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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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31 Easy
31 of 287
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.
32 Easy
32 of 287
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.
33 Easy
33 of 287
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.
34 Easy
34 of 287
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.
35 Easy
35 of 287
Application +4 | -1
If the area of a △ABC\triangle A B C be λ\lambda, then a2sin⁡2B+b2sin⁡2Aa^2 \sin 2 B+b^2 \sin 2 A is equal to
View answer and explanation
Correct option: C
4λ4 \lambda
Option C (4λ4 \lambda) is correct because using trigonometric identities and the sine rule, the expression simplifies to four times the area of the triangle.
36 Easy
36 of 287
Application +4 | -1
In any ΔABC\Delta ABC under usual notation, a(bcos⁡C−ccos⁡B)a\left( {b\cos C - c\cos B} \right) is equal to
View answer and explanation
Correct option: A
b2−c2{b^2} - {c^2}
Option A (b2−c2{b^2} - {c^2}) is correct because using the cosine rule, the expression simplifies to b2−c2{b^2} - {c^2}.
37 Easy
37 of 287
Application +4 | -1
If the sides a,  ba,\,\,b and cc of a Δ ABC\Delta \,ABC are in A.P., then
(tan⁡  A2+tan⁡  C2):cot⁡  B2\left( {\tan \,\,\frac{A}{2} + \tan \,\,\frac{C}{2}} \right):\cot \,\,\frac{B}{2}
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the ratio simplifies to 2/3, which is not among the given options.
38 Easy
38 of 287
Application +4 | -1
A △ABC\triangle A B C is right angled at BB. Then, the diameter of the incircle of the triangle is
View answer and explanation
Correct option: C
c+a−bc+a-b
Option C (c+a−bc+a-b) is correct because it represents the diameter (2r) of the incircle of a right-angled triangle.
39 Easy
39 of 287
Application +4 | -1
In Δ  ABC,1+cos⁡(A−B)cos⁡C1+cos⁡(A−C)cos⁡B\Delta \;ABC,\frac{{1 + \cos \left( {A - B} \right)\cos C}}{{1 + \cos \left( {A - C} \right)\cos B}} is equal to
View answer and explanation
Correct option: D
a2+b2a2+c2\frac{{{a^2} + {b^2}}}{{{a^2} + {c^2}}}
Option D ((a²+b²)/(a²+c²)) is correct because the given expression simplifies to (sin²A + sin²B)/(sin²A + sin²C), which is equal to (a²+b²)/(a²+c²) by the sine rule.
40 Easy
40 of 287
Analyzing +4 | -1
In Δ  ABC,ifsin⁡2A2,sin⁡2B2,sin⁡2C2\Delta \;ABC,\mathrm{if}{\sin ^2}\frac{A}{2},{\sin ^2}\frac{B}{2},{\sin ^2}\frac{C}{2} be in HP. Then, a,b,ca,b,c will be in
View answer and explanation
Correct option: C
HP
Option C (HP) is correct because the reciprocals of the given terms are in AP, leading to a, b, c being in HP.
41 Easy
41 of 287
Application +4 | -1
If in a ΔABC,  a=6\Delta ABC,\;a = 6 cm, b=8b = 8 cm, c=10c = 10 cm, then the value of sin⁡2A\sin 2A is
View answer and explanation
Correct option: D
24/2524/25
Option D (24/25) is correct because using the cosine rule to find cos A, then sin A, and finally sin 2A gives 24/25.
42 Easy
42 of 287
Application +4 | -1
The circumradius of the triangle whose sides are 13, 12 and 5, is
View answer and explanation
Correct option: B
132\frac{{13}}{2}
Option B (132\frac{{13}}{2}) is correct because the triangle is right-angled, and the circumradius is half the length of the hypotenuse.
43 Easy
43 of 287
Application +4 | -1
In a Δ  ABC,  a(bcos⁡C−ccos⁡B)\Delta \;ABC,\;a\left( {b\cos C - c\cos B} \right) is equal to
View answer and explanation
Correct option: B
b2−c2{b^2} - {c^2}
Option B b2−c2{b^2} - {c^2} is correct because it is derived from applying the cosine rule and simplifying the expression.
