Practice Start Test

Properties of Triangle

JEE Mathematics · 68 questions · Page 6 of 7 · Click an option or "Show Solution" to reveal answer

Q51
The triangle of maximum area that can be inscribed in a given circle of radius 'r' is :
A An equilateral triangle having each of its side of length 3\sqrt 3 r.
B An equilateral triangle of height 2r3{{2r} \over 3}.
C A right angle triangle having two of its sides of length 2r and r.
D An isosceles triangle with base equal to 2r.
Correct Answer
Option A
Solution

Area of triangle ABC

A=12×BC×AMA = {1 \over 2} \times BC \times AM
=12×2r2−x2×(r+x)= {1 \over 2} \times 2\sqrt {{r^2} - {x^2}} \times (r + x)
A=(r+x)r2−x2A = (r + x)\sqrt {{r^2} - {x^2}}
dAdx=r2−x2−xr2−x2×(r+x){{dA} \over {dx}} = \sqrt {{r^2} - {x^2}} - {x \over {\sqrt {{r^2} - {x^2}} }} \times (r + x)
=r2−x2−rx−x2r2−x2=r2−rx−2x2r2−x2=−(x+r)(2x−r)r2−x2= {{{r^2} - {x^2} - rx - {x^2}} \over {\sqrt {{r^2} - {x^2}} }} = {{{r^2} - rx - 2{x^2}} \over {\sqrt {{r^2} - {x^2}} }} = {{ - (x + r)(2x - r)} \over {\sqrt {{r^2} - {x^2}} }}
dAdx=0⇒x=r2{{dA} \over {dx}} = 0 \Rightarrow x = {r \over 2}

Sign change of

dAdx{{dA} \over {dx}}

at

x=r2x = {r \over 2}

⇒\Rightarrow A has maximum at

x=r2x = {r \over 2}
BC=2r2−x2=3rBC = 2\sqrt {{r^2} - {x^2}} = \sqrt 3 r

,

AM=r+12rAM = r + {1 \over 2}r

=

32r{3 \over 2}r
⇒AB=AC=3r\Rightarrow AB = AC = \sqrt 3 r
Q52
If in a triangle ABC, AB = 5 units, ∠B=cos⁡−1(35)\angle B = {\cos ^{ - 1}}\left( {{3 \over 5}} \right) and radius of circumcircle of Δ\DeltaABC is 5 units, then the area (in sq. units) of Δ\DeltaABC is :
A 10+6210 + 6\sqrt 2
B 8+228 + 2\sqrt 2
C 6+836 + 8\sqrt 3
D 4+234 + 2\sqrt 3
Correct Answer
Option C
Solution

As,

cos⁡B=35⇒B=53∘\cos B = {3 \over 5} \Rightarrow B = 53^\circ

As,

R=5⇒csin⁡c=2RR = 5 \Rightarrow {c \over {\sin c}} = 2R
⇒510=sin⁡c⇒C=30∘\Rightarrow {5 \over {10}} = \sin c \Rightarrow C = 30^\circ

Now,

bsin⁡B=2R⇒b=2(5)(45)=8{b \over {\sin B}} = 2R \Rightarrow b = 2(5)\left( {{4 \over 5}} \right) = 8

Now, by cosine formula

cos⁡B=a2+c2−b22ac\cos B = {{{a^2} + {c^2} - {b^2}} \over {2ac}}
⇒35=a2+25−642(5)a\Rightarrow {3 \over 5} = {{{a^2} + 25 - 64} \over {2(5)a}}
⇒a2−6a−3g=0\Rightarrow {a^2} - 6a - 3g = 0

∴\therefore

a=6±1922=6±832a = {{6 \pm \sqrt {192} } \over 2} = {{6 \pm 8\sqrt 3 } \over 2}
⇒3+43\Rightarrow 3 + 4\sqrt 3

(Reject

a=3−43a = 3 - 4\sqrt 3

) Now,

Δ=abc4R=(3+43)(8)(5)4(5)=2(3+43)\Delta = {{abc} \over {4R}} = {{(3 + 4\sqrt 3 )(8)(5)} \over {4(5)}} = 2(3 + 4\sqrt 3 )
⇒Δ=(6+83)\Rightarrow \Delta = (6 + 8\sqrt 3 )

⇒\Rightarrow Option (3) is correct.

