Find the area of a rectangle whose vertices are
A (-2, 6), B (5, 3), C (-1,-11) and D (-8,-8)
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Answers
ANSWER:-
Given:
Find the area of a rectangle whose vertices are A(-2,6), B(5,3), C(-1,-11)&
D(-8,-8).
SOLUTION:
Let the vertices of quadrilateral ABCD be;
⏺️A(-2,6)
⏺️B(5,3)
⏺️C(-1,-11)
⏺️D(-8,-8)
Therefore,
AB² = [5-(-2)]² + (3-6)²
AB² = (5+2)² + (-3)²
AB² = (7)² + (-3)²
AB² = 49 + 9
AB² = 58
AB = √58
&
BC² = (-1-5)² + (-11-3)²
BC² = (-6)² + (-14)²
BC² = 36 + 196
BC² = 232
BC = √232
&
CD² = (-8+1)² +(-8+11)²
CD² = (-7)² + (3)²
CD² = 49 + 9
CD² = 58
CD √58
&
AD² = (-8+2)² + (-8-6)²
AD² = (-6)² + (-14)²
AD² = 36 + 196
AD² = 232
AD = √232
Therefore,
AB =CD=√58 units & BC= AD= √232 units
So,
AC² = (-1+2)² +(-11-6)²
AC² = (1)² + (-17)²
AC² = 1 + 289
AC² = 290
AC = √290
&
BD² = (-8-5)² + (-8-3)²
BD² = (-13)² + (-11)²
BD² = 169 + 121
BD² = 290
BD = √290
So,
Diagonal AC= diagonal BD.
Hence,
ABCD is a quadrilateral whose opposite sides and diagonals are equal. So, quadrilateral. &
ABCD is a rectangle.