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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 241-270 of 434 questions

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241 Easy
241 of 434
Application +4 | -1
In a triangle ABCa=6,b=3&cos⁡(A−B)=4/5A B C a=6, b=3 \& \cos (A-B)=4 / 5. Find area of Δ\Delta.
View answer and explanation
Correct option: A
9
Option A (9) is correct because using the given information and trigonometric identities, we can find that the triangle is right-angled, and then calculate the area.
242 Medium
242 of 434
Application +4 | -1
In any △ABC\triangle A B C, acos⁡A+bcos⁡B+ccos⁡Ca \cos A+b \cos B+c \cos C is equal to
View answer and explanation
Correct option: D
4Rsin⁡Asin⁡Bsin⁡C4 R \sin A \sin B \sin C
Option D (4Rsin⁡Asin⁡Bsin⁡C4 R \sin A \sin B \sin C) is correct because by using the sine rule and trigonometric identities, the expression simplifies to 4Rsin⁡Asin⁡Bsin⁡C4 R \sin A \sin B \sin C.
243 Medium
243 of 434
Application +4 | -1
If the angles of a Δ  ABC\Delta \;ABC be in AP, then
View answer and explanation
Correct option: B
b2=a2+c2−ac{b^2} = {a^2} + {c^2} - ac
Option B (b2=a2+c2−ac{b^2} = {a^2} + {c^2} - ac) is correct because when the angles are in AP, one angle is 60 degrees, and the cosine rule simplifies to this form.
244 Easy
244 of 434
Analyzing +4 | -1
In a Δ ABC,  cot⁡  A2+cot⁡  B2+cot⁡  C2cot⁡  A+cot⁡  B+cot⁡  C=\Delta \,ABC,\,\,\frac{{\cot \,\,\frac{A}{2} + \cot \,\,\frac{B}{2} + \cot \,\,\frac{C}{2}}}{{\cot \,\,A + \cot \,\,B + \cot \,\,C}} =
View answer and explanation
Correct option: A
(a+b+c)2a2+b2+c2\frac{{{{\left( {a + b + c} \right)}^2}}}{{{a^2} + {b^2} + {c^2}}}
The correct answer is A ((a+b+c)2a2+b2+c2\frac{{(a + b + c)^2}}{{a^2 + b^2 + c^2}}) because using trigonometric identities and the properties of triangles, the given expression simplifies to this ratio.
245 Medium
245 of 434
Application +4 | -1
If in a triangle, RR and rr are the circumradius and inradius respectively, then the HM of the exradii of the triangle is
View answer and explanation
Correct option: A
3r3 r
Option A (3r3r) is correct because the harmonic mean of the exradii is equal to three times the inradius.
246 Medium
246 of 434
Application +4 | -1
If the angle A\mathrm{A} of a triangle ABC\mathrm{ABC} is given by the equation 5cos⁡A+3=05 \cos A+3=0, then sin⁡A\sin A and tan⁡A\tan A are the roots of the equation
View answer and explanation
Correct option: D
15x2+8x−16=015 x^2+8 x-16=0
Option D (15x2+8x−16=015 x^2+8 x-16=0) is correct because it is the quadratic equation with roots sin A and tan A, derived from the given cosine value.
247 Easy
247 of 434
Application +4 | -1
In triangle ABC,  A=30∘,  b=8,  a=6ABC,\,\,A = 30^\circ ,\,\,b = 8,\,\,a = 6, then B=sin⁡−1xB = {\sin ^{ - 1}}x, where x=x =
View answer and explanation
Correct option: C
2/32/3
Option C (2/32/3) is correct because by using the sine rule, the value of sin⁡B\sin B is calculated to be 2/32/3.
248 Medium
248 of 434
Application +4 | -1
In a △ABC\triangle \mathrm{ABC}, if angle C\mathrm{C} is obtuse, then
View answer and explanation
Correct option: A
tan⁡Atan⁡B<1\tan A \tan B < 1
Option A (tan⁡Atan⁡B<1\tan A \tan B < 1) is correct because when angle C is obtuse, the product of tan A and tan B must be less than 1.
249 Easy
249 of 434
Analyzing +4 | -1
In a Δ  ABC,  if  r1=2r2=3r3\Delta \;ABC,\;\mathrm{if}\;{r_1} = 2{r_2} = 3{r_3}, then
View answer and explanation
Correct option: B
ab=54\frac{a}{b} = \frac{5}{4}
Option B (ab=54\frac{a}{b} = \frac{5}{4}) is correct because by using the given relationship between the exradii and the formula for exradii, we can derive the relationship between sides a and b.
