Skip to main content
Back to PROPERTIES OF TRIANGLES

Your Progress

0 / 287 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • PROPERTIES OF TRIANGLES

Properties of triangles practice · Easy

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →
Showing 181-210 of 287 questions

Preparing practice and saved progress…

181 Easy
181 of 287
Application +4 | -1
In a triangle ABC,(a+b+c)(b+c−a)=kbcA B C,(a+b+c)(b+c-a)=k b c if
View answer and explanation
Correct option: C
0<k<40 < {k} < 4
Option C (0<k<40 < {k} < 4) is correct because the given equation can be transformed using trigonometric identities to show that kk lies between 0 and 4.
182 Easy
182 of 287
Application +4 | -1
In any Δ  ABC\Delta \;ABC, if cot⁡  A2  cot⁡  B2,  cot⁡C2\cot \,\,\frac{A}{2}\,\,\cot \,\,\frac{B}{2},\,\,\cot \frac{C}{2} are in A.P., then a,  b,  ca,\,\,b,\,\,c are in
View answer and explanation
Correct option: A
A.P.\mathrm{A}.\mathrm{P}.
Option A is correct because the condition cot⁡A2 cot⁡B2, cot⁡C2\cot \frac{A}{2}\,\cot \frac{B}{2},\,\cot \frac{C}{2} are in A.P. implies that a,b,ca, b, c are in A.P.
183 Easy
183 of 287
Application +4 | -1
In △ABC\triangle A B C, the value of r1+r21+cos⁡C\frac{r_1+r_2}{1+\cos C} is always equal to
View answer and explanation
Correct option: B
2R2 R
Option B (2R2R) is correct because the expression simplifies to 2R2R using trigonometric identities and the properties of exradii.
184 Easy
184 of 287
Application +4 | -1
In Δ  ABC,  (a−b)2cos⁡2C2+(a+b)2sin⁡2C2\Delta \;ABC,\;{\left( {a - b} \right)^2}{\cos ^2}\frac{C}{2} + {\left( {a + b} \right)^2}{\sin ^2}\frac{C}{2} is equal to
View answer and explanation
Correct option: C
c2{c^2}
Option C (c2{c^2}) is correct because the expression simplifies to c2c^2 using trigonometric identities and the cosine rule.
185 Easy
185 of 287
Analyzing +4 | -1
In △ABC,(a+b−c)(b+c−a)(c+a−b)−abc\triangle A B C,(a+b-c)(b+c-a)(c+a-b)-a b c is always
View answer and explanation
Correct option: B
non-positive
Option B (non-positive) is correct because the expression (a+b−c)(b+c−a)(c+a−b)−abc(a+b-c)(b+c-a)(c+a-b)-abc is always less than or equal to zero.
186 Easy
186 of 287
Analyzing +4 | -1
If twice the square of the diameter of a circle is equal to half the sum of the squares of the sides of incribed triangle ABCABC, then sin⁡2A+sin⁡2B+sin⁡2C{\sin ^2}A + \sin^2 B+ {\sin ^2}C is equal to
View answer and explanation
Correct option: C
4
Option C (4) is correct because the given condition leads to the relationship sin⁡2A+sin⁡2B+sin⁡2C=4{\sin ^2}A + {\sin ^2}B+ {\sin ^2}C = 4.
187 Easy
187 of 287
Application +4 | -1
If in a Δ  ABC,  acos⁡2(C2)+ccos⁡2(A2)=3b2\Delta \;ABC,\;a{\cos ^2}\left( {\frac{C}{2}} \right) + c{\cos ^2}\left( {\frac{A}{2}} \right) = \frac{{3b}}{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 the given condition simplifies to 2b=a+c2b = a + c, which is the condition for a, b, and c to be in arithmetic progression.
188 Easy
188 of 287
Analyzing +4 | -1
Which of the following pieces of data does not uniquely determine acute angled Δ  ABC\Delta \;ABC(R=R =circumradius)?
View answer and explanation
Correct option: D
