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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 241-270 of 287 questions

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241 Easy
241 of 287
Application +4 | -1
The angles A,  BA,\;B and CC of a Δ  ABC\Delta \;ABC are in AP. if b:c=3:2b:c = \sqrt 3 :\sqrt 2, then the angle AA is
View answer and explanation
Correct option: C
75∘75^\circ
Option C (75°) is correct because, with angles in AP and b:c = √3:√2, solving for the angles gives A = 75°.
242 Easy
242 of 287
Understanding +4 | -1
In an equilateral triangle, the ratio of the inradius, circumradius and exradius are
View answer and explanation
Correct option: D
1:2:31: 2: 3
This question is incomplete and contains errors. The explanation provided is not relevant to the question. The correct answer should relate to the ratio of inradius, circumradius, and exradius in an equilateral triangle.
243 Easy
243 of 287
Application +4 | -1
In a triangle ABCA B C, let ∠C=π2\angle C=\frac{\pi}{2}. If rr is the inradius and RR is the circumradius of the triangle ABCA B C, then 2(r+R)2(r+R) equals
View answer and explanation
Correct option: B
a+ba+b
Option B (a+b) is correct because in a right-angled triangle, 2(r+R) simplifies to a+b.
244 Easy
244 of 287
Application +4 | -1
If α,β\alpha, \beta and γ\gamma are the altitudes of a △ABC\triangle A B C and 2s2 s denotes its perimeter, then α−1+β−1+γ−1\alpha^{-1}+\beta^{-1}+\gamma^{-1} is equal to
View answer and explanation
Correct option: B
sΔ\frac{s}{\Delta}
Option B (sΔ\frac{s}{\Delta}) is correct because it is derived by expressing the reciprocals of altitudes in terms of sides and area of the triangle.
245 Easy
245 of 287
Application +4 | -1
In the angles A,BA,B and CC of a triangular are in the arithmetic progression and if a,ba,b and cc denotes the lengths of the sides opposite to A,BA,B and CC respectively, then the value of the expression acsin⁡  2C+casin⁡  2A\frac{a}{c}\sin \;2C + \frac{c}{a}\sin \;2A is
View answer and explanation
Correct option: D
3\sqrt 3
Option D (3\sqrt{3}) is correct because the expression simplifies to 2sinB, and since B = 60 degrees, the value is 3\sqrt{3}.
246 Easy
246 of 287
Application +4 | -1
If in a triangle ABC,  b+c=3aABC,\,\,b + c = 3a, then tan⁡ (B2)tan⁡(C2)\tan \,\left( {\frac{B}{2}} \right)\tan \left( {\frac{C}{2}} \right) is equal to
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the correct value of tan⁡(B2)tan⁡(C2)\tan \left( {\frac{B}{2}} \right)\tan \left( {\frac{C}{2}} \right) is 12\frac{1}{2}, which is not among the given options.
247 Easy
247 of 287
Application +4 | -1
Angles A,BA,B and CC of a Δ  ABC\Delta \;ABC are in AP. If bc=32\frac{b}{c} = \frac{{\sqrt 3 }}{{\sqrt 2 }}, then angle AA is equal to
View answer and explanation
Correct option: C
5π12\frac{{5\pi }}{{12}}
Option C (5π12\frac{{5\pi }}{{12}}) is correct because by using the sine rule and the given ratio of sides, we can find angle A to be 75 degrees, which is 5π12\frac{{5\pi }}{{12}} radians.
248 Easy
248 of 287
Analyzing +4 | -1
If in a ΔABC,  r1<r2<r3\Delta ABC,\;{r_1} < {r_2} < {r_3}, then
View answer and explanation
Correct option: A
a<b<ca < b < c
Option A (a < b < c) is correct because the given inequality r1<r2<r3{r_1} < {r_2} < {r_3} implies that (s−a)>(s−b)>(s−c)\left( {s - a} \right) > \left( {s - b} \right) > \left( {s - c} \right), which leads to a<b<ca < b < c.
249 Easy
249 of 287
Analyzing +4 | -1
In Δ  ABC\Delta\; ABC, if sin⁡2A2,sin⁡2B2,sin⁡2C2{\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 if the sines squared are in HP, then the sides a, b, c are also in HP.
250 Easy
250 of 287
Analyzing +4 | -1
If a,ba, b and cc are the sides of a triangle such that b⋅c=λ2b \cdot c=\lambda^2, for some positive λ\lambda, then
View answer and explanation
Correct option: C
a≥2λsin⁡A2a \geq 2 \lambda \sin \frac{A}{2}
Option C (a≥2λsin⁡A2a \geq 2 \lambda \sin \frac{A}{2}) is correct because it is derived from the AM-GM inequality and the sine rule applied to the triangle.
251 Easy
251 of 287
Application +4 | -1
If in a ΔABC,  4sin⁡A=4sin⁡B=3sin⁡C\Delta ABC,\;4\sin A = 4\sin B = 3\sin C, then cos⁡C\cos C is equal to
View answer and explanation
Correct option: B
1/91/9
Option B (19\frac{1}{9}) is correct because it is the value of cos⁡C\cos C calculated using the given ratios of sines and the cosine rule.
