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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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91 Easy
91 of 434
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.
92 Easy
92 of 434
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 }}.
93 Easy
93 of 434
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.
94 Medium
94 of 434
Analyzing +4 | -1
In a Δ  ABC,(cot⁡A2+cot⁡B2)(asin⁡2B2+bsin⁡2A2)\Delta \;ABC,\left( {\cot \frac{A}{2} + \cot \frac{B}{2}} \right)\left( {a{{\sin }^2}\frac{B}{2} + b{{\sin }^2}\frac{A}{2}} \right) is equal to
View answer and explanation
Correct option: D
ccot⁡C2c\cot \frac{C}{2}
Option D (ccot⁡C2c\cot \frac{C}{2}) is correct because after simplification using trigonometric identities, the expression reduces to this form.
95 Easy
95 of 434
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.
96 Easy
96 of 434
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
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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.
97 Easy
97 of 434
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°.
98 Easy
98 of 434
Analyzing +4 | -1
If in a △ABC,atan⁡A+btan⁡B=(a+b)tan⁡(A+B)2\triangle A B C, a \tan A+b \tan B=(a+b) \tan \frac{(A+B)}{2}, then
View answer and explanation
Correct option: C
A=BA=B
Option C (A=BA=B) is correct because the given equation simplifies to tan⁡A=tan⁡B\tan A = \tan B, implying A=BA = B.
99 Easy
99 of 434
Application +4 | -1
In Δ  ABC,(a+b+c)(tan⁡A2+tan⁡B2)\Delta \;ABC,\left( {a + b + c} \right)\left( {\tan \frac{A}{2} + \tan \frac{B}{2}} \right) is equal to
View answer and explanation
Correct option: A
2c  cot⁡C22c\;\cot \frac{C}{2}
Option A (2c  cot⁡C22c\;\cot \frac{C}{2}) is correct because by using trigonometric identities and the area of a triangle formula, the expression simplifies to this form.
100 Easy
100 of 434
Application +4 | -1
Three poles whose feet lie on a circle subtend angle α,β,γ\alpha, \beta, \gamma respectively at the centre of the circle. If the height of the poles are in A.P. then cot⁡α,cot⁡β,cot⁡γ\cot \alpha, \cot \beta, \cot \gamma are in
View answer and explanation
Correct option: C
H.P.
Option C (H.P.) is correct because if the heights are in A.P., then their cotangents are in H.P. due to the relationship between height, radius, and angle.
101 Easy
101 of 434
Application +4 | -1
If A,  A1,A2,  A3A,\;{A_1},{A_2},\;{A_3} be the areas of the incircle and excircles, then 1A1+1A2+1A3\frac{1}{{\sqrt {{A_1}} }} + \frac{1}{{\sqrt {{A_2}} }} + \frac{1}{{\sqrt {{A_3}} }} is equal to
View answer and explanation
Correct option: A
1A\frac{1}{{\sqrt A }}
Option A 1A\frac{1}{{\sqrt A }} is correct because it is derived from the relationship between the inradius, exradii, and the area of the triangle.
102 Easy
102 of 434
Application +4 | -1
In a △ABC,2casin⁡(A−B+C2)\triangle A B C, 2 c a \sin \left(\frac{A-B+C}{2}\right) is equal to
View answer and explanation
Correct option: B
c2+a2−b2c^2+a^2-b^2
Option B (c2+a2−b2c^2+a^2-b^2) is correct because by simplifying the trigonometric expression and using the Law of Cosines, we can show that 2casin⁡(A−B+C2)=c2+a2−b22 c a \sin \left(\frac{A-B+C}{2}\right) = c^2+a^2-b^2.
103 Easy
103 of 434
Application +4 | -1
If a=22,  b=6,  A=45∘a = 2\sqrt 2 ,\;b = 6,\;A = 45^\circ, then
View answer and explanation
Correct option: A
No triangle is possible
Option A (No triangle is possible) is correct because the calculated value of sin B is greater than 1, which is not possible.
104 Easy
104 of 434
Application +4 | -1
In a ΔABC\Delta ABC, if 2s=a+b+c2s = a + b + c and (s−b)(s−c)=x  sin⁡2A2\left( {s - b} \right)\left( {s - c} \right) = x\;{\sin ^2}\frac{A}{2}, then the value of xx is
View answer and explanation
Correct option: A
bcbc
Option A (bcbc) is correct because by using the formula for sin⁡A2\sin \frac{A}{2} and comparing it with the given equation, we find that x=bcx = bc.
