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JEE • MATHEMATICS • PROPERTIES OF TRIANGLES

Properties of triangles practice · Mixed difficulty

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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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Showing 31-60 of 434 questions
31 Easy
31 of 434
Understanding +4 | -1
In a triangle, if r1>r2>r3r_1>r_2>r_3, then
View answer and explanation
Correct option: A
a>b>ca > b > c
Option B (a<b<ca < b < c) is correct because in a triangle, if r1>r2>r3r_1 > r_2 > r_3, it implies that the sides opposite to the angles are in the reverse order, i.e., a<b<ca < b < c.
32 Easy
32 of 434
Application +4 | -1
If the length of the sides of a triangle are 3, 4 and 5 unit, then RR is
View answer and explanation
Correct option: D
2.5
Option D (2.5) is correct because a triangle with sides 3, 4, and 5 is a right-angled triangle, and the circumradius of a right-angled triangle is half the length of the hypotenuse.
33 Easy
33 of 434
Application +4 | -1
If λ\lambda be the perimeter of the Δ  ABC\Delta \;ABC, then b  cos⁡2(C2)+ccos⁡2(B2)b\;{\cos ^2}\left( {\frac{C}{2}} \right) + c{\cos ^2}\left( {\frac{B}{2}} \right) is equal to
View answer and explanation
Correct option: C
λ2\frac{\lambda }{2}
Option C (λ2\frac{\lambda }{2}) is correct because by using trigonometric identities and the definition of the semi-perimeter, the expression bcos⁡2(C2)+ccos⁡2(B2)b\cos ^2(\frac{C}{2}) + c\cos ^2(\frac{B}{2}) simplifies to the semi-perimeter, which is half the perimeter.
34 Easy
34 of 434
Application +4 | -1
If in a Δ  ABC\Delta \;ABC, sides a,b,c are in A.P., then tan⁡  A2  tan⁡  C2=\tan \,\,\frac{A}{2}\,\,\tan \,\,\frac{C}{2} =
View answer and explanation
Correct option: B
1/31/3
Option B (1/31/3) is correct because when the sides of a triangle are in A.P., a relationship between tan⁡ A2\tan \,\frac{A}{2} and tan⁡ C2\tan \,\frac{C}{2} can be derived using the properties of A.P. and trigonometric identities.
35 Easy
35 of 434
Analyzing +4 | -1
In a triangle (1−r1r2)(1−r1r3)=2\left( {1 - \frac{{{r_1}}}{{{r_2}}}} \right)\left( {1 - \frac{{{r_1}}}{{{r_3}}}} \right) = 2, then the triangle is
View answer and explanation
Correct option: A
Right angled
The correct answer is A (Right angled) because the given condition simplifies to a2=b2+c2a^2 = b^2 + c^2, which is the Pythagorean theorem, indicating a right-angled triangle.
36 Easy
36 of 434
Application +4 | -1
In a △ABC,a=4,b=3,∠A=60∘\triangle A B C, a=4, b=3, \angle A=60^{\circ}, then cc is the root of the equation
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Correct option: A
c2−3c−7=0c^2-3 c-7=0
Option A (c2−3c−7=0c^2-3c-7=0) is correct because by applying the Law of Cosines and substituting the given values, we can derive the quadratic equation c2−3c−7=0c^2-3c-7=0.
37 Medium
37 of 434
Analyzing +4 | -1
In △ABC\triangle A B C, Σ(b+c)tan⁡A2tan⁡(B−C2)\Sigma(b+c) \tan \frac{A}{2} \tan \left(\frac{B-C}{2}\right) is equal to
View answer and explanation
Correct option: D
0
Option D (0) is correct because the expression simplifies to b-c + c-a + a-b, which equals zero due to the tangent half-angle and difference identities.
38 Medium
38 of 434
Application +4 | -1
In a Δ ABC\Delta \,ABC, if A=30∘,  b=2,  c=3+1A = 30^\circ ,\,\,b = 2,\,\,c = \sqrt 3 + 1, then C−B2=\frac{{C - B}}{2} =
View answer and explanation
Correct option: B
30∘30^\circ
Option B (30∘30^\circ) is correct because applying the tangent half-angle formula and solving for C−B2\frac{{C - B}}{2} yields 30∘30^\circ.
