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JEE • MATHEMATICS • PROPERTIES OF TRIANGLES

Properties of triangles practice · Easy

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Practice format and chapter quiz

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 1-30 of 287 questions
1 Easy
1 of 287
Application +4 | -1
If p1,p2,p3{p_1},{p_2},{p_3} are altitudes of a Δ  ABC\Delta \;ABC drawn from the vertices A,B,CA,B,C and Δ\Delta the area of the triangle, then p1−2+p2−2+p3−2p_1^{ - 2} + p_2^{ - 2} + p_3^{ - 2} is equal to
View answer and explanation
Correct option: B
a2+b2+c24Δ2\frac{{{a^2} + {b^2} + {c^2}}}{{4{\Delta ^2}}}
Option B a2+b2+c24Δ2\frac{{{a^2} + {b^2} + {c^2}}}{{4{\Delta ^2}}} is correct because it is derived from the relationship between altitudes and the area of the triangle.
2 Easy
2 of 287
Application +4 | -1
If the sides of the triangles are 5k,  6k,  5k5k,\;6k,\;5k and radius of incircle is 6, the value of kk is equal to
View answer and explanation
Correct option: A
4
Option A (4) is correct because by using Heron's formula to find the area of the triangle and then using the formula for the inradius, we can solve for kk.
3 Easy
3 of 287
Application +4 | -1
In a Δ ABC\Delta \,ABC, if a=(b−c)sec⁡θa = \left( {b - c} \right)\sec \theta, then 2 bcb−c  sin⁡  A2=\frac{{2\,\sqrt {bc} }}{{b - c}}\,\,\sin \,\,\frac{A}{2} =
View answer and explanation
Correct option: C
tan⁡  θ\tan \,\,\theta
Option C (tan⁡  θ\tan \,\,\theta) is correct because by substituting a=(b−c)sec⁡θa = (b - c)\sec \theta into the cosine rule and simplifying, we find that 2 bcb−c  sin⁡  A2=tan⁡  θ\frac{{2\,\sqrt {bc} }}{{b - c}}\,\,\sin \,\,\frac{A}{2} = \tan \,\,\theta.
4 Easy
4 of 287
Application +4 | -1
If Δ=a2−(b−c)2\Delta = {a^2} - {\left( {b - c} \right)^2}, where Δ\Delta is the area of Δ  ABC\Delta \;ABC, then tan⁡A\tan A is equal to
View answer and explanation
Correct option: C
815\frac{8}{{15}}
Option C (815\frac{8}{{15}}) is correct because by using the given condition Δ=a2−(b−c)2\Delta = {a^2} - {\left( {b - c} \right)^2} and trigonometric identities, we can find that tan⁡A=815\tan A = \frac{8}{{15}}.
5 Easy
5 of 287
Application +4 | -1
In a triangle ABCA B C, angles are in AP and b:c=3:2b: c=\sqrt{3}: \sqrt{2}. Then the angle AA is
View answer and explanation
Correct option: B
75∘75^{\circ}
Option B (75∘75^{\circ}) is correct because using the given ratio of sides and the fact that angles are in AP allows us to determine angle C as 45 degrees, leading to angle A being 75 degrees.
6 Easy
6 of 287
Application +4 | -1
If in a Δ  ABC,1a+c+1b+c=3a+b+c\Delta \;ABC,\frac{1}{{a + c}} + \frac{1}{{b + c}} = \frac{3}{{a + b + c}}, then the value of the angle CC is
View answer and explanation
Correct option: A
60∘60^\circ
Option A (60∘60^\circ) is correct because the given equation leads to cos⁡C=12\cos C = \frac{1}{2}, which implies C=60∘C = 60^\circ.
7 Easy
7 of 287
Application +4 | -1
If PQPQ and PRPR are the two sides of a triangle, then the angle between them which gives maximum area of the triangle, is
View answer and explanation
Correct option: D
π2\frac{\pi }{2}
Option D (2a2a) is correct because the length of the subnormal of the parabola y2=4ax{y^2} = 4ax is constant and equal to 2a2a.
8 Easy
8 of 287
Application +4 | -1
In a triangle, 1−tan⁡A2tan⁡B2=1-\tan \frac{A}{2} \tan \frac{B}{2}=
View answer and explanation
Correct option: B
2ca+b+c\frac{2 c}{a+b+c}
Option B (2ca+b+c\frac{2 c}{a+b+c}) is correct because it is the simplified form of the given expression using the properties of a triangle.
9 Easy
9 of 287
Application +4 | -1
If the sides of a triangle are 7  cm,  4 3  cm7\;\mathrm{cm},\,\,4\,\sqrt 3 \mathrm{\;cm} and 13    cm\sqrt {13} \,\,\;\mathrm{cm}, then the smallest angle of the triangle is
View answer and explanation
Correct option: C
30∘30^\circ
Option C (30∘30^\circ) is correct because it is the smallest angle calculated using the Law of Cosines with the given side lengths.
10 Easy
10 of 287
Application +4 | -1
In a Δ  ABC,b−ccos⁡Ac−bcos⁡A\Delta \;ABC,\frac{{b - c\cos A}}{{c - b\cos A}} is equal to
View answer and explanation
Correct option: B
cos⁡Ccos⁡B\frac{{\cos C}}{{\cos B}}