44 Easy
44 of 287
Application +4 | -1
The greatest angle of a triangle whose sides are x2+x+1,2x+1x^2+x+1,2 x+1 and x2−1x^2-1, is
View answer and explanation
Correct option: C
120∘120^{\circ}
Option A (120∘120^{\circ}) is correct because the greatest angle is found using the cosine rule, with x2+x+1x^2+x+1 being the longest side.
45 Easy
45 of 287
Application +4 | -1
If the angles of a triangle are in the ratio 1  :2  :31\;:2\;:3, the corresponding sides are in the ratio

View answer and explanation
Correct option: D
1  :3  :21\;:\sqrt 3 \;:2
Option D (1:3:21:\sqrt 3 :2) is correct because the ratio of sides is the same as the ratio of sines of the angles, and the angles are 30, 60, and 90 degrees.
46 Easy
46 of 287
Application +4 | -1
In a ΔABC,acot⁡A+bcot⁡B+ccot⁡C\Delta ABC,a\cot A + b\cot B + c\cot C is equal to
View answer and explanation
Correct option: C
2(r+R)2\left( {r + R} \right)
Option C (2(r+R)2\left( {r + R} \right)) is correct because acot⁡A+bcot⁡B+ccot⁡Ca\cot A + b\cot B + c\cot C can be simplified to 2R(cos⁡A+cos⁡B+cos⁡C)2R(\cos A + \cos B + \cos C), which equals 2(r+R)2(r+R).
47 Easy
47 of 287
Application +4 | -1
In Δ  ABC,if(a+b+c)(a−b+c)=3ac\Delta \;ABC,\mathrm{if}\left( {a + b + c} \right)\left( {a - b + c} \right) = 3ac, then
View answer and explanation
Correct option: A
∠B=60∘\angle B = 60^\circ
Option A (∠B=60∘\angle B = 60^\circ) is correct because by simplifying the given equation and using the cosine rule, we find that angle B is 60 degrees.
48 Easy
48 of 287
Application +4 | -1
In a Δ  ABC\Delta \;ABC, if b2+c2=3a2{b^2} + {c^2} = 3{a^2}, then cot⁡B+cot⁡C−cot⁡A\cot B + \cot C - \cot A is equal to
View answer and explanation
Correct option: C
0
Option C (0) is correct because substituting the given condition b² + c² = 3a² into the expression and simplifying leads to 0.
49 Easy
49 of 287
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, then 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 leads to one angle being 90 degrees.
50 Easy
50 of 287
Application +4 | -1
The vertices of a triangle are A(1,1),B(4,5)A(1,1), B(4,5) and C(6,13)C(6,13). Then, the value of cos⁡A\cos A is
View answer and explanation
Correct option: A
6365\frac{63}{65}
Option A (6365\frac{63}{65}) is correct because it is the cosine of angle A, calculated using the cosine rule with the given vertices of the triangle.
51 Easy
51 of 287
Analyzing +4 | -1
The ex-radii of a triangle r1,  r2,  r3{r_1},\,\,{r_2},\,\,{r_3} are in harmonic progression, then the sides a,  b,  ca,\,\,b,\,\,c are
View answer and explanation
Correct option: B
In A.P.
Option B (In A.P.) is correct because the harmonic progression of ex-radii implies an arithmetic progression of the sides.
52 Easy
52 of 287
Application +4 | -1
If x,y,zx,y,z are perpendicular drawn from the vertices of triangle having sides a,ba,b and cc, then the value of bxc+cya+azb\frac{{bx}}{c} + \frac{{cy}}{a} + \frac{{az}}{b} will be
View answer and explanation
Correct option: A
a2+b2+c22R\frac{{{a^2} + {b^2} + {c^2}}}{{2R}}
Option A (a2+b2+c22R\frac{{{a^2} + {b^2} + {c^2}}}{{2R}}) is correct because it is derived from the area of the triangle and the circumradius.
53 Easy
53 of 287
Application +4 | -1
In a △ABC,R=\triangle A B C, R= circumradius and r=r= inradius. The value of acos⁡A+bcos⁡B+ccos⁡Ca+b+c\frac{a \cos A+b \cos B+c \cos C}{a+b+c} is equal to
View answer and explanation
Correct option: C
rR\frac{r}{R}
Option C (r/Rr/R) is correct because by using trigonometric identities and the definitions of inradius and circumradius, the given expression simplifies to r/Rr/R.