39 Medium
39 of 434
Application +4 | -1
In a △ABC,a=5,b=4\triangle A B C, a=5, b=4 and tan⁡C2=79\tan \frac{C}{2}=\sqrt{\frac{7}{9}}. The side cc is
View answer and explanation
Correct option: A
6
Option A (6) is correct because it is the solution obtained by applying the tangent half-angle formula and solving for side c.
40 Easy
40 of 434
Application +4 | -1
In ΔABC,  a=2,  b=4\Delta ABC,\;a = 2,\;b = 4 and ∠C=60∘,\angle C = 60^\circ ,, then ∠A\angle A and ∠B\angle B are equal to
View answer and explanation
Correct option: C
30∘,  90∘30^\circ ,\;90^\circ
Option C (30∘,  90∘30^\circ ,\;90^\circ) is correct because the angles A and B can be calculated using the Law of Cosines and the Law of Sines, given the sides a, b, and angle C.
41 Easy
41 of 434
Application +4 | -1
In a Δ  ABC\Delta \;ABC, if 2s=a+b+c2s = a + b + c, then the value of the s(s−a)bc−(s−b)(s−c)bc\frac{{s\left( {s - a} \right)}}{{bc}} - \frac{{\left( {s - b} \right)\left( {s - c} \right)}}{{bc}} is equal to
View answer and explanation
Correct option: B
cos⁡A\cos A
Option B (cos A) is correct because the given expression simplifies to cos²(A/2) - sin²(A/2), which is equal to cos A using trigonometric identities.
42 Easy
42 of 434
Application +4 | -1
If p1,p2,p3{p_1},{p_2},{p_3} are respectively the perpendiculars from the vertices of a triangle to the opposite sides, then cos⁡  Ap1  +  cos⁡  Bp2  +  cos⁡  Cp3\frac{{\cos \,\,A}}{{{p_1}}}\,\, + \,\,\frac{{\cos \,\,B}}{{{p_2}}}\,\, + \,\,\frac{{\cos \,\,C}}{{{p_3}}}&nbsp; is equal to
View answer and explanation
Correct option: B
1R\frac{1}{R}
Option B (1R\frac{1}{R}) is correct because the expression cos⁡Ap1+cos⁡Bp2+cos⁡Cp3\frac{{\cos A}}{{{p_1}}} + \frac{{\cos B}}{{{p_2}}} + \frac{{\cos C}}{{{p_3}}} simplifies to 1R\frac{1}{R} using trigonometric identities and the area of the triangle.
43 Easy
43 of 434
Application +4 | -1
If in a Δ  ABC,  2b2=a2+c2\Delta \;ABC,\;2{b^2} = {a^2} + {c^2}, then sin⁡3Bsin⁡B\frac{{\sin 3B}}{{\sin B}} is equal to
View answer and explanation
Correct option: D
(c2−a22ca)2{\left( {\frac{{{c^2} - {a^2}}}{{2ca}}} \right)^2}
Option D ((c2−a22ca)2{\left( {\frac{{{c^2} - {a^2}}}{{2ca}}} \right)^2}) is correct because it is the simplified form of sin⁡3Bsin⁡B\frac{{\sin 3B}}{{\sin B}} using the given condition and trigonometric identities.
44 Hard
44 of 434
Application +4 | -1
If a,b,ca,b,c are the sides of a tringle, then ab+c−a+bc+a−b+ca+b−c  \frac{a}{{b + c - a}} + \frac{b}{{c + a - b}} + \frac{c}{{a + b - c}}\; is
View answer and explanation
Correct option: B
≥3\ge 3
Option B ≥3\ge 3 is correct because it is derived using the AM-GM inequality after a suitable substitution.
45 Easy
45 of 434
Application +4 | -1
Two sides of a triangle are given by the roots of the equation x2−5x+6=0{x^2} - 5x + 6 = 0 and the angle between the sides is π3\frac{\pi }{3}. Then, the perimeter of the triangle is
View answer and explanation
Correct option: D
5+75 + \sqrt 7
Option D (5+75 + \sqrt{7}) is correct because it is the perimeter calculated using the cosine rule to find the third side of the triangle.

