Properties of Kites 8-5,6 and Trapezoids Warm Up Lesson Presentation Lesson Quiz Holt Geometry

Objectives Use properties of kites to solve problems. Use properties of trapezoids to solve problems.

Vocabulary kite trapezoid base of a trapezoid leg of a trapezoid base angle of a trapezoid isosceles trapezoid midsegment of a trapezoid

A kite is a quadrilateral with exactly two pairs of congruent consecutive sides.

Example 2A: Using Properties of Kites In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mBCD. Kite  cons. sides  ∆BCD is isos. 2  sides isos. ∆ CBF  CDF isos. ∆ base s  mCBF = mCDF Def. of   s mBCD + mCBF + mCDF = 180° Polygon  Sum Thm.

Example 2A Continued mBCD + mCBF + mCDF = 180° Substitute mCDF for mCBF. mBCD + mCBF + mCDF = 180° Substitute 52 for mCBF. mBCD + 52° + 52° = 180° Subtract 104 from both sides. mBCD = 76°

Example 2B: Using Properties of Kites In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mABC. ADC  ABC Kite  one pair opp. s  mADC = mABC Def. of  s Polygon  Sum Thm. mABC + mBCD + mADC + mDAB = 360° Substitute mABC for mADC. mABC + mBCD + mABC + mDAB = 360°

Example 2B Continued mABC + mBCD + mABC + mDAB = 360° mABC + 76° + mABC + 54° = 360° Substitute. 2mABC = 230° Simplify. mABC = 115° Solve.

Example 2C: Using Properties of Kites In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mFDA. CDA  ABC Kite  one pair opp. s  mCDA = mABC Def. of  s mCDF + mFDA = mABC  Add. Post. 52° + mFDA = 115° Substitute. mFDA = 63° Solve.

A trapezoid is a quadrilateral with exactly one pair of parallel sides A trapezoid is a quadrilateral with exactly one pair of parallel sides. Each of the parallel sides is called a base. The nonparallel sides are called legs. Base angles of a trapezoid are two consecutive angles whose common side is a base.

If the legs of a trapezoid are congruent, the trapezoid is an isosceles trapezoid. The following theorems state the properties of an isosceles trapezoid.

Example 3A: Using Properties of Isosceles Trapezoids Find mA. mC + mB = 180° Same-Side Int. s Thm. 100 + mB = 180 Substitute 100 for mC. mB = 80° Subtract 100 from both sides. A  B Isos. trap. s base  mA = mB Def. of  s mA = 80° Substitute 80 for mB

Example 3B: Using Properties of Isosceles Trapezoids KB = 21.9m and MF = 32.7. Find FB. Isos.  trap. s base  KJ = FM Def. of  segs. KJ = 32.7 Substitute 32.7 for FM. KB + BJ = KJ Seg. Add. Post. 21.9 + BJ = 32.7 Substitute 21.9 for KB and 32.7 for KJ. BJ = 10.8 Subtract 21.9 from both sides.

Example 3B Continued Same line. KFJ  MJF Isos. trap.  s base  Isos. trap.  legs  ∆FKJ  ∆JMF SAS CPCTC BKF  BMJ FBK  JBM Vert. s 

Example 3B Continued Isos. trap.  legs  ∆FBK  ∆JBM AAS CPCTC FB = JB Def. of  segs. FB = 10.8 Substitute 10.8 for JB.

Check It Out! Example 3a Find mF. mF + mE = 180° Same-Side Int. s Thm. E  H Isos. trap. s base  mE = mH Def. of  s mF + 49° = 180° Substitute 49 for mE. mF = 131° Simplify.

The midsegment of a trapezoid is the segment whose endpoints are the midpoints of the legs. In Lesson 5-1, you studied the Triangle Midsegment Theorem. The Trapezoid Midsegment Theorem is similar to it.

Review: Triangle Midsegment Theorem If a segment joins the midpoints of two sides of a triangle, then the segment is parallel to the third side, and is half of its length. Midsegment x 2x

Example 5: Finding Lengths Using Midsegments Find EF. Trap. Midsegment Thm. Substitute the given values. EF = 10.75 Solve.

Substitute the given values. Check It Out! Example 5 Find EH. Trap. Midsegment Thm. 1 16.5 = (25 + EH) 2 Substitute the given values. Simplify. 33 = 25 + EH Multiply both sides by 2. 8 = EH Subtract 25 from both sides.

8-6 Identify Special Quadrilaterals Warm Up Lesson Presentation Lesson Quiz Holt Geometry

Objective Prove that a given quadrilateral is a rectangle, rhombus, or square.

When you are given a parallelogram with certain properties, you can use the theorems below to determine whether the parallelogram is a rectangle.

Example 1: Carpentry Application A manufacturer builds a mold for a desktop so that , , and mABC = 90°. Why must ABCD be a rectangle? Both pairs of opposites sides of ABCD are congruent, so ABCD is a . Since mABC = 90°, one angle ABCD is a