The area of circle circumscribing abc is-Turito

Maths-

General

Easy

Question

The area of circle circumscribing ΔABC is

$\frac{π}{8}$
$\frac{π}{4}$
$\frac{π}{2}$
$π$

The correct answer is: $pi over 4$

Book A Free Demo

Name*

Mobile*

+91

Email*

Grade*

I agree to get WhatsApp notifications & Marketing updates

Related Questions to study

General

maths-

Statement-I : The statement that circumradius and inradius of a triangle are 12 and 8 respectively can not be correct. Statement-II : Circumradius 2 (inradius)

maths-General

General

maths-

The complete solution set of the equation $open vertical bar x to the power of 2 end exponent minus x close vertical bar plus vertical line x plus 3 vertical line equals open vertical bar x to the power of 2 end exponent minus 2 x minus 3 close vertical bar$ is

table row cell open parentheses x to the power of 2 end exponent minus x close parentheses left parenthesis x plus 3 right parenthesis less or equal than 0 end cell row cell x left parenthesis x minus 1 right parenthesis left parenthesis x plus 3 right parenthesis less or equal than 0 end cell row cell x element of left parenthesis negative infinity comma negative 3 right square bracket union left square bracket 0 , 1 right square bracket end cell end table

The complete solution set of the equation $open vertical bar x to the power of 2 end exponent minus x close vertical bar plus vertical line x plus 3 vertical line equals open vertical bar x to the power of 2 end exponent minus 2 x minus 3 close vertical bar$ is

maths-General

table row cell open parentheses x to the power of 2 end exponent minus x close parentheses left parenthesis x plus 3 right parenthesis less or equal than 0 end cell row cell x left parenthesis x minus 1 right parenthesis left parenthesis x plus 3 right parenthesis less or equal than 0 end cell row cell x element of left parenthesis negative infinity comma negative 3 right square bracket union left square bracket 0 , 1 right square bracket end cell end table

General

Maths-

The solution(s) of the equation cos2x sin6x = cos3x sin5x in the interval [0, $straight pi$ ] is/are –

Maths-General

General

maths-

The equation $4 s i n to the power of 2 end exponent invisible function application x minus 2 left parenthesis square root of 3 plus 1 right parenthesis s i n invisible function application x plus square root of 3 equals 0$ has

sin to the power of 4 end exponent invisible function application x minus cos to the power of 2 end exponent invisible function application xsin invisible function application x plus 2 sin to the power of 2 end exponent invisible function application x plus sin invisible function application x equals 0

table row cell s i n invisible function application x open square brackets sin to the power of 3 end exponent invisible function application x minus cos to the power of 2 end exponent invisible function application x plus 2 sin invisible function application x plus 1 close square brackets equals 0 end cell row cell s i n invisible function application x open square brackets sin to the power of 3 end exponent invisible function application x minus 1 plus sin to the power of 2 end exponent invisible function application x plus 2 sin invisible function application x plus 1 close square brackets equals 0 end cell row cell s i n invisible function application x open square brackets sin to the power of 3 end exponent invisible function application x plus sin to the power of 2 end exponent invisible function application x plus 2 sin invisible function application x close square brackets equals 0 end cell row cell s i n to the power of 2 end exponent invisible function application x equals 0 text end text text o end text text r end text text end text s i n to the power of 2 end exponent invisible function application x plus s i n invisible function application x plus 2 equals 0 end cell end table

(not possible for real x)

text or  end text sin invisible function application x equals 0

Hence, the solutions are x = 0, p, 2p, 3p.

