Download Geometry Triangle Questions for MAH-CET<\/a><\/p>\n Enroll to MAH-CET 2022 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>If the height of a triangle is decreased by 40% and it\u2019s base is increased by 40% , what will be the effect on its area?<\/p>\n a)\u00a0No change<\/p>\n b)\u00a016 % increase<\/p>\n c)\u00a08% decrease<\/p>\n d)\u00a016% decrease<\/p>\n e)\u00a0None of these<\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n area of triangle = 1\/2(base*height)<\/p>\n after decreasing height by 40%, new height = 0.6H<\/p>\n after increasing base by 40%, new base = 1.4B<\/p>\n now, area of triangle = 1\/2(1.4B*0.6H) = 0.84(1\/2*B*H)<\/p>\n Area decreased by 100-84%= 16%<\/p>\n so the answer is option D.<\/p>\n Question 2:\u00a0<\/b>If the height of a triangle is decreased by 40% and it\u2019s base is increased by 40% , what will be the effect on its area?<\/p>\n a)\u00a0No change<\/p>\n b)\u00a016 % increase<\/p>\n c)\u00a08% decrease<\/p>\n d)\u00a016% decrease<\/p>\n e)\u00a0None of these<\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let base and height of the triangle be 10 cm each<\/p>\n Area of original triangle, $\\triangle = \\frac{1}{2} \\times 10 \\times 10 = 50$ $cm^2$<\/p>\n If the height of a triangle is decreased by 40%, => New height = $10-\\frac{40}{100} \\times 10=6$ cm<\/p>\n and it\u2019s base is increased by 40%, => New base =\u00a0$10+\\frac{40}{100} \\times 10=14$ cm<\/p>\n Area of new triangle, $\\triangle’ = \\frac{1}{2} \\times 6 \\times 14 = 42$ $cm^2$<\/p>\n $\\therefore$ Area decreases<\/strong> by = $\\frac{\\triangle – \\triangle’}{\\triangle} \\times 100$<\/p>\n = $\\frac{50-42}{50} \\times 100 = 8 \\times 2=16\\%$<\/p>\n => Ans – (D)<\/p>\n Question 3:\u00a0<\/b>The area of a right angled triangle is 80 sq. cm. The ratio of the base and the height of the triangle is 4 : 5. Find the length of hypotenuse.<\/p>\n a)\u00a082cm<\/p>\n b)\u00a02 82cm<\/p>\n c)\u00a02 41cm<\/p>\n d)\u00a03 82cm<\/p>\n e)\u00a0None of these<\/p>\n 3)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let base and height be $4x$ and $5x$ respectively.<\/p>\n Area of triangle = $\\frac{1}{2} \\times$ base $\\times$ height = 80<\/p>\n => $\\frac{1}{2} \\times 4x \\times 5x = 80$<\/p>\n => $x^2 = \\frac{80}{10} = 8$<\/p>\n => $x = \\sqrt{8} = 2 \\sqrt{2}$<\/p>\n => Base = $8 \\sqrt{2}$ and Height = $10 \\sqrt{2}$<\/p>\n $\\therefore$ Hypotenuse = $\\sqrt{(8 \\sqrt{2})^2 + (10 \\sqrt{2})^2}$<\/p>\n = $\\sqrt{128 + 200} = \\sqrt{328}$<\/p>\n $\\approx 18$ cm<\/p>\n Question 4:\u00a0<\/b>The area of a triangle is half the area of square. The perimeter of the square is 224 cms. What is the area of the triangle?<\/p>\n a)\u00a0$1856 cm^{2}$<\/p>\n b)\u00a0$1658 cm^{2}$<\/p>\n c)\u00a0$1558 cm^{2}$<\/p>\n d)\u00a0$1586 cm^{2}$<\/p>\n e)\u00a0None of these<\/p>\n 4)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Perimeter of square = 224 cms<\/p>\n => Side of square = $\\frac{Perimeter}{4} = \\frac{224}{4} = 56$ cm<\/p>\n Area of square = $(56)^2 = 3136 cm^2$<\/p>\n $\\therefore$ Area of triangle = $\\frac{3136}{2} cm^2$<\/p>\n = $1568 cm^2$<\/p>\n Question 5:\u00a0<\/b>The height and base of a triangle are equal to the length and breadth of a rectangle respectively. If the perimeter of the rectangle is 86m