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Triangles — PYPs

12 solved Triangles previous year questions (PYQs) from past year papers — attempt each and check the answer.

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Q1:ipmat indore 2024QA › TrianglesEasySA · TITA
The number of triangles with integer sides and with perimeter 15 is:
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The Setup: We need to calculate the total number of triangles that can be formed with integer sides and a fixed perimeter of 1515. Step 1: Define the basic parameters and constraints. Let the integer sides of the triangle be aa, bb, and cc, strictly ordered such that a≤b≤ca \le b \le c. The perimeter condition dictates: a+b+c=15a + b + c = 15 Step 2: Apply the triangle inequality theorem to bound the longest side cc. The sum of the two shorter sides must strictly exceed the longest side: a+b>ca + b > c Substitute a+b=15−ca + b = 15 - c into the inequality: 15−c>c⇒2c<15⇒c≤7.515 - c > c \Rightarrow 2c < 15 \Rightarrow c \le 7.5 Since cc is an integer, the maximum valid dimension for cc is 77. Furthermore, cc is the maximum side, so it must be at least the average length of the perimeter: c≥153=5c \ge \frac{15}{3} = 5 Thus, c∈{5,6,7}c \in \{5, 6, 7\}. Step 3: Systematically evaluate integer pairs (a,b)(a, b) for each possible value of cc, maintaining a≤b≤ca \le b \le c. Case 1: c=5c = 5 Requires a+b=10a + b = 10. The only integer pair satisfying a≤b≤5a \le b \le 5 is (5,5)(5, 5). (Yields 11 triangle) Case 2: c=6c = 6 Requires a+b=9a + b = 9. The pairs satisfying a≤b≤6a \le b \le 6 are (3,6)(3, 6) and (4,5)(4, 5). (Yields 22 triangles) Case 3: c=7c = 7 Requires a+b=8a + b = 8. The pairs satisfying a≤b≤7a \le b \le 7 are (1,7)(1, 7), (2,6)(2, 6), (3,5)(3, 5), and (4,4)(4, 4). (Yields 44 triangles) Step 4: Aggregate the valid triangle formations. Total Triangles=1+2+4=7\text{Total Triangles} = 1 + 2 + 4 = 7 Final Answer: 7
Q2:ipmat indore 2024QA › TrianglesEasySA · TITA
Let △ABC\triangle ABC be a triangle right-angled at BB with AB=BC=18AB = BC = 18. The area of the largest rectangle that can be inscribed in this triangle and has BB as one of the vertices is:
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The Setup: A right-angled triangle has vertex BB at the right angle, with legs AB=BC=18AB = BC = 18. We need to find the area of the largest inscribed rectangle sharing vertex BB. Step 1: Establish the coordinate system. Place the right-angled vertex BB at the origin (0,0)(0,0). Given leg lengths AB=BC=18AB = BC = 18, set vertex AA at (0,18)(0,18) and vertex CC at (18,0)(18,0). The hypotenuse ACAC forms a line passing through (0,18)(0,18) and (18,0)(18,0). Its linear equation is: x+y=18⇒y=18−xx + y = 18 \Rightarrow y = 18 - x Step 2: Define the area function of the inscribed rectangle. A rectangle sharing vertex BB at (0,0)(0,0) and bounded by the triangle will have its opposing vertex (x,y)(x, y) strictly on the hypotenuse ACAC. The Area (AA) is the product of its length and width: Arect=x×y=x(18−x)=18x−x2A_{\text{rect}} = x \times y = x(18 - x) = 18x - x^2 Step 3: Maximize the quadratic area function. The function is a downward-opening parabola. We find its maximum via differentiation (or the vertex formula x=−b2ax = \frac{-b}{2a}): dArectdx=18−2x=0⇒x=9\frac{d A_{\text{rect}}}{dx} = 18 - 2x = 0 \Rightarrow x = 9 Step 4: Calculate the maximum area footprint. Substitute x=9x = 9 into the bounding equation to find yy: y=18−9=9y = 18 - 9 = 9 Max Area=9×9=81\text{Max Area} = 9 \times 9 = 81 Final Answer: 81
