Enroll to TISSNET 2023 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>ABCD is a trapezoid where BC is parallel to AD and perpendicular to AB. Kindly note that BC< AD. Pis a point on AD such that CPD is an equilateral triangle. Q is a point on BC such that AQ is\u00a0parallel to PC. If the area of the triangle CPD is $4\\sqrt{\\ 3}$, find the area of the triangle ABQ.<\/p>\n a)\u00a0$2\\sqrt{3}$<\/p>\n b)\u00a0$4\\sqrt{3}$<\/p>\n c)\u00a04<\/p>\n d)\u00a0$8\\sqrt{3}$<\/p>\n e)\u00a0None of the above<\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given that,\u00a0CPD is an equilateral triangle<\/p>\n $\\angle\\ CPD=\\angle\\ PDC=\\angle\\ DCP=60^{\\circ\\ }$<\/p>\n As AQ is parallel to PC\u00a0$\\angle\\ CPD=\\angle\\ QAP=60^{\\circ\\ }$<\/p>\n As BC is parallel to AD\u00a0$\\angle\\ QAP=\\angle\\ AQB=60^{\\circ\\ }$ ( alternate interior angles).<\/p>\n Given,\u00a0Area of equilateral triangle CPD is $4\\sqrt{\\ 3}$<\/p>\n This implies,\u00a0$\\ \\frac{\\ \\sqrt{3}a^2}{4}=4\\sqrt{3}$<\/p>\n Solving, we get\u00a0$a=\\ 4$.<\/p>\n From figure, length of $AB$ = Height of\u00a0$\\triangle\\ CPD=\\frac{\\sqrt{\\ 3}a}{2}=2\\sqrt{\\ 3}$.<\/p>\n From $\\triangle\\ ABQ,\\ BQ=\\ \\frac{\\ AB}{\\tan\\ 60^{\\circ\\ }}=\\frac{2\\sqrt{\\ 3}}{\\sqrt{\\ 3}}=2$<\/p>\n Area of\u00a0$\\triangle\\ ABQ=\\frac{1}{2}\\times\\ AB\\times\\ BQ=\\ \\ \\frac{\\ 2\\times\\ 2\\sqrt{\\ 3}}{2}=2\\sqrt{\\ 3}\\ $<\/p>\n Option A is correct.<\/p>\n Question 2:\u00a0<\/b>A non-flying ant wants to travel from the bottom corner to the diagonally opposite top corner of a cubical room. The side of the room is 2 meters. What will be the minimum distance that the ant needs to travel?<\/p>\n a)\u00a06 meters<\/p>\n b)\u00a0$(2\\surd{2} + 2)$ meters<\/p>\n c)\u00a0$2\\surd{3}$ meters<\/p>\n d)\u00a0$2\\surd{6}$ meters<\/p>\n e)\u00a0$2\\surd{5}$ meters<\/p>\n 2)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n The shortest route ant takes to travel from the bottom corner to the diagonally opposite top corner is\u00a0 shown below.<\/p>\n The route goes through lateral faces (vertical faces) of the cubical room.<\/p>\n The length of route, r =\u00a0$\\sqrt{\\ \\left(2+2\\right)^2+2^2}$ =\u00a0$\\sqrt{\\ 20}$ =\u00a0$2\\sqrt{5}$<\/p>\n Question 3:\u00a0<\/b>ABC is a triangle and the coordinates of A, B and C are (a, b-2c), (a, b+4c) and (-2a,3c) respectively where a, b and c are positive numbers. a)\u00a06abc<\/p>\n b)\u00a09abc<\/p>\n c)\u00a06bc<\/p>\n d)\u00a09ac<\/p>\n e)\u00a0None of the above<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b> The length of AB =\u00a0$\\left(b+4c\\right)-\\left(b-2c\\right)=6c$\u00a0 (X-coordinates of A&B are same).<\/p>\n The altitude of triangle ABC, CD\u00a0= a-(-2a)=3a.<\/p>\n Area of triangle ABC =\u00a0$\\ \\frac{\\ AB\\ \\times\\ CD}{2}$ =\u00a0$\\ \\frac{6c\\ \\times\\ 3a\\ }{2}=9ac$<\/p>\n Option (D) is correct.