Q53
Let sin⁡Asin⁡B=sin⁡(A−C)sin⁡(C−B){{\sin A} \over {\sin B}} = {{\sin (A - C)} \over {\sin (C - B)}}, where A, B, C are angles of triangle ABC. If the lengths of the sides opposite these angles are a, b, c respectively, then :
A b2 −- a2 = a2 + c2
B b2, c2, a2 are in A.P.
C c2, a2, b2 are in A.P.
D a2, b2, c2 are in A.P.
Correct Answer
Option B
Solution
sin⁡Asin⁡B=sin⁡(A−C)sin⁡(C−B){{\sin A} \over {\sin B}} = {{\sin (A - C)} \over {\sin (C - B)}}

As A, B, C are angles of triangle.

A + B + C = π\pi A = π\pi −- (B + C) ......

(1) Similarly sinB = sin(A + C) .....

(2) From (1) and (2)

sin⁡(B+C)sin⁡(A+C)=sin⁡(A−C)sin⁡(C−B){{\sin (B + C)} \over {\sin (A + C)}} = {{\sin (A - C)} \over {\sin (C - B)}}
sin⁡(C+B).sin⁡(C−B)=sin⁡(A−C)sin⁡(A+C)\sin (C + B).\sin (C - B) = \sin (A - C)\sin (A + C)
sin⁡2C−sin⁡2B=sin⁡2A−sin⁡2C{\sin ^2}C - {\sin ^2}B = {\sin ^2}A - {\sin ^2}C

∵\because

{sin⁡(x+y)sin⁡(x−y)=sin⁡2x−sin⁡2y}\{ \sin (x + y)\sin (x - y) = {\sin ^2}x - {\sin ^2}y\}
2sin⁡2C=sin⁡2A+sin⁡2B2{\sin ^2}C = {\sin ^2}A + {\sin ^2}B

By sine rule

2c2=a2+b22{c^2} = {a^2} + {b^2}

⇒\Rightarrow b2, c2 and a2 are in A.P.

Q54
The lengths of the sides of a triangle are 10 + x2, 10 + x2 and 20 −- 2x2. If for x = k, the area of the triangle is maximum, then 3k2 is equal to :
A 5
B 8
C 10
D 12
Correct Answer
Option C
Solution
CD=(10+x2)2−(10−x2)2=210∣x∣CD = \sqrt {{{(10 + {x^2})}^2} - {{(10 - {x^2})}^2}} = 2\sqrt {10} |x|

Area

=12×CD×AB=12×210∣x∣(20−2x2)= {1 \over 2} \times CD \times AB = {1 \over 2} \times 2\sqrt {10} |x|(20 - 2{x^2})
A=10∣x∣(10−x2)A = \sqrt {10} |x|(10 - {x^2})
dAdx=10∣x∣x(10−x2)+10∣x∣(−2x)=0{{dA} \over {dx}} = \sqrt {10} {{|x|} \over x}(10 - {x^2}) + \sqrt {10} |x|( - 2x) = 0
⇒10−x2=2x2\Rightarrow 10 - {x^2} = 2{x^2}
3x2=103{x^2} = 10
x=kx = k
3k2=103{k^2} = 10
Q55
Let a, b and c be the length of sides of a triangle ABC such that a+b7=b+c8=c+a9{{a + b} \over 7} = {{b + c} \over 8} = {{c + a} \over 9}. If r and R are the radius of incircle and radius of circumcircle of the triangle ABC, respectively, then the value of Rr{R \over r} is equal to :
A 52{5 \over 2}
B 2
C 32{3 \over 2}
D 1
Correct Answer
Option A
Solution
a+b7=b+c8=c+a9=λ{{a + b} \over 7} = {{b + c} \over 8} = {{c + a} \over 9} = \lambda
a+b=7λa + b = 7\lambda
b+c=8λb + c = 8\lambda
c+a=9λc + a = 9\lambda
a+b+c=12λa + b + c = 12\lambda

∴\therefore

a=4λ, b=3λ, c=5λa = 4\lambda ,\,b = 3\lambda ,\,c = 5\lambda
S=4λ+3λ+5λ2=6λS = {{4\lambda + 3\lambda + 5\lambda } \over 2} = 6\lambda
Δ=S(s−a)(s−b)(s−c)=(6λ)(2λ)(3λ)(λ)=6λ2\Delta = \sqrt {S(s - a)(s - b)(s - c)} = \sqrt {(6\lambda )(2\lambda )(3\lambda )(\lambda )} = 6{\lambda ^2}
R=abc4Δ=(4λ)(3λ)(5λ)4(6λ2)=52λR = {{abc} \over {4\Delta }} = {{(4\lambda )(3\lambda )(5\lambda )} \over {4(6{\lambda ^2})}} = {5 \over 2}\lambda
r=Δs=6λ26λ=λr = {\Delta \over s} = {{6{\lambda ^2}} \over {6\lambda }} = \lambda
Rr=52λλ=52{R \over r} = {{{5 \over 2}\lambda } \over \lambda } = {5 \over 2}
Q56
For a triangle ABCABC, the value of cos⁡2A+cos⁡2B+cos⁡2C\cos 2A + \cos 2B + \cos 2C is least. If its inradius is 3 and incentre is M, then which of the following is NOT correct?
A MA→ . MB→=−18\overrightarrow {MA} \,.\,\overrightarrow {MB} = - 18
B sin⁡2A+sin⁡2B+sin⁡2C=sin⁡A+sin⁡B+sin⁡C\sin 2A + \sin 2B + \sin 2C = \sin A + \sin B + \sin C
C perimeter of ΔABC\Delta ABC is 183\sqrt3
D area of ΔABC\Delta ABC is 2732{{27\sqrt 3 } \over 2}
Correct Answer
Option D
Solution