Q3:ipmat indore 2020QA › TrianglesHardMCQ · MCQ
The number of acute angled triangles whose sides are three consecutive positive integers and whose perimeter is at most 100 is
  • A28
  • B29
  • C31
  • D33
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The Setup: This geometry problem is a triple filter. The sides must pass the triangle inequality vibe check, then survive the acute-angle condition, then fit under the perimeter cap. Whichever filter is strictest wins. Step 1: Set up the side lengths. Let the three consecutive integer sides be x−1x-1, xx and x+1x+1, with x−1≥1x-1\geq 1. For any triangle the two smaller sides must beat the largest: (x−1)+x>x+1  ⟹  x>2(x-1)+x>x+1\implies x>2. Step 2: Enforce the acute angle rule. A triangle is acute when *every* angle is acute - but we only have to test one. The largest angle always sits opposite the longest side, so if that angle is acute the two smaller ones are smaller still and acute automatically. The longest side here is x+1x+1, so the single condition to check is a2+b2>c2a^2+b^2>c^2: (x−1)2+x2>(x+1)2(x-1)^2+x^2>(x+1)^2 x2−2x+1+x2>x2+2x+1x^2-2x+1+x^2>x^2+2x+1 x2>4xx^2>4x Since xx is positive we may divide by xx, giving x>4x>4. As xx is an integer, the lowest valid value is x=5x=5 (the smallest qualifying triangle being 4,5,64,5,6). Note this condition is strictly tighter than Step 1's x>2x>2, so the triangle inequality is absorbed and adds nothing further. Step 3: Apply the perimeter cap. The perimeter is at most 100: (x−1)+x+(x+1)=3x≤100  ⟹  x≤33.33…(x-1)+x+(x+1)=3x\leq 100 \implies x\leq 33.33\ldots Since xx is an integer, the maximum is x=33x=33, giving sides 32,33,3432,33,34 and a perimeter of 9999. Step 4: Count the surviving squad. The valid range for xx runs from 5 to 33 inclusive: 33−5+1=2933-5+1=29 Final Answer: 29
Q4:ipmat indore 2025QA › TrianglesMediumMCQ · MCQ
In triangle ABCABC, AB=AC=xAB=AC=x, ∠ABC=θ\angle ABC=\theta and the circumradius is equal to yy. Then xy\frac{x}{y} equals
  • Asin⁡(θ)\sin(\theta)
  • Bcos⁡(θ)\cos(\theta)
  • C2cos⁡(θ)2\cos(\theta)
  • D2sin⁡(θ)2\sin(\theta)
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The Setup: We have an isosceles triangle stepping into the arena, and we need to link its side lengths, angles, and circumradius. The ultimate cheat code for any circle circumscribing a triangle is the Extended Law of Sines. It is basically the universal API bridging triangle sides, their opposite angles, and the circumcircle's radius. Step 1: Decode the triangle's geometry. We are given that AB=AC=xAB=AC=x, making triangle ABCABC isosceles. In geometry lore, the angle opposite to side ACAC is ∠ABC\angle ABC, which we are told is θ\theta. Step 2: Deploy the Extended Law of Sines. The law dictates that the ratio of any side length to the sine of its opposite angle is exactly equal to the circumcircle's diameter (twice the circumradius, or 2R2R). The general framework is: asin⁡(A)=bsin⁡(B)=csin⁡(C)=2R\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} = 2R Step 3: Plug in the stats. Let's target side ACAC (which has a length of xx) and its corresponding opposite angle, ∠ABC\angle ABC (which is θ\theta). The circumradius RR is given to us as yy. Substitute these into the formula: xsin⁡(θ)=2y\frac{x}{\sin(\theta)} = 2y Step 4: Isolate the target ratio. The prompt is asking for the exact value of xy\frac{x}{y}. We just need to execute a quick algebraic shuffle—divide both sides by yy and multiply both sides by sin⁡(θ)\sin(\theta) to get the variables aligned perfectly. xy=2sin⁡(θ)\frac{x}{y} = 2\sin(\theta) The math literally solves itself. Final Answer: 2sin⁡(θ)2\sin(\theta)
Q5:ipmat indore 2022QA › TrianglesEasyMCQ · MCQ
The lengths of the sides of a triangle are x,21x, 21 and 4040, where xx is the shortest side. A possible value of xx is:
  • A18
  • B20
  • C19
  • D16
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The Setup: A triangle is composed of side lengths x,21x, 21 and 4040, where xx is stated to be the shortest side. We must determine a viable integer value for xx bounded by the triangle inequality. Step 1: Apply the triangle inequality theorem. For three lengths to successfully form a closed triangle, the sum of the two shorter sides must strictly exceed the longest side. x+21>40  ⟹  x>19x + 21 > 40 \implies x > 19 Step 2: Apply the specific length constraint. The problem states that xx is the shortest side. Therefore, xx must be less than or equal to 2121. Combining both strict inequalities: 19<x≤2119 < x \le 21 Step 3: Evaluate the given options. The mathematically possible integer values for xx are 2020 and 2121. Reviewing the multiple-choice options provided (18,20,19,1618, 20, 19, 16), the only valid candidate listed is 2020. Final Answer: 20
Q6:ipmat indore 2024QA › TrianglesHardMCQ · MCQ
The side AB of a triangle ABC is c. The median BD is of length k. If ∠BDA=θ\angle BDA = \theta and θ<90∘\theta < 90^\circ, then the area of triangle ABC is
  • Ak2sin⁡θ2+ksin⁡θc2+k2sin⁡2θ\dfrac{k^2 \sin \theta}{2} + k \sin \theta \sqrt{c^2 + k^2 \sin^2 \theta}
  • Bk2sin⁡2θ2+ksin⁡θc2−k2sin⁡2θ\dfrac{k^2 \sin 2\theta}{2} + k \sin \theta \sqrt{c^2 - k^2 \sin^2 \theta}
  • Ck2cos⁡2θ2+ksin⁡θc2−k2sin⁡2θ\dfrac{k^2 \cos 2\theta}{2} + k \sin \theta \sqrt{c^2 - k^2 \sin^2 \theta}
  • Dk2cos⁡θ2+ksin⁡θc2+k2sin⁡2θ\dfrac{k^2 \cos \theta}{2} + k \sin \theta \sqrt{c^2 + k^2 \sin^2 \theta}
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The Setup: In triangle ABCABC, the side ABAB is cc, the median BDBD has length kk, and the angle ∠BDA=θ\angle BDA = \theta where θ<90∘\theta < 90^\circ. We need to find the area of the entire triangle ABCABC. Step 1: Establish the area formula using the median. A median divides a triangle into two smaller triangles of identical area. Area(ABC)=2×Area(ABD)\text{Area}(ABC) = 2 \times \text{Area}(ABD) Using the base ADAD and the altitude from BB to ADAD (which is ksin⁡θk\sin\theta): Area(ABD)=12×AD×(ksin⁡θ)\text{Area}(ABD) = \frac{1}{2} \times AD \times (k\sin\theta) Area(ABC)=2(12×AD×ksin⁡θ)=AD⋅ksin⁡θ\text{Area}(ABC) = 2 \left( \frac{1}{2} \times