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JEE • MATHEMATICS • PROPERTIES OF TRIANGLES

Properties of triangles practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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91 Medium
91 of 140
Application +4 | -1
The area of the triangle whose sides are 6,  5,  136,\;5,\;\sqrt {13} (in square unit) is
View answer and explanation
Correct option: B
9
Option B (9) is correct because using Heron's formula or trigonometric methods, the area of the triangle with sides 6, 5, and √13 is 9 square units.
92 Medium
92 of 140
Application +4 | -1
In a right-angled triangle, the sides are a,  ba,\;b and cc, with cc as hypotenuse, and c−b≠1,  c+b≠1c - b \ne 1,\;c + b \ne 1. Then the value of (log⁡c+ba+log⁡c−ba)/(2log⁡c+ba×log⁡c−ba)\left( {{{\log }_{c + b}}a + {{\log }_{c - b}}a} \right)/\left( {2{{\log }_{c + b}}a \times {{\log }_{c - b}}a} \right) will be
View answer and explanation
Correct option: D
1
Option D (1) is correct because the expression simplifies to log⁡a2log⁡a2\frac{{\log {a^2}}}{{\log {a^2}}}, which equals 1.
93 Medium
93 of 140
Analyzing +4 | -1
In a triangle, the sum of two sides is xx and the product of the same two sides is yy. If x2−c2=yx^2-c^2=y, where cc is the third side of the triangle, then the ratio of the inradius to the circumradius of the triangle is
View answer and explanation
Correct option: A
3y2x(x+c)\frac{3 y}{2 x(x+c)}
Option B is the correct answer because it correctly calculates the ratio of inradius to circumradius using the given conditions and trigonometric relationships.
94 Medium
94 of 140
Application +4 | -1
If in a △ABC,∠B=90∘\triangle A B C, \angle B=90^{\circ}, then tan⁡2(A2)\tan ^2\left(\frac{A}{2}\right) is equal to
View answer and explanation
Correct option: A
b−cb+c\frac{b-c}{b+c}
Option A (b−cb+c\frac{b-c}{b+c}) is correct because it is derived by using the definition of cosine and the half-angle formula for tangent in a right-angled triangle.
95 Medium
95 of 140
Analyzing +4 | -1
If A,  B,  C,  DA,\;B,\;C,\;D are the angles of a quadrilateral, then ∑tan⁡A∑cot⁡A\frac{{\sum \tan A}}{{\sum \cot A}} is equal to
View answer and explanation
Correct option: A
∏tan⁡A\prod \tan A
Option A (∏tan⁡A\prod \tan A) is correct because using the condition A+B+C+D=2πA + B + C + D = 2\pi, we can derive the given expression to be equal to the product of tangents.
96 Medium
96 of 140
Analyzing +4 | -1
In a △ABC,a,b,A\triangle A B C, a, b, A are given and b1,b2b_1, b_2 are two values of the third side bb such that b2=2b1b_2=2 b_1. Then, sin⁡A\sin A is equal to
View answer and explanation
Correct option: B
9a2−c28c2\sqrt{\frac{9 a^2-c^2}{8 c^2}}
Option B (9a2−c28c2\sqrt{\frac{{9 a^2-c^2}}{{8 c^2}}}) is correct because using the properties of quadratic equations and trigonometric identities, we can derive the expression for sin⁡A\sin A.
97 Medium
97 of 140
Application +4 | -1
In a Δ  ABC\Delta \;ABC, angle AA is greater than angle BB. If the measures of angles AA and BB satisfy the equation 3sin⁡x−4sin⁡3x−k=0,    0<k<13\sin x - 4{\sin ^3}x - k = 0,\;\;0 < k < 1, then the measure of angle CC is
View answer and explanation
Correct option: C
2π3\frac{{2\pi }}{3}
Option C (2π3\frac{{2\pi }}{3}) is correct because the given equation implies sin⁡3A=sin⁡3B\sin 3A = \sin 3B, and since A > B, we have A+B=60∘A + B = 60^\circ, leading to C=120∘=2π3C = 120^\circ = \frac{{2\pi }}{3}.