If cos⁡2 A+cos⁡2 B+cos⁡2C\cos 2 \mathrm{~A}+\cos 2 \mathrm{~B}+\cos 2 \mathrm{C} is minimum then A=\mathrm{A}= B=C=60∘\mathrm{B}=\mathrm{C}=60^{\circ} So △ABC\triangle \mathrm{ABC} is equilateral Now in-radias r=3r=3 So in △MBD\triangle \mathrm{MBD} we have

tan⁡30∘=MDBD=ra/2=6a1/3=1a=a=63\begin{aligned} & \operatorname{tan} 30^{\circ}=\frac{M D}{B D}=\frac{r}{a / 2}=\frac{6}{a} \\\\ & 1 / \sqrt{3}=\frac{1}{a}=a=6 \sqrt{3} \end{aligned}

Perimeter of △ABC=183\triangle \mathrm{ABC}=18 \sqrt{3} Area of △ABC=34a2=273\triangle \mathrm{ABC}=\dfrac{\sqrt{3}}{4} a^2=27 \sqrt{3}

Q57
A straight line cuts off the intercepts OA=a\mathrm{OA}=\mathrm{a} and OB=b\mathrm{OB}=\mathrm{b} on the positive directions of xx-axis and yy axis respectively. If the perpendicular from origin OO to this line makes an angle of π6\dfrac{\pi}{6} with positive direction of yy-axis and the area of △OAB\triangle \mathrm{OAB} is 9833\dfrac{98}{3} \sqrt{3}, then a2−b2\mathrm{a}^{2}-\mathrm{b}^{2} is equal to :
A 3923\dfrac{392}{3}
B 98
C 196
D 1963\dfrac{196}{3}
Correct Answer
Option A
Solution
12ab=9833{1 \over 2}ab = {{98\sqrt 3 } \over 3}
⇒3ab=196\Rightarrow \sqrt 3 ab = 196

..... (i)

OP=OBcos⁡30∘=OAcos⁡60∘OP = OB\cos 30^\circ = OA\cos 60^\circ
⇒b32=a2\Rightarrow {{b\sqrt 3 } \over 2} = {a \over 2}
⇒3b=a\Rightarrow \sqrt 3 b = a

..... (ii) By (i) and (ii)

