\circ$<\/p>\n $\\angle$ADB =\u00a0180$^\\circ$ – x<\/p>\n =\u00a0180$^\\circ$ –\u00a0116$^\\circ$<\/p>\n = 64$^\\circ$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 8:\u00a0<\/b>The area of a triangular plot having sides 12 m, 35 m and 37 m is equal to the area of a rectangular field whose sides are in the ratio 7 : 3. The perimeter (in m) of the field is:<\/p>\n a)\u00a0$20\\sqrt{10}$<\/p>\n b)\u00a0$20\\sqrt{5}$<\/p>\n c)\u00a0$24\\sqrt{10}$<\/p>\n d)\u00a0$24\\sqrt{5}$<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Perimeter of triangular plot = 12 + 35 + 37 = 84 m<\/p>\n Half of the perimeter = s = 42 m<\/p>\n Area of the\u00a0triangular plot =\u00a0$\\sqrt{42\\left(42-12\\right)\\left(42-35\\right)\\left(42-37\\right)}$<\/p>\n =\u00a0$\\sqrt{42\\left(30\\right)\\left(7\\right)\\left(5\\right)}$<\/p>\n = 210 cm$^2$<\/p>\n The area of a triangular plot is equal to the area of a rectangular field.<\/p>\n Area of a rectangular field = 210 cm$^2$<\/p>\n Sides of rectangular filed are in the ratio 7:3.<\/p>\n Let the sides of the rectangular field are 7p and 3p respectively.<\/p>\n 7p x 3p = 210<\/p>\n p$^2$ = 10<\/p>\n p =\u00a0$\\sqrt{10}$<\/p>\n Perimeter of the\u00a0rectangular field = 2(7p + 3p)<\/p>\n = 20p<\/p>\n = 20$\\sqrt{10}$ cm<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 9:\u00a0<\/b>In triangle PQR, points E and F are on sides PQ and PR respectively such that EF is parallel to QR. If PE = 2 cm and EQ = 3 cm, then area($\\triangle$ PQR) : area($\\triangle$ PEF) is equal to:<\/p>\n a)\u00a03 : 2<\/p>\n b)\u00a025 : 4<\/p>\n c)\u00a05 : 2<\/p>\n d)\u00a09 : 4<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b> $\\triangle$PQR and\u00a0$\\triangle$PEF are similar triangles.<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\frac{\\text{Area of triangle PQR}}{\\text{Area of triangle PEF}}$ = $\\left(\\frac{PQ}{PE}\\right)^2$<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\frac{\\text{Area of triangle PQR}}{\\text{Area of triangle PEF}}$ = $\\left(\\frac{PE+EQ}{PE}\\right)^2$<\/p>\n $\\Rightarrow$ $\\frac{\\text{Area of triangle PQR}}{\\text{Area of triangle PEF}}$ = $\\left(\\frac{2+3}{2}\\right)^2$<\/p>\n $\\Rightarrow$ $\\frac{\\text{Area of triangle PQR}}{\\text{Area of triangle PEF}}$ =\u00a0$\\frac{25}{4}$<\/p>\n $\\Rightarrow$\u00a0 Area of triangle PQR :\u00a0Area of triangle PEF = 25:4<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 10:\u00a0<\/b>The angles of a triangle are in AP (arithmetic progression). If measure of the smallest angle is $50^\\circ$ less than that of the largest angle, then find the largest angle (in degrees).<\/p>\n a)\u00a090<\/p>\n b)\u00a085<\/p>\n c)\u00a080<\/p>\n d)\u00a075<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n The angles of triangle are in AP (arithmetic progression).<\/p>\n Let the angles are a, a+r, a+2r.<\/p>\n Measure of the smallest angle is $50^\\circ$ less than that of the largest angle.