AD \times k\sin\theta \right) = AD \cdot k\sin\theta Step 2: Use the Cosine Rule to express ADAD. In △ABD\triangle ABD, apply the Law of Cosines: c2=AD2+k2−2(AD)(k)cos⁡θc^2 = AD^2 + k^2 - 2(AD)(k)\cos\theta Rearrange into a quadratic equation in terms of ADAD: AD2−(2kcos⁡θ)AD+(k2−c2)=0AD^2 - (2k\cos\theta)AD + (k^2 - c^2) = 0 Solve for ADAD via the quadratic formula (taking the positive valid geometric root): AD=2kcos⁡θ+4k2cos⁡2θ−4(k2−c2)2AD = \frac{2k\cos\theta + \sqrt{4k^2\cos^2\theta - 4(k^2 - c^2)}}{2} AD=kcos⁡θ+c2−k2(1−cos⁡2θ)=kcos⁡θ+c2−k2sin⁡2θAD = k\cos\theta + \sqrt{c^2 - k^2(1 - \cos^2\theta)} = k\cos\theta + \sqrt{c^2 - k^2\sin^2\theta} Step 3: Substitute ADAD back into the Area equation. Area(ABC)=(kcos⁡θ+c2−k2sin⁡2θ)ksin⁡θ\text{Area}(ABC) = \left( k\cos\theta + \sqrt{c^2 - k^2\sin^2\theta} \right) k\sin\theta Area(ABC)=k2sin⁡θcos⁡θ+ksin⁡θc2−k2sin⁡2θ\text{Area}(ABC) = k^2\sin\theta\cos\theta + k\sin\theta\sqrt{c^2 - k^2\sin^2\theta} Using the double-angle identity 2sin⁡θcos⁡θ=sin⁡(2θ)2\sin\theta\cos\theta = \sin(2\theta): Area(ABC)=k2sin⁡2θ2+ksin⁡θc2−k2sin⁡2θ\text{Area}(ABC) = \frac{k^2\sin 2\theta}{2} + k\sin\theta\sqrt{c^2 - k^2\sin^2\theta} Final Answer: k2sin⁡2θ2+ksin⁡θc2−k2sin⁡2θ\dfrac{k^{2}\sin 2\theta}{2}+k\sin\theta\sqrt{c^{2}-k^{2}\sin^{2}\theta}
Q7:ipmat indore 2024QA › TrianglesMediumMCQ · MCQ
Let △ABC\triangle ABC be a triangle with AB=ACAB = AC and DD be a point on BCBC such that ∠BAD=30∘\angle BAD = 30^\circ. If EE is a point on ACAC such that AD=AEAD = AE, then ∠CDE\angle CDE equals
  • A60∘60^\circ
  • B30∘30^\circ
  • C10∘10^\circ
  • D15∘15^\circ
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The Setup: In a triangle ABCABC with AB=ACAB=AC, DD is a point on BCBC creating ∠BAD=30∘\angle BAD=30^\circ, and EE is a point on ACAC creating AD=AEAD=AE. We need to determine the angle ∠CDE\angle CDE. Step 1: Assign variables to the base angles of the primary triangle. Let ∠B=∠C=α\angle B = \angle C = \alpha (since AB=ACAB = AC). The vertex angle of the large triangle is ∠BAC=180∘−2α\angle BAC = 180^\circ - 2\alpha. The remaining upper angle inside is ∠DAC=∠BAC−∠BAD=(180∘−2α)−30∘=150∘−2α\angle DAC = \angle BAC - \angle BAD = (180^\circ - 2\alpha) - 30^\circ = 150^\circ - 2\alpha. Step 2: Determine the base angles of the internal isosceles triangle. In △ADE\triangle ADE, since AD=AEAD = AE, the base angles are equal: ∠ADE=∠AED=180∘−∠DAC2\angle ADE = \angle AED = \frac{180^\circ - \angle DAC}{2} ∠ADE=180∘−(150∘−2α)2=30∘+2α2=15∘+α\angle ADE = \frac{180^\circ - (150^\circ - 2\alpha)}{2} = \frac{30^\circ + 2\alpha}{2} = 15^\circ + \alpha Step 3: Use the exterior angle theorem to map ∠CDE\angle CDE. Look at △ABD\triangle ABD. The exterior angle at DD is ∠ADC\angle ADC, equaling the sum of the remote interior angles: ∠ADC=∠B+∠BAD=α+30∘\angle ADC = \angle B + \angle BAD = \alpha + 30^\circ Geometrically, angle ∠ADC\angle ADC is comprised of two adjacent components: ∠ADC=∠ADE+∠CDE\angle ADC = \angle ADE + \angle CDE Substitute the known expressions: α+30∘=(15∘+α)+∠CDE\alpha + 30^\circ = (15^\circ + \alpha) + \angle CDE ∠CDE=30∘−15∘=15∘\angle CDE = 30^\circ - 15^\circ = 15^\circ Final Answer: 15∘15^{\circ}
Q8:ipmat indore 2025QA › TrianglesMediumMCQ · MCQ
