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Must Do Problems Of Properties of Triangles For NDA

Must Do Problems Of Properties of Triangles

Must Do Problems
Q 2680512417

In `Delta ABC`, least value of `e^A/A + e^B/B + e^C/C ` is equal to


BCECE Mains 2015
(A)

` 9/pi e^(pi//3)`

(B)

`pi/3 e^(pi//3)`

(C)

`pi/9 e^(pi//3)`

(D)

None of these

Solution:

We have,

` e^A/A + e^B/B + e^C/C >= 3 (e^(A + B + C)/(ABC) )^(1//3)`

[ using `AM >= GM`]

` => e^A/A + e^B/B + e^C/C >= 3 ( e^pi /(ABC) )^(1//3)`

Also, `A + B + C >= 3(ABC)^(1//3) ` [using `AM >= GM`]

` => pi/3 >= (ABC)^(1//3)`

` => ( 1/(ABC) )^(1//2) >= 3/(11)`

` => ( e^pi/(ABC))^(1//3) >= 3/pi e^(pi//3)`

` => 3 ( e^pi/(ABC))^(1//3) >= 9/pi e^(pi//3)`

` => e^A/A + e^B/B + e^C/C >= 9/pi e^(pi//3)`
Correct Answer is `=>` (A) ` 9/pi e^(pi//3)`
Q 2335278162

If in a `Delta ABC, cos B = (sin C)/(2 sin A)`, then the triangle is
BITSAT Mock
(A)

right-angled

(B)

equilateral

(C)

isosceles

(D)

right-angled isosceles

Solution:

`sin C = 2 sin A cos B`

`= sin (A + B) + sin (A - B)`

`= sin C + sin (A - B)`

`=> sin (A - B) = 0 => A = B`

i.e., the triangle is isosceles.
Correct Answer is `=>` (C) isosceles
Q 1563880745

The angles of a triangle are in `AP` and the least angle is `30^o`. The greatest angle in radians is:
BITSAT 2012
(A)

`\frac{7\pi}{12}`

(B)

`\frac{2\pi}{3}`

(C)

`\frac{5\pi}{6}`

(D)

`\frac{\pi}{2}`

Solution:

Let triangle be `ABC` and we have

`A=30^o=\ frac{\pi}{6}`

`\therefore B =\ frac{\pi}{6}+d, C =\ frac{\pi}{6}+2d`

Where, `d` is the common difference of the `A.P.`

Now, sum of all the angles is `\pi`

`\Rightarrow \ frac{\pi}{2}+3d=\pi\Rightarrow d=\ frac{\pi}{6}`

Hence, greatest angle `=\ frac{\pi}{6}+2d=\ frac{\pi}{2}`
Correct Answer is `=>` (D) `\frac{\pi}{2}`
Q 2268467305

If in a `DeltaABC, a^2 + b^2 + c^2 = ca + ab sqrt(3)`,

then the triangle
BITSAT Mock
(A)

is equilateral

(B)

is right-angled isosceles

(C)

has angles `30^(circ), 60^(circ)` and `90^(circ) `

(D)

None of these

Solution:

`a^2 + b^2 + c^2 - ca - ab sqrt(3) = 0`

Regrouping,

`(b- (a sqrt(3))/2)^2 + (c- a/2)^2 =0`

`:. a : b : c = 2 : sqrt(3) : 1`.