<\/p>\n a = a + 2r –\u00a0$50^\\circ$<\/p>\n 2r =\u00a0$50^\\circ$<\/p>\n r =\u00a0$25^\\circ$<\/p>\n Sum of the angles of triangle =\u00a0$180^\\circ$<\/p>\n a + a + r + a + 2r =\u00a0$180^\\circ$<\/p>\n 3a + 3r =\u00a0$180^\\circ$<\/p>\n 3a +\u00a0$75^\\circ$ =\u00a0$180^\\circ$<\/p>\n 3a =\u00a0$105^\\circ$<\/p>\n a =\u00a0$35^\\circ$<\/p>\n Largest angle of triangle = a + 2r =\u00a0$35^\\circ$ +\u00a0$50^\\circ$ =\u00a0$85^\\circ$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 11:\u00a0<\/b>Triangle ABC is an equilateral triangle. D and E are points on AB and AC respectively such that DE is parallel to BC and is equal to half the length of BC. If AD +\u00a0CE + BC = 30 cm, then find the perimeter (in cm) of the quadrilateral BCED.<\/p>\n a)\u00a037.5<\/p>\n b)\u00a025<\/p>\n c)\u00a045<\/p>\n d)\u00a035<\/p>\n 11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b> Triangle ABC is an equilateral triangle.<\/p>\n Let the length of BC = 2p<\/p>\n BC = AB = AC = 2p<\/p>\n DE is equal to half the length of BC.<\/p>\n Triangle ABC and triangle ADE are similar triangles.<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\frac{AD}{AB}=\\frac{DE}{BC}$<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\frac{AD}{AB}=\\frac{p}{2p}$<\/p>\n $\\Rightarrow$\u00a0 $AD=\\frac{1}{2}AB$<\/p>\n $\\Rightarrow$\u00a0 $AD=\\frac{1}{2}\\times2p$<\/p>\n $\\Rightarrow$\u00a0 AD = p<\/p>\n Similarly, AE = p<\/p>\n and EC = AC – AE = 2p – p = p<\/p>\n AD + CE + BC = 30 cm<\/p>\n p + p + 2p = 30<\/p>\n 4p = 30<\/p>\n p =\u00a0$\\frac{15}{2}$ cm<\/p>\n Perimeter of\u00a0the quadrilateral BCED = BD + DE + CE + BC<\/p>\n = p + p + p + 2p<\/p>\n = 5p<\/p>\n =\u00a0$5\\times\\frac{15}{2}$<\/p>\n = 37.5 cm<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 12:\u00a0<\/b>The side BC of a triangle ABC is extended to the point D. If $\\angle$ACD = $132^\\circ$ and $\\angle$B = $\\frac{4}{7} \\angle$A, then the measure of $\\angle$A is<\/p>\n a)\u00a0$60^\\circ$<\/p>\n b)\u00a0$50^\\circ$<\/p>\n c)\u00a0$84^\\circ$<\/p>\n d)\u00a0$80^\\circ$<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b> Given,\u00a0$\\angle$ACD = $132^\\circ$\u00a0 and \u00a0$\\angle$B = $\\frac{4}{7} \\angle$A<\/p>\n In\u00a0$\\triangle$ABC,<\/p>\n $\\angle$ACD is the external angle at C which is equal to the sum of opposite angles A and B.<\/p>\n $\\Rightarrow$ \u00a0$\\angle$ACD = $\\angle$A +\u00a0$\\angle$B<\/p>\n $\\Rightarrow$ \u00a0$132^\\circ$ =\u00a0$\\angle$A +\u00a0$\\frac{4}{7} \\angle$A<\/p>\n $\\Rightarrow$ \u00a0$132^\\circ$ =\u00a0$\\frac{11}{7} \\angle$A<\/p>\n $\\Rightarrow$ \u00a0$\\angle$A =\u00a0$84^\\circ$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 13:\u00a0<\/b>In a $\\triangle$ABC, $\\angle$BAC = $90^\\circ$ and AD is perpendicular to BC where D is a point on BC. If BD = 4 cm and CD = 5 cm,then the length of AD is equal to:<\/p>\n a)\u00a0$5 \\sqrt 2$ cm<\/p>\n b)\u00a0$2 \\sqrt 5$ cm<\/p>\n c)\u00a06 cm<\/p>\n d)\u00a04.5 cm<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b> In\u00a0$\\triangle$ABC,<\/p>\n $\\angle$A = 90$^\\circ$<\/p>\n $\\Rightarrow$ \u00a0$\\angle$B =\u00a090$^\\circ$ –\u00a0$\\angle$C<\/p>\n AD is perpendicular to BC<\/p>\n $\\Rightarrow$ \u00a0$\\angle$ADC =\u00a090$^\\circ$<\/p>\n In $\\triangle$ADC,<\/p>\n $\\angle$ADC = 90$^\\circ$<\/p>\n $\\Rightarrow$ \u00a0$\\angle$DAC =\u00a090$^\\circ$ – $\\angle$C<\/p>\n In\u00a0$\\triangle$ADB and\u00a0$\\triangle$CDA,<\/p>\n $\\angle$B = $\\angle$DAC =\u00a090$^\\circ$ – $\\angle$C<\/p>\n $\\angle$BDA = $\\angle$ADC = 90$^\\circ$<\/p>\n Two angles are equal for both the triangles, $\\triangle$ADB is similar to $\\triangle$CDA<\/p>\n Ratio of respective sides are equal in both the triangles<\/p>\n $\\Rightarrow$ \u00a0$\\frac{BD}{AD}=\\frac{AD}{CD}$<\/p>\n $\\Rightarrow$\u00a0 AD$^2$ = BD.CD<\/p>\n $\\Rightarrow$\u00a0\u00a0AD$^2$ = 4 x 5<\/p>\n $\\Rightarrow$ \u00a0AD$^2$ =\u00a0 20<\/p>\n $\\Rightarrow$\u00a0 AD =\u00a0$2 \\sqrt 5$ cm<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 14:\u00a0<\/b>ABC is a right angled triangle, right angled at A. A circle is inscribed in it. The lengths of two sides containing the right angle are 48 cm and 14 cm. The radius of the inscribed circle is:<\/p>\n a)\u00a04 cm<\/p>\n b)\u00a08 cm<\/p>\n c)\u00a06 cm<\/p>\n d)\u00a05 cm<\/p>\n 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b> Using the pythagoras theorem,<\/p>\n BC$^2$ = AB$^2$ + AC$^2$<\/p>\n $\\Rightarrow$ \u00a0BC$^2$ = 48$^2$ + 14$^2$<\/p>\n $\\Rightarrow$ \u00a0BC$^2$ = 2304 + 196<\/p>\n $\\Rightarrow$ \u00a0BC$^2$ = 2500<\/p>\n $\\Rightarrow$\u00a0 BC = 50 cm<\/p>\n Let the radius of the circle = r<\/p>\n $\\Rightarrow$\u00a0 OD = OI = OJ = r<\/p>\n AB, BC, AC are tangents to the circle<\/p>\n Area of $\\triangle$ABC = Area of $\\triangle$OAC + Area of $\\triangle$OBC + Area of $\\triangle$OAB<\/p>\n $\\Rightarrow$ \u00a0$\\frac{1}{2}$ x 48 x 14 =\u00a0$\\frac{1}{2}$ x AC x OD +\u00a0$\\frac{1}{2}$ x BC x OI +\u00a0$\\frac{1}{2}$ x AB x OJ<\/p>\n $\\Rightarrow$ \u00a0336 =\u00a0$\\frac{1}{2}$ x 14 x r +\u00a0$\\frac{1}{2}$ x 50 x r +\u00a0$\\frac{1}{2}$ x 48 x r<\/p>\n $\\Rightarrow$\u00a0 336 = 7r + 25r + 24r<\/p>\n $\\Rightarrow$\u00a0 56r = 336<\/p>\n $\\Rightarrow$\u00a0 r = 6 cm<\/p>\n $\\therefore\\ $Radius of the inscribed circle = 6 cm<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 15:\u00a0<\/b>In an isosceles triangle ABC with AB = AC and AD is perpendicular to BC, if AD = 6 cm and the perimeter of $\\triangle$ABC is 36 cm, then the area of $\\triangle$ABC is:<\/p>\n a)\u00a054 $cm^2$<\/p>\n b)\u00a064 $cm^2$<\/p>\n c)\u00a045 $cm^2$<\/p>\n d)\u00a048 $cm^2$<\/p>\n 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b> Given, AB = AC<\/p>\n Let AB = AC = p<\/p>\n Perimeter of $\\triangle$ABC = 36 cm<\/p>\n $\\Rightarrow$\u00a0 AB + AC + BC = 36<\/p>\n $\\Rightarrow$\u00a0 p + p + BC = 36<\/p>\n $\\Rightarrow$\u00a0 BC = 36 – 2p<\/p>\n Since AB = AC and AD is perpendicular to BC<\/p>\n AD will be the perpendicular bisector which bisects BC<\/p>\n $\\Rightarrow$\u00a0 BD = CD = $\\frac{36-2p}{2}$ = 18 – p<\/p>\n