250 Easy
250 of 434
Application +4 | -1
In any triangle ABC,  c2sin⁡2B+b2sin⁡2CABC,\;{c^2}\sin 2B + {b^2}\sin 2C is equal to
View answer and explanation
Correct option: D
4Δ4\Delta
Option D (4Δ4\Delta) is correct because by using the sine rule and area of triangle formula, the expression c2sin⁡2B+b2sin⁡2C{c^2}\sin 2B + {b^2}\sin 2C simplifies to 4Δ4\Delta.
251 Easy
251 of 434
Application +4 | -1
In a Δ ABC,  b=2,  C=60∘,  c=6\Delta \,ABC,\,\,b = 2,\,\,C = 60^\circ ,\,\,c = \sqrt 6, then a=a =
View answer and explanation
Correct option: C
3+1\sqrt 3 + 1
Option C (3+1\sqrt 3 + 1) is correct because it is obtained by applying the sine rule and solving for side 'a' using the given values.
252 Medium
252 of 434
Application +4 | -1
The inradius of the triangle whose sides are 3, 5, 6 is
View answer and explanation
Correct option: A
87\sqrt {\frac{8}{7}}
Option A (87\sqrt{\frac{8}{7}}) is correct because it is the inradius calculated using the formula r=Δsr = \frac{\Delta}{s}, where Δ\Delta is the area and ss is the semi-perimeter.
253 Medium
253 of 434
Analyzing +4 | -1
If A,BA, B and CC are angles of triangle such that ∠A\angle A is obtuse, then tan⁡Btan⁡C\tan B \tan C will be less than
View answer and explanation
Correct option: C
1
Option C (1) is correct because, given that angle A is obtuse, the product of tan B and tan C must be less than 1.
254 Medium
254 of 434
Application +4 | -1
If RR and R′R^{\prime} are the resultant of two forces PQ\frac{P}{Q} and QP(P>Q)\frac{{Q}}{{P}}({P}>{Q}) according as they are like or unlike such that R:R′=25:7{R}: {R}^{\prime}=25: 7, then P:Q={P}: {Q}=
View answer and explanation
Correct option: C
4:34: 3
Option C (4:34: 3) is correct because it is derived by setting up a ratio of the resultants and solving for the ratio of the forces P and Q.
255 Medium
255 of 434
Application +4 | -1
In any △ABC,2acos⁡B2cos⁡C2\triangle A B C, 2 a \cos \frac{B}{2} \cos \frac{C}{2} is equal to
View answer and explanation
Correct option: B
(a+b+c)sin⁡A2(a+b+c) \sin \frac{A}{2}
Option B ((a+b+c)sin⁡A2(a+b+c) \sin \frac{A}{2}) is correct because it is derived using trigonometric identities and the properties of a triangle.
256 Easy
256 of 434
Application +4 | -1
In a triangle vertex angles are A,  B,  CA,\;B,\;C and side BCBC are given. The area of ΔABC\Delta ABC is
View answer and explanation
Correct option: D
12.a2sin⁡Bsin⁡Csin⁡A\frac{1}{2}.\frac{{{a^2}\sin B\sin C}}{{\sin A}}
Option D (12.a2sin⁡Bsin⁡Csin⁡A\frac{1}{2}.\frac{{{a^2}\sin B\sin C}}{{\sin A}}) is correct because it represents the area of the triangle in terms of one side and all three angles, derived using the sine rule.
257 Easy
257 of 434
Application +4 | -1
If the lengths of the sides of a triangle are in AP and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle is
View answer and explanation
Correct option: B
04:05:06
Option B (04:05:06) is correct because when the sides are in AP and the greatest angle is double the smallest, the ratio of the sides is 4:5:6.
258 Medium
258 of 434
Application +4 | -1
The area of the triangle whose sides are 6,  5,  136,\;5,\;\sqrt {13} (in square unit) is
View answer and explanation
Correct option: B
9
Option B (9) is correct because using Heron's formula or trigonometric methods, the area of the triangle with sides 6, 5, and √13 is 9 square units.
259 Easy
259 of 434
Application +4 | -1
If the altitudes of a triangle are in AP, then the sides of the triangle are in
View answer and explanation
Correct option: C
H.P.
Option C (H.P.) is correct because the sides are inversely proportional to the altitudes, and if altitudes are in A.P., the sides are in H.P.
260 Medium
260 of 434
Application +4 | -1
In a right-angled triangle, the sides are a,  ba,\;b and cc, with cc as hypotenuse, and c−b≠1,  c+b≠1c - b \ne 1,\;c + b \ne 1. Then the value of (log⁡c+ba+log⁡c−ba)/(2log⁡c+ba×log⁡c−ba)\left( {{{\log }_{c + b}}a + {{\log }_{c - b}}a} \right)/\left( {2{{\log }_{c + b}}a \times {{\log }_{c - b}}a} \right) will be
View answer and explanation
Correct option: D
1
Option D (1) is correct because the expression simplifies to log⁡a2log⁡a2\frac{{\log {a^2}}}{{\log {a^2}}}, which equals 1.