Hence the triangle has angles

`30^(circ), 60^(circ)` and `90^(circ)`.
Correct Answer is `=>` (C) has angles `30^(circ), 60^(circ)` and `90^(circ) `
Q 2539080812

In `Delta ABC`, if ` a/(b^2 - c^2) + c/(b^2 - a^2) = 0`, then `B` is equal to


BCECE Mains 2015
(A)

`pi/2`

(B)

`pi//4`

(C)

`2pi//3`

(D)

`pi//3`

Solution:

We have,

`a/(sin A) = b/(sin B) = c/(sin C) = k`

`:. a/(b^2 - c^2) + c/(b^2 - a^2) = 0`

`=> (k sin A)/(k^2 (sin^2 B - B sin^2 C)`

`+ (k sin C)/(k^2 (sin^2 B - B sin^2 A) = 0`

` => (sin A)/(sin (B + C) sin (B - C))`

` + ( sin C)/(sin (B + A) sin (B - A)) = 0`

` => 1/(sin (B -C)) + 1/(sin (B -A)) = 0`

` => sin (B - A) + sin (B -C) = 0`

`=> sin (A - B) = sin (B -C)`

`=> A - B = B - C`

`=> A + C= 2B => B = 60^0`
Q 1510434319

In a `triangle ABC` if the sides are `a=3, b = 5` and `c = 4`, then `sin (B/2) + cos (B/2)` is equal to :
BITSAT 2006
(A)

`sqrt2`

(B)

`(sqrt3 + 1)/2`

(C)

`(sqrt3 - 1)/2`

(D)

`1`

Solution:

We know,

`cos B = (a^2 + c^2 - b^2)/(2ac)`

`cos B = (3^2 + 4^2 - 5^2)/(2(3)(4)) = (9+16-25)/(2(3)
(4)) = 0`

=> `B = 90^0`

`therefore sin(B/2) + cos(B/2) = sin 45^0 + cos 45^0`

= `1/sqrt2 + 1/sqrt2 = sqrt2`
Correct Answer is `=>` (A) `sqrt2`
Q 2542212133

Let p, q and r be the sides opposite to the angles P, Q and R, respectively in a `Delta PQR`. If `r^2 sin P sin Q = pq`, then the triangle is
WBJEE 2012
(A)

equilateral

(B)

acute angled but not equilateral

(C)

obtuse angled

(D)

right angled

Solution:

We know that in `Delta ABC`

`a/(sin A) = b/(sin B) = c/(sin C) = 2R`

`:. r^2 sin P sin Q = pq`

`=> r^2 · p/(2R_1) · q/(2 R_1) = pq`, where `R_1` is circumradius

of `Delta PQR`.

`=> r^2 = 4R_1^2`

`=> r = 2R_1`

` => 2R_1 sin R = 2R_1`

`=> R_1 = 90°`

`:. Delta PQR` is right angled.
Correct Answer is `=>` (D) right angled
Q 2476267176

If in `DeltaABC, cos A+ cosB + cosC = 3/2`, then
triangle `Delta` is
UPSEE 2013
(A)

right angled

(B)

isosceles

(C)

acute

(D)

equilateral

Solution:

`:. cos A+ cos B + cos C = 3/2`

`=> (b^2 +c^2 -a^2)/(2bc) + (a^2 + c^2 -b^2)/(2ac)+ (a^2 + b^2 -c^2)/(2ab) =3/2`

`=> (a+b -c) (a-b)^2 + (b+c -a) (b-c)^2`

`+ (c+a-b) (c-a)^2 =0`

As we know, `a+ b > c, b + c >a, c + a > b`


`=> a = b = c`

Hence, triangle is an equilateral.
Correct Answer is `=>` (D) equilateral
Q 1665478365

Angles `A, B, C` of a `DeltaABC` are in `AP` and
`b: c = sqrt(3) : sqrt(2)`, then the `angleA` is given by
GGSIPU 2014
(A)

`45^(0)`

(B)

`60^(0)`

(C)

`75^(0)`

(D)

`90^(0)`

Solution:

Since, `A, B, C` are in `AP`.