Consider a triangle with side lengths 44 meters, 66 meters, and 99 meters. A dog runs around the triangle in such a way that the shortest distance of the dog from the triangle is exactly 11 meter. The total distance covered (in meters) by the dog in one round is
  • A22+2π22+2\pi
  • B19+2π19+2\pi
  • C2222
  • D22−2π22-2\pi
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The Setup: This is a locus problem disguised as a dog-walking simulator. The dog maintains a strict 11-meter social distance from the triangle at all times. When you trace this path, the dog's route is built from two types of geometry: straight paths parallel to the triangle's sides, and circular arcs pivoting around the corners. It is basically Mario Kart drifting mechanics. Step 1: Analyze the straightaways. For every side of the triangle, the dog runs a parallel straight line exactly 11 meter away. The lengths of these straight paths match the triangle's base stats perfectly. Straight distance=4+6+9=19\text{Straight distance} = 4 + 6 + 9 = 19 Step 2: Calculate the corner drifts. When the dog reaches a corner, it has to swing around the vertex to stay exactly 11 meter away from the point. This movement creates a circular arc. The angle of this arc is exactly equal to the exterior angle of the triangle at that specific vertex. Step 3: Piece the arcs together. In Euclidean geometry lore, the sum of the exterior angles of *any* convex polygon is an automatic 360∘360^\circ (or 2π2\pi radians). This means if you take those three corner arcs and stitch them together, they form one flawless, complete circle with a radius of r=1r = 1 meter. Let's run the circumference formula: Arc distance=2πr=2π(1)=2π\text{Arc distance} = 2\pi r = 2\pi(1) = 2\pi Step 4: Sum the total distance. Combine the straightaways and the corner drifts to get the final run length. The total distance covered by the dog in one round is just the sum of our two parts. Total=19+2π\text{Total} = 19 + 2\pi Final Answer: 19+2π19+2\pi
Q9:ipmat indore 2023QA › TrianglesHardMCQ · MCQ
In a triangle ABC, let D be the midpoint of BC, and AM be the altitude on BC. If the lengths of AB, BC and CA are in the ratio of 2:4:3, then the ratio of the lengths of BM and AD would be
  • A11:41011:4\sqrt{10}
  • B12:1112:11
  • C11:1211:12
  • D410:114\sqrt{10}:11
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The Setup: In △ABC\triangle ABC, DD is the midpoint of BCBC, and AMAM is the altitude on BCBC. With side ratios AB:BC:CA=2:4:3AB:BC:CA = 2:4:3, we must find the length ratio BM:ADBM:AD. Step 1: Assign algebraic lengths to the sides. Let the sides be c=AB=2xc = AB = 2x, a=BC=4xa = BC = 4x, and b=CA=3xb = CA = 3x. Since DD is the midpoint of BCBC, BD=BC2=2xBD = \frac{BC}{2} = 2x. Step 2: Calculate the length of BMBM. Use the Law of Cosines to find cos⁡B\cos B in △ABC\triangle ABC: cos⁡B=a2+c2−b22ac=(4x)2+(2x)2−(3x)22(4x)(2x)\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{(4x)^2 + (2x)^2 - (3x)^2}{2(4x)(2x)} cos⁡B=16x2+4x2−9x216x2=11x216x2=1116\cos B = \frac{16x^2 + 4x^2 - 9x^2}{16x^2} = \frac{11x^2}{16x^2} = \frac{11}{16} In the right-angled △ABM\triangle ABM, the segment BMBM represents the adjacent side to