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bhai thanks boalne ki jarurat nahi h
aaur thanks kal dedunga mere pre bord hone vale h to aabhi chalta hu bye
stay blessed
stay safe
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Answer:
★\huge\underline\bold\orange{Given}
Given
★
\bold{Measures \: of \: triangle \: (-1,-2),\: (1,0),}Measuresoftriangle(−1,−2),(1,0),
\bold{(-1,3), \: (-3,0)}(−1,3),(−3,0)
★\huge\underline\bold\orange{Find\:out}
Findout
★
\textbf{Name the type of quadrilateral and }Name the type of quadrilateral and \textbf{give reason for your answer}give reason for your answer
★\huge\underline\bold\orange{Formula \: used}
Formulaused
★
\textbf{Distance \: formula}Distance formula
\sqrt{(x2-x1)^2+(y2-y1)^2}
(x2−x1)
2
+(y2−y1)
2
★\huge\underline\bold\orange{Solution}
Solution
★
\large\bold\blue{Distance \: of \: AB}DistanceofAB
\bold{AB \: =}AB= \sqrt{(x2-x1)^2+(y2-y1)^2}
(x2−x1)
2
+(y2−y1)
2
\bold{AB \: =}AB= \sqrt{(1-(-1))^2+(0-(-2))^2}
(1−(−1))
2
+(0−(−2))
2
\bold{AB \: =}AB= \sqrt{(1+1)^2+(0+2)^2}
(1+1)
2
+(0+2)
2
\bold{AB \: =}AB= \sqrt{(2)^2+(2)^2}
(2)
2
+(2)
2
\bold{AB \: =}AB= \sqrt{4+4}
4+4
\bold{AB \: =}AB= \sqrt{8}
8
\bold{AB \: =}AB= {2}2 \sqrt{2}
2
unit
\large\bold\blue{Distance \: of \: BC}DistanceofBC
\bold{BC \: =}BC= \sqrt{(-1-1)^2+(2-0)^2}
(−1−1)
2
+(2−0)
2
\bold{BC \: =}BC= \sqrt{(-2)^2+(2)^2}
(−2)
2
+(2)
2
\bold{BC \: =}BC= \sqrt{4+4}
4+4
\bold{BC \: =}BC= \sqrt{8}
8
\bold{BC \: =}BC= {2}2 \sqrt{2}
2
unit
\large\bold\blue{Distance \: of \: CD}DistanceofCD
\bold{CD \: =}CD= \sqrt{(-3-(-1))^2+(0-2)^2}
(−3−(−1))
2
+(0−2)
2
\bold{CD \: =}CD= \sqrt{(-3+1)^2+(-2)^2}
(−3+1)
2
+(−2)
2
\bold{CD \: =}CD= \sqrt{(-2)^2+4}
(−2)
2
+4
\bold{CD \: =}CD= \sqrt{4+4}
4+4
\bold{CD \: =}CD= \sqrt{8}
8
\bold{CD \: =}CD= {2}2 \sqrt{2}
2
unit
\large\bold\blue{Distance \: of \: DA}DistanceofDA
\bold{DA \: =}DA= \sqrt{(-1-(-3))^2+(-2-0)^2}
(−1−(−3))
2
+(−2−0)
2
\bold{DA \: =}DA= \sqrt{(-1+3)^2+(-2)^2}
(−1+3)
2
+(−2)
2
\bold{DA \: =}DA= \sqrt{(-2)^2+(-2)^2}
(−2)
2
+(−2)
2
\bold{DA \: =}DA= \sqrt{4+4}
4+4
\bold{DA \: =}DA= \sqrt{8}
8
\bold{DA \: =}DA= {2}2 \sqrt{2}
2
unit
\large\bold\blue{Distance \: of \: AC}DistanceofAC
\bold{AC \: =}AC= \sqrt{(-1-(-1))^2+(-2-2)^2}
(−1−(−1))
2
+(−2−2)
2
\bold{AC \: =}AC= \sqrt{(0)^2+(-4)^2}
(0)
2
+(−4)
2
\bold{AC \: =}AC= \sqrt{16}
16
\bold{AC \: =}AC= {4}4 unit
\large\bold\blue{Distance \: of \: BD}DistanceofBD
\bold{BD \: =}BD= \sqrt{(-3-1)^2+(0-0)^2}
(−3−1)
2
+(0−0)
2
\bold{BD \: =}BD= \sqrt{(-4)^2+(0)^2}
(−4)
2
+(0)
2
\bold{BD \: =}BD= \sqrt{16}
16
\bold{BD \: = \: 4}BD=4 unit
\textbf\red{Hence, AB=BC=CD=DA}Hence, AB=BC=CD=DA
\textbf\red{AC=BD}AC=BD
.•.\textbf\red{ABCD \: is \: a \: square}ABCD is a square
★\textbf\red{All \: sides \: are \: equal \: in \: square}All sides are equal in square ★
★\textbf\red{Diagonals \: are \: equal \: in \: square}Diagonals are equal in square ★