:. We have, `2B =A+ C`

Also, since `A+ B + C = 180^0` ............(1)

`=> 3B = 180^0`

`=> B =60^0` ................(2)

Now, by using sine rule. we have

` a/ (sin A) = b/ (sin B) =c/ (sin C)`

`=> sin C = c/b sin B`

` = sqrt(2)/ sqrt(3) sin 60^0 = sqrt(2)/ sqrt(3) xx sqrt(3)/ 2 = 1/sqrt(2) = sin 45^0`

` => C = 45^0` .............(3)

` => A = 75^0`

[using Eqs. (i), (ii) and (iii))
Correct Answer is `=>` (C) `75^(0)`
Q 1113378240

In a triangle `Delta ABC`, `C=90^0;` then `(a^2-b^2)/(a^2+b^2) =` ?
EAMCET 2010
(A)

`sin(A+B)`

(B)

`sin(A-B)`

(C)

`cos(A+B)`

(D)

`cos(A-B)`

Solution:

We know the value of `C` which is `90^0.`Using sine law of triangles,

`a/sin(A) = b/sin(B) = c/sin(C)`

`a/sin(A) = b/sin(B) = c/sin(90^0)`

`=> a=csin(A) and b=csin(B)`

Also from cosine law of triangles,we know that:

`c^2 = a^2 + b^2 - 2ab cos(C)`

Put `C=90^0` and value of a and `b` in the cosine law,

we get `c^2=a^2+b^2= (csinA)^2+(csinB)^2= c^2(sin^2A+sin^2B) => sin^2A+sin^2B=1`

`(a^2-b^2)/(a^2+b^2) = ((c^2)(sin^2 (A) - sin^2 (B))/(c^2))`
`=> (a^2-b^2)/(a^2+b^2) = (sin^2 (A) - sin^2 (B))`

`B=90-A`

`=>(a^2-b^2)/(a^2+b^2) = (sin^2(A)-cos^2(A))=1-2cos^2(A)`

`=-cos(2A)=-cos(A+90-B)=sin(A-B)`
Correct Answer is `=>` (B) `sin(A-B)`
Q 2573601546

If in a `Delta ABC, sin A, sin B, sin C` are in AP,
then
WBJEE 2011
(A)

the altitudes are in AP

(B)

the altitudes are in HP

(C)

the angles are in AP

(D)

the angles are in HP

Solution:

If `p_1, p_2` and `p_3` are the altitudes of a triangle.

Then, `Delta 1/2 ap_1 =1/2 bp_2 = 1/2 c p_3`

`=> a= (2 Delta)/p_1 , b =(2 Delta)/p_2, c= (2 Delta)/p_3`

Since, `sin A, sin B` and `sin C` are in AP.

`:. sin B = (sin A + sin C)/2`

`=> b= (a+c)/2`

`=> (2 Delta)/(p_2) = ( ( (2 Delta)/(p_1) ) + ((2 Delta)/(p_3)) )/2`

`=> 2/p_2 = (p_3 +p_1)/(p_1 p_3)`

`=> p_2 = (2p_1 p_3)/(p_1 + p_3)`

Hence, altitudes are in AP.
Correct Answer is `=>` (A) the altitudes are in AP
Q 2583501447

In a `DeltaABC , 2ac sin ((A-B+C)/2) ` is equal to
WBJEE 2010
(A)

`a^2 + b^2 - c^2`

(B)

`c^2 + a^2 - b^2`

(C)

`b^2 - a^2 - c^2`

(D)

`c^2 - a^2 - b^2`

Solution:

`2ac sin ((A-B+C)/2) = 2ac sin ((A+C)/2 -B/2)`


` = 2ac sin (pi/2-B/2-B/2)` `[because (A+C)/2 = pi/2-B/2]`



` = 2ac sin (pi/2-B)`


` = 2ac cosB`

` = a^2+c^2-b^2 [because cosB = (a^2+c^2-b^2)/(2ac)]`
Correct Answer is `=>` (B) `c^2 + a^2 - b^2`
Q 2562312235

Let `p, q` and `r` be the side4s opposite to the angles P,Q and R, respectively in a `Delta PQR`. Then, `2pr sin ( (P - Q + R)/2)` equals
WBJEE 2012
(A)