<\/p>\n Question 4:\u00a0<\/b>ABC is a triangle with BC=5. D is the foot of the perpendicular from A on BC. E is a point on CD such that BE=3. The value of $AB^2 – AE^2 + 6CD$ is:<\/p>\n a)\u00a05<\/p>\n b)\u00a010<\/p>\n c)\u00a014<\/p>\n d)\u00a018<\/p>\n e)\u00a021<\/p>\n 4)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given that,\u00a0$AD\\bot\\ BC$<\/p>\n $BE=3$<\/p>\n $BC=5$<\/p>\n Using Pythagoras’ theorem,<\/p>\n $AD^2+BD^2=AB^2$ …..(1)<\/p>\n $AD^2+DE^2=AE^2$……(2)<\/p>\n $AD^2+DC^2=AC^2$……(3)<\/p>\n (1) – (2) gives<\/p>\n $BD^2+DE^2=AB^2-AE^2$<\/p>\n $x^2-\\left(3-x\\right)^2=AB^2-AE^2$<\/p>\n $AB^2-AE^2=6x-9$<\/p>\n $AB^2-AE^2+6CD=6x-9+6\\left(5-x\\right)$\u00a0 \u00a0 \u00a0 \u00a0($CD=\\left(5-x\\right)$)<\/p>\n $AB^2-AE^2+6CD=6x-9+30-6x$<\/p>\n $AB^2-AE^2+6CD=21$<\/p>\n Question 5:\u00a0<\/b>Find the value of a)\u00a0$\\sin 15^{\\circ} + \\sin 75^{\\circ}$<\/p>\n b)\u00a0$\\frac{6}{5}$<\/p>\n c)\u00a01<\/p>\n d)\u00a0$\\sin 15^{\\circ} \\cos 15^{\\circ}$<\/p>\n e)\u00a0None of the above<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $Let\\ \\sin^215^{\\circ}\\ =a\\ and\\ \\sin^275^{\\circ}\\ =\\ b\\ $<\/p>\n Then the given equation becomes,<\/p>\n $\\ \\frac{\\ a^3+b^3+6ab}{a^2+b^2+5ab}$<\/p>\n =$\\ \\frac{\\ \\left(a+b\\right)^3-3ab\\left(a+b\\right)+6ab}{\\left(a+b\\right)^2-2ab+5ab}$\u00a0 \u00a0 \u00a0$\\ \\therefore\\ \\ a^3+b^3=\\left(a+b\\right)^3-3ab\\left(a+b\\right)$ ,\u00a0$\\ \\therefore\\ \\ a^2+b^2=\\left(a+b\\right)^2-2ab$<\/p>\n Using\u00a0$\\sin\\left(\\theta\\right)\\ =\\cos\\left(90-\\theta\\ \\ \\right)$<\/p>\n we get,\u00a0$\\sin^275^{\\circ\\ }=\\cos^215^{\\circ\\ }$<\/p>\n now, a+b = $\\sin^215^{\\circ\\ }+\\sin^275^{\\circ\\ }$ = $\\sin^215^{\\circ\\ }+\\cos^215^{\\circ\\ }$ = 1.<\/p>\n The equation becomes,\u00a0$=\\ \\frac{\\ 1-3ab+6ab}{1-2ab+5ab}\\ =\\ \\frac{\\ 1+3ab}{1+3ab}\\ =1\\ $<\/p>\n Question 6:\u00a0<\/b>In a triangle ABC, AB = AC = 8 cm. A circle drawn with BC as diameter passes through A. Another circle drawn with center at A passes through Band C. Then the area, in sq. cm, of the overlapping region between the two circles is<\/p>\n a)\u00a0$16\\pi$<\/p>\n b)\u00a0$16(\\pi – 1)$<\/p>\n c)\u00a0$32(\\pi – 1)$<\/p>\n d)\u00a0$32\\pi$<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> BC is the diameter of circle C2 so we can say that\u00a0$\\angle BAC=90^{\\circ\\ }$ as angle in the semi circle is\u00a0$90^{\\circ\\ }$<\/p>\n Therefore overlapping area =\u00a0$\\frac{1}{2}$(Area of circle C2) + Area of the minor sector made be BC in C1<\/p>\n AB= AC = 8 cm and as\u00a0$\\angle BAC=90^{\\circ\\ }$, so we can conclude that BC=\u00a0$8\\sqrt{2}$ cm<\/p>\n Radius of C2 = Half of length of\u00a0BC =\u00a0$4\\sqrt{2}$ cm<\/p>\n Area of C2 =\u00a0$\\pi\\left(4\\sqrt{2}\\right)^2=32\\pi$\u00a0$cm^2$<\/p>\n A is the centre of C1 and C1 passes through B, so AB is the radius of C1 and is equal to 8 cm<\/p>\n Area of the minor sector made be BC in