The equation $4 s i n to the power of 2 end exponent invisible function application x minus 2 left parenthesis square root of 3 plus 1 right parenthesis s i n invisible function application x plus square root of 3 equals 0$ has

maths-General

sin to the power of 4 end exponent invisible function application x minus cos to the power of 2 end exponent invisible function application xsin invisible function application x plus 2 sin to the power of 2 end exponent invisible function application x plus sin invisible function application x equals 0

table row cell s i n invisible function application x open square brackets sin to the power of 3 end exponent invisible function application x minus cos to the power of 2 end exponent invisible function application x plus 2 sin invisible function application x plus 1 close square brackets equals 0 end cell row cell s i n invisible function application x open square brackets sin to the power of 3 end exponent invisible function application x minus 1 plus sin to the power of 2 end exponent invisible function application x plus 2 sin invisible function application x plus 1 close square brackets equals 0 end cell row cell s i n invisible function application x open square brackets sin to the power of 3 end exponent invisible function application x plus sin to the power of 2 end exponent invisible function application x plus 2 sin invisible function application x close square brackets equals 0 end cell row cell s i n to the power of 2 end exponent invisible function application x equals 0 text end text text o end text text r end text text end text s i n to the power of 2 end exponent invisible function application x plus s i n invisible function application x plus 2 equals 0 end cell end table

(not possible for real x)

text or  end text sin invisible function application x equals 0

Hence, the solutions are x = 0, p, 2p, 3p.

General

maths-

$s i n to the power of 2 end exponent invisible function application x minus c o s invisible function application 2 x equals 2 minus s i n invisible function application 2 x$ if

table row cell s i n to the power of 2 end exponent invisible function application x minus c o s invisible function application 2 x equals 2 minus s i n invisible function application 2 x end cell row cell rightwards double arrow s i n to the power of 2 end exponent invisible function application x minus open parentheses 1 minus 2 s i n to the power of 2 end exponent invisible function application x close parentheses equals 2 minus 2 s i n invisible function application x c o s invisible function application x end cell row cell rightwards double arrow 3 s i n to the power of 2 end exponent invisible function application x plus 2 s i n invisible function application x c o s invisible function application x equals 3 end cell row cell text end text text C end text text a end text text s end text text e end text text-end text text 1 end text text end text colon c o s invisible function application x not equal to 0 end cell row cell therefore 3 t a n to the power of 2 end exponent invisible function application x plus 2 t a n invisible function application x equals 3 open parentheses 1 plus t a n to the power of 2 end exponent invisible function application x close parentheses rightwards double arrow t a n invisible function application x equals fraction numerator 3 over denominator 2 end fraction end cell row cell text end text text C end text text a end text text s end text text e end text text-end text text I end text text I end text text end text colon c o s invisible function application x equals 0 end cell row cell therefore 3 left parenthesis 1 right parenthesis plus 2 left parenthesis plus-or-minus 1 right parenthesis left parenthesis 0 right parenthesis equals 3 text end text text w end text text h end text text i end text text c end text text h end text text end text text i end text text s end text text end text text t end text text r end text text u end text text e end text text end text therefore x equals left parenthesis 2 n plus 1 right parenthesis fraction numerator pi over denominator 2 end fraction end cell end table

$s i n to the power of 2 end exponent invisible function application x minus c o s invisible function application 2 x equals 2 minus s i n invisible function application 2 x$ if

maths-General

table row cell s i n to the power of 2 end exponent invisible function application x minus c o s invisible function application 2 x equals 2 minus s i n invisible function application 2 x end cell row cell rightwards double arrow s i n to the power of 2 end exponent invisible function application x minus open parentheses 1 minus 2 s i n to the power of 2 end exponent invisible function application x close parentheses equals 2 minus 2 s i n invisible function application x c o s invisible function application x end cell row cell rightwards double arrow 3 s i n to the power of 2 end exponent invisible function application x plus 2 s i n invisible function application x c o s invisible function application x equals 3 end cell row cell text end text text C end text text a end text text s end text text e end text text-end text text 1 end text text end text colon c o s invisible function application x not equal to 0 end cell row cell therefore 3 t a n to the power of 2 end exponent invisible function application x plus 2 t a n invisible function application x equals 3 open parentheses 1 plus t a n to the power of 2 end exponent invisible function application x close parentheses rightwards double arrow t a n invisible function application x equals fraction numerator 3 over denominator 2 end fraction end cell row cell text end text text C end text text a end text text s end text text e end text text-end text text I end text text I end text text end text colon c o s invisible function application x equals 0 end cell row cell therefore 3 left parenthesis 1 right parenthesis plus 2 left parenthesis plus-or-minus 1 right parenthesis left parenthesis 0 right parenthesis equals 3 text end text text w end text text h end text text i end text text c end text text h end text text end text text i end text text s end text text end text text t end text text r end text text u end text text e end text text end text therefore x equals left parenthesis 2 n plus 1 right parenthesis fraction numerator pi over denominator 2 end fraction end cell end table