`p^2 + q^2 + r^2`

(B)

`p^2 + r^2 - q^2`

(C)

`q^2 + r^2 - p^2`

(D)

`p^2 + q^2 - r^2`

Solution:

In `Delta PQR, P + Q + R = 180°`

`:. 2pr sin ( (P-Q+R)/2) = 2pr sin (180°-Q-Q)/2`

`= 2pr sin (90°- Q)`

`= 2pr cos Q`

`( In Delta PQR cos Q = (p^2 + r^2 - q^2)/(2pr))`

`= 2pr ((p^2 + r^2- q^2)/(2pr)) = p^2 + r^2 - q^2`
Correct Answer is `=>` (B) `p^2 + r^2 - q^2`
Q 2448178003

If in a `Delta ABC`, the altitudes from the
vertices `A, B, C` on opposite sides are in HP
then `sin A, sin B, sin C` are in
BCECE Stage 1 2016
(A)

HP

(B)

Arithmetico-Geometric Progression

(C)

AP

(D)

GP

Solution:

In `Delta BAD`,

`cos (90-B)=(AD)/c`

`AD=c sin B`

Similarly, `BE= a sin C` and `CF = bsin A`

Since, `AD, BE, CF` are in HP

`:. csin B, a sin C, bsin A` are in HP

`1/(sin C sin B), 1/(sin A sin C), 1/(sin B sin A)` are in AP

`sin A, sin B, sin C` are in AP
Correct Answer is `=>` (C) AP
Q 1666578475

Consider a `Delta ABC` satisfying

`2a sin^2 (C/2) + 2c sin^2 ( A/2) = 2a + 2c - 3b`

`sin A, sin B, sin C` are in

DSSB Paper 1 2015
(A)

GP

(B)

AP

(C)

HP

(D)

Neither in GP nor in AP nor in HP

Solution:

Since, `a, b`, care in `AP`.

We know that if each term of an AP is multiplied or

divide by a non-zero constant K, then the resulting

sequence is also an `AP`.

So, `K sin A, K sin B, K sin C` also in `AP`.

`:. sin A sin B sin C` are in `AP`.
Correct Answer is `=>` (B) AP
Q 1184723657

In any `DeltaABC,((a+b+c)(b+c-a)(c+a-b)(a+b-c))/(4b^2c^2)=`
EAMCET 2014
(A)

`sin^2B`

(B)

`cos^2A`

(C)

` cos^2B`

(D)

`sin^2A`

Solution:

As, `s=(a+b+c)/2=>2s=a+b+c`

And, `b+c-a=a+b+c-2a=2s-2a=2(s-a)`

Similarly, `c+a-b=2s-2b=2(s-b)`

And, `a+b-c=2s-2c=2(s-c)`

`=16*(s(s-a)(s-b)(s-c))/(4b^2c^2)`

`=((2sqrt(s(s-a)(s-b)(s-c)))/(bc))^2`

Also, we know that `Delta= sqrt(s(s-a)(s-b)(s-c))`

`=((2Delta)/(bc))^2`

`=sin^2A`
Q 1133878742

In `DeltaABC`, if `1/(b+c)+1/(c+a)=3/(a+b+c)`, then angle `C` is equal to
EAMCET 2008
(A)

`90^@`

(B)

`60^@`

(C)

`45^@`

(D)

`30^@`

Solution:

Given,

`1/(b+c)+1/(c+a)=3/(a+b+c)`

=>`1+b/(a+c)+1+a/(b+c)=3`

=>`b(b+c)+a(a+c)=(a+c)(b+c)`

=>`b^2+bc+a^2+ac=ab+ac+bc+c^2`

=>`a^2+b^2-c^2=ab`

=>Now ,`cosC=(a^2+b^2-c^2)/(2ab)=(ab)/(2ab)=1/2`

=>`angleC=60^@`
Correct Answer is `=>` (B) `60^@`
Q 2542212133

Let p, q and r be the sides opposite to the angles P, Q and R, respectively in a `Delta PQR`. If `r^2 sin P sin Q = pq`, then the triangle is
WBJEE 2012
(A)

equilateral

(B)

acute angled but not equilateral

(C)

obtuse angled

(D)

right angled

Solution:

