Let P(2,1), Q(4,-1) and R(3,2) be the vertices or triangle PQR. If through P and R lines parallel to opposite sides are drawn to intersect in S, then find the area of square PQRS, in sq. units.

Question

Let P(2,1), Q(4,-1) and R(3,2) be the vertices or triangle PQR. If through P and R lines parallel to opposite sides are drawn to intersect in S, then find the area of square PQRS, in sq. units. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/7d6c347d7ace51994239a5e2d0ced74e-.png" /> | PQ |=√(4-2)2+(1+1)2=2 √2 | PR |=√(3-2)2+(2-1)2=√2 ⇒ Area of square PQRS =2 area of triangle PQR =2 × (1/2) × 2 √2 × √2 =4

Q. Let $P (2, 1), Q (4, - 1)$ and $R (3, 2)$ be the vertices or $△ PQR$ . If through $P$ and $R$ lines parallel to opposite sides are drawn to intersect in $S$ , then find the area of $□ PQRS$ , in sq. units.

$∣ PQ ∣ = (4 - 2)^{2} + (1 + 1)^{2} = 22$
$∣ PR ∣ = (3 - 2)^{2} + (2 - 1)^{2} = 2$
$\Rightarrow$ Area of $□ PQRS = 2$ area of $△ PQR$
$= 2 \times \frac{1}{2} \times 22 \times 2$
$= 4$

Q. Let P(2,1),Q(4,−1) and R(3,2) be the vertices or △PQR. If through P and R lines parallel to opposite sides are drawn to intersect in S, then find the area of □PQRS, in sq. units.

∣PQ∣=(4−2)2+(1+1)2​=22​ ∣PR∣=(3−2)2+(2−1)2​=2​ ⇒ Area of □PQRS=2 area of △PQR =2×21​×22​×2​ =4

Q. Let $P (2, 1), Q (4, - 1)$ and $R (3, 2)$ be the vertices or $△ PQR$ . If through $P$ and $R$ lines parallel to opposite sides are drawn to intersect in $S$ , then find the area of $□ PQRS$ , in sq. units.

$∣ PQ ∣ = (4 - 2)^{2} + (1 + 1)^{2} = 22$
$∣ PR ∣ = (3 - 2)^{2} + (2 - 1)^{2} = 2$
$\Rightarrow$ Area of $□ PQRS = 2$ area of $△ PQR$
$= 2 \times \frac{1}{2} \times 22 \times 2$
$= 4$