General

maths-

$fraction numerator 3 x squared plus x plus 1 over denominator left parenthesis x minus 1 right parenthesis to the power of 4 end fraction equals fraction numerator a over denominator left parenthesis x minus 1 right parenthesis end fraction plus fraction numerator b over denominator left parenthesis x minus 1 right parenthesis squared end fraction plus fraction numerator c over denominator left parenthesis x minus 1 right parenthesis cubed end fraction plus fraction numerator d over denominator left parenthesis x minus 1 right parenthesis to the power of 4 end fraction text then end text open square brackets table attributes columnalign left left columnspacing 1em end attributes row cell a b end cell row cell c d end cell end table close square brackets equals$

maths-General

General

maths-

If $vertical line x vertical line less than 1 fifth$ , the coefficient of $x cubed$ in the expansion of $fraction numerator 1 over denominator left parenthesis 1 minus 5 x right parenthesis left parenthesis 1 minus 4 x right parenthesis end fraction$ is

maths-General

General

maths-

If $a subscript K equals fraction numerator 1 over denominator K left parenthesis K plus 1 right parenthesis end fraction$ for $K equals 1 comma 2 comma 3 horizontal ellipsis.. n$ then $open parentheses sum from K equals 1 to n of a subscript K close parentheses squared equals$

Because of the greater size of Br, Br exert greater repulsive force. Hence Br in the equitorial position add maximum stability.

If $a subscript K equals fraction numerator 1 over denominator K left parenthesis K plus 1 right parenthesis end fraction$ for $K equals 1 comma 2 comma 3 horizontal ellipsis.. n$ then $open parentheses sum from K equals 1 to n of a subscript K close parentheses squared equals$

maths-General

Because of the greater size of Br, Br exert greater repulsive force. Hence Br in the equitorial position add maximum stability.

General

maths-

The number of partial fractions of $fraction numerator 2 over denominator x to the power of 4 plus x squared plus 1 end fraction$ is

maths-General

General

maths-

The number of partial fractions of $fraction numerator x cubed minus 3 x squared plus 3 x over denominator left parenthesis x minus 1 right parenthesis to the power of 5 end fraction$ is

maths-General

General

maths-

Coefficient of $x cubed y to the power of 4 z squared$ is $left parenthesis 2 x minus 3 y plus 4 z right parenthesis to the power of 9$ is

6 A g N O subscript 3 end subscript plus A s H subscript 3 end subscript plus 3 H subscript 2 end subscript O ⟶

6 A g stack stack downwards arrow with _ below plus H subscript 3 end subscript A s O subscript 3 end subscript with _ below plus 6 H N O subscript 3 end subscript

Coefficient of $x cubed y to the power of 4 z squared$ is $left parenthesis 2 x minus 3 y plus 4 z right parenthesis to the power of 9$ is

maths-General

6 A g N O subscript 3 end subscript plus A s H subscript 3 end subscript plus 3 H subscript 2 end subscript O ⟶

6 A g stack stack downwards arrow with _ below plus H subscript 3 end subscript A s O subscript 3 end subscript with _ below plus 6 H N O subscript 3 end subscript

General

maths-

Coefficient of $x to the power of r$ is $1 plus left parenthesis 1 plus x right parenthesis plus left parenthesis 1 plus x right parenthesis squared plus$ $horizontal ellipsis plus left parenthesis 1 plus x right parenthesis to the power of n$ is

Because of the partial positive charge on adjacent N-atoms there is a repulsion along N  N bond, hence N  N bond distance become higher.

Coefficient of $x to the power of r$ is $1 plus left parenthesis 1 plus x right parenthesis plus left parenthesis 1 plus x right parenthesis squared plus$ $horizontal ellipsis plus left parenthesis 1 plus x right parenthesis to the power of n$ is

maths-General

Because of the partial positive charge on adjacent N-atoms there is a repulsion along N  N bond, hence N  N bond distance become higher.