In\u00a0$\\triangle$ADB,<\/p>\n AB$^2$ = BD$^2$ + AD$^2$<\/p>\n $\\Rightarrow$\u00a0 p$^2$ = (18 – p)$^2$ + 6$^2$<\/p>\n $\\Rightarrow$ \u00a0p$^2$ = 324 +\u00a0p$^2$ – 36p + 36<\/p>\n $\\Rightarrow$\u00a0 36p = 360<\/p>\n $\\Rightarrow$\u00a0 p = 10<\/p>\n BC = 36 – 2p = 36 – 20 = 16<\/p>\n $\\therefore\\ $Area of the triangle =\u00a0$\\frac{1}{2}$ x AD x BC =\u00a0$\\frac{1}{2}$ x 6 x 16 =\u00a048 $cm^2$<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 16:\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 16)\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 17:\u00a0<\/b>The side MN of $\\triangle$LMN is produced to X. If $\\angle$LNX = $117^\\circ$ and $\\angle$M = $\\frac{1}{2} \\angle$L, then $\\angle$L is:<\/p>\n a)\u00a0$78^\\circ$<\/p>\n b)\u00a0$76^\\circ$<\/p>\n c)\u00a0$77^\\circ$<\/p>\n d)\u00a0$75^\\circ$<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b> Given,\u00a0$\\angle$LNX = $117^\\circ$ and\u00a0$\\angle$M = $\\frac{1}{2} \\angle$L<\/p>\n In\u00a0$\\triangle$LMN,<\/p>\n The exterior angle LNX is equal to the sum of the opposite interior angles L and M.<\/p>\n $\\Rightarrow$ \u00a0$\\angle$LNX =\u00a0$\\angle$L +\u00a0$\\angle$M<\/p>\n $\\Rightarrow$ \u00a0$117^\\circ$ =\u00a0$\\angle$L +\u00a0$\\frac{1}{2} \\angle$L<\/p>\n $\\Rightarrow$ \u00a0$117^\\circ$ =\u00a0$\\frac{3}{2} \\angle$L<\/p>\n $\\Rightarrow$ \u00a0$\\angle$L = \u00a0$78^\\circ$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 18:\u00a0<\/b>The centroid of an equilateral $\\triangle$XYZ is L. If XY = 12 cm, then the length of XL (in cm), is:<\/p>\n a)\u00a0$5 \\sqrt 3$<\/p>\n b)\u00a0$2 \\sqrt 3$<\/p>\n c)\u00a0$4 \\sqrt 3$<\/p>\n d)\u00a0$3 \\sqrt 3$<\/p>\n 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b> In the equilateral traingle XYZ, \u00a0XY = 12 cm<\/p>\n $\\Rightarrow$\u00a0 YZ = XZ = 12 cm<\/p>\n The median XP bisects YZ at P and also perpendicular to YZ in the equilateral triangle.<\/p>\n $\\Rightarrow$\u00a0 YP = PZ = 6 cm \u00a0 and\u00a0 XP$\\bot\\ $YZ<\/p>\n The centroid L divides the median XP in the ratio of 2 : 1<\/p>\n $\\Rightarrow$ XL : PL = 2 : 1<\/p>\n $\\Rightarrow$ XL = 2PL ………….(1)<\/p>\n In\u00a0$\\triangle$XPZ,<\/p>\n XP$^2$ + PZ$^2$ = XZ$^2$<\/p>\n $\\Rightarrow$ XP$^2$ + 6$^2$ = 12$^2$<\/p>\n $\\Rightarrow$\u00a0XP$^2$ + 36 = 144<\/p>\n $\\Rightarrow$\u00a0 XP$^2$ = 108<\/p>\n $\\Rightarrow$\u00a0 XP =\u00a0$6\\sqrt{3}$<\/p>\n $\\Rightarrow$\u00a0 XL + PL =\u00a0$6\\sqrt{3}$<\/p>\n $\\Rightarrow$\u00a0 2PL + PL = $6\\sqrt{3}$<\/p>\n $\\Rightarrow$\u00a0 3PL =\u00a0$6\\sqrt{3}$<\/p>\n $\\Rightarrow$\u00a0 PL =\u00a0$2\\sqrt{3}$<\/p>\n $\\Rightarrow$\u00a0 XL = 2PL =\u00a0$4\\sqrt{3}$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 19:\u00a0<\/b>The measure of one of the exterior angles of a triangle is twice one of the interior opposite angles and the measure of the other interior opposite angles is 60$^\\circ$. The triangle is a\/an:<\/p>\n a)\u00a0equilateral triangle<\/p>\n b)\u00a0right triangle<\/p>\n c)\u00a0scalene triangle<\/p>\n d)\u00a0isosceles triangle<\/p>\n 19)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, two interior angles of the triangle is\u00a060$^\\circ$.