Question

Two triangles of a rectangle differ by 3.5cm. Find dimension of rectangle if its Perimeter is 67cm.

Answer

Let,

• one side of rectangle = 3.5 cm
• other side of rectangle = 67 cm

According to the statement,

• x - y = 3.5 ( e.q 1 )
• 2( x + y ) = 67
• 2x - 2y = 67 ( e.q 2 )

Write in matrix form,

See the achievement( 1 )

By using Cramer rule,

See the achievement

( 2 , 3 )

Now Put values in the formula,

.

Answer:

Question

Two triangles of a rectangle differ by 3.5cm. Find dimension of rectangle if its Perimeter is 67cm.

Answer

Let,

one side of rectangle = 3.5 cm

other side of rectangle = 67 cm

According to the statement,

x - y = 3.5 ( e.q 1 )

2( x + y ) = 67

2x - 2y = 67 ( e.q 2 )

Write in matrix form,

See the achievement( 1 )

By using Cramer rule,

\begin{gathered} \implies \sf{Ax = \large\frac{ |Ax| }{ |A| } } \\ \\ \implies \sf{Ay = \large\frac{ |Ay| }{ |A| } }\end{gathered}

⟹Ax=

∣A∣

∣Ax∣

⟹Ay=

∣A∣

∣Ay∣

See the achievement

( 2 , 3 )

Now Put values in the formula,

\begin{gathered} \implies \sf{Ax = \large\frac{74 }{ 4 } } \\ \\ \implies \sf{Ax = \large\frac{ \cancel{74}^{37} }{ \cancel{ 4 }^{2} } } \\ \\ \implies \boxed { \red{\sf{Ax = \large\frac{37}{2} }} }\\ \\ \implies \sf{Ay = \large\frac{ 60 }{ 4} } \\ \\ \implies \sf{Ay = \large\frac{ \cancel{ 60 }^{15} }{ \cancel{4}^{1} } } \\ \\ \implies \boxed{ \red{{\sf{Ay = 15 }}}}\end{gathered}

⟹Ax=

4

74

⟹Ax=

4

2

74

37

⟹

Ax=

2

37

⟹Ay=

4

60

⟹Ay=

4

1

60

15

⟹

Ay=15

.Hope it is helpful.

Ziddi girl here.

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