General

maths-

If a,b,c,d are consective binomial coefficients of $left parenthesis 1 plus x right parenthesis to the power of n$ then $fraction numerator a plus b over denominator a end fraction comma fraction numerator b plus c over denominator b end fraction comma fraction numerator c plus d over denominator c end fraction$ are is

maths-General

General

maths-

No of terms whose value depend on 'x' is $left parenthesis x squared minus 2 plus 1 divided by x squared right parenthesis to the power of n$ is

B a open parentheses N subscript 3 end subscript close parentheses subscript 2 end subscript stack ⟶ with capital delta on top B a plus 3 N subscript 2 end subscript

No of terms whose value depend on 'x' is $left parenthesis x squared minus 2 plus 1 divided by x squared right parenthesis to the power of n$ is

maths-General

B a open parentheses N subscript 3 end subscript close parentheses subscript 2 end subscript stack ⟶ with capital delta on top B a plus 3 N subscript 2 end subscript

General

Maths-

If $T subscript 0 comma T subscript 1 comma T subscript 2 comma horizontal ellipsis. T subscript n$ represent the terms is $left parenthesis x plus a right parenthesis to the power of n$ then $left parenthesis T subscript 0 minus T subscript 2 plus T subscript 4 minus T subscript 6 plus midline horizontal ellipsis right parenthesis squared plus$ $left parenthesis T subscript 1 minus T subscript 3 plus T subscript 5 minus horizontal ellipsis right parenthesis squared$ is

Maths-General

Have a query? Ask a Tutor!!

Can’t find what you’re looking for?

Question

The area of circle circumscribing ΔABC is

$\frac{π}{8}$
$\frac{π}{4}$
$\frac{π}{2}$
$π$

Answer

The correct answer is: $pi over 4$

Book A Free Demo

Related Questions to study

Statement-I : The statement that circumradius and inradius of a triangle are 12 and 8 respectively can not be correct. Statement-II : Circumradius 2 (inradius)

Statement-I : The statement that circumradius and inradius of a triangle are 12 and 8 respectively can not be correct. Statement-II : Circumradius 2 (inradius)

The complete solution set of the equation $open vertical bar x to the power of 2 end exponent minus x close vertical bar plus vertical line x plus 3 vertical line equals open vertical bar x to the power of 2 end exponent minus 2 x minus 3 close vertical bar$ is

The complete solution set of the equation $open vertical bar x to the power of 2 end exponent minus x close vertical bar plus vertical line x plus 3 vertical line equals open vertical bar x to the power of 2 end exponent minus 2 x minus 3 close vertical bar$ is

The solution(s) of the equation cos2x sin6x = cos3x sin5x in the interval [0, $straight pi$ ] is/are –

The solution(s) of the equation cos2x sin6x = cos3x sin5x in the interval [0, $straight pi$ ] is/are –

The equation $4 s i n to the power of 2 end exponent invisible function application x minus 2 left parenthesis square root of 3 plus 1 right parenthesis s i n invisible function application x plus square root of 3 equals 0$ has

The equation $4 s i n to the power of 2 end exponent invisible function application x minus 2 left parenthesis square root of 3 plus 1 right parenthesis s i n invisible function application x plus square root of 3 equals 0$ has

$s i n to the power of 2 end exponent invisible function application x minus c o s invisible function application 2 x equals 2 minus s i n invisible function application 2 x$ if

$s i n to the power of 2 end exponent invisible function application x minus c o s invisible function application 2 x equals 2 minus s i n invisible function application 2 x$ if

If $vertical line x vertical line less than 1 fifth$ , the coefficient of $x cubed$ in the expansion of $fraction numerator 1 over denominator left parenthesis 1 minus 5 x right parenthesis left parenthesis 1 minus 4 x right parenthesis end fraction$ is

If $vertical line x vertical line less than 1 fifth$ , the coefficient of $x cubed$ in the expansion of $fraction numerator 1 over denominator left parenthesis 1 minus 5 x right parenthesis left parenthesis 1 minus 4 x right parenthesis end fraction$ is

If $a subscript K equals fraction numerator 1 over denominator K left parenthesis K plus 1 right parenthesis end fraction$ for $K equals 1 comma 2 comma 3 horizontal ellipsis.. n$ then $open parentheses sum from K equals 1 to n of a subscript K close parentheses squared equals$

If $a subscript K equals fraction numerator 1 over denominator K left parenthesis K plus 1 right parenthesis end fraction$ for $K equals 1 comma 2 comma 3 horizontal ellipsis.. n$ then $open parentheses sum from K equals 1 to n of a subscript K close parentheses squared equals$

The number of partial fractions of $fraction numerator 2 over denominator x to the power of 4 plus x squared plus 1 end fraction$ is

The number of partial fractions of $fraction numerator 2 over denominator x to the power of 4 plus x squared plus 1 end fraction$ is

The number of partial fractions of $fraction numerator x cubed minus 3 x squared plus 3 x over denominator left parenthesis x minus 1 right parenthesis to the power of 5 end fraction$ is

The number of partial fractions of $fraction numerator x cubed minus 3 x squared plus 3 x over denominator left parenthesis x minus 1 right parenthesis to the power of 5 end fraction$ is

Coefficient of $x cubed y to the power of 4 z squared$ is $left parenthesis 2 x minus 3 y plus 4 z right parenthesis to the power of 9$ is

Coefficient of $x cubed y to the power of 4 z squared$ is $left parenthesis 2 x minus 3 y plus 4 z right parenthesis to the power of 9$ is

Coefficient of $x to the power of r$ is $1 plus left parenthesis 1 plus x right parenthesis plus left parenthesis 1 plus x right parenthesis squared plus$ $horizontal ellipsis plus left parenthesis 1 plus x right parenthesis to the power of n$ is

Coefficient of $x to the power of r$ is $1 plus left parenthesis 1 plus x right parenthesis plus left parenthesis 1 plus x right parenthesis squared plus$ $horizontal ellipsis plus left parenthesis 1 plus x right parenthesis to the power of n$ is

No of terms whose value depend on 'x' is $left parenthesis x squared minus 2 plus 1 divided by x squared right parenthesis to the power of n$ is

No of terms whose value depend on 'x' is $left parenthesis x squared minus 2 plus 1 divided by x squared right parenthesis to the power of n$ is

Courses

Quick Links

Follow Us at

Support

Download App

Courses

Quick Links

Have a query? Ask a Tutor!!

Can’t find what you’re looking for?

Question

The area of circle circumscribing ΔABC is

Answer

The correct answer is:

Book A Free Demo

Related Questions to study

Statement-I : The statement that circumradius and inradius of a triangle are 12 and 8 respectively can not be correct. Statement-II : Circumradius 2 (inradius)

Statement-I : The statement that circumradius and inradius of a triangle are 12 and 8 respectively can not be correct. Statement-II : Circumradius 2 (inradius)

The complete solution set of the equation is

The complete solution set of the equation is

The solution(s) of the equation cos2x sin6x = cos3x sin5x in the interval [0, ] is/are –

The solution(s) of the equation cos2x sin6x = cos3x sin5x in the interval [0, ] is/are –

The equation has

The equation has

if

if

If , the coefficient of in the expansion of is

If , the coefficient of in the expansion of is

If for then

If for then

The number of partial fractions of is

The number of partial fractions of is

The number of partial fractions of is

The number of partial fractions of is

Coefficient of is is

Coefficient of is is

Coefficient of is is

Coefficient of is is

If a,b,c,d are consective binomial coefficients of then are is

If a,b,c,d are consective binomial coefficients of then are is

No of terms whose value depend on 'x' is is

No of terms whose value depend on 'x' is is

If represent the terms is then is

If represent the terms is then is

Courses

Quick Links

The correct answer is: $pi over 4$

The complete solution set of the equation $open vertical bar x to the power of 2 end exponent minus x close vertical bar plus vertical line x plus 3 vertical line equals open vertical bar x to the power of 2 end exponent minus 2 x minus 3 close vertical bar$ is