<\/p>\n $\\Rightarrow$ The third interior angle must be\u00a060$^\\circ$ as the sum of the interior angles is 180$^\\circ$.<\/p>\n $\\therefore$The triangle is an equilateral triangle.<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 20:\u00a0<\/b>In $\\triangle$ABC, AB = AC, and $\\angle$BAC is 50$^\\circ$. Then $\\angle$ABC and $\\angle$BCA are, respectively:<\/p>\n a)\u00a0$70^\\circ$ and $75^\\circ$<\/p>\n b)\u00a0$65^\\circ$ and $65^\\circ$<\/p>\n c)\u00a0$50^\\circ$ and $55^\\circ$<\/p>\n d)\u00a0$55^\\circ$ and $55^\\circ$<\/p>\n 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n In $\\triangle$ABC, \u00a0AB = AC<\/p>\n Angles opposite to equal sides in triangle are equal.<\/p>\n $\\Rightarrow$\u00a0$\\angle$BCA =\u00a0$\\angle$ABC<\/p>\n Let\u00a0$\\angle$ABC = $\\angle$BCA = x<\/p>\n In $\\triangle$ABC,<\/p>\n $\\angle$ABC +\u00a0$\\angle$BCA +\u00a0$\\angle$BAC =\u00a0180$^\\circ$<\/p>\n $\\Rightarrow$\u00a0 x + x +\u00a050$^\\circ$ =\u00a0180$^\\circ$<\/p>\n $\\Rightarrow$\u00a0 2x =\u00a0130$^\\circ$<\/p>\n $\\Rightarrow$\u00a0 x =\u00a065$^\\circ$<\/p>\n $\\therefore\\ $\u00a0$\\angle$ABC = $\\angle$BCA =\u00a065$^\\circ$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 21:\u00a0<\/b>In $\\triangle$ABC, D is a point on BC. If $\\frac{AB}{AC} = \\frac{BD}{DC}$, $\\angle$B = $75^\\circ$ and $\\angle$C = $45^\\circ$ then $\\angle$BAD is equal to:<\/p>\n a)\u00a0$50^\\circ$<\/p>\n b)\u00a0$45^\\circ$<\/p>\n c)\u00a0$30^\\circ$<\/p>\n d)\u00a0$60^\\circ$<\/p>\n 21)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b> Given,\u00a0$\\angle$B = $75^\\circ$\u00a0 and \u00a0$\\angle$C = $45^\\circ$<\/p>\n In\u00a0$\\triangle$ABC,<\/p>\n $\\angle$A + $\\angle$B + $\\angle$C = $180^\\circ$<\/p>\n $\\Rightarrow$ \u00a0$\\angle$A + $75^\\circ$ + $45^\\circ$ = $180^\\circ$<\/p>\n $\\Rightarrow$ \u00a0$\\angle$A +\u00a0$120^\\circ$ =\u00a0$180^\\circ$<\/p>\n $\\Rightarrow$ \u00a0$\\angle$A =\u00a0$60^\\circ$<\/p>\n Given, $\\frac{AB}{AC} = \\frac{BD}{DC}$<\/p>\n AD divides the side BC in the ratio of other two sides so AD is the angular bisector $\\angle$A.<\/p>\n $\\Rightarrow$ \u00a0$\\angle$BAD =\u00a0$\\frac{1}{2}\\angle$A<\/p>\n $\\Rightarrow$ \u00a0$\\angle$BAD = $\\frac{60^{\\circ}}{2}$<\/p>\n $\\Rightarrow$ \u00a0$\\angle$BAD =\u00a0$30^\\circ$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 22:\u00a0<\/b>In $\\triangle$ABC, E and D are points on sides AB and AC, respectively, such that $\\angle$ABC = $\\angle$ADE. If AE = 6 cm, AD = 4 cm a)\u00a08 cm<\/p>\n b)\u00a09.5 cm<\/p>\n c)\u00a010 cm<\/p>\n d)\u00a011 cm<\/p>\n 22)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b> In\u00a0$\\triangle$ABC and\u00a0$\\triangle$ADE,<\/p>\n $\\angle$ABC = $\\angle$ADE<\/p>\n $\\angle$BAC = $\\angle$DAE<\/p>\n So $\\triangle$ABC is similar to $\\triangle$ADE<\/p>\n $\\Rightarrow$ \u00a0$\\frac{AB}{AC}=\\frac{AD}{AE}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{AE+EB}{AD+DC}=\\frac{AD}{AE}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{6+4}{4+DC}=\\frac{4}{6}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{10}{4+DC}=\\frac{2}{3}$<\/p>\n $\\Rightarrow$\u00a0 4 + DC = 15<\/p>\n $\\Rightarrow$\u00a0 DC = 11 cm<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 23:\u00a0<\/b>Triangle ABC is right angled at B and D is a point of BC such that BD = 5 cm, AD = 13 cm and AC = 37 cm. then find the length of DC in cm.<\/p>\n a)\u00a025<\/p>\n b)\u00a030<\/p>\n c)\u00a05<\/p>\n d)\u00a035<\/p>\n 23)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b> From right angled triangle ABD,<\/p>\n AD$^2$ = AB$^2$ + BD$^2$<\/p>\n 13$^2$ = AB$^2$ + 5$^2$<\/p>\n 169 = AB$^2$ + 25<\/p>\n AB$^2$ = 144<\/p>\n AB = 12 cm<\/p>\n From right angled triangle ABC,<\/p>\n AC$^2$ = AB$^2$ + BC$^2$<\/p>\n 37$^2$ = 12$^2$ + BC$^2$<\/p>\n 1369 = 144 + BC$^2$<\/p>\n BC$^2$ = 1225<\/p>\n BC = 35<\/p>\n BD + DC = 35<\/p>\n 5 + DC = 35<\/p>\n DC = 30 cm<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 24:\u00a0<\/b>In $\\triangle$ABC, $\\angle$A = 50$^\\circ$. If the bisectors of the angle B and angle C, meet at a point O, then $\\angle$BOC is equal to:<\/p>\n a)\u00a0$130^\\circ$<\/p>\n b)\u00a0$65^\\circ$<\/p>\n c)\u00a0$50^\\circ$<\/p>\n d)\u00a0$115^\\circ$<\/p>\n 24)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b> BO is the bisector of angle B.<\/p>\n Let $\\angle$OBC =\u00a0$\\angle$OBA = y<\/p>\n CO is the bisector of angle C.<\/p>\n Let\u00a0$\\angle$OCA = $\\angle$OCB = x<\/p>\n From\u00a0$\\triangle$ABC,<\/p>\n $\\angle$A +\u00a0$\\angle$B +\u00a0$\\angle$C =\u00a0180$^\\circ$<\/p>\n 50$^\\circ$ + 2y + 2x =\u00a0180$^\\circ$<\/p>\n 2(x+y) =\u00a0130$^\\circ$<\/p>\n x + y =\u00a065$^\\circ$……(1)<\/p>\n From $\\triangle$OBC,<\/p>\n $\\angle$OBC + $\\angle$OCB + $\\angle$BOC = 180$^\\circ$<\/p>\n y + x +\u00a0$\\angle$BOC =\u00a0180$^\\circ$<\/p>\n 65$^\\circ$ + $\\angle$BOC =\u00a0180$^\\circ$ [From (1)]<\/p>\n $\\angle$BOC\u00a0 =\u00a0115$^\\circ$<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 25:\u00a0<\/b>The perimeter of an isosceles triangle is 125 cm. If the base is 33 cm, find the length of the equal sides.<\/p>\n a)\u00a046 cm<\/p>\n b)\u00a034 cm<\/p>\n c)\u00a032 cm<\/p>\n d)\u00a042 cm<\/p>\n 25)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b> Let length of the equal sides = a<\/p>\n Length of the base = 33 cm<\/p>\n Perimeter of isosceles triangle = 125 cm<\/p>\n $\\Rightarrow$\u00a0 a + a + 33 = 125<\/p>\n $\\Rightarrow$\u00a0 2a = 92<\/p>\n $\\Rightarrow$\u00a0 a = 46 cm<\/p>\n $\\therefore\\ $Length of equal sides = 46 cm<\/p>\n Hence, the correct answer is Option A<\/p>\n
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\nthen the area of triangle BDE is equal to:<\/p>\n
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\nand EB = 4 cm, then the length of DC is:<\/p>\n
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