C1 =\u00a0$\\frac{1}{4}$(Area of circle C1) – Area of triangle ABC =\u00a0$\\frac{1}{4}\\pi\\left(8\\right)^2-\\left(\\frac{1}{2}\\times8\\times8\\right)=16\\pi-32$\u00a0$cm^2$<\/p>\n Therefore,<\/p>\n Overlapping area between the two circles= $\\frac{1}{2}$(Area of circle C2) + Area of the minor sector made be BC in C1<\/p>\n =\u00a0$\\frac{1}{2}\\left(32\\pi\\right)\\ +\\left(16\\pi-32\\right)=32\\left(\\pi-1\\right)$\u00a0$cm^2$<\/p>\n Question 7:\u00a0<\/b>The lengths of all four sides of a quadrilateral are integer valued. If three of its sides are of length 1 cm, 2 cm and 4 cm, then the total number of possible lengths of the fourth side is<\/p>\n a)\u00a03<\/p>\n b)\u00a04<\/p>\n c)\u00a06<\/p>\n d)\u00a05<\/p>\n 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Sum of the three sides of a quadrilateral is greater than the fourth side.<\/p>\n Therefore, let the fourth side be<\/p>\n 1+2+4>d or d<7<\/p>\n 1+2+d>4 or d>1<\/p>\n Possible values of d are 2, 3, 4, 5 and 6.<\/p>\n Question 8:\u00a0<\/b>Suppose the medians BD and CE of a triangle ABC intersect at a point O. If area of triangle ABC is 108 sq. cm., then, the area of the triangle EOD, in sq. cm., is<\/p>\n 8)\u00a0Answer:\u00a09<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> Area of ABD : Area of BDC = 1:1<\/p>\n Therefore, area of ABD = 54<\/p>\n Area of ADE : Area of EDB = 1:1<\/p>\n Therefore, area of ADE = 27<\/p>\n O is the centroid and it divides the medians in the ratio of 2:1<\/p>\n Area of BEO : Area of EOD = 2:1<\/p>\n Area of EOD = 9<\/p>\n Question 9:\u00a0<\/b>The length of each side of an equilateral triangle ABC is 3 cm. Let D be a point on BC such that the area of triangle ADC is half the area of triangle ABD. Then the length of AD, in cm, is<\/p>\n a)\u00a0$\\sqrt{8}$<\/p>\n b)\u00a0$\\sqrt{6}$<\/p>\n c)\u00a0$\\sqrt{7}$<\/p>\n d)\u00a0$\\sqrt{5}$<\/p>\n 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> Area of triangle ABD is twice the area of triangle ACD<\/p>\n $\\angle\\ ADB=\\theta\\ $<\/p>\n $\\frac{1}{2}\\left(AD\\right)\\left(BD\\right)\\sin\\theta\\ =2\\left(\\frac{1}{2}\\left(AD\\left(CD\\right)\\sin\\left(180-\\theta\\ \\right)\\right)\\right)$<\/p>\n $BD\\ =2CD$<\/p>\n Therefore, BD = 2 and CD = 1<\/p>\n $\\angle\\ ABC=\\angle\\ ACB=60^{\\circ\\ }$<\/p>\n Applying cosine rule in triangle ADC, we get<\/p>\n $\\cos\\angle\\ ACD=\\ \\frac{\\ AC^2+CD^2-AD^2}{2\\left(AC\\right)\\left(CD\\right)}$<\/p>\n $\\frac{1}{2}=\\ \\frac{\\ 9+1-AD^2}{6}$<\/p>\n $AD^2=7$<\/p>\n $AD=\\sqrt{\\ 7}$<\/p>\n The answer is option C.<\/p>\n Question 10:\u00a0<\/b>Regular polygons A and B have number of sides in the ratio 1 : 2 and interior angles in the ratio 3 : 4. Then the number of sides of B equals<\/p>\n 10)\u00a0Answer:\u00a010<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the number of sides of polygons A and B be n and 2n, respectively.