98 Medium
98 of 140
Application +4 | -1
In a triangle the length of the two larger sides are 24\mathrm{24} and 22\mathrm{22}, respectively. If the angles are in AP, then the third side is
View answer and explanation
Correct option: A
12+2 1312 + 2\,\sqrt {13}
Option A (12+2 1312 + 2\,\sqrt {13}) is correct because it is the value of the third side calculated using the properties of triangles with angles in A.P. and the sine rule.
99 Medium
99 of 140
Application +4 | -1
ABCA B C is an equilateral triangle of each side a (>0)(>0). The inradius of the triangle is
View answer and explanation
Correct option: D
a23\frac{a}{2 \sqrt{3}}
Option D (a23\frac{a}{2 \sqrt{3}}) is correct because the inradius of an equilateral triangle is given by the formula a23\frac{a}{2 \sqrt{3}}, where a is the side length.
100 Medium
100 of 140
Application +4 | -1
If the angles of a right angled triangle are in A.P., then the ratio of the in-radius and the perimeter is
View answer and explanation
Correct option: C
(2−3)  :2 3\left( {2 - \sqrt 3 } \right)\;:2\,\sqrt 3
Option C is correct because it accurately calculates the ratio of the in-radius to the perimeter of the right-angled triangle with angles in A.P.
101 Medium
101 of 140
Analyzing +4 | -1
In △ABC,a=5,b=4\triangle A B C, a=5, b=4 and cos⁡(A−B)=3132\cos (A-B)=\frac{31}{32}, then side cc is
View answer and explanation
Correct option: A
6
Option A (6) is correct because applying trigonometric identities and the cosine rule leads to the side c being equal to 6.
102 Medium
102 of 140
Application +4 | -1
Let a,ba, b and cc be the sides of a triangle and a2+b2+c2ab+bc+ca=P\frac{a^2+b^2+c^2}{a b+b c+c a}=P. Then
View answer and explanation
Correct option: D
1≤P<21 \leq P < 2
Option D (1≤P<21 \leq P < 2) is correct because the ratio a2+b2+c2ab+bc+ca\frac{a^2+b^2+c^2}{ab+bc+ca} is always greater than or equal to 1, and less than 2 for a triangle.
103 Medium
103 of 140
Analyzing +4 | -1
If in △ABC,∠A=90∘\triangle A B C, \angle A=90^{\circ} and c,sin⁡B,cos⁡Bc, \sin B, \cos B are rational numbers, then
View answer and explanation
Correct option: A
aa and bb are rational
Option A (aa and bb are rational) is correct because given the conditions, it can be shown that both a and b are rational numbers.
104 Medium
104 of 140
Application +4 | -1
If the ex-radii of a triangle are in H.P., then the corresponding sides are in
View answer and explanation
Correct option: A
A.P.
Option A is correct because if the ex-radii are in H.P., the corresponding sides are in A.P.
105 Medium
105 of 140
Application +4 | -1
In any △ABC\triangle A B C, if C=90∘C=90^{\circ}, then tan⁡B2\tan \frac{B}{2} is equal to
View answer and explanation
Correct option: A
c−ac+a\sqrt{\frac{c-a}{c+a}}
Option A (c−ac+a\sqrt{\frac{c-a}{c+a}}) is correct because it is derived from Napier's analogy and the given condition C=90∘C=90^{\circ}.
106 Medium
106 of 140
Application +4 | -1
In a triangle ABC,2a2+4b2+c2=4ab+2ac\mathrm{ABC}, 2 a^2+4 b^2+c^2=4 a b+2 a c, then cos⁡B\cos \mathrm{B} is equal to
View answer and explanation
Correct option: A
0.875
Option A (0.875) is correct because by manipulating the given equation, we can find the relationship between a, b, and c, and then calculate cos B using the Law of Cosines.
