Lesson Introduction to properties of a rectangle
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A rectangle is one of the most commonly known <A HREF=http://en.wikipedia.org/wiki/Quadrilateral>quadrilaterals</A>. In this lesson we are going to deal with rectangles and their basic properties. Further we are going to build a deep understanding of their properties and will prove them simultaneously. <B><A HREF=http://www.algebra.com/algebra/homework/Rectangles/Rectangle.wikipedia>Rectangle</A></B> A parallelogram in which each angle is 90 degrees is called a rectangle. Hence a rectangle has all the properties of a parallelogram. {{{drawing( 160, 144, -10, 10, -10, 8, line( -6, -6, 4, -6) , line( -6, -6, -6, 1.5), line( -6, 1.5, 4, 1.5), line( 4, 1.5, 4, -6), line( -6, -6, 4, 1.5), line( 4, -6, -6, 1.5), locate( -6.5, -6.5, A), locate( 4.5, -6.5, B), locate( 4.5, 2.5, C), locate( -7, 3, D) )}}} <B>The properties of a <A HREF=http://www.algebra.com/algebra/homework/Rectangles/Parallelogram.wikipedia>Parallelogram</A> common to rectangle are:</B> 1. Opposite sides of a parallelogram are equal. 2. Opposite angles of a parallelogram are equal. 3. Diagonals of a parallelogram bisect each other in their intersection point. <B>The rectangle has following special properties</B>: 1. The corner angles all are right angles (90°). 2. Diagonals of a rectangle are equal. 3. A square of a diagonal length is equal to a sum of squares of its sides lengths. <B>Proof of properties of a Rectangle</B> <B>Proof 1</B>. If in a parallelogram one angle is 90 degrees then all angles are 90 degrees. Consider a parallelogram <B>ABCD</B>, where it is given that <I>L</I><B>BCD</B> = 90°. {{{drawing( 160, 144, -10, 10, -10, 8, line( -6, -6, 4, -6), line( -6, -6, -2, 2), line( -2, 2, 8, 2), line( 8, 2, 4, -6), line( -6, -6, 8, 2), line( -2, 2, 4, -6), locate( -6.5, -6.5, A), locate( 4.5, -6.5, B), locate( 8.5, 3.5, C), locate( -2.5, 4, D), locate( -0, -2.4, O) )}}} Since opposite angles in a parallelogram are equal, we have <I>L</I><B>DAB</B> = <I>L</I><B>BCD</B> = 90°. (1) Now, <B>AB</B> || <B>DC</B>. Hence, by the <B>Sum of angles</B> property in a parallel lines, <I>L</I><B>DAB</B> + <I>L</I><B>ADC</B>=180°, 90° + <I>L</I><B>ADC</B> = 180°, <I>L</I><B>ADC</B> = 180°-90° = 90°. (2) Again, by the property of parallelogram that opposite angles are equal, <I>L</I><B>ABC</B> = <I>L</I><B>ADC</B> = 90°. (3) From equations (1), (2) and (3) <I>L</I><B>BCD</B> = <I>L</I><B>DAB</B> = <I>L</I><B>ADC</B> = <I>L</I><B>ABC</B> = 90°. Hence, all <A HREF=http://www.algebra.com/algebra/homework/Triangles/Angle.wikipedia>angles</A> in a rectangle are equal to 90 degrees. <B>Proof 2</B>. Diagonals of a rectangle are equal. Consider two <A HREF=http://www.algebra.com/algebra/homework/Triangles/Triangles-and-its-basic-properties.lesson>Triangles</A> <B>ABD</B> and <B>ADC</B> containing two diagonals <B>BD</B> and <B>AC</B> respectively. Triangles <B>ABD</B> and <B>ADC</B> are Right Angled Triangles right angled at <I>L</I><B>DAC</B> and <I>L</I><B>ADC</B> respectively. From Pythagorean theorem, In Triangle <B>ABD</B>, {{{BD^2=(AB^2+AD^2)}}}. (4) In Triangle <B>ADC</B>, {{{AC^2=AD^2+DC^2}}}. (5) From equations (5) and (6) {{{BD^2=AC^2=(AD^2+DC^2)=(AB^2+AD^2)}}}, {{{BD^2=AC^2}}}, {{{BD=AC}}}. Hence, diagonals of a rectangle are of equal length. From the property of a parallelogram the diagonals of a rectangle bisect each other at their intersection point. Thus diagonals of a rectangle are of equal length and bisect each other. <B>Proof 3</B>. A square of a diagonal length is equal to a sum of squares of its sides lengths. Consider two Triangles <B>ABD</B> and <B>ADC</B> containing two diagonals BD and AC respectively. Triangles <B>ABD</B> and <B>ADC</B> are <A HREF=http://www.algebra.com/algebra/homework/Triangles/Right-Triangles.lesson>Right Angled Triangles</A> right angled at <I>L</I><B>DAC</B> and <I>L</I><B>ADC</B> respectively. From <A HREF=http://www.algebra.com/algebra/homework/Pythagorean-theorem/proof-of-pythagorean-theorem.lesson>Pythagorean theorem</A>, In Triangle <B>ABD</B>, {{{BD^2=(AB^2+AD^2)}}}. (7) In Triangle <B>ADC</B>, {{{AC^2=AD^2+DC^2}}}. (8) From equations (7) and (8) {{{BD^2=AC^2=(AD^2+DC^2)=(AB^2+AD^2)}}}, {{{BD^2=AC^2}}}. QED