a,sin⁡A,  Ra,\sin A,\;R
Option D (a,sin⁡A,  Ra,\sin A,\;R) is correct because knowing aa, sin⁡A\sin A, and RR only gives the ratio bsin⁡B\frac{b}{{\sin B}} or csin⁡(A+B)\frac{c}{{\sin (A + B)}}, and we cannot determine the values of bb, cc, sin⁡B\sin B, and sin⁡C\sin C separately.
189 Easy
189 of 287
Analyzing +4 | -1
The base of a triangle is 80  cm\mathrm{80}\,\,\mathrm{cm} and one of the base angles is 60∘60^\circ. If the sum of the lengths of the other two sides is 90  cm\mathrm{90}\,\,\mathrm{cm}, then the length of the shortest side is
View answer and explanation
Correct option: D
17  cm\mathrm{17}\,\,\mathrm{cm}
Option D (17 cm) is correct because by using the cosine rule and the given conditions, we can find the length of the shortest side to be 17 cm.
190 Easy
190 of 287
Application +4 | -1
If in a Δ  ABC,b+c11=c+a12=a+b13\Delta \;ABC,\frac{{b + c}}{{11}} = \frac{{c + a}}{{12}} = \frac{{a + b}}{{13}}, then cos⁡A\cos A is equal to
View answer and explanation
Correct option: A
15\frac{1}{5}
Option B (-4) is correct because the midpoint of the rectangle's diagonal lies on the given line, allowing us to solve for c.
191 Easy
191 of 287
Application +4 | -1
In triangle ABC\mathrm{ABC} given 9a2+9b2−17c2=09 a^2+9 b^2-17 c^2=0.If cot⁡A+cot⁡Bcot⁡C=mn\frac{\cot \mathrm{A}+\cot \mathrm{B}}{\cot \mathrm{C}}=\frac{m}{n}, then the value of (m+n)(m+n) equals
View answer and explanation
Correct option: A
13
Option A (13) is correct because by using the Law of Cosines and trigonometric identities, the given equation simplifies to (cot A + cot B) / cot C = 9/4, and thus m+n = 13.
192 Easy
192 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+b2a2+c2\frac{{{a^2} + {b^2}}}{{{a^2} + {c^2}}}) is correct because by using trigonometric identities and the sine rule, the given expression can be simplified to a2+b2a2+c2\frac{{{a^2} + {b^2}}}{{{a^2} + {c^2}}}.
193 Easy
193 of 287
Application +4 | -1
In a Δ ABC\Delta \,ABC, if a=8,  b=10a = 8,\,\,b = 10 and c=12c = 12, then CC is equal to
View answer and explanation
Correct option: B
2A2A
Option B (2A2A) is correct because by using the Law of Cosines, we find that cos⁡C=18\cos C = \frac{1}{8} and cos⁡2A=18\cos 2A = \frac{1}{8}, thus C=2AC = 2A.
194 Easy
194 of 287
Application +4 | -1
Let a,ba, b and cc be real numbers such that 4a+2b+c4 a+2 b+c =0=0 and ab>0a b>0. Then the quadratic equation ax2+bxa x^2+b x +c=0+c=0 has
View answer and explanation
Correct option: C
Real roots
Option C (Real roots) is correct because given the condition 4a+2b+c=04a + 2b + c = 0, x=2x=2 is a real root of the quadratic equation ax2+bx+c=0ax^2 + bx + c = 0. Since one root is real, the other must also be real.
195 Easy
195 of 287
Application +4 | -1
If the sides of a triangle are the roots of the equation x3−2x2−x−16=0{x^3} - 2{x^2} - x - 16 = 0, then the product of the in-radius and circum-radius of the triangle is
View answer and explanation
Correct option: C
4\mathrm{4}
Option C (4) is correct because the product of the in-radius and circum-radius is calculated using the relationship between the sides of the triangle and the roots of the given equation.