angle BB. Thus, BM=c⋅cos⁡BBM = c \cdot \cos B: BM=2x(1116)=11x8BM = 2x \left(\frac{11}{16}\right) = \frac{11x}{8} Step 3: Calculate the length of the median ADAD. Apply Apollonius's Theorem to the median ADAD: AB2+AC2=2(AD2+BD2)AB^2 + AC^2 = 2(AD^2 + BD^2) (2x)2+(3x)2=2(AD2+(2x)2)(2x)^2 + (3x)^2 = 2(AD^2 + (2x)^2) 4x2+9x2=2AD2+8x24x^2 + 9x^2 = 2AD^2 + 8x^2 13x2−8x2=2AD213x^2 - 8x^2 = 2AD^2 5x2=2AD2  ⟹  AD2=5x22  ⟹  AD=x52=x1025x^2 = 2AD^2 \implies AD^2 = \frac{5x^2}{2} \implies AD = \frac{x\sqrt{5}}{\sqrt{2}} = \frac{x\sqrt{10}}{2} Step 4: Compute the ratio BM:ADBM : AD. Ratio=11x8x102=118×210=22810=11410\text{Ratio} = \frac{\frac{11x}{8}}{\frac{x\sqrt{10}}{2}} = \frac{11}{8} \times \frac{2}{\sqrt{10}} = \frac{22}{8\sqrt{10}} = \frac{11}{4\sqrt{10}} Final Answer: 11:41011:4\sqrt{10}
Q10:ipmat indore 2022QA › TrianglesMediumMCQ · MCQ
In a right-angled triangle ABC, the hypotenuse AC is of length 13 cm. A line drawn connecting the midpoints D and E of sides AB and AC is found to be 6 cm in length. The length of BC is
  • A1212 cm
  • B55 cm
  • C232 \sqrt{3} cm
  • D88 cm
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The Setup: In a right-angled triangle ABCABC, the hypotenuse ACAC is 13 cm13\text{ cm}. The segment DEDE joining the midpoint of the leg ABAB to the midpoint of the hypotenuse ACAC measures 6 cm6\text{ cm}. We must find the length of BCBC. Step 1: Map the geometry using the Midpoint Theorem. The problem defines a line segment drawn connecting the midpoints (DD and EE) of two sides of a triangle (ABAB and ACAC). According to the Triangle Midpoint Theorem, the line segment connecting the midpoints of two sides of a triangle is perfectly parallel to the third side and exactly half of its length. DE=12BCDE = \frac{1}{2} BC Step 2: Calculate the base. We are explicitly given that DE=6 cmDE = 6\text{ cm}. 6=12BC  ⟹  BC=12 cm6 = \frac{1}{2} BC \implies BC = 12\text{ cm} *(The hypotenuse measurement of 13 is consistent data: with BC=12BC = 12 and AC=13AC = 13, the remaining leg AB=5AB = 5, forming the standard 5-12-13 Pythagorean triple.)* Final Answer: 1212 cm
Q11:ipmat indore 2024QA › TrianglesHardMCQ · MCQ
Let ABC be an equilateral triangle, with each side of length kk. If a circle is drawn with diameter AB, then the area of the portion of the triangle lying inside the circle is
  • A(33+π)k224\left(3\sqrt{3} + \pi\right) \frac{k^2}{24}
  • B(33+π)k26\left(3\sqrt{3} + \pi\right) \frac{k^2}{6}
  • C(33−π)k224\left(3\sqrt{3} - \pi\right) \frac{k^2}{24}
  • D(33+π)k28\left(3\sqrt{3} + \pi\right) \frac{k^2}{8}
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The Setup: An equilateral triangle is partially overlapped by a circle whose diameter is one of its sides. We must calculate the area of the geometric intersection of the two figures. Step 1: Establish the geometry of the intersection. Let the equilateral triangle be ABCABC with side length kk. The circle has diameter ABAB, meaning its center OO is the midpoint of ABAB, and its radius is k/2k/2. The circle intersects side ACAC at a point DD and side BCBC at a point EE. According to Thales's Theorem, any angle inscribed in a semicircle is a right angle. Since ABAB is the diameter, ∠ADB=90∘\angle