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46 Medium
46 of 434
Application +4 | -1
The area of a △ABC\triangle A B C is a2−(b−c)2a^2-(b-c)^2. Then, tan⁡A\tan A is equal to&nbsp;
View answer and explanation
Correct option: C
815\frac{8}{15}
Option C (815\frac{{8}}{{15}}) is correct because by using the area formula and trigonometric identities, we can find the value of tan A.
47 Easy
47 of 434
Application +4 | -1
If the sines of the angles of a triangle are in the ratio 4:5:64: 5: 6, then their cosines are in the ratio
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Correct option: B
12:9:212: 9: 2
Option B (12: 9: 2) is correct because using the cosine rule and the sine rule, the ratio of cosines can be found to be 12:9:2.
48 Medium
48 of 434
Application +4 | -1
If p1,p2,  p3{p_1},{p_2},\;{p_3} are respectively the perpendicular from the vertices of a triangle to the opposite sides, then p1p2p3{p_1}{p_2}{p_3} is equal to
View answer and explanation
Correct option: D
a2b2c28R2\frac{{{a^2}{b^2}{c^2}}}{{8{R^2}}}
Option D (a2b2c28R2\frac{{{a^2}{b^2}{c^2}}}{{8{R^2}}}) is correct because the product of the perpendiculars from the vertices to the opposite sides can be expressed in terms of the sides and the circumradius of the triangle.
49 Medium
49 of 434
Analyzing +4 | -1
The sides of a triangle are 3x+4y,  4x+3y3x + 4y,\,\,4x + 3y and 5x+5y5x + 5y, where, x, y>0x,\,y > 0 then the triangle is
View answer and explanation
Correct option: B
Obtuse angled
Option B (Obtuse angled) is correct because by using the cosine rule, we find that cos⁡C<0\cos C < 0, which means angle C is obtuse.
50 Medium
50 of 434
Application +4 | -1
In a Δ ABC\Delta \,ABC if a=5,  b=4a = 5,\,\,b = 4 and tan⁡C2=73\tan \frac{C}{2} = \frac{{\sqrt 7 }}{3}, then c=c =
View answer and explanation
Correct option: C
6
Option C (6) is correct because by using the Law of Cosines and the given value of tan⁡C2\tan \frac{C}{2}, we can find the value of c.
51 Easy
51 of 434
Application +4 | -1
In a △ABC,ab=2+3\triangle \mathrm{ABC}, \frac{\mathrm{a}}{\mathrm{b}}=2+\sqrt{3} and ∠C=60∘\angle \mathrm{C}=60^{\circ}. Then the ordered pair (∠A,∠B)(\angle \mathrm{A}, \angle \mathrm{B}) is equal to :
View answer and explanation
Correct option: B
(105∘,15∘)\left(105^{\circ}, 15^{\circ}\right)
Option B ((105°, 15°)) is correct because using the sine rule and the given ratio, the angles A and B can be found to be 105° and 15° respectively.
52 Easy
52 of 434
Analyzing +4 | -1
In a Δ  ABC\Delta \;ABC, among the following which one is true?
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Correct option: D
(b−c)cos⁡A2=asin⁡(B−C2)\left( {b - c} \right)\cos \frac{A}{2} = a\sin \left( {\frac{{B - C}}{2}} \right)
Option D ((b−c)cos⁡A2=asin⁡(B−C2)\left( {b - c} \right)\cos \frac{A}{2} = a\sin \left( {\frac{{B - C}}{2}} \right)) is correct because it is derived from sine rule and trigonometric identities.
53 Medium
53 of 434
Analyzing +4 | -1
If in a Δ  ABC,r3=r1+r2+r\Delta \;ABC,{r_3} = {r_1} + {r_2} + r, then ∠A+∠B\angle A + \angle B is equal to
View answer and explanation
Correct option: C
90∘90^\circ
Option C (90∘90^\circ) is correct because the given condition r3=r1+r2+r{r_3} = {r_1} + {r_2} + r leads to C=90∘C = 90^\circ, and therefore A+B=90∘A + B = 90^\circ.