105 Easy
105 of 434
Application +4 | -1
The radii r1,r2,r3r_1, r_2, r_3 of escribed circles of the △ABC\triangle A B C are in HP. If its area is 24sqcm24 \mathrm{sq} \mathrm{cm} and its perimeter is 24 cm24 \mathrm{~cm}, then the lengths of its sides are
View answer and explanation
Correct option: C
6,8,106,8,10
Option C (6,8,10) is correct because these side lengths form an arithmetic progression and satisfy the area and perimeter conditions.
106 Easy
106 of 434
Application +4 | -1
In a triangle ABC,tan⁡A−B2cot⁡A+B2A B C, \tan \frac{A-B}{2} \cot \frac{A+B}{2} is equal to
View answer and explanation
Correct option: C
a−ba+b\frac{a-b}{a+b}
Option C (a−ba+b\frac{a-b}{a+b}) is correct because using Napier's analogy, tan⁡A−B2=a−ba+bcot⁡C2\tan \frac{A-B}{2} = \frac{a-b}{a+b} \cot \frac{C}{2}, and since A+B+C=πA+B+C = \pi, cot⁡A+B2=tan⁡C2\cot \frac{A+B}{2} = \tan \frac{C}{2}.
107 Easy
107 of 434
Understanding +4 | -1
If in Δ   ABC,   a  sin⁡  A=b  sin⁡  B\Delta \,\;ABC,\;\,a\,\,\sin \,\,A = b\,\,\sin \,\,B, then the triangle is

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Correct option: A
Isosceles
Option A (Isosceles) is correct because if a sin A = b sin B, then by the sine rule, a²k = b²k, implying a = b, which defines an isosceles triangle.
108 Medium
108 of 434
Application +4 | -1
In a triangle ABC,  cos⁡  A+cos⁡  B+cos⁡  C=ABC,\,\,\cos \,\,A + \cos \,\,B + \cos \,\,C =
View answer and explanation
Correct option: A
1+rR1 + \frac{r}{R}
Option A (1+rR1 + \frac{r}{R}) is correct because the expression cos⁡A+cos⁡B+cos⁡C\cos A + \cos B + \cos C can be rewritten in terms of the inradius rr and circumradius RR as 1+rR1 + \frac{r}{R}.
109 Easy
109 of 434
Application +4 | -1
In a Δ ABC,  a=2b\mathrm{\Delta }\,ABC,\,\,a = 2b and A=3BA = 3B, the A=A =
View answer and explanation
Correct option: A
90∘90^\circ
Option A (90∘90^\circ) is correct because by applying the sine rule and trigonometric identities, we find that A = 90 degrees.
110 Medium
110 of 434
Analyzing +4 | -1
If in a △ABC\triangle A B C, cos⁡Acos⁡B+sin⁡Asin⁡Bsin⁡C=1\cos A \cos B+\sin A \sin B \sin C=1, then a:b:ca: b: c is equal to
View answer and explanation
Correct option: A
1:1:21: 1: \sqrt{2}
Option A (1:1:21: 1: \sqrt{{2}}) is correct because the given equation implies that the triangle is right-angled and isosceles, leading to the specified side ratio.
111 Medium
111 of 434
Application +4 | -1
The area of the triangle inscribed in a circle of radius of 4 and the measures of whose angles are in the ratio 5:4:35:4:3 is
View answer and explanation
Correct option: A
4(3+3)4(3+\sqrt{3})
Option A (4(3+3)4(3+\sqrt{3})) is correct because the area of the triangle can be calculated using the formula 12absin⁡C\frac{1}{2}ab\sin C after finding the angles and sides.
112 Medium
112 of 434
Application +4 | -1
In a △ABC,cos⁡C+cos⁡Ac+a+cos⁡Bb\triangle A B C, \frac{\cos C+\cos A}{c+a}+\frac{\cos B}{b} is equal to
View answer and explanation
Correct option: B
1b\frac{1}{b}
Option B (1b\frac{1}{b}) is correct because it simplifies to 1b\frac{1}{b} using projection formulas after substituting the cosine values.