<\/p>\n $\\ \\dfrac{\\ \\ \\dfrac{\\ \\left(n-2\\right)180}{n}}{\\ \\dfrac{\\ \\left(2n-2\\right)180}{2n}}=\\dfrac{3}{4}$<\/p>\n $\\ \\dfrac{\\ \\ \\ n-2}{\\ \\ n-1}=\\dfrac{3}{4}$<\/p>\n $\\ \\ \\ \\ 4n-8=3n-3$<\/p>\n $n=5$<\/p>\n The number of sides of polygon B is 2*5, i.e. 10.<\/p>\n Question 11:\u00a0<\/b>In triangle ABC, altitudes AD and BE are drawn to the corresponding bases. If $\\angle BAC = 45^{\\circ}$ and $\\angle ABC=\\theta\\ $, then $\\frac{AD}{BE}$ equals<\/p>\n a)\u00a0$\\sqrt{2} \\cos \\theta$<\/p>\n b)\u00a0$\\frac{(\\sin \\theta + \\cos \\theta)}{\\sqrt{2}}$<\/p>\n c)\u00a01<\/p>\n d)\u00a0$\\sqrt{2} \\sin \\theta$<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> It is given, Angle BAE = 45 degrees<\/p>\n This implies AE = BE<\/p>\n Let AE = BE = x<\/p>\n In right-angled triangle ABD, it is given $\\angle ABC=\\theta\\ $<\/p>\n $\\sin\\theta=\\frac{AD}{AB}\\ $<\/p>\n $\\sin\\theta=\\frac{AD}{x\\sqrt{\\ 2}}\\ $<\/p>\n $\\sqrt{\\ 2}\\sin\\theta=\\frac{AD}{BE}\\ $<\/p>\n The answer is option D.<\/p>\n Question 12:\u00a0<\/b>Let ABCD be a parallelogram such that the coordinates of its three vertices A, B, C are (1, 1), (3, 4) and (\u22122, 8), respectively. Then, the coordinates of the vertex D are<\/p>\n a)\u00a0(0, 11)<\/p>\n b)\u00a0(4, 5)<\/p>\n c)\u00a0(\u22123, 4)<\/p>\n d)\u00a0(\u22124, 5)<\/p>\n 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n In a parallelogram, two diagonals of parallelogram bisects each other, which concludes that mid-point of both diagonals are the same.<\/p>\n Midpoint of AC =\u00a0$\\left(\\ \\frac{\\ 1-2}{2},\\ \\frac{\\ 1+8}{2}\\right)$<\/p>\n Let the coordinates of vertex D be (x,y)<\/p>\n $\\left(\\ \\frac{\\ x+3}{2},\\ \\frac{\\ y+4}{2}\\right)=\\left(\\ \\frac{\\ 1-2}{2},\\ \\frac{\\ 1+8}{2}\\right)$<\/p>\n x = -4 and y = 5<\/p>\n The answer is option D.<\/p>\n Question 13:\u00a0<\/b>All the vertices of a rectangle lie on a circle of radius R. If the perimeter of the rectangle is P, then the area of the rectangle is<\/p>\n a)\u00a0$\\frac{P^2}{16} – R^2$<\/p>\n b)\u00a0$\\frac{P^2}{8} – 2R^2$<\/p>\n c)\u00a0$\\frac{P^2}{2} – 2PR$<\/p>\n d)\u00a0$\\frac{P^2}{8} – \\frac{R^2}{2}$<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> $A=lb$<\/p>\n $l^2+b^2=4r^2$<\/p>\n $P\\ =2\\left(l+b\\right)$<\/p>\n $\\frac{P}{2}=l+b$<\/p>\n Squaring on both the sides, we get<\/p>\n $\\frac{P^2}{4}=l^2+b^2+2lb$<\/p>\n $\\frac{P^2}{4}=4r^2+2lb$<\/p>\n $\\frac{P^2}{8}-2r^2=lb$<\/p>\n The answer is option B.<\/p>\n Question 14:\u00a0<\/b>A trapezium $ABCD$ has side $AD$ parallel to $BC, \\angle BAD = 90^\\circ, BC = 3$ cm and $AD= 8$ cm. If the perimeter of this trapezium is 36 cm, then its area, in sq. cm, is<\/p>\n 14)\u00a0Answer:\u00a066<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b> CD =\u00a0$\\sqrt{\\ y^2+25}$<\/p>\n $11+y+\\sqrt{y^2+25}=36$<\/p>\n $\\sqrt{y^2+25}=25-y$<\/p>\n $y^2+25=25^2+y^2-50y$<\/p>\n 2y = 24<\/p>\n y = 12<\/p>\n Area