Explore Other Chapters in MATHEMATICS

View All →
SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
196 Easy
196 of 287
Application +4 | -1
If A=30∘,a=100,c=1002A=30^{\circ}, a=100, c=100 \sqrt{2}, find the number of triangles that can be formed.
View answer and explanation
Correct option: A
2
Option A (2) is correct because two triangles are possible when a<ca < c and a>csin⁡Aa > c \sin A.
197 Easy
197 of 287
Application +4 | -1
The smallest angle of the triangle whose sides are 6+12, 48,246 + \sqrt {12} ,\,\sqrt {48} ,\sqrt {24} is
View answer and explanation
Correct option: C
π/6\pi /6
Option C (π/6\pi /6) is correct because it represents the smallest angle of the triangle, calculated using the law of cosines with the given side lengths.
198 Easy
198 of 287
Application +4 | -1
If a ΔABC,  2casin⁡A−B+C2\Delta ABC,\;2ca\sin \frac{{A - B + C}}{2} is equal to
View answer and explanation
Correct option: B
c2+a2−b2{c^2} + {a^2} - {b^2}
Option B (c2+a2−b2{c^2} + {a^2} - {b^2}) is correct because the expression simplifies to 2accos⁡B2ac \cos B, which equals c2+a2−b2{c^2} + {a^2} - {b^2} by the cosine rule.
199 Easy
199 of 287
Application +4 | -1
In a triangle with one angle if 120∘120^\circ, the length of the sides forms an AP. If the length of the greatest sides is 7 cm, then area of triangle is
View answer and explanation
Correct option: B
1534  cm2\frac{{15\sqrt 3 }}{4}\mathrm{\;c}{\mathrm{m}^2}
Option C (6 cm) is correct because the radius of the incircle is calculated as the area of the triangle divided by its semi-perimeter, which gives 6 cm.
200 Easy
200 of 287
Application +4 | -1
If in a Δ  ABC,cot⁡A2=b+ca\Delta \;ABC,\cot \frac{A}{2} = \frac{{b + c}}{a}, then the Δ  ABC\Delta \;ABC is
View answer and explanation
Correct option: C
Right angled
Option C (Right angled) is correct because given cot⁡A2=b+ca\cot \frac{A}{2} = \frac{{b + c}}{a}, it can be shown that angle B is equal to π2\frac{\pi}{2}.
201 Easy
201 of 287
Application +4 | -1
In a Δ ABC,  cos⁡2  A2  +  cos⁡2  B2 + cos⁡2  C2=\Delta \,ABC,\,\,{\cos ^2}\,\,\frac{A}{2}\,\, + \,\,{\cos ^2}\,\,\frac{B}{2}\, + \,{\cos ^2}\,\,\frac{C}{2} =
View answer and explanation
Correct option: C
2+r2R2 + \frac{r}{{2R}}
Option C (2+r2R2 + \frac{r}{{2R}}) is correct because the expression cos⁡2 A2 + cos⁡2 B2 + cos⁡2 C2{\cos ^2}\,\frac{A}{2}\, + \,{\cos ^2}\,\frac{B}{2}\, + \,{\cos ^2}\,\frac{C}{2} simplifies to 2+r2R2 + \frac{r}{{2R}} using trigonometric identities and the relationship between inradius and circumradius.
202 Easy
202 of 287
Application +4 | -1
In a Δ  ABC\Delta \;ABC, if cos⁡Aa=cos⁡Bb=cos⁡Cc\frac{{\cos A}}{a} = \frac{{\cos B}}{b} = \frac{{\cos C}}{c} and the side a=2a = 2, then area of the triangle is
View answer and explanation
Correct option: D
3\sqrt 3 sq unit
Option D (3\sqrt{3}) is correct because the given condition implies that the triangle is equilateral, and the area can be calculated using the side length.
203 Easy
203 of 287
Understanding +4 | -1
If RR is the radius of the circumcircle of the △ABC\triangle A B C and its area is Δ\Delta, then
View answer and explanation
Correct option: C
R=abc4ΔR=\frac{a b c}{4 \Delta}
Option C (R=abc4ΔR=\frac{{a b c}}{{4 \Delta}}) is correct because it is the standard formula relating the circumradius (R) to the sides (a, b, c) and area (Δ\Delta) of a triangle.
204 Easy
204 of 287
Application +4 | -1
The area of the circle and the area of a regular polygon of nn sides and of perimeter equal to that of the circle are in the ratio of
View answer and explanation
Correct option: A
tan⁡  (πn)  :  πn\tan \,\,\left( {\frac{\pi }{n}} \right)\;:\,\,\frac{\pi }{n}
Option A (tan⁡,(πn);:,πn\tan \\,\left( {\frac{\pi }{n}} \right)\\;:\\,\frac{\pi }{n}) is correct because the ratio of the area of the circle to the area of the regular polygon can be expressed in terms of tan⁡(πn)\tan \left( {\frac{\pi }{n}} \right) and πn\frac{\pi }{n}.
205 Easy
205 of 287
Application +4 | -1
If in a △ABC,r1=2,r2=3\triangle A B C, r_1=2, r_2=3 and r3=6r_3=6, then aa equals
View answer and explanation
Correct option: D
3
Option D (3) is correct because it is the value of side 'a' calculated using the given exradii and the properties of triangles.
206 Easy
206 of 287
Application +4 | -1
If sides of a triangle are 18,24,30 cm18,24,30 \mathrm{~cm}, then radius of circumcircle is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the circumradius is calculated as 15 cm, which is not among the given options.
207 Easy
207 of 287
Understanding +4 | -1
In a ∆ABC, the correct formulae among the following are