We know that in `Delta ABC`

`a/(sin A) = b/(sin B) = c/(sin C) = 2R`

`:. r^2 sin P sin Q = pq`

`=> r^2 · p/(2R_1) · q/(2 R_1) = pq`, where `R_1` is circumradius

of `Delta PQR`.

`=> r^2 = 4R_1^2`

`=> r = 2R_1`

` => 2R_1 sin R = 2R_1`

`=> R_1 = 90°`

`:. Delta PQR` is right angled.
Correct Answer is `=>` (D) right angled
Q 1580578417

The circumradius of the triangle whose sides are `13, 12` and `5`, is :
BITSAT 2006
(A)

`15`

(B)

`13/2`

(C)

`15/2`

(D)

`6`

Solution:

Let sides are `a= 13, b = 12, c = 5`

Now, `a^2 = b^2 + c^2`

`=> (13)^2 = (12)^2 + 5^2`

`=> 169 = 169`

`=> angleA = 90^0`

we know, `R = a/(2sinA)`

`R = 13/(2sin90^0) =13/2`
Correct Answer is `=>` (B) `13/2`
Q 1550723614

In a triangle, if ` r_1 = 2r_2 =3r_3` , then `a/b + b/c + c/a` is equal to
BITSAT 2008
(A)

`170/60`

(B)

`181/60`

(C)

`190/60`

(D)

`191/60`

Solution:

Given that, `r_1=2r_2=3r_3`

`Delta/(s-a) = (2 Delta)/(s-b) =( 3 Delta)/(s-c) = Delta/k`

Then , `s – a = k, s – b = 2 k, s – c = 3 k`

`3 s – (a + b + c) = 6 k` ,

`s = 6 k`

`a/5 = b/4 = c/3 =k`

Now, `a/b + b/c +c/a =5/4 + 4/3 + 3/5`

`(75+80+36)/60` = `191/60`
Correct Answer is `=>` (D) `191/60`
Q 1545580463

If in a `\triangle ABC, r_{1} = 2r_{2} = 3r_{3}`, then the perimeter of the triangle is equal to
EAMCET 2015
(A)

`3a`

(B)

`3b`

(C)

`2c`

(D)

`3(a + b + c)`

Solution:

We have, `\triangle ABC, r_{1} = 2r_{2} = 3r_{3}`

`\Rightarrow \ frac {\Delta}{s - a} = \ frac {2\Delta}{s - b} = \ frac {3\Delta}{s - c} = \ frac {1}{k},`

`\Rightarrow s=a+k\Delta`......(1),

` s=b+2k\Delta` .....(2)

and `s=c+3k\Delta`........(3)

add all above equations

`\Rightarrow 3s=2s+6k\Delta`

`\Rightarrow s =6k\Delta`

`\Rightarrow k\Delta =\ frac{s}{6}`

Now using (3) `s=c+\ frac{s}{2}`

` \Rightarrow s=2c`
Correct Answer is `=>` (C) `2c`
Q 1629223111

The area of the largest triangle that can be inscribed in a semi-circle of radius `r` is
CDS 2015
(A)

`r^2`

(B)

`2r^2`

(C)

`3 r^2`

(D)

`4 r^2`

Solution:

Let the `r` be the radius of semi-circle.

In `Delta ABC, AO = OB = OC = r`

`:.` Area of triangle `=1/2 xx` Base x Height

`=1/2 xx AB xx OC`

`=1/2 xx 2 r xx r =r^2`

Hence, the area of the largest triangle is `r^ 2`
Correct Answer is `=>` (A) `r^2`
 
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