and the difference between its length and breadth is 5m, what is the area of the triangle ? (in m^{2} )<\/p>\n a)\u00a0224<\/p>\n b)\u00a0228<\/p>\n c)\u00a0216<\/p>\n d)\u00a0242<\/p>\n e)\u00a0256<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let length of rectangle = $x$ m<\/p>\n Breadth = $(x – 5)$ m<\/p>\n => Perimeter of rectangle = $2 (x + x – 5) = 86$<\/p>\n => $2x – 5 = \\frac{86}{2} = 43$<\/p>\n => $2x = 43 + 5 = 48$<\/p>\n => $x = \\frac{48}{2} = 24$<\/p>\n => Breadth = 24 – 5 = 19 m<\/p>\n => Height of triangle = 24 m and Base of triangle = 19 m<\/p>\n $\\therefore$ Area of triangle = $\\frac{1}{2} \\times 24 \\times 19$<\/p>\n = $12 \\times 19 = 228 m^2$<\/p>\n Take Free MAH-CET mock tests here<\/a><\/strong><\/span><\/p>\n Enroll to 5 MAH CET Latest Mocks For Just Rs. 299<\/a><\/strong><\/span><\/p>\n Question 6:\u00a0<\/b>The height and base of a triangle are equal to the length and breadth of a rectangle respectively. If the perimeter of the rectangle is 90m and the difference between its length and breadth is 7m, what is the area of the triangle ? (in m )<\/p>\n a)\u00a0239<\/p>\n b)\u00a0253<\/p>\n c)\u00a0241<\/p>\n d)\u00a0257<\/p>\n e)\u00a0247<\/p>\n 6)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let length of rectangle = $x$ m<\/p>\n Breadth = $(x – 7)$ m<\/p>\n => Perimeter of rectangle = $2 (x + x – 7) = 90$<\/p>\n => $2x – 7 = \\frac{90}{2} = 45$<\/p>\n => $2x = 45 + 7 = 52$<\/p>\n => $x = \\frac{52}{2} = 26$<\/p>\n => Breadth = 26 – 7 = 19 m<\/p>\n => Height of triangle = 26 m and Base of triangle = 19 m<\/p>\n $\\therefore$ Area of triangle = $\\frac{1}{2} \\times 26 \\times 19$<\/p>\n = $13 \\times 19 = 247 m^2$<\/p>\n Question 7:\u00a0<\/b>The side of a square is equal to height of a triangle. If the area of the triangle is 294 m\u00b2 and the respective ratio of its height and base is 3 : 4, what is the perimeter of the square (in m) ?<\/p>\n a)\u00a0108<\/p>\n b)\u00a096<\/p>\n c)\u00a084<\/p>\n d)\u00a072<\/p>\n e)\u00a060<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let height of triangle = $3x$<\/p>\n => Base of triangle = $4x$<\/p>\n => Area of triangle = $\\frac{1}{2} \\times 3x \\times 4x = 294$<\/p>\n => $6x^2 = 294$<\/p>\n => $x^2 = \\frac{294}{6} = 49$<\/p>\n => $x = \\sqrt{49} = 7$<\/p>\n => Height of triangle = $3 \\times 7 = 21$ m<\/p>\n Also, side of square = height of triangle = 21 m<\/p>\n $\\therefore$ Perimeter of square = $4 \\times 21$<\/p>\n = $84$ m<\/p>\n Question 8:\u00a0<\/b>A triangle has two of its angles in the ratio of 1 : 2. If the measure of one of its angles is 30 degrees, what is the measure of the largest angle of the triangle in degrees ?<\/p>\n a)\u00a0100<\/p>\n b)\u00a090<\/p>\n c)\u00a0135<\/p>\n d)\u00a0Cannot be determined<\/p>\n e)\u00a0None of these<\/p>\n 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Case 1\u00a0: Let the first angle be 30\u00b0<\/p>\n So according to the ratio of 1\u00a0: 2, the other angle = 60\u00b0<\/p>\n => Third angle = $180^{\\circ} – (30^{\\circ} + 60^{\\circ}) = 180^{\\circ} – 90^{\\circ} = 90^{\\circ}$<\/p>\n Case 2\u00a0: Let the third angle be 30\u00b0<\/p>\n => First and second angles respectively are = $x , 2x$<\/p>\n => $x + 2x + 30^{\\circ} = 180^{\\circ}$<\/p>\n => $3x = 180^{\\circ} – 30^{\\circ} = 150^{\\circ}$<\/p>\n => $x = \\frac{150^{\\circ}}{3} = 50^{\\circ}$<\/p>\n Thus, largest angle = $2 \\times 50^{\\circ} = 100^{\\circ}$<\/p>\n $\\therefore$ Largest angle can be either 90 or 100\u00b0<\/p>\n => Ans – (D)<\/p>\n Instructions<\/b><\/p>\n In the given questions, two quantities are given, one as Quantity I and another as Quantity II. You have to determine relationship between two quantities and choose the appropriate option.