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JEE • MATHEMATICS • PROPERTIES OF TRIANGLES

Properties of triangles practice · Mixed difficulty

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Showing 181-210 of 434 questions

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181 Easy
181 of 434
Analyzing +4 | -1
If in a Δ  ABC,tan⁡(A2),tan⁡(B2),tan⁡(C2)\Delta \;ABC,\tan \left( {\frac{A}{2}} \right),\tan \left( {\frac{B}{2}} \right),\tan \left( {\frac{C}{2}} \right) are in HP, then the sides a,  b,  ca,\;b,\;c are in
View answer and explanation
Correct option: A
AP
Option A (AP) is correct because if tan⁡A2,tan⁡B2,tan⁡C2\tan \frac{A}{2},\tan \frac{B}{2},\tan \frac{C}{2} are in HP, then cot⁡A2,cot⁡B2,cot⁡C2\cot \frac{A}{2},\cot \frac{B}{2},\cot \frac{C}{2} are in AP, which leads to a,b,ca, b, c being in AP.
182 Easy
182 of 434
Application +4 | -1
In a triangle ABC,a2b2c2(sin⁡2A+sin⁡2B+sin⁡2C)=A B C, a^2 b^2 c^2(\sin 2 A+\sin 2 B+\sin 2 C)=
View answer and explanation
Correct option: D
32Δ332 \Delta^3
Option D (32Δ332 \Delta^3) is correct because the expression simplifies to 32Δ332 \Delta^3 using trigonometric identities and the formula for the area of a triangle.
183 Easy
183 of 434
Understanding +4 | -1
In an equilateral triangle,
View answer and explanation
Correct option: C
r1=r2=r3=3rr_1=r_2=r_3=3 r
Option C (r1=r2=r3=3rr_1=r_2=r_3=3 r) is correct because in an equilateral triangle, all exradii are equal and each is three times the inradius.
184 Medium
184 of 434
Understanding +4 | -1
In a △ABC\triangle A B C, the inradius and three exradii are r,r1,r2r, r_1, r_2 and r3r_3, respectively. In usual notations, the value of r⋅r1⋅r2⋅r3r \cdot r_1 \cdot r_2 \cdot r_3 is equal to
View answer and explanation
Correct option: B
Δ2\Delta^2
Option B (Δ2\Delta^2) is correct because the product of the inradius and exradii is equal to the square of the area of the triangle.
185 Medium
185 of 434
Application +4 | -1
A pole indined from vertical at 15∘15^{\circ} towards the sun. Length of the shadow of the pole is 3+1\sqrt{3}+1 meter at the time when angle of elevation of the sun is 30∘30^{\circ}, then length of pole is
View answer and explanation
Correct option: B
23−1m\frac{\sqrt{2}}{\sqrt{3}-1} m
Option B (23−1m\frac{\sqrt{2}}{\sqrt{3}-1} m) is correct because it applies the sine rule to the triangle formed by the pole, its shadow, and the sun's rays, accounting for the pole's inclination.
186 Easy
186 of 434
Application +4 | -1
In any △ABC\triangle A B C,