Option B (cos⁡Ccos⁡B\frac{{\cos C}}{{\cos B}}) is correct because the expression can be simplified using the Law of Cosines and the Law of Sines.
11 Easy
11 of 287
Application +4 | -1
In a △ABC,2s=\triangle A B C, 2 s= perimeter and R=R= circumradius. Then, SR\frac{S}{R} is equal to
View answer and explanation
Correct option: A
sin⁡A+sin⁡B+sin⁡C\sin A+\sin B+\sin C
Option A (sin⁡A+sin⁡B+sin⁡C\sin A + \sin B + \sin C) is correct because using the sine rule, s/Rs/R can be expressed as the sum of sines of the angles.
12 Easy
12 of 287
Application +4 | -1
If two angles of Δ  ABC\Delta \;ABC are 45∘45^\circ and 60∘60^\circ, then the ratio of the smallest and the greatest sides are
View answer and explanation
Correct option: A
(3−1):1\left( {\sqrt 3 - 1} \right):1
Option A ((3−1):1\left( {\sqrt 3 - 1} \right):1) is correct because it represents the ratio of the smallest to the greatest side, calculated using the sine rule.
13 Easy
13 of 287
Analyzing +4 | -1
Let sin⁡Asin⁡B=sin⁡(A−C)sin⁡(C−B)\frac{{\sin A}}{{\sin B}} = \frac{{\sin (A - C)}}{{\sin (C - B)}}, where A,BA,B and CC are angles of a ΔABC\Delta ABC. If the lengths of the sides opposite these angles are a,ba,b and cc repectively, then
View answer and explanation
Correct option: B
b2,c2   and   a2{b^2},{c^2}\,\,\,\mathrm{and}\,\,\,{a^2} are in AP
Option C (b2,c2   and   a2{b^2},{c^2}\,\,\,\mathrm{and}\,\,\,{a^2} are in AP) is correct because the given condition leads to the relationship 2c2=a2+b22c^2 = a^2 + b^2, which defines an arithmetic progression.
14 Easy
14 of 287
Application +4 | -1
If in a triangle ABC,  cos⁡  Aa  =cos⁡  Bb=cos⁡  CcABC,\,\,\frac{{\cos \,\,A}}{a}\,\, = \frac{{\cos \,\,B}}{b} = \frac{{\cos \,\,C}}{c}, then the triangle is
View answer and explanation
Correct option: C
Equilateral
Option C (Equilateral) is correct because if cos⁡ Aa =cos⁡ Bb=cos⁡ Cc\frac{{\cos \,A}}{a}\, = \frac{{\cos \,B}}{b} = \frac{{\cos \,C}}{c}, then A = B = C, which implies the triangle is equilateral.
15 Easy
15 of 287
Application +4 | -1
Two sides of a triangle are 2 2  cm2\,\sqrt 2 \;\mathrm{cm} and 2 3   cm2\,\sqrt 3 \;\,\mathrm{cm} and the angle opposite to the shorter side of the two is π4\frac{\pi }{4}. The largest possible length of the third side is
View answer and explanation
Correct option: A
(6+2)  cm\left( {\sqrt 6 + \sqrt 2 } \right)\;\mathrm{cm}
The correct answer is None of these because the expression simplifies to cos⁡ 11∘\cos \,11^\circ, which is not among the given options.

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16 Easy
16 of 287
Application +4 | -1
If a=2,  b=3,  c=5a = 2,\,\,b = 3,\,\,c = 5 in Δ ABC\Delta \,ABC, then C=C =
View answer and explanation
Correct option: D
None of these
The correct answer is D (None of these) because the calculated value of C is \π\pi, which is impossible in a triangle.
17 Easy
17 of 287
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.
18 Easy
18 of 287
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.
19 Easy
19 of 287
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.
20 Easy
20 of 287
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.
21 Easy
21 of 287
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.
22 Easy
22 of 287
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
View answer and explanation
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.
23 Easy
23 of 287
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.
24 Easy
24 of 287
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.
25 Easy
25 of 287
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.
26 Easy
26 of 287
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.
27 Easy
27 of 287
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.
28 Easy
28 of 287
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
View answer and explanation
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.
29 Easy
29 of 287
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.
30 Easy
30 of 287
Analyzing +4 | -1
In a Δ  ABC\Delta \;ABC, among the following which one is true?
View answer and explanation
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.
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