<\/p>\n a: If quantity I \u2265 quantity II Question 9:\u00a0<\/b>A right-angled triangle is inscribed in a given circle. What is the area of the given circle (in cm2) ? a)\u00a0III and either omly I or only II<\/p>\n b)\u00a0ALL I,II and III<\/p>\n c)\u00a0Only II and III<\/p>\n d)\u00a0Only I<\/p>\n e)\u00a0Either I or II<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n I : $x^2 – 23x + 120 = 0$<\/p>\n => $x^2 – 8x – 15x + 120 = 0$<\/p>\n => $x (x – 8) – 15 (x – 8) = 0$<\/p>\n => $(x – 8) (x – 15) = 0$<\/p>\n => $x = 8 , 15$<\/p>\n Thus, base = 8 cm and height = 15 cm (or vice versa)<\/p>\n => Hypotenuse of right angled triangle = $\\sqrt{(8)^2 + (15)^2}$<\/p>\n = $\\sqrt{64 + 225} = \\sqrt{289} = 17 cm$<\/p>\n Since, triangle is inscribed in circle, => Radius of circle = half of hypotenuse<\/p>\n => $r = \\frac{17}{2} = 8.5$ cm<\/p>\n $\\therefore$ Area of circle = $\\pi r^2$<\/p>\n = $\\frac{22}{7} \\times 8.5 \\times 8.5 \\approx 227 cm^2$<\/p>\n Thus, I alone is sufficient.<\/strong><\/p>\n Clearly, we cannot find base and height from statements II or III. Thus, they are insufficient.<\/p>\n Question 10:\u00a0<\/b>The ratio of angles of a triangle is 3 : 5 : 7. What is the difterence between double the smallest angle and half of the largest angle ?<\/p>\n a)\u00a025<\/p>\n b)\u00a036<\/p>\n c)\u00a038<\/p>\n d)\u00a030<\/p>\n e)\u00a0None of these<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n We know that the sum of interior angles of a triangle is equal to $180^{o}$. Question 11:\u00a0<\/b>The angles of 1 triangle are in ratio ol 3 : 5 : 1 respectively. What is the difference between twice the smallest angle and the second largest arigle of the triangle<\/p>\n a)\u00a025\u00b0<\/p>\n b)\u00a010\u00b0<\/p>\n c)\u00a045\u00b0<\/p>\n d)\u00a030\u00b0<\/p>\n e)\u00a0None of these<\/p>\n 11)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n We know that the sum of interior angles of a triangle is equal to $180^{o}$. Question 12:\u00a0<\/b>The perimeter of a rectangle whose length is 6rn more than its breadth is 84 m. What would be the area of a triangle whose base is equal to the diagonal of the rectangle and whose height is equal to the length of the rectangle ? (in m )<\/p>\n a)\u00a0324<\/p>\n b)\u00a0372<\/p>\n c)\u00a0360<\/p>\n d)\u00a0364<\/p>\n e)\u00a0348<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Breadth = b meters Perimeter = 4b+12 = 84, i.e. b = 18 Triangle – base = 30, height = 24 Question 13:\u00a0<\/b>The perimeter of a rectangle whose length is 6 metre more than its breadth is 84 metre. What is the area of the triangle whose base is equal to the diagonal of the rectangle and height is equal to the length of the rectangle ?