113 Easy
113 of 434
Analyzing +4 | -1
In a ΔABC,(a+b+c)(b+c−a)=kbc\Delta ABC,\left( {a + b + c} \right)\left( {b + c - a} \right) = kbc, if
View answer and explanation
Correct option: C
0<k<40 < k < 4
Option C (0<k<40 < k < 4) is correct because it is derived from the trigonometric identity relating the given expression to cos⁡2(A/2)\cos^2(A/2) and the range of cosine squared.
114 Easy
114 of 434
Application +4 | -1
In a △ABC,a=8 cm,b=10 cm,c=12 cm\triangle A B C, a=8 \mathrm{~cm}, b=10 \mathrm{~cm}, c=12 \mathrm{~cm}. Then, relation between angles of the triangle is
View answer and explanation
Correct option: C
C=2AC=2 A
Option C (C=2AC=2A) is correct because by using the cosine rule, we find that cos(C) = cos(2A), which implies C = 2A.
115 Easy
115 of 434
Application +4 | -1
The sides of a triangle are three consecutive natural numbers and its largest angle is twice the smallest one. Then, the sides of the triangle are
View answer and explanation
Correct option: D
4, 5, 6
Option D (4, 5, 6) is correct because by using the sine rule and the given condition C = 2A, we can derive a relationship between the sides and solve for the consecutive natural numbers.
116 Medium
116 of 434
Application +4 | -1
In a △ABC,a,b,c,A\triangle A B C, a, b, c, A are given and c1,c2c_1, c_2 are two values of the third side cc. The sum of the areas of two triangles with sides a,b,c1a, b, c_1 and a,b,c2a, b, c_2 is
View answer and explanation
Correct option: A
12b2sin⁡2A\frac{1}{2} b^2 \sin 2 A
Option A (12b2sin⁡2A\frac{1}{2} b^2 \sin 2A) is correct because the sum of the areas can be expressed in terms of bb and AA using trigonometric identities.
117 Easy
117 of 434
Application +4 | -1
In a Δ ABC\Delta \,ABC if a=26,  b=30a = 26,\,\,b = 30 and cos⁡  C=6365\cos \,\,C = \frac{{63}}{{65}}, then r2={r_2} =
View answer and explanation
Correct option: C
48\mathrm{48}
Option C (48) is correct because by using the cosine rule to find c and then using Heron's formula to find the area, r2r_2 can be calculated as Δs−b\frac{\Delta}{s-b}.
118 Easy
118 of 434
Application +4 | -1
In a Δ  ABC\Delta \;ABC, if a,  b,  ca,\;b,\;c are in AP, then the value of sin⁡A2sin⁡C2sin⁡B2\frac{{\sin \frac{A}{2}\sin \frac{C}{2}}}{{\sin \frac{B}{2}}} is
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B 12\frac{1}{2} is correct because it is the simplified value of the given expression using the sine rule and the condition that a, b, c are in AP.
119 Easy
119 of 434
Application +4 | -1
In triangle ABC, the value of cot⁡A2cot⁡B2−1cot⁡A2cot⁡B2\frac{{\cot \frac{A}{2}\cot \frac{B}{2} - 1}}{{\cot \frac{A}{2}\cot \frac{B}{2}}} is
View answer and explanation
Correct option: D
2ca+b+c\frac{{2c}}{{a + b + c}}
Option D (2ca+b+c\frac{{2c}}{{a + b + c}}) is correct because simplifying the given expression using trigonometric identities and half-angle formulas leads to the result cs\frac{c}{s}, where ss is the semi-perimeter of the triangle, which is equal to 2ca+b+c\frac{{2c}}{{a + b + c}}.
120 Easy
120 of 434
Application +4 | -1
The angle of a right angled triangle are in A.P. The ratio of the in-radius and the perimeter is
View answer and explanation
Correct option: A
(2−3)  :23\left( {2 - \sqrt 3 } \right)\;:2\sqrt 3
Option A ((2−3) :23\left( {2 - \sqrt 3 } \right)\,:2\sqrt 3) is correct because the ratio of the in-radius and the perimeter of the right-angled triangle with angles in A.P. is (2−3) :23\left( {2 - \sqrt 3 } \right)\,:2\sqrt 3.
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