of trapezium =\u00a0$3y+\\frac{5y}{2}=\\frac{11y}{2}=\\frac{11}{2}\\left(12\\right)=66$<\/p>\n Question 15:\u00a0<\/b>If a 30 meter ladder is placed against a wall such that it just reaches the top of the wall, if the horizontal distance between the wall and the base of the ladder is 1\/3rd of the length of ladder, then the height of wall is :<\/p>\n a)\u00a025 meter<\/p>\n b)\u00a0$20\\sqrt{2}$ meter<\/p>\n c)\u00a0$20\\sqrt{3}$ meter<\/p>\n d)\u00a020 meter<\/p>\n 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b> AC is ladder, C is base of ladder, AB is wall<\/p>\n Given AC = 30 m and BC =\u00a0$\\frac{1}{3}AC$ =\u00a0$\\frac{1}{3}\\left(30\\right)$ = 10 m<\/p>\n Applying pythagoras theorem,<\/p>\n $AB^2+\\ BC^2=\\ AC^2$<\/p>\n $AB^2$ =\u00a0$30^2-10^2$ = 800<\/p>\n AB =\u00a0$20\\sqrt{\\ 2}$<\/p>\n Answer is option B.<\/p>\n Question 16:\u00a0<\/b>A copper wire having length of 243m and diameter 4 mm was melted to form a sphere. Find the diameter of the sphere :<\/p>\n a)\u00a017 m<\/p>\n b)\u00a018 cm<\/p>\n c)\u00a015 cm<\/p>\n d)\u00a020 cm<\/p>\n 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Volume of copper wire melted will be equal to volume of sphere.<\/p>\n Volume of copper wire = Area\u00a0$\\times\\ $ length =\u00a0$\\pi\\ \\left(\\frac{2}{10}\\right)^2\\times\\ 24300\\ cm^3$<\/p>\n Let the radius of sphere be r<\/p>\n $\\frac{4}{3}\\pi\\ r^{3\\ \\ }=\\ \\pi\\ \\left(\\frac{2}{10}\\right)^2\\times\\ 24300\\ cm^3$<\/p>\n $r^3$ = 729<\/p>\n r = 9 cm<\/p>\n Diameter of sphere = 2(9) = 18 cm<\/p>\n Answer is option B.<\/p>\n Question 17:\u00a0<\/b>If a cuboidal box has height, length and width as 20 cm, 15 cm and 10 cm respectively.\u00a0Then its total surface area is__________.<\/p>\n a)\u00a01300$cm^{2}$<\/p>\n b)\u00a01400$cm^{2}$<\/p>\n c)\u00a01200$cm^{2}$<\/p>\n d)\u00a01100$cm^{2}$<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Total surface area = 2(lb\u00a0+ bh + hl) = 2(150 + 200 + 300) = 1300\u00a0$cm^2$<\/p>\n Answer is option A.<\/p>\n Question 18:\u00a0<\/b>Angle bisectors of a parallelogram form a _____.<\/p>\n a)\u00a0parallelogram<\/p>\n b)\u00a0square<\/p>\n c)\u00a0rhombus<\/p>\n d)\u00a0rectangle<\/p>\n 18)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b> Angular bisectors from A and B intersect at P, B and D intersect at Q, C and D intersect at R, and A and C intersect at S. Answer is option D.<\/p>\n Question 19:\u00a0<\/b>A tree bends due to heavy storm and now its peak touches the ground making an\u00a0angle of $30^{\\circ}$ with it. If the bent part is 20 metre long, find the original height of the tree in metres.<\/p>\n a)\u00a010<\/p>\n b)\u00a020\/\u221a3<\/p>\n c)\u00a040<\/p>\n d)\u00a030<\/p>\n 19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b> BC is ground and it is given that\u00a0$\\angle\\ ACB\\ =30^{\\circ\\ }$ Answer is option D.