ADB = 90^\circ and ∠AEB=90∘\angle AEB = 90^\circ. Thus, BDBD and AEAE are altitudes of the equilateral triangle ABCABC. In an equilateral triangle, altitudes also act as medians, meaning DD and EE exactly bisect sides ACAC and BCBC. Step 2: Partition the target area. The portion of the triangle lying inside the circle is the shape bounded by segments ADAD, EBEB, ABAB, and the circular arc DEDE. We can decompose this bounded region by connecting the center OO to points DD and EE. The area consists of three non-overlapping geometric shapes: △AOD\triangle AOD, △BOE\triangle BOE, and the circular sector DOEDOE. Step 3: Calculate the component areas. Because OO is the midpoint of ABAB and DD is the midpoint of ACAC, AD=k/2AD = k/2 and AO=k/2AO = k/2. Since DD is on the circle, OD=k/2OD = k/2. Thus, △AOD\triangle AOD is equilateral with side length k/2k/2. Area(△AOD)=34(k2)2=316k2\text{Area}(\triangle AOD) = \frac{\sqrt{3}}{4} \left(\frac{k}{2}\right)^2 = \frac{\sqrt{3}}{16} k^2 By identical symmetry, △BOE\triangle BOE is also equilateral with the same area. Because ∠AOD=60∘\angle AOD = 60^\circ and ∠BOE=60∘\angle BOE = 60^\circ, the central angle of the sector DOEDOE is 180∘−60∘−60∘=60∘180^\circ - 60^\circ - 60^\circ = 60^\circ. Area(Sector DOE)=60∘360∘×π(k2)2=16×πk24=π24k2\text{Area}(\text{Sector } DOE) = \frac{60^\circ}{360^\circ} \times \pi \left(\frac{k}{2}\right)^2 = \frac{1}{6} \times \frac{\pi k^2}{4} = \frac{\pi}{24} k^2 Step 4: Sum the components to find total area. Total Area=2(316k2)+π24k2\text{Total Area} = 2 \left( \frac{\sqrt{3}}{16} k^2 \right) + \frac{\pi}{24} k^2 Total Area=38k2+π24k2=(3324+π24)k2=33+π24k2\text{Total Area} = \frac{\sqrt{3}}{8} k^2 + \frac{\pi}{24} k^2 = \left( \frac{3\sqrt{3}}{24} + \frac{\pi}{24} \right) k^2 = \frac{3\sqrt{3} + \pi}{24} k^2 Final Answer: (33+π)k224\left(3\sqrt{3}+\pi\right)\dfrac{k^{2}}{24}
Q12:ipmat indore 2019QA › TrianglesMediumMCQ · MCQ
The number of points, having both coordinates as integers, that lie in the interior of the triangle with vertices (0,0),(0,31),(0, 0), (0, 31), and (31,0)(31, 0) is
  • A435
  • B465
  • C450
  • D464
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The Setup: We are hunting for lattice points strictly inside the triangle. The boundary lines are the axes (x=0,y=0x=0, y=0) and the hypotenuse (x+y=31x+y=31). To stay *interior*, our coordinates must be strictly positive integers that sum to less than 31. Boundary points fail the vibe check and get instantly disqualified. Step 1: Set the Constraints We need integers x>0x > 0 and y>0y > 0 such that x+y<31x + y < 31. This means the absolute max limit for the sum is x+y≤30x + y \le 30. Step 2: Sum the Rows If x=1x = 1, yy can range from 11 to 2929 (giving us 29 valid points). If x=2x = 2, yy can range from 11 to 2828 (28 points). This sequence cascades linearly all the way down to x=29x = 29, where yy can only be 11 (1 point). Step 3: Calculate the Total We just need to sum the first 29 positive integers. We plug this into the classic arithmetic series sum formula S=n(n+1)2S = \frac{n(n+1)}{2}. S=29(30)2S = \frac{29(30)}{2} S=29×15=435S = 29 \times 15 = 435 Final Answer: 435

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