252 Easy
252 of 287
Application +4 | -1
In a Δ ABC,  A=2  π3,  b−c=3 3  cm\Delta \,ABC,\,\,A = \frac{{2\;\pi }}{3},\,\,b - c = 3\,\sqrt 3 \;\mathrm{cm} and Δ=9 32  cm2\Delta = \frac{{9\,\sqrt 3 }}{2}\,\,\mathrm{c}{\mathrm{m}^2}. Then, a=a =
View answer and explanation
Correct option: B
9   cm9\;\,\mathrm{cm}
Option B, 9   cm9\;\,\mathrm{cm}, is correct because it is the length of side 'a' calculated using the cosine rule and the given area and angle.
253 Easy
253 of 287
Application +4 | -1
In a △ABC,tan⁡A2=56,tan⁡C2=25\triangle A B C, \tan \frac{A}{2}=\frac{5}{6}, \tan \frac{C}{2}=\frac{2}{5}, then
View answer and explanation
Correct option: B
a,b,ca, b, c are in AP
Option B (a,b,ca, b, c are in AP) is correct because the given trigonometric ratios lead to the condition a+c=2ba + c = 2b, which defines an arithmetic progression.
254 Easy
254 of 287
Understanding +4 | -1
In a triangle r1>r2>r3{r_1} > {r_2} > {r_3}, then
View answer and explanation
Correct option: A
a>b>ca > b > c
Option A (a>b>ca > b > c) is correct because the inradii r1,r2,r3r_1, r_2, r_3 are inversely proportional to s−a,s−b,s−cs-a, s-b, s-c respectively.
255 Easy
255 of 287
Analyzing +4 | -1
If the data given to construct a triangle ABCABC are a=5,  b=7,  sin⁡A=3/4a = 5,\,\,b = 7,\,\,\sin A = 3/4, then it is possible to construct
View answer and explanation
Correct option: D
No triangles
Option D (No triangles) is correct because the value of sin B is greater than 1, which is impossible for any angle in a triangle.
256 Easy
256 of 287
Application +4 | -1
The longest side of a triangle is 3 times the shortest side and the third side is 2 cm2 \mathrm{~cm} shorter than the longest side. If the perimeter of the triangle is at least 61 cm61 \mathrm{~cm}, find the minimum length of the shortest side.
View answer and explanation
Correct option: A
9
Option A (9) is correct because setting up the inequality based on the problem statement and solving for the shortest side (x) results in x≥9x \geq 9.
257 Easy
257 of 287
Application +4 | -1
In ΔABC,2(a  sin⁡2C2+c  sin⁡2A2)\Delta ABC,2\left( {a\;{{\sin }^2}\frac{C}{2} + c\;{{\sin }^2}\frac{A}{2}} \right) is equal to
View answer and explanation
Correct option: B
c+a−bc + a - b
Option B (c+a−bc + a - b) is correct because the expression simplifies to this using trigonometric identities and the semi-perimeter formula.
258 Easy
258 of 287
Application +4 | -1
If c2=a2+b2,  2  s=a+b+c{c^2} = {a^2} + {b^2},\,\,2\;s = a + b + c, then 4s  (s−a)  (s−b)  (s−c)=4s\,\,\left( {s - a} \right)\,\,\left( {s - b} \right)\,\,\left( {s - c} \right) =
View answer and explanation
Correct option: C
c2 a2{c^2}\,{a^2}
Option D (a2 b2{a^2}\,{b^2}) is correct because using the given condition c2=a2+b2{c^2} = {a^2} + {b^2} and the area of the triangle, we can deduce that 4s(s−a)(s−b)(s−c)=a2b24s(s-a)(s-b)(s-c) = a^2b^2.
259 Easy
259 of 287
Application +4 | -1
In △ABC,1−tan⁡B2⋅tan⁡C2\triangle A B C, 1-\tan \frac{B}{2} \cdot \tan \frac{C}{2} is equal to
View answer and explanation
Correct option: A
2aa+b+c\frac{2 a}{a+b+c}
Option A (2aa+b+c\frac{2 a}{a+b+c}) is correct because it is the simplified form of the given expression using trigonometric identities and triangle properties.
260 Easy
260 of 287
Application +4 | -1
If r1=2r2=3r3r_1=2 r_2=3 r_3 then a+b+ca+b+c is equal to
View answer and explanation
Correct option: A
3b3 b
Option B (3b) is correct because given the relationship between the ex-radii, we can express a, b, and c in terms of each other and find that a+b+c=3ba+b+c = 3b.
261 Easy
261 of 287
Application +4 | -1
If in a △ABC,acos⁡2(C2)+ccos⁡2(A2)=3b2\triangle A B C, a \cos ^2\left(\frac{C}{2}\right)+c \cos ^2\left(\frac{A}{2}\right)=\frac{3 b}{2}, then the sides a,ba, b and cc
View answer and explanation
Correct option: A
are in AP
Option A (are in AP) is correct because by substituting the half-angle formula into the given equation and simplifying, we can show that 2b=a+c2b = a+c, which means a,b,ca, b, c are in arithmetic progression.