<\/p>\n Question 20:\u00a0<\/b>$ABC$ is a triangle, $\\angle B=60^\\circ, \\angle C=45^\\circ$, BC is produced to \/ extended till D so that $\\angle ADB=30^\\circ$, then given $\\sqrt{3}=\\frac{19}{11}$ and $BC \\times CD=3^{4}\\times 2^{3}\\times\\frac{11}{19}$, what will be the square of the altitude from A to BC?<\/p>\n a)\u00a0144<\/p>\n b)\u00a0324<\/p>\n c)\u00a0484<\/p>\n d)\u00a01254<\/p>\n 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b> AE is altitude from A on BC.\u00a0$\\angle\\ AEC\\ =\\ 90^{\\circ\\ }$<\/p>\n It is given,<\/p>\n $\\angle\\ ABC\\ =\\ 60^{\\circ\\ },\\ \\angle\\ ACB\\ =\\ 45^{\\circ\\ }$<\/p>\n In triangle ABE,<\/p>\n AE =\u00a0$\\sqrt{\\ 3}x$<\/p>\n As $\\angle\\ EAC\\ =\\angle\\ ECA$, EC =\u00a0$\\sqrt{\\ 3}x$<\/p>\n In triangle AED,\u00a0$\\angle\\ EAD\\ =\\ 60^{\\circ\\ }$<\/p>\n Therefore, CD =\u00a0$3x-\\sqrt{\\ 3}x$<\/p>\n It is given,<\/p>\n $BC\\ \\cdot\\ CD\\ =\\ \\frac{3^4.2^3.11}{19}$<\/p>\n $x\\left(\\sqrt{\\ 3}+1\\right)\\sqrt{\\ 3}\\left(\\sqrt{\\ 3}-1\\right)x\\ =\\ \\frac{3^4.2^3.11}{19}$<\/p>\n $x\\ =\\ \\frac{3^2.2.11}{19}$<\/p>\n AE = 9*2 = 18<\/p>\n $AE^2$ = 324<\/p>\n Answer is option B.<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/image_p2dHrjq.png\")
<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/image_wL6BMpl.png\")
<\/p>\n
\nThe area of the triangle ABC is:<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/image_jYtSOuE.png\")
<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/image_qKF7w6x.png\")
<\/p>\n
\n$\\frac{\\sin^{6}15^{\\circ} + \\sin^{6}75^{\\circ} + 6\\sin^{2}15^{\\circ}\\sin^{2}75^{\\circ}}{\\sin^{4}15^{\\circ} + \\sin^{4}75^{\\circ} + 5\\sin^{2}15^{\\circ}\\sin^{2}75^{\\circ}}$<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/Screenshot_20221207_092834.png\")
<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/Screenshot_20221206_172342.png\")
<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/image_pEtGtAZ.png\")
<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/image_BeCpLbp.png\")
<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/image_ijAStIe.png\")
<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/image_xs8Ka8E.png\")
<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/image_i9fQr5T.png\")
<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/image_BfIYXwm.png\")
<\/p>\n
\nIn a parallelogram ABCD, AC is parallel to BD.
\n$\\angle\\ CAB\\ +\\angle\\ ACD\\ =\\ 180^{\\circ\\ }$
\n$\\angle\\ CAS\\ +\\angle\\ ACS\\ =\\ 90^{\\circ\\ }$
\nIn triangle CAS,
\n$\\angle\\ ASC\\ =\\ 180^{\\circ\\ }-90^{\\circ\\ }=90^{\\circ\\ }$
\nAs all the angles of PQRS are $90^{\\circ\\ }$,\u00a0PQRS is a rectangle.<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/image_zdZ9JtW.png\")
<\/p>\n
\nActual length of the tree = AB + AC
\n$\\sin30^{\\circ\\ }=\\frac{AB}{AC}$
\nAB = 20\/2 = 10 cm
\nLength of the tree = AB + AC = 10 + 20 = 30 cm<\/p>\n
\n![](\"https:\/\/cracku.in\/media\/uploads\/image_QJTEFKR.png\")
<\/p>\n