are: Opposite sides of a rectangle are parallel. 1. Opposite sides - - PowerPoint PPT Presentation

DAY 114 – DIAGONALS OF A

RECTANGLE

INTRODUCTION

We have already discussed the properties of a rectangle. Among them, is that a rectangle is a parallelogram with right angles at all the four

• vertices. Unlike the other parallelograms whose

diagonals are not equal, a rectangle has equal diagonals. In this lesson, we will prove that the diagonals of a rectangle are equal. We will also show that at the intersection of these diagonals, the opposite sides are equal and adjacent angles are supplementary.

VOCABULARY

Supplementary angles These are two angles whose sum is 180°.

In this lesson, there are important properties of rectangles we need to remember. These properties are:

1.

Opposite sides of a rectangle are parallel.

2.

Opposite sides of a rectangle are equal.

3.

A rectangle has right angles at all the four corners. We would also like to mention a result that we will use in this lesson, that when two triangles are congruent, their corresponding sides are equal.

Theorem: Diagonals of a rectangle are equal. To prove this theorem, we consider the rectangle below. We want to prove that 𝑁𝐿 = 𝐾𝑀.

J K M L

The two diagonals divide rectangle JKLM into ∆𝐾𝐿𝑁 and ∆𝐾𝑀𝑁. Since opposite sides of a rectangle are equal, JK in ∆𝐾𝐿𝑁 is equal to ML in ∆𝐾𝑀𝑁. KL is shared by the two triangles. Since a rectangle has right angles at its corners, ∠𝑁𝐾𝐿 = ∠𝐾𝑁𝑀 = 90°. By S.A.S postulate, ∆𝐾𝐿𝑁 and ∆𝐿𝑀𝑁 are congruent. JL and MK are corresponding sides. Since corresponding sides of triangles are equal, 𝑵𝑳 = 𝑲𝑴. Therefore, diagonals of a rectangle are equal.

Example 1 Show that NP = MO.

M N P O

Solution In ∆𝑁𝑃𝑄 and ∆𝑁𝑄𝑂 we have, MN = PO(opposite sides of a rectangle are equal) ∠𝑁𝑄𝑃 = ∠𝑄𝑁𝑂 (each equal to 90°) MP is common in both triangles. By S.A.S postulate ∆𝑁𝑃𝑄 ≅ ∆𝑁𝑄𝑂 NP and MO are corresponding sides and thus NP = MO.

Showing that opposite angles at the intersection of the diagonals are equal. By S.S.S postulate, ∆𝐾𝑁𝑀, ∆𝐾𝐿𝑀, ∆𝐾𝐿𝑁 and ∆𝐿𝑀𝑁 are all congruent, since all these triangles share the diagonals, one longer side and a shorter side. ∠𝐾𝑀𝑁, ∠𝐾𝐿𝑀, ∠𝐾𝐿𝑁 and ∠𝐿𝑁𝑀 are corresponding angles and therefore equal. Similarly, ∠𝑀𝐾𝑁, ∠𝐿𝑁𝐾, ∠𝐾𝑀𝐿 and ∠𝑁𝐿𝑀 are corresponding angles and therefore equal.

J K M L O

By A.S.A postulate, ∆𝐾𝑁𝑃 ≅ ∆𝐿𝑀𝑃.(Two adjacent sides are congruent and the included sides are equal) ∠𝐾𝑃𝑁 and ∠𝐿𝑃𝑀 are corresponding angles and therefore equal. By A.S.A postulate, ∆𝑁𝑃𝑀 ≅ ∆𝐾𝑃𝐿. ∠𝐾𝑃𝐿 and ∠𝑁𝑃𝑀 are corresponding angles and therefore equal. Thus, opposite angles at the point of intersection of diagonals of a triangles are equal. The distance from each corner to the point of intersection is equal.

∠𝑁𝑃𝑀 and ∠𝐿𝑃𝑀 lie on a straight line and therefore their sum is 180°, thus they are supplementary. ∠𝐾𝑃𝑁 and ∠𝑁𝑃𝑀 lie on a straight line and therefore their sum is 180°, thus they are supplementary. ∠𝐿𝑃𝑀 and ∠𝐾𝑃𝐿 lie on a straight line and therefore their sum is 180°, thus they are supplementary. ∠𝐾𝑃𝑁 and ∠𝐾𝑃𝐿 lie on a straight line and therefore their sum is 180°, thus they are supplementary.

Example Find the value of 𝑦 in the figure below. Solution Since opposite angles at the point of the intersection of the diagonals are equal, 2𝑦 + 40 = 𝑦 + 80 2𝑦 − 𝑦 = 80 − 40 𝑦 = 40

C D A B 2𝑦 + 40

°

𝑦 + 80 °

HOMEWORK Find the value of the angle marked with letter a.

29° a

ANSWERS TO HOMEWORK

29°

THE END