107 Medium
107 of 140
Analyzing +4 | -1
In Δ  ABC,a2(cos⁡2B−cos⁡2C)+b2(cos⁡2C−cos⁡2A)+c2(cos⁡2A−cos⁡2B)\Delta \;ABC,{a^2}\left( {{{\cos }^2}B - {{\cos }^2}C} \right) + {b^2}\left( {{{\cos }^2}C - {{\cos }^2}A} \right) + {c^2}\left( {{{\cos }^2}A - {{\cos }^2}B} \right) is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because by using the sine rule and trigonometric identities, the expression simplifies to zero.
108 Medium
108 of 140
Analyzing +4 | -1
In a Δ ABC\Delta \,ABC, angles A,  B,  CA,\,\,B,\,\,C are in A.P., then
lim⁡A→C  3−4  sin⁡  A  sin⁡C∣ A−C ∣\lim\limits_{A \to C} \,\,\frac{{\sqrt {3 - 4\,\,\sin \,\,A\,\,\sin C} }}{{\left| {\,A - C\,} \right|}} is equal to
View answer and explanation
Correct option: A
1\mathrm{1}
Option A (1) is correct because after applying the A.P. condition and trigonometric identities, the limit simplifies to 1.
109 Medium
109 of 140
Application +4 | -1
The angles of elevation of a vertical tower standing inside a triangular field at the vertices of the field are each equal to θ\theta. If the length of the sides of the field are 30 m,50 m30 \mathrm{~m}, 50 \mathrm{~m} and 70 m70 \mathrm{~m}, the height of the tower is
View answer and explanation
Correct option: B
(70/3)tan⁡θm(70 / \sqrt{3}) \tan \theta \mathrm{m}
Option B ((70/3)tan⁡θm(70 / \sqrt{3}) \tan \theta \mathrm{m}) is correct because it is derived by finding the circumradius of the triangle and relating it to the height of the tower and the angle of elevation.
110 Medium
110 of 140
Application +4 | -1
In a △ABC,A:B:C=3:5:4\triangle \mathrm{ABC}, \mathrm{A}: \mathrm{B}: \mathrm{C}=3: 5: 4.Then a+b+c2\mathrm{a}+\mathrm{b}+\mathrm{c} \sqrt{2} is equal to
View answer and explanation
Correct option: C
3 b3 \mathrm{~b}
Option C (3 b3 \mathrm{~b}) is correct because after substituting the values of a, b, and c in terms of b, the expression simplifies to 3b.
111 Medium
111 of 140
Application +4 | -1
The sines of two angles of a triangle are equal to 5/135 / 13 and 99/10199 / 101. The cosine of the third angle is
View answer and explanation
Correct option: C
2551313\frac{255}{1313}
Option C (2551313\frac{255}{1313}) is correct because it is the result of applying trigonometric identities to find the cosine of the third angle.
112 Medium
112 of 140
Analyzing +4 | -1
A triangle be such that cos⁡2A+cos⁡2B+cos⁡2C\cos 2 A+\cos 2 B+\cos 2 C is minimum. If in radius of the triangle is 3, then which of the following is CORRECT?
View answer and explanation
Correct option: B
Perimeter of Δ\Delta is 18318 \sqrt{3} Units
Option B (Perimeter of Δ\Delta is 18318 \sqrt{3} Units) is correct because when cos⁡2A+cos⁡2B+cos⁡2C\cos 2 A+\cos 2 B+\cos 2 C is minimum, the triangle is equilateral, and the perimeter can be calculated using the inradius.
113 Medium
113 of 140
Application +4 | -1
In any △ABC\triangle A B C, if sin⁡A,sin⁡B\sin A, \sin B and sin⁡C\sin C are in APA P, then the maximum value of tan⁡B2\tan \frac{B}{2} is
View answer and explanation
Correct option: B
13\frac{1}{\sqrt{3}}
Option B (13\frac{1}{\sqrt{3}}) is correct. The maximum value of tan⁡B2\tan \frac{B}{2} is found by expressing it in terms of sin⁡A\sin A and cos⁡A\cos A, then using trigonometric identities and inequalities.
