Squares

Problem Setter & Dataset: Md. Albin Hossain

Reviewer: Saiful Islam Ramim

Alternate Writer: Siam Ahmed

Problem Category: Geometry

Contest Solve Count: $150/500$ (Expected) (To be updated after the contest).

Problem Analysis:

Given $PQ$. To find the area of the square $ABCD$ we need to find a side $AB$. Then we can find the area of $ABCD={AB}^2$.

Sides of square $ABCD$ are equal, therefore $AB = BC$ and $P, Q$ are midpoints of $AB$, and $BC$ respectively. So $PB=BQ$ and $AB= 2PB$.

Consider the triangle $\bigtriangleup PBQ$. $\angle PBQ$ is $90°$ as $ABCD$ is a square by definition.

We can get by Pythagorean theorem,

$PB^2 + BQ^2 = PQ^2$

$PB^2 + PB^2 = PQ^2 \quad [ \because PB = BQ]$

$2PB^2 = PQ^2$

$4PB^2 = 2PQ^2 \quad$ [Multiply both side by 2]

$(2PB)^2 = 2PQ^2$

$AB^2 = 2PQ^2 \quad [ \because AB = 2PB]$

$AB^2=$Area of $ABCD$ square.

Source Code:

#include<stdio.h>

int main() {

    int pq;

    scanf("%d", &pq);

    int area = 2 * pq * pq;

    printf("%d\n", area);

    return 0;

}