261 Easy
261 of 434
Application +4 | -1
If the radius of the incircle of a triangle with its sides 5 k,  6 k5\,k,\,\,6\,k, and 5k5k is 6\mathrm{6}, then kk is equal to
View answer and explanation
Correct option: B
4\mathrm{4}
Option B, 4\mathrm{4}, is correct because it is the value of k calculated by equating the area of the triangle using Heron's formula and the formula involving inradius.
262 Easy
262 of 434
Application +4 | -1
The area of a triangle ABCA B C, where a=2(3+1),B=45∘,C=60∘a=2(\sqrt{3}+1), B=45^{\circ}, C=60^{\circ} is
View answer and explanation
Correct option: C
23(3+1)2 \sqrt{3}(\sqrt{3}+1) square unit
Option C (23(3+1)2 \sqrt{3}(\sqrt{3}+1) square unit) is correct because it is the area calculated using the sine rule and the given values of aa, BB, and CC.
263 Easy
263 of 434
Application +4 | -1
In a ΔABC,a:b:c=4:5:6\Delta ABC,a:b:c = 4:5:6. The ratio of the radius of the circumcircle to that of the incircle is
View answer and explanation
Correct option: B
167\frac{{16}}{7}
Option B 167\frac{{16}}{7} is correct because it is derived by using the formulas for the circumradius and inradius of a triangle and the given ratio of sides.
264 Easy
264 of 434
Analyzing +4 | -1
If a2,b2,c2{a^2},{b^2},{c^2} are in AP, then which of the following are also in AP?
View answer and explanation
Correct option: C
cot⁡A,cot⁡B,cot⁡C\cot A,\cot B,\cot C
Option C (cot⁡A,cot⁡B,cot⁡C\cot A,\cot B,\cot C) is correct because if a2,b2,c2{a^2},{b^2},{c^2} are in AP, then cot⁡A,cot⁡B,cot⁡C\cot A,\cot B,\cot C are also in AP.
265 Easy
265 of 434
Application +4 | -1
In a triangle ABCA B C, if (1+ab+cb)(1+bc−ac)=3\left(1+\frac{a}{b}+\frac{c}{b}\right)\left(1+\frac{b}{c}-\frac{a}{c}\right)=3, then the angle AA is equal to
View answer and explanation
Correct option: A
π3\frac{\pi}{3}
Option A (π3\frac{\pi}{3}) is correct because solving the given equation using trigonometric identities and the cosine rule leads to angle A being 60 degrees.
266 Easy
266 of 434
Analyzing +4 | -1
If cos⁡2A+cos⁡2C=sin⁡2B{\cos ^2}A + {\cos ^2}C = {\sin ^2}B, then ΔABC\Delta ABC is
View answer and explanation
Correct option: B
Right angled
Option B (Right angled) is correct because the given trigonometric identity holds true for a right-angled triangle, specifically when one of the angles is 90 degrees.
267 Easy
267 of 434
Application +4 | -1
The radius of the incircle of triangle when sides are 18, 24 and 30 cm is
View answer and explanation
Correct option: C
6 cm
Option C (6 cm) is correct because the inradius is calculated by dividing the area of the triangle by its semi-perimeter.
268 Medium
268 of 434
Analyzing +4 | -1
In a triangle, the sum of two sides is xx and the product of the same two sides is yy. If x2−c2=yx^2-c^2=y, where cc is the third side of the triangle, then the ratio of the inradius to the circumradius of the triangle is
View answer and explanation
Correct option: A
3y2x(x+c)\frac{3 y}{2 x(x+c)}
Option B is the correct answer because it correctly calculates the ratio of inradius to circumradius using the given conditions and trigonometric relationships.
269 Easy
269 of 434
Application +4 | -1
A triangle ABCA B C is inscribed in a circle of radius 4 cm.∠A=60∘AC−AB=4 cm4 \mathrm{~cm} . \angle A=60^{\circ} A C-A B=4 \mathrm{~cm}, then area of △ABC\triangle \mathrm{ABC} will be?
View answer and explanation
Correct option: B
838 \sqrt{3}
Option B (838 \sqrt{3}) is correct because it is the area of the triangle calculated using the given conditions and trigonometric relationships.
270 Easy
270 of 434
Analyzing +4 | -1
In any △ABC,rr1+r2r3\triangle A B C, r r_1+r_2 r_3 is equal to
View answer and explanation
Correct option: C
bcb c
Option C (bcbc) is correct because the expression rr1+r2r3rr_1 + r_2r_3 simplifies to bcbc using the properties of inradius and exradii.
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