If the extremities of the base of an isoscele | vedclass

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

Question

(B) Let the vertices of the base be $A(2a, 0)$ and $B(0, a)$.Since one side is $x = 2a$,the third vertex $C$ must lie on this line. Let $C = (2a, t)$.

Vedclass · Accepted Answer

$\frac{5}{2}a^2$ sq. units

Vedclass · Answer

(B) Let the vertices of the base be $A(2a, 0)$ and $B(0, a)$.Since one side is $x = 2a$,the third vertex $C$ must lie on this line. Let $C = (2a, t)$.For an isosceles triangle,the distance from the third vertex to the base vertices must be equal,or the vertex must lie on the perpendicular bisector.Given $AC = BC$,we have $AC^2 = BC^2$.$(2a - 2a)^2 + (t - 0)^2 = (2a - 0)^2 + (t - a)^2$$t^2 = 4a^2 + t^2 - 2at + a^2$$2at = 5a^2 \Rightarrow t = \frac{5a}{2}$.Thus,the third vertex is $C(2a, \frac{5a}{2})$.The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Area $= \frac{1}{2} |2a(0 - \frac{5a}{2}) + 2a(\frac{5a}{2} - 0) + 0(0 - 0)| = \frac{1}{2} |-5a^2 + 5a^2| = 0$ (This implies the points are collinear).Re-evaluating: The side $x=2a$ connects $(2a, 0)$ and $(2a, t)$. The base is the segment connecting $(2a, 0)$ and $(0, a)$.Length of base $AB = \sqrt{(2a-0)^2 + (0-a)^2} = \sqrt{4a^2 + a^2} = a\sqrt{5}$.Height $h$ from $C(2a, t)$ to line $AB$ $(x + 2y - 2a = 0)$: $h = \frac{|2a + 2t - 2a|}{\sqrt{1^2 + 2^2}} = \frac{2t}{\sqrt{5}}$.Since $AC = BC$,$C$ lies on the perpendicular bisector of $AB$. The slope of $AB$ is $m = \frac{a-0}{0-2a} = -1/2$. The perpendicular bisector has slope $2$ and passes through midpoint $(a, a/2)$.Equation: $y - a/2 = 2(x - a) \Rightarrow y = 2x - 3a/2$.Intersection with $x = 2a$: $y = 2(2a) - 3a/2 = 4a - 1.5a = 2.5a = \frac{5a}{2}$.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (a\sqrt{5}) 	imes \frac{|2a + 2(5a/2) - 2a|}{\sqrt{5}} = \frac{1}{2} 	imes a\sqrt{5} 	imes \frac{5a}{\sqrt{5}} = \frac{5a^2}{2}$.

If the extremities of the base of an isosceles triangle are the points $(2a, 0)$ and $(0, a)$ and the equation of one of the sides is $x = 2a$,then the area of the triangle is

Similar Questions

Find the area of the quadrilateral formed by the lines $4y - 3x = 1, 4y - 3x - 3 = 0, 3y - 4x + 1 = 0,$ and $3y - 4x + 2 = 0$.

The area of the quadrilateral formed by the lines $x+2y+3=0$,$2x+4y+9=0$,$x-2y+3=0$,and $3x-6y+11=0$ is

The circumcentre of the triangle formed by the lines $x+y+2=0, 2x+y+8=0$ and $x-y-2=0$ is

If the lines $3x + y - 4 = 0$,$x - \alpha y + 10 = 0$,$\beta x + 2y + 4 = 0$,and $3x + y + k = 0$ represent the sides of a square,then $\alpha \beta (k + 4)^2 = $

In a quadrilateral $ABCD$,which is not a trapezium,it is known that $\angle DAB = \angle ABC = 60^{\circ}$. Moreover,$\angle CAB = \angle CBD$. Then,

Vedclass Test Series

Exam Paper Generator

Online Exam Module