a2=196{a^2} = 196
a=14a = 14
b2=a23{b^2} = {{{a^2}} \over 3}
a2−b2=2a23=3923{a^2} - {b^2} = {{2{a^2}} \over 3} = {{392} \over 3}
Q58
In a triangle ABC, if cos⁡A+2cos⁡B+cos⁡C=2\cos \mathrm{A}+2 \cos \mathrm{B}+\cos C=2 and the lengths of the sides opposite to the angles A and C are 3 and 7 respectively, then cos⁡A−cos⁡C\mathrm{\cos A-\cos C} is equal to
A 37\dfrac{3}{7}
B 97\dfrac{9}{7}
C 107\dfrac{10}{7}
D 57\dfrac{5}{7}
Correct Answer
Option C
Solution
cos⁡A+2cos⁡B+cos⁡C=2cos⁡A+cos⁡C=2(1−cos⁡B)2cos⁡A+C2cos⁡(A−C2)=2×2sin⁡2B2cos⁡A−C2=2sin⁡B22cos⁡B2cos⁡A−C2=4sin⁡B2cos⁡B22sin⁡(A+C2)cos⁡(A−C2)=2sin⁡Bsin⁡A+sin⁡C=2sin⁡Ba+c=2b(∵a=3,c=7)⇒b=5\begin{aligned} & \cos A+2 \cos B+\cos C=2 \\\\ & \cos A+\cos C=2(1-\cos B) \\\\ & 2 \cos \frac{A+C}{2} \cos \left(\frac{A-C}{2}\right)=2 \times 2 \sin ^2 \frac{B}{2} \\\\ & \cos \frac{A-C}{2}=2 \sin \frac{B}{2} \\\\ & 2 \cos \frac{B}{2} \cos \frac{A-C}{2}=4 \sin \frac{B}{2} \cos \frac{B}{2} \\\\ & 2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right)=2 \sin B \\\\ & \sin A+\sin C=2 \sin B \\\\ & a+c=2 b \quad(\because a=3, c=7) \\\\ & \Rightarrow b=5 \quad \end{aligned}
cos⁡A−cos⁡C=b2+c2−a22bc−a2+b2−c22ab=25+49−970−9+25−4930=6570+12=2014=107\begin{aligned} \cos A & -\cos C=\frac{b^2+c^2-a^2}{2 b c}-\frac{a^2+b^2-c^2}{2 \mathrm{ab}} \\\\ & =\frac{25+49-9}{70}-\frac{9+25-49}{30} \\\\ & =\frac{65}{70}+\frac{1}{2}=\frac{20}{14}=\frac{10}{7} \end{aligned}
Q59
Let (5,a4)\left(5, \dfrac{a}{4}\right) be the circumcenter of a triangle with vertices A(a,−2),B(a,6)\mathrm{A}(a,-2), \mathrm{B}(a, 6) and C(a4,−2)C\left(\dfrac{a}{4},-2\right). Let α\alpha denote the circumradius, β\beta denote the area and γ\gamma denote the perimeter of the triangle. Then α+β+γ\alpha+\beta+\gamma is
A 60
B 62
C 53
D 30
Correct Answer
Option C
Solution
A(a,−2),B(a,6),C(a4,−2),O(5,a4)AO=BO(a−5)2+(a4+2)2=(a−5)2+(a4−6)2a=8AB=8,AC=6,BC=10α=5,β=24,γ=24\begin{aligned} & A(a,-2), B(a, 6), C\left(\frac{a}{4},-2\right), O\left(5, \frac{a}{4}\right) \\ & A O=B O \\ & (a-5)^2+\left(\frac{a}{4}+2\right)^2=(a-5)^2+\left(\frac{a}{4}-6\right)^2 \\ & a=8 \\ & A B=8, A C=6, B C=10 \\ & \alpha=5, \beta=24, \gamma=24 \end{aligned}
Q60
Two vertices of a triangle ABC\mathrm{ABC} are A(3,−1)\mathrm{A}(3,-1) and B(−2,3)\mathrm{B}(-2,3), and its orthocentre is P(1,1)\mathrm{P}(1,1). If the coordinates of the point C\mathrm{C} are (α,β)(\alpha, \beta) and the centre of the of the circle circumscribing the triangle PAB\mathrm{PAB} is (h,k)(\mathrm{h}, \mathrm{k}), then the value of (α+β)+2( h+k)(\alpha+\beta)+2(\mathrm{~h}+\mathrm{k}) equals
A 81
B 15
C 51
D 5
Correct Answer
Option D
Solution
mPA=2−2=−1∴mBC=1\begin{aligned} & m_{P A}=\frac{2}{-2}=-1 \\ & \therefore \quad m_{B C}=1 \end{aligned}
BC:y=x+5mBP=2−3=−23∴mAC=32AC:y=32x−112⇒2y=3x−11∴C:(21,26)\begin{aligned} & B C: y=x+5 \\ & m_{B P}=\frac{2}{-3}=\frac{-2}{3} \\ & \therefore m_{A C}=\frac{3}{2} \\ & A C: y=\frac{3}{2} x-\frac{11}{2} \quad \Rightarrow 2 y=3 x-11 \\ & \therefore \quad C:(21,26) \end{aligned}

Let the circumcentre be

(h,k)(h, k)
(h−1)2+(k−1)2=(h+2)2+(k−3)2... (i)(h−1)2+(k−1)2=(h−3)2+(k+1)2... (ii)\begin{aligned} & (h-1)^2+(k-1)^2=(h+2)^2+(k-3)^2 \quad \text{... (i)}\\ & (h-1)^2+(k-1)^2=(h-3)^2+(k+1)^2 \quad \text{... (ii)} \end{aligned}

Solving (i) and (ii)

h=−192,k=−232α+β+2(h+k)=21+26−19−23=2+3=5\begin{aligned} & h=\frac{-19}{2}, k=\frac{-23}{2} \\ & \alpha+\beta+2(h+k) \\ & =21+26-19-23 \\ & =2+3=5 \end{aligned}
Ready for a full JEE mock test? Timed · full syllabus · instant results
Take a Mock Test →