114 Medium
114 of 140
Application +4 | -1
If kk is the perimeter of △ABC\triangle A B C, then bcos⁡2C2+ccos⁡2B2b \cos ^2 \frac{C}{2}+c \cos ^2 \frac{B}{2} is equal to
View answer and explanation
Correct option: C
k2\frac{k}{2}
Option C (k2\frac{{k}}{{2}}) is correct because by using trigonometric identities and the cosine rule, the expression simplifies to half the perimeter of the triangle.
115 Medium
115 of 140
Application +4 | -1
If two forces acting at right angles have their resultant 10 kg\sqrt{10} {~kg}. wt. and when they act at an angle of 60∘60^{\circ}, the resultant is 13 kg\sqrt{13} {~kg}. wt., then the forces are
View answer and explanation
Correct option: A
3 kg3 {~kg}. wt., 1 kg1 {~kg}. wt.
Option A (3 kg. wt., 1 kg. wt.) is correct because these forces satisfy the given conditions for the resultant at right angles and at 60 degrees.
116 Medium
116 of 140
Application +4 | -1
In any triangle ABCA B C, if cos⁡A=sin⁡B2sin⁡C\cos A=\frac{\sin B}{2 \sin C}, then
View answer and explanation
Correct option: B
c=ac=a
Option B (c=ac=a) is correct because by using the Law of Cosines and the Law of Sines, the given equation simplifies to c = a.
117 Medium
117 of 140
Application +4 | -1
If in a Δ ABC,  A=π3\Delta \,ABC,\,\,A = \frac{\pi }{3} and ADAD is the median, then
View answer and explanation
Correct option: A
2   AD2=b2+c2+  bc2\;\,A{D^2} = {b^2} + {c^2} + \,\,bc
Option A (2   AD2=b2+c2+  bc2\;\,A{D^2} = {b^2} + {c^2} + \,\,bc) is correct because it is derived from applying the cosine rule and the median formula.
118 Medium
118 of 140
Application +4 | -1
If A1 A2 A3 A4 A5{A_1}\,{A_2}\,{A_3}\,{A_4}\,{A_5} be a regular pentagon inscribed in a unit circle. Then (A1A2)  (A1A3)\left( {{A_1}{A_2}} \right)\,\,\left( {{A_1}{A_3}} \right) is equal to
View answer and explanation
Correct option: D
5\sqrt 5
Option D (5\sqrt 5) is correct because the product of the lengths A1A2{A_1}{A_2} and A1A3{A_1}{A_3} in a regular pentagon inscribed in a unit circle equals 5\sqrt 5.
119 Medium
119 of 140
Application +4 | -1
If a,ba, b and AA are given in aa triangle and c1,c2c_1, c_2 are the possible values of the third side, then c12+c22−2c1c2cos⁡2Ac_1^2+c_2^2-2 c_1 c_2 \cos 2 A is equal to
View answer and explanation
Correct option: C
4a2cos⁡2A4 a^2 \cos ^2 A
Option C (4a2cos⁡2A4 a^2 \cos ^2 A) is correct because by using the cosine rule and Vieta's formulas, the expression simplifies to 4a2cos⁡2A4 a^2 \cos ^2 A.
120 Medium
120 of 140
Application +4 | -1
If in a △ABC,a2+b2+c2=8R2\triangle A B C, a^2+b^2+c^2=8 R^2, where R=R= circumradius, the the triangle is
View answer and explanation
Correct option: C
right angled
Option C (right angled) is correct because the given condition a2+b2+c2=8R2a^2 + b^2 + c^2 = 8R^2 implies that sin⁡2A+sin⁡2B+sin⁡2C=2\sin^2 A + \sin^2 B + \sin^2 C = 2, which leads to one of the angles being π2\frac{{\pi}}{{2}}.
Revision notes and key takeaways

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