<\/p>\n a)\u00a0360 sq. metre<\/p>\n b)\u00a0380 sq. metre<\/p>\n c)\u00a0360 metre<\/p>\n d)\u00a0400 sq. metre<\/p>\n e)\u00a0None of these<\/p>\n 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the base of rectangle be y , then its length will be = y+6<\/p>\n perimeter of triangle = 84<\/p>\n 2(length + breadth) = 84<\/p>\n 2(2y + 6) = 84<\/p>\n 2y + 6 = 42<\/p>\n y = 18 m<\/p>\n length of rectangle = height of triangle = 18 + 6 = 24 m<\/p>\n Diagonal of rectangle = base of Triangle = $\\sqrt{(24)^2 + (18)^2}$ = 30<\/p>\n Area of Triangle = $\\frac{1}{2}\\times30\\times24$<\/p>\n Area of Triangle = 360 sq m<\/p>\n Question 14:\u00a0<\/b>In the following figure, ABC is an equilateral triangle and BCDE is a square whose each side is 8 cm long. Find the area of pentagon ABDEC in square cm. a)\u00a018(4 + \u221a3 )<\/p>\n b)\u00a016 (4 +\u221a3)<\/p>\n c)\u00a08 (4 + \u221a3 )<\/p>\n d)\u00a016 (2 + \u221a3 )<\/p>\n e)\u00a0None of these<\/p>\n 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n side of square = 8 cm<\/p>\n As the side of square is coinciding with side of equilateral triangle so side of equilateral teiangle = 8 cm<\/p>\n Area of equilateral triangle = (\u221a3\/4 ) side^2<\/p>\n Area of square = side ^2<\/p>\n Total area = area of square + area of traingle<\/p>\n = ((\u221a3\/4) x64) + 64<\/p>\n =16(\u221a3+4) sqr cm<\/p>\n Question 15:\u00a0<\/b>What will be the effect on the area of an isosceles triangle if all the sides are doubled in length ?<\/p>\n a)\u00a0Increase to four times the area of original triangle<\/p>\n b)\u00a0Increase to two times the area of original triangle<\/p>\n c)\u00a0Increased by 150%<\/p>\n d)\u00a0Increased by 200%<\/p>\n e)\u00a0None of these<\/p>\n 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n If the sides are all doubled, the semi preimeter gets doubled.<\/p>\n Area = $ \\sqrt {s(s-a)(s-b)(s-c)}$<\/p>\n New area =\u00a0$ \\sqrt {2s(2s-2a)(2s-2b)(2s-2c)}$<\/p>\n 4 times the original area.<\/p>\n
\nb: If quantity I > quantity II
\nc: If quantity I < quantity II
\nd: If quantity I = quantity II or the relationship cannot be established from the information that is given
\ne: If quantity quantity II<\/p>\n
\nI. The base and height of the triangle (in cm) are both the roots of the equation $x^{2}-23x+120=0$
\nII. The sum of the base and height of the triangle is 23 cm.
\nIII. The height of the right-angled triangle is greater than the base of the same.<\/p>\n
\nThe total sum of the ratios must be equal to $180^{o}$.
\nHence each angle of the triangle is $\\frac{3\\times180}{15}=36^{o}$ , $\\frac{5\\times180}{15}=60^{o}$ and $\\frac{7\\times180}{15}=84^{o}$
\nTwice the smallest angle $=2\\times36 = 72.$
\nHalf the largest angle $=\\frac{84}{2} = 42.$
\nDifference $=72-42 = 30.$
\nHence Option D is the correct answer.<\/span><\/span><\/p>\n
\nThe total sum of the ratios must be equal to $180^{o}$.
\nHence each angle of the triangle is $\\frac{3\\times180}{9}=60^{o}$ , $\\frac{5\\times180}{9}=100^{o}$ and $\\frac{1\\times180}{9}=20^{o}$<\/span>
\nTwice the smallest angle = $2 \\times 20^{o} = 40^{o}$
\nSecond largest angle =$60^{o}$
\nDifference = $60^{o} – 40^{0} = 20^{o}$
\nHence Option E is the correct answer.
\n<\/span><\/p>\n
\nLength = (b+6) meters<\/p>\n
\nb = 18, L = 24
\nDiagonal = 30 (3:4:5 Pythagoras triplet)<\/p>\n
\nArea = 12*30 = 360 square meters<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/quant_1234.JPG\")
<\/p>\n