Two medians drawn from acute angles vertices of a right angled triangle intersects at an angle (π /6) . If the length of the hypotenuse of the triangle is 3 units, then area of the triangle (in sq. units) is

Question

Two medians drawn from acute angles vertices of a right angled triangle intersects at an angle (π /6) . If the length of the hypotenuse of the triangle is 3 units, then area of the triangle (in sq. units) is (A) √3 (B) 3 (C) √2 (D) 9

Tardigrade Admin · Accepted Answer

Slope of A G = - (b/2 a) , tan 30o = ((3 b/2 a)/1 + (b2)a2) <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-dxbodttb72o1.png" /> a2+b2=9⇒ (1/√3)=(3 b . a/2 (a2 + b2)) ⇒ (1/2) ab=((a2 + b2/3 √3))=(9/3 √3)=√3

Q. Two medians drawn from acute angles vertices of a right angled triangle intersects at an angle $\frac{π}{6}$ . If the length of the hypotenuse of the triangle is $3$ units, then area of the triangle (in sq. units) is

$3$

$3$

$2$

$9$

Slope of $A G = - \frac{b}{2 a}, t an 3 0^{o} = \frac{\frac{3 b}{2 a}}{1 + \frac{b ^{2}}{a ^{2}}}$

$a^{2} + b^{2} = 9 \Rightarrow \frac{1}{3} = \frac{3 b . a}{2 ( a ^{2} + b ^{2} )}$
$\Rightarrow \frac{1}{2} ab = (\frac{a ^{2} + b ^{2}}{3 3}) = \frac{9}{3 3} = 3$

Q. Two medians drawn from acute angles vertices of a right angled triangle intersects at an angle 6π​ . If the length of the hypotenuse of the triangle is 3 units, then area of the triangle (in sq. units) is

3​

3

2​

9

Slope of AG=−2ab​,tan30o=1+a2b2​2a3b​​ a2+b2=9⇒3​1​=2(a2+b2)3b.a​ ⇒21​ab=(33​a2+b2​)=33​9​=3​

Q. Two medians drawn from acute angles vertices of a right angled triangle intersects at an angle $\frac{π}{6}$ . If the length of the hypotenuse of the triangle is $3$ units, then area of the triangle (in sq. units) is

$3$

$3$

$2$

$9$

Slope of $A G = - \frac{b}{2 a}, t an 3 0^{o} = \frac{\frac{3 b}{2 a}}{1 + \frac{b ^{2}}{a ^{2}}}$

$a^{2} + b^{2} = 9 \Rightarrow \frac{1}{3} = \frac{3 b . a}{2 ( a ^{2} + b ^{2} )}$
$\Rightarrow \frac{1}{2} ab = (\frac{a ^{2} + b ^{2}}{3 3}) = \frac{9}{3 3} = 3$