(a+b+c)(b+c−a)(c+a−b)(a+b−c)4b2c2\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4 b^2 c^2} equals
View answer and explanation
Correct option: D
sin⁡2A\sin ^2 A
Option D (sin⁡2A\sin ^2 A) is correct because using Heron's formula and the area of a triangle formula, the given expression simplifies to sin⁡2A\sin^2 A.
187 Easy
187 of 434
Application +4 | -1
Angles of a triangle are in the ratio 4:1:14:1:1. The ratio between its greatest side and perimeter is
View answer and explanation
Correct option: C
33+2\frac{{\sqrt 3 }}{{\sqrt 3 + 2}}
Option C (33+2\frac{{\sqrt 3 }}{{\sqrt 3 + 2}}) is correct because, with angles in the ratio 4:1:1, the ratio of the greatest side to the perimeter simplifies to 33+2\frac{{\sqrt 3 }}{{\sqrt 3 + 2}}.
188 Medium
188 of 434
Application +4 | -1
The abscissa of the point on the curve 9y29 y^2 =x3=x^3, the normal at which cuts off equal intercepts on the coordinate axes is
View answer and explanation
Correct option: B
4
Option B (4) is correct because the abscissa of the point is 4, which is found by setting the intercepts of the normal equal and solving for x.
189 Easy
189 of 434
Application +4 | -1
In a ΔABC,  if  b+c=2a\Delta ABC,\;\mathrm{if}\;b + c = 2a and ∠A=60∘\angle A = 60^\circ, then ΔABC\Delta ABC is
View answer and explanation
Correct option: A
Equilateral
Option A (Equilateral) is correct because given the condition b+c=2ab + c = 2a and ∠A=60∘\angle A = 60^\circ, the triangle must be equilateral.
190 Easy
190 of 434
Application +4 | -1
If rr and RR are respectively the radii of the inscribed and circumscribed circles of a regular polygon of nn sides such that Rr=5−1\frac{R}{r}=\sqrt{5}-1, then nn is equal to
View answer and explanation
Correct option: A
5
Option A (5) is correct because by using the formulas for the radii of inscribed and circumscribed circles of a regular polygon, and the given ratio, we can solve for nn and find that n=5n=5.
191 Easy
191 of 434
Application +4 | -1
If in a Δ  ABC,  CD\Delta \;ABC,\;CD is the angular bisector of the ∠  ACB\angle \;ACB, then CDCD is equal to
View answer and explanation
Correct option: C
2aba+bcos⁡C2\frac{{2ab}}{{a + b}}\cos \frac{C}{2}
Option C (2aba+bcos⁡C2\frac{{2ab}}{{a + b}}\cos \frac{C}{2}) is correct because the length of the angular bisector CD is derived using the area of the triangle.
192 Medium
192 of 434
Application +4 | -1
If in a ΔABC,a=15,b=36,c=39\Delta ABC,a = 15,b = 36,c = 39, then sin⁡C2\sin \frac{C}{2} is equal to
View answer and explanation
Correct option: C
12\frac{1}{{\sqrt 2 }}
Option C 12\frac{1}{{\sqrt 2 }} is correct because it is the value of sin⁡C2\sin \frac{C}{2} calculated using the given sides of the triangle.
193 Easy
193 of 434
Application +4 | -1
In a triangle ABCABC, if sin⁡A  sin⁡B=abc2\sin A\;\sin B = \frac{{ab}}{{{c^2}}}, then the triangle is
View answer and explanation
Correct option: C
Right angled
Option C (Right angled) is correct because the given condition leads to sin⁡2C=1\sin^2 C = 1, implying that angle C is 90 degrees, hence a right-angled triangle.
194 Easy
194 of 434
Analyzing +4 | -1
In a triangle, if r1=2r2=3r3{r_1} = 2{r_2} = 3{r_3}, then ab+bc+ca\frac{a}{b} + \frac{b}{c} + \frac{c}{a} is equal to
View answer and explanation
Correct option: D
19160\frac{{191}}{{60}}
Option D (19160\frac{{191}}{{60}}) is correct because it's the result of calculating ab+bc+ca\frac{a}{b} + \frac{b}{c} + \frac{c}{a} after finding the ratios of a, b, and c using the given relationship between the exradii.
195 Easy
195 of 434
Application +4 | -1
In ΔABC\Delta ABC, if ∠A=45∘,∠B=75∘\angle A = 45\mathrm{^\circ },\angle B = 75\mathrm{^\circ }, then a+c2a + c\sqrt 2 is equal to
View answer and explanation
Correct option: D
2b2b
Option D (2b2b) is correct because, using the sine rule and given angles, the expression a+c2a + c\sqrt{2} simplifies to 2b2b.

