Two sides of a triangle have lengths a text{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the two sides of the triangle be $a$ and $b$,and the angle between them be $	heta$.The area of the triangle is given by $A = \frac{1}{2}ab \s

Vedclass · Accepted Answer

$\frac{1}{2}\sqrt{a^2 + b^2}$

Vedclass · Answer

(A) Let the two sides of the triangle be $a$ and $b$,and the angle between them be $	heta$.The area of the triangle is given by $A = \frac{1}{2}ab \sin	heta$.For the area to be maximum,$\sin	heta$ must be maximum,which occurs when $	heta = 90^\circ$.Thus,the triangle is a right-angled triangle with sides $a$ and $b$ forming the right angle.Let the vertex containing sides $a$ and $b$ be at the origin $(0,0)$ in the Cartesian plane.The coordinates of the vertices are $(0,0)$,$(a,0)$,and $(0,b)$.The midpoint of the hypotenuse (the side opposite to the right angle) is $M = (\frac{a}{2}, \frac{b}{2})$.The median from the vertex $(0,0)$ to the hypotenuse is the distance from $(0,0)$ to $M$.Length of the median $m = \sqrt{(\frac{a}{2} - 0)^2 + (\frac{b}{2} - 0)^2} = \sqrt{\frac{a^2}{4} + \frac{b^2}{4}} = \frac{1}{2}\sqrt{a^2 + b^2}$.

Two sides of a triangle have lengths $a \text{ cm}$ and $b \text{ cm}$. If the triangle is to have the maximum area,then the length of the median from the vertex containing the sides $a$ and $b$ is

Similar Questions

Given two points $A(3, 4)$ and $B(5, -2)$. If $PA = PB$ and the area of $\Delta PAB = 10$ square units,find the coordinates of point $P$.

The centroid of the triangle formed by the lines $x+y-1=0$,$x-y-1=0$,and $x-3y+3=0$ is

The vertex of the right angle of a right-angled triangle lies on the straight line $2x + y - 10 = 0$. If the two other vertices are at points $(2, -3)$ and $(4, 1)$,then the area of the triangle in sq. units is:

The area of the parallelogram formed by the lines $x \cos \alpha + y \sin \alpha = p$,$x \cos \alpha + y \sin \alpha = q$,$x \cos \beta + y \sin \beta = r$,and $x \cos \beta + y \sin \beta = s$ is:

If in a parallelogram $ABDC$,the coordinates of $A, B$ and $C$ are respectively $(1, 2), (3, 4)$ and $(2, 5)$,then the equation of the diagonal $AD$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module