Let R be the point (3,7) and let P and Q be two points on the line x+y=5 such that P Q R is an equilateral triangle. Then the area of triangle PQR is :

Question

Let R be the point (3,7) and let P and Q be two points on the line x+y=5 such that P Q R is an equilateral triangle. Then the area of triangle PQR is : (A) (25/4 √3) (B) (25 √3/2) (C) (25/√3) (D) (25/2 √3)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/896791e6a8de52c6f3094df04d103617-.png" /> sin 60°=(5 / √2/ a ) a =(5 √2/3) Area of triangle PQR =(√3/4) a 2=(25/2 √3)

Q. Let $R$ be the point $(3, 7)$ and let $P$ and $Q$ be two points on the line $x + y = 5$ such that $PQR$ is an equilateral triangle. Then the area of $△ PQR$ is :

$\frac{25}{4 3}$

$\frac{25 3}{2}$

$\frac{25}{3}$

$\frac{25}{2 3}$

$sin 6 0^{\circ} = \frac{5/ 2}{a}$
$a = \frac{5 2}{3}$
Area of $△ PQR = \frac{3}{4} a^{2} = \frac{25}{2 3}$

Q. Let R be the point (3,7) and let P and Q be two points on the line x+y=5 such that PQR is an equilateral triangle. Then the area of △PQR is :

43​25​

2253​​

3​25​

23​25​