III.     r=4Rsin⁡A2sin⁡B2sin⁡C2\;\;r = 4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}

IV.     r1=(s−a)tan⁡A2\;\;{r_1} = \left( {s - a} \right)\tan \frac{A}{2}

V.     r3=Δs−c\;\;{r_3} = \frac{\Delta }{{s - c}}
View answer and explanation
Correct option: C
Only I, III
Option C (Only I, III) is correct because these are the standard formulae for inradius and exradius of a triangle.
208 Easy
208 of 287
Application +4 | -1
If, in a Δ ABC,  (a+b+c)  (b+c−a)=λ   bc\Delta \,ABC,\,\,\left( {a + b + c} \right)\,\,\left( {b + c - a} \right) = \lambda \;\,bc, then
View answer and explanation
Correct option: D
0<λ<40 < \lambda < 4
The correct answer is D (0<λ<40 < \lambda < 4) because the given equation can be rewritten in terms of cos⁡2(A2)\cos^2(\frac{A}{2}), which must be between 0 and 1.
209 Easy
209 of 287
Application +4 | -1
Angles A,BA,B and CC of a triangle are in AP with common difference 15 degree, then angle AA is equal to
View answer and explanation
Correct option: B
60∘60^\circ
Option A (60∘60^\circ) is incorrect. The correct answer is B (60∘60^\circ) because the angles are in AP and their sum is 180 degrees.
210 Easy
210 of 287
Application +4 | -1
In a ΔABC\Delta ABC, if tan⁡A2=56,tan⁡C2=25\tan \frac{A}{2} = \frac{5}{6},\tan \frac{C}{2} = \frac{2}{5}, then
View answer and explanation
Correct option: D
a,b,ca,b,c are in AP
Option A (a,b,ca,b,c are in AP) is correct because given the values of tan⁡A2\tan \frac{A}{2} and tan⁡C2\tan \frac{C}{2}, we can derive that 2b=a+c2b = a + c, which means a,b,ca, b, c are in arithmetic progression.
Revision notes and key takeaways

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

Showing 181-210 of 287 questions

Continue Your Practice

📚 More Sets in This Chapter

Try different difficulty levels to master this topic

→ ENGINEERING MATHEMATICS - PROPERTIES OF TRIANGLES MCQs (All) 434 MCQs → ENGINEERING MATHEMATICS - PROPERTIES OF TRIANGLES MCQs (Hard) 7 MCQs → ENGINEERING MATHEMATICS - PROPERTIES OF TRIANGLES MCQs (Medium) 140 MCQs
🎯 Next Chapters to Practice

Continue building your subject mastery

→ SEQUENCE OF SERIES 1764 MCQs → DEFINITE INTEGRATION 1443 MCQs → MATRICES 1363 MCQs → INDEFINITE INTEGRATION 1179 MCQs

Your practice

0 / 287 answered · 287 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →