Consider four triangles with side lengths (5, | vedclass

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

Question

(C) For a triangle with sides $a, b, c$,the area $A$ is given by Heron's formula: $A = \sqrt{s(s-a)(s-b)(s-c)}$,where $s = \frac{a+b+c}{2}$.For all tr

Vedclass · Accepted Answer

$(5, 12, 13)$

Vedclass · Answer

(C) For a triangle with sides $a, b, c$,the area $A$ is given by Heron's formula: $A = \sqrt{s(s-a)(s-b)(s-c)}$,where $s = \frac{a+b+c}{2}$.For all triangles,two sides are fixed at $5$ and $12$. Let the third side be $x$. Then $s = \frac{5+12+x}{2} = \frac{17+x}{2}$.The area $A(x) = \sqrt{\frac{17+x}{2} \cdot \frac{17-x}{2} \cdot \frac{x+7}{2} \cdot \frac{x-7}{2}} = \frac{1}{4} \sqrt{(17^2 - x^2)(x^2 - 7^2)}$.Let $u = x^2$. The function $f(u) = (289 - u)(u - 49) = -u^2 + 338u - 14161$. This is a downward parabola with a maximum at $u = \frac{-338}{2(-1)} = 169$.Thus,$x^2 = 169$,which means $x = 13$.For $x = 13$,the sides are $(5, 12, 13)$,which satisfy $5^2 + 12^2 = 13^2$,forming a right-angled triangle with area $\frac{1}{2} 	imes 5 	imes 12 = 30$.Comparing areas for $x \in \{9, 11, 13, 15\}$,the maximum area occurs at $x = 13$.

Consider four triangles with side lengths $(5, 12, 9)$,$(5, 12, 11)$,$(5, 12, 13)$,and $(5, 12, 15)$. Among these,the triangle with the maximum area has sides:

Similar Questions

The area of the parallelogram formed by the lines $a_1x + b_1y + c_1 = 0$,$a_1x + b_1y + d_1 = 0$,$a_2x + b_2y + c_2 = 0$,and $a_2x + b_2y + d_2 = 0$ is:

The equations of two altitudes of an equilateral triangle are $\sqrt{3}x - y + 8 - 4\sqrt{3} = 0$ and $\sqrt{3}x + y - 12 - 4\sqrt{3} = 0$. The equation of the third altitude is

The triangle with vertices at $(-2, 2)$,$(2, -2)$,and $(1, 1)$ is:

Straight lines $x-y=7$ and $x+4y=2$ intersect at $B$. Points $A$ and $C$ are so chosen on these two lines such that $AB=AC$. The equation of line $AC$ passing through $(2,-7)$ is

The area of the triangle formed by the lines $y = m_1x + c_1$,$y = m_2x + c_2$ and $x = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module