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196 Easy
196 of 434
Application +4 | -1
In any △ABC,b2sin⁡2C+c2sin⁡2B\triangle A B C, b^2 \sin 2 C+c^2 \sin 2 B is equal to
View answer and explanation
Correct option: D
4Δ4 \Delta
Option D (4Δ4 \Delta) is correct because the expression simplifies to 4Δ4 \Delta using trigonometric identities and the area formula of a triangle.
197 Easy
197 of 434
Application +4 | -1
In a △ABC,a[bcos⁡C−ccos⁡B]\triangle A B C, a[b \cos C-c \cos B] is equal to
View answer and explanation
Correct option: C
b2−c2b^2-c^2
Option C (b2−c2b^2-c^2) is correct because by applying the cosine rule and simplifying, the expression reduces to b2−c2b^2 - c^2.
198 Easy
198 of 434
Analyzing +4 | -1
If a,ba,b and cc are the sides of a triangle such that a4+b4+c4=2c2(a2+b2){a^4} + {b^4} + {c^4} = 2{c^2}\left( {{a^2} + {b^2}} \right), then the angles opposite to the side CC is
View answer and explanation
Correct option: C
45∘or  135∘45\mathrm{^\circ or}\;135\mathrm{^\circ }
Option C (45∘or  135∘45\mathrm{^\circ or}\;135\mathrm{^\circ }) is correct because by using the cosine rule and the given equation, we can derive that cos⁡2C=12\cos^2 C = \frac{1}{2}, which implies that CC is either 45 or 135 degrees.
199 Easy
199 of 434
Application +4 | -1
In a ABCABC, if b=2,∠B=30∘b = 2,\angle B = 30^\circ, then the area of the circumcircle of ΔABC\Delta ABC in square unit is
View answer and explanation
Correct option: C
4π4\pi
Option C (4π4\pi) is correct because the area of the circumcircle is calculated using the sine rule to find the radius.
200 Hard
200 of 434
Application +4 | -1
The least perimeter of a cyclic quadrilateral of given area AA square units is
View answer and explanation
Correct option: D
4A4\sqrt A
The correct answer is D (4A4\sqrt A) because using AM-GM inequality on the area formula of a cyclic quadrilateral gives the least perimeter.
201 Easy
201 of 434
Application +4 | -1
In Δ  ABC\Delta \;ABC, if 8R2=a2+b2+c28{R^2} = {a^2} + {b^2} + {c^2}, then the triangle is
View answer and explanation
Correct option: A
Right angled
Option A (Right angled) is correct because the given condition leads to one of the angles being equal to 90 degrees.
202 Medium
202 of 434
Analyzing +4 | -1
If ABCA B C is a triangle whose internal angles are A,B,CA, B, C and △DEF\triangle D E F is pedal triangle of △ABC\triangle A B C, then the ratio of the area of △DEF\triangle D E F to area of △ABC\triangle A B C is
View answer and explanation
Correct option: A
2cos⁡Acos⁡Bcos⁡C2 \cos A \cos B \cos C
The correct answer is A (2cos⁡Acos⁡Bcos⁡C2 \cos A \cos B \cos C). The ratio of the area of the pedal triangle to the area of the original triangle is 2cos⁡Acos⁡Bcos⁡C2 \cos A \cos B \cos C.
203 Easy
203 of 434
Application +4 | -1
In a Δ  ABC,  a=5,  a=\Delta \;ABC,\;a = 5,\;a = and cos⁡(A+B)=3132\cos \left( {A + B} \right) = \frac{{31}}{{32}}. In this triangle, cc is equal to
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the calculated value of c, 3192\frac{{\sqrt {319} }}{2}, does not match any of the provided options.
204 Medium
204 of 434
Application +4 | -1
A stone is projected vertically upward moves such that its distance feet in time t sec is given by s=112t−16t2s=112 t-16 t^2 then the maximum height reached by the stone in (feet) is
View answer and explanation
Correct option: A
196
Option A (196) is correct because it is the maximum height reached by the stone, calculated by finding the time when the velocity is zero and substituting it into the distance equation.
205 Medium
205 of 434
Application +4 | -1
In △ABC,a:b:c=(1+x):1:(1−x)\triangle A B C, a: b: c=(1+x): 1:(1-x), where x∈(0,1)x \in(0,1). If ∠A=π2+∠C\angle A=\frac{\pi}{2}+\angle C, then xx is equal to
View answer and explanation
Correct option: C
17\frac{1}{\sqrt{7}}
Option C (17\frac{1}{\sqrt{7}}) is correct because it is the value of x that satisfies the given conditions and trigonometric relationships in the triangle.
206 Easy
206 of 434
Application +4 | -1
If OO is the origin and P(2,3)P\left( {2,3} \right) and Q(4,5)Q\left( {4,5} \right) are two points, then OP.OQcos⁡∠POQ=OP.OQ\cos \angle POQ =
View answer and explanation
Correct option: D
23
Option D ((c2−a22ca)2{\left( {\frac{{{c^2} - {a^2}}}{{2ca}}} \right)^2}) is correct because by using the trigonometric identity for sin⁡3B\sin 3B and the given condition 2b2=a2+c22b^2 = a^2 + c^2, the expression sin⁡3Bsin⁡B\frac{{\sin 3B}}{{\sin B}} simplifies to (c2−a22ca)2{\left( {\frac{{{c^2} - {a^2}}}{{2ca}}} \right)^2}.
207 Medium
207 of 434
Application +4 | -1
In Δ  ABC\Delta \;ABC, if tan⁡A2tan⁡C2=12\tan \frac{A}{2}\tan \frac{C}{2} = \frac{1}{2}, then a,  b,  ca,\;b,\;c are in
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the given condition tan⁡A2tan⁡C2=12\tan \frac{A}{2}\tan \frac{C}{2} = \frac{1}{2} leads to the relation a+c=3ba + c = 3b, which does not satisfy the conditions for AP, GP, or HP.
208 Easy
208 of 434
Application +4 | -1
In a Δ ABC\Delta \,ABC, if b=20, c=21b = 20,\,c = 21 and sin⁡A=35\sin A = \frac{3}{5}, then a=a =
View answer and explanation
Correct option: B
13
Option B (13) is correct because applying the cosine rule with the given values results in a = 13.
209 Easy
209 of 434
Application +4 | -1
In a △ABC\triangle A B C, let ∠C=π2\angle C=\frac{\pi}{2}, if rr is the inradius and RR is the circumradius of the △ABC\triangle A B C, then 2(r+R)2(r+R) equals
View answer and explanation
Correct option: C
a+ba+b
Option C (a+ba+b) is correct because in a right-angled triangle, 2(r+R)=a+b2(r+R) = a+b can be derived using the relationships between inradius, circumradius, and side lengths.
210 Medium
210 of 434
Application +4 | -1
In a △ABC,A=2π3,b−c=33\triangle A B C, A=\frac{2 \pi}{3}, b-c=3 \sqrt{3} and ar (△ABC)=932 cm2(\triangle A B C)=\frac{9 \sqrt{3}}{2} \mathrm{~cm}^2. Then a is
View answer and explanation
Correct option: B
9 cm9 \mathrm{~cm}.
Option B (9 cm9 \mathrm{~cm}.) is correct because by using the area and the given angle, we can find bc, and then using the cosine rule, we can find a.
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