The complete solution set of the equation $open vertical bar x to the power of 2 end exponent minus x close vertical bar plus vertical line x plus 3 vertical line equals open vertical bar x to the power of 2 end exponent minus 2 x minus 3 close vertical bar$ is

The solution(s) of the equation cos2x sin6x = cos3x sin5x in the interval [0, $straight pi$ ] is/are –

The solution(s) of the equation cos2x sin6x = cos3x sin5x in the interval [0, $straight pi$ ] is/are –

The equation $4 s i n to the power of 2 end exponent invisible function application x minus 2 left parenthesis square root of 3 plus 1 right parenthesis s i n invisible function application x plus square root of 3 equals 0$ has

The equation $4 s i n to the power of 2 end exponent invisible function application x minus 2 left parenthesis square root of 3 plus 1 right parenthesis s i n invisible function application x plus square root of 3 equals 0$ has

$s i n to the power of 2 end exponent invisible function application x minus c o s invisible function application 2 x equals 2 minus s i n invisible function application 2 x$ if

$s i n to the power of 2 end exponent invisible function application x minus c o s invisible function application 2 x equals 2 minus s i n invisible function application 2 x$ if

If $vertical line x vertical line less than 1 fifth$ , the coefficient of $x cubed$ in the expansion of $fraction numerator 1 over denominator left parenthesis 1 minus 5 x right parenthesis left parenthesis 1 minus 4 x right parenthesis end fraction$ is

If $vertical line x vertical line less than 1 fifth$ , the coefficient of $x cubed$ in the expansion of $fraction numerator 1 over denominator left parenthesis 1 minus 5 x right parenthesis left parenthesis 1 minus 4 x right parenthesis end fraction$ is

If $a subscript K equals fraction numerator 1 over denominator K left parenthesis K plus 1 right parenthesis end fraction$ for $K equals 1 comma 2 comma 3 horizontal ellipsis.. n$ then $open parentheses sum from K equals 1 to n of a subscript K close parentheses squared equals$

If $a subscript K equals fraction numerator 1 over denominator K left parenthesis K plus 1 right parenthesis end fraction$ for $K equals 1 comma 2 comma 3 horizontal ellipsis.. n$ then $open parentheses sum from K equals 1 to n of a subscript K close parentheses squared equals$

The number of partial fractions of $fraction numerator 2 over denominator x to the power of 4 plus x squared plus 1 end fraction$ is

The number of partial fractions of $fraction numerator 2 over denominator x to the power of 4 plus x squared plus 1 end fraction$ is

The number of partial fractions of $fraction numerator x cubed minus 3 x squared plus 3 x over denominator left parenthesis x minus 1 right parenthesis to the power of 5 end fraction$ is

The number of partial fractions of $fraction numerator x cubed minus 3 x squared plus 3 x over denominator left parenthesis x minus 1 right parenthesis to the power of 5 end fraction$ is

Coefficient of $x cubed y to the power of 4 z squared$ is $left parenthesis 2 x minus 3 y plus 4 z right parenthesis to the power of 9$ is

Coefficient of $x cubed y to the power of 4 z squared$ is $left parenthesis 2 x minus 3 y plus 4 z right parenthesis to the power of 9$ is

Coefficient of $x to the power of r$ is $1 plus left parenthesis 1 plus x right parenthesis plus left parenthesis 1 plus x right parenthesis squared plus$ $horizontal ellipsis plus left parenthesis 1 plus x right parenthesis to the power of n$ is

Coefficient of $x to the power of r$ is $1 plus left parenthesis 1 plus x right parenthesis plus left parenthesis 1 plus x right parenthesis squared plus$ $horizontal ellipsis plus left parenthesis 1 plus x right parenthesis to the power of n$ is

No of terms whose value depend on 'x' is $left parenthesis x squared minus 2 plus 1 divided by x squared right parenthesis to the power of n$ is

No of terms whose value depend on 'x' is $left parenthesis x squared minus 2 plus 1 divided by x squared right parenthesis to the power of n$ is