54 Easy
54 of 434
Analyzing +4 | -1
If a,  b,  ca,\;b,\;c be the sides of a Δ  ABC\Delta \;ABC and if roots of the equation a(b−c)x2+b(c−a)x+c(a−b)=0a\left( {b - c} \right){x^2} + b\left( {c - a} \right)x + c\left( {a - b} \right) = 0 are equal, then sin⁡2(A2),sin⁡2(B2),sin⁡2(C2){\sin ^2}\left( {\frac{A}{2}} \right),{\sin ^2}\left( {\frac{B}{2}} \right),{\sin ^2}\left( {\frac{C}{2}} \right) are in
View answer and explanation
Correct option: C
HP
Option C (HP) is correct because if the roots of the given equation are equal, then a, b, and c are in HP, which leads to sin⁡2(A2),sin⁡2(B2),sin⁡2(C2){\sin ^2}\left( {\frac{A}{2}} \right),{\sin ^2}\left( {\frac{B}{2}} \right),{\sin ^2}\left( {\frac{C}{2}} \right) being in HP.
55 Medium
55 of 434
Analyzing +4 | -1
In a △ABC,r1bc+r2ca+r3ab\triangle A B C, \frac{r_1}{b c}+\frac{r_2}{c a}+\frac{r_3}{a b} is equal to
View answer and explanation
Correct option: D
1r−12R\frac{1}{r}-\frac{1}{2 R}
Option D (1r−12R\frac{1}{r}-\frac{1}{2 R}) is correct because by using trigonometric identities and the definitions of inradius and circumradius, the expression can be simplified to 1r−12R\frac{1}{r}-\frac{1}{2 R}.
56 Medium
56 of 434
Application +4 | -1
If in a Δ  ABC,cos⁡A+2cos⁡B+cos⁡C=2\Delta \;ABC,\cos A + 2\cos B + \cos C = 2, then a,  b,  ca,\;b,\;c are in
View answer and explanation
Correct option: A
AP
Option A (AP) is correct because the given condition leads to the relationship sin⁡A+sin⁡C=2sin⁡B\sin A + \sin C = 2\sin B, which implies that a, b, and c are in arithmetic progression.
57 Medium
57 of 434
Application +4 | -1
In △ABC(a+b)(a−b)=c(b+c)\triangle A B C(a+b)(a-b)=c(b+c), the measure of angle AA, is-
View answer and explanation
Correct option: D
120∘120^{\circ}
Option D (120∘120^{\circ}) is correct because by applying the law of cosines and simplifying the given equation, we find that cos⁡A=−1/2\cos A = -1/2, which corresponds to an angle of 120∘120^{\circ}.
58 Easy
58 of 434
Application +4 | -1
In Δ ABC\Delta \,ABC, if s−aΔ=18,  s−bΔ=112\frac{{s - a}}{\Delta } = \frac{1}{8},\,\,\frac{{s - b}}{\Delta } = \frac{1}{{12}} and s−cΔ=124\frac{{s - c}}{\Delta } = \frac{1}{{24}}, then b=b =
View answer and explanation
Correct option: A
16\mathrm{16}
Option A (16) is correct because it is derived from the given ratios involving the semi-perimeter, inradius, and area of the triangle.
59 Easy
59 of 434
Application +4 | -1
Given b=2,c=3,∠A=30∘b=2, c=\sqrt{3}, \angle A=30^{\circ}. Then the in-radius of △ABC\triangle A B C is
View answer and explanation
Correct option: A
3−12\frac{\sqrt{3}-1}{2}
Option A (3−12\frac{\sqrt{3}-1}{2}) is correct because the in-radius can be calculated using the formula r=Δsr = \frac{\Delta}{s} after finding the area and semi-perimeter.
60 Easy
60 of 434
Application +4 | -1
In a Δ  ABC,(b+c)(bc)cos⁡A+(a+c)(ac)cos⁡B+(a+b)(ab)cos⁡C\Delta \;ABC,\left( {b + c} \right)\left( {bc} \right)\cos A + \left( {a + c} \right)\left( {ac} \right)\cos B + \left( {a + b} \right)\left( {ab} \right)\cos C is
View answer and explanation
Correct option: B
a3+b3+c3{a^3} + {b^3} + {c^3}
Option B (a3+b3+c3{a^3} + {b^3} + {c^3}) is correct because the given expression simplifies to this form using the cosine rule.
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