262 Easy
262 of 287
Analyzing +4 | -1
A quadrilateral ABCDABCD in which AB=a,  BC=b,  CD=cAB = a,\;BC = b,\;CD = c and DA=dDA = d is such that one circle can be inscribed in it and another circle can be circumscribed about it, then cos⁡A\cos A is equal to
View answer and explanation
Correct option: B
ad−bcad+bc\frac{{ad - bc}}{{ad + bc}}
Option B (ad−bcad+bc\frac{{ad - bc}}{{ad + bc}}) is correct because it is the expression derived from the given conditions of a cyclic quadrilateral with an inscribed circle.
263 Easy
263 of 287
Application +4 | -1
In a △ABC\triangle A B C, if A=60∘A=60^{\circ}, then bc+a+ca+b\frac{b}{c+a}+\frac{c}{a+b} is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because by using the Law of Cosines and simplifying, we can show that bc+a+ca+b=1\frac{b}{c+a} + \frac{c}{a+b} = 1.
264 Easy
264 of 287
Application +4 | -1
In a Δ  ABC,  a,  c  ,  A\Delta \;ABC,\;a,\;c\;,\;A are given and b1,b2{b_1},{b_2} are two values of third side bb such that b=2b1b = 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 it is derived from the quadratic equation formed using the Law of Cosines and the given relationship between the two possible values of the third side.
265 Easy
265 of 287
Application +4 | -1
In a Δ ABC\Delta \,ABC,

a(b2+c2)  cos⁡  A+b(c2+a2)  cos⁡  B+c(a2+b2)cos⁡  Ca\left( {{b^2} + {c^2}} \right)\,\,\cos \,\,A + b\left( {{c^2} + {a^2}} \right)\,\,\cos \,\,B + c\left( {{a^2} + {b^2}} \right)\cos \,\,C is equal to
View answer and explanation
Correct option: C
3 abc3\,abc
The correct answer is C (3abc3abc) because by using the projection rule, the given expression simplifies to 3abc3abc.
266 Easy
266 of 287
Analyzing +4 | -1
In a Δ ABC\Delta \,ABC if c=(a+b)sin⁡  θc = \left( {a + b} \right)\sin \,\,\theta and cos⁡  θ=k aba+b\cos \,\,\theta = \frac{{k\,\sqrt {ab} }}{{a + b}}, then k=k =
View answer and explanation
Correct option: A
2  cos⁡  C22\,\,\cos \,\,\frac{C}{2}
Option A (2cos⁡C22 \cos \frac{C}{2}) is correct because using sine rule and half angle formula, we can derive that k=2cos⁡C2k = 2 \cos \frac{C}{2}.
267 Easy
267 of 287
Application +4 | -1
If A=30∘,  a=7,  b=8A = 30^\circ ,\,\,a = 7,\,\,b = 8 in Δ ABC\Delta \,ABC, then BB has
View answer and explanation
Correct option: B
Two solutions
Option B (Two solutions) is correct because the condition b>a>bsin⁡Ab > a > b \sin A is satisfied, indicating two possible triangles.
268 Easy
268 of 287
Analyzing +4 | -1
If a Δ  ABC,if  a4+b4+c4=2c2(a2+b2)\Delta \;ABC,\mathrm{if}\;{a^4} + {b^4} + {c^4} = 2{c^2}\left( {{a^2} + {b^2}} \right), then ∠C\angle C is equal to
View answer and explanation
Correct option: B
135∘135^\circ
Option B (135∘135^\circ) is correct because it satisfies the given equation when substituting into the cosine rule.
269 Easy
269 of 287
Application +4 | -1
The horizontal distance between two tower is 60 m60 \mathrm{~m}. The angular elevation of the top of the taller tower as seen from the top of the shorter one is 30∘30^{\circ}. If the height of the taller tower is 150 m150 \mathrm{~m}, the height of the shorter one is
View answer and explanation
Correct option: A
(150−203)m(150-20 \sqrt{3}) \mathrm{m}
Option A ((150−203)m(150-20 \sqrt{3}) \mathrm{m}) is correct because it is derived using trigonometry and the given heights and distance.
270 Easy
270 of 287
Application +4 | -1
If p1,  p2,  p3{p_1},\,\,{p_2},\,\,{p_3} are altitudes of a triangle ABCABC from the vertices A,  B,  CA,\,\,B,\,\,C and Δ\Delta, the area of the triangle, then p1−1+p2−1−p3−1p_1^{ - 1} + p_2^{ - 1} - p_3^{ - 1} is equal to
View answer and explanation
Correct option: C
s−cΔ\frac{{s - c}}{\Delta }
Option C (s−cΔ\frac{{s - c}}{\Delta }) is correct because by using the area formula and substituting, the expression simplifies to s−cΔ\frac{{s - c}}{\Delta }.
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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