54 Easy
54 of 287
Application +4 | -1
In a Δ ABC\Delta \,ABC, if a,  c,  ba,\,\,c,\,\,b are in A.P., then the value of a  cos⁡  B−b  cos⁡  Aa−b\frac{{a\,\,\cos \,\,B - b\,\,\cos \,\,A}}{{a - b}}, is
View answer and explanation
Correct option: B
2\mathrm{2}
Option B (2) is correct because the value of the expression simplifies to 2 using the cosine rule and the given A.P. condition.
55 Easy
55 of 287
Application +4 | -1
In a ΔABC\Delta ABC, (a+b+c)(b+c−a)(c+a−b)(a+b−c)4b2c2\frac{{\left( {a + b + c} \right)\left( {b + c - a} \right)\left( {c + a - b} \right)\left( {a + b - c} \right)}}{{4{b^2}{c^2}}} equals
View answer and explanation
Correct option: C
sin⁡2A{\sin ^2}A
Option C (sin⁡2A{\sin ^2}A) is correct because the given expression can be simplified using the semi-perimeter formula and trigonometric identities to obtain sin⁡2A{\sin ^2}A.
56 Easy
56 of 287
Application +4 | -1
In a triangle ABC,cos⁡Aa=cos⁡Bb=cos⁡CcABC,\frac{{\cos A}}{a} = \frac{{\cos B}}{b} = \frac{{\cos C}}{c}. if a=16a = \frac{1}{{\sqrt 6 }}, then the area of the triangle (in square unit) is
View answer and explanation
Correct option: B
3/24\sqrt 3 /24
Option B (3/24\sqrt 3 /24) is correct because if cos⁡Aa=cos⁡Bb=cos⁡Cc\frac{{\cos A}}{a} = \frac{{\cos B}}{b} = \frac{{\cos C}}{c}, then the triangle is equilateral, and the area can be calculated using the side length a=16a = \frac{1}{{\sqrt 6 }}.
57 Easy
57 of 287
Application +4 | -1
Consider a △ABC\triangle A B C and let a,ba, b and cc denote the lengths of the sides opposite to vertices A,BA, B and CC, respectively. Suppose a=6,b=10a=6, b=10 and the area of the triangle is 15315 \sqrt{3}. If ∠ACB\angle A C B is obtuse and if rr denotes the radius of the incircle of the triangle, then r2r^2 is equal to
View answer and explanation
Correct option: A
3
Option A (3) is correct because by using the area of the triangle and the given side lengths, we can find the third side and then calculate the inradius rr and r2r^2.
58 Easy
58 of 287
Application +4 | -1
If ΔABC,ifa=3,b=4,c=5\Delta ABC,\mathrm{if}a = 3,b = 4,c = 5, then the value of sin⁡2B\sin 2B is
View answer and explanation
Correct option: C
24/2524/25
Option C (2425\frac{{24}}{{25}}) is correct because it is the result of correctly applying the double angle formula for sine, sin⁡2B=2sin⁡Bcos⁡B\sin 2B = 2 \sin B \cos B, after finding sin⁡B\sin B and cos⁡B\cos B using the given side lengths.
59 Easy
59 of 287
Analyzing +4 | -1
If the sides of a triangle are x2+x+1,  x2−1,  2x+1{x^2} + x + 1,\,\,{x^2} - 1,\,\,2x + 1, where x>1x > 1, then the largest angle is
View answer and explanation
Correct option: D
30∘\mathrm{30^\circ }
Option A (120°) is correct because by using the cosine rule, the largest angle opposite the largest side is found to be 120 degrees.
60 Easy
60 of 287
Application +4 | -1
In a Δ ABC\Delta \,ABC, if B=90∘B = 90^\circ, then the value of tan⁡A2\tan \frac{A}{2} in terms of the sides is
View answer and explanation
Correct option: B
b−cb+c\sqrt {\frac{{b - c}}{{b + c}}}
Option B (b−cb+c\sqrt {\frac{{b - c}}{{b + c}}}) is correct because it is derived from trigonometric identities and the given condition B = 90°.
Revision notes and key takeaways

Review PROPERTIES OF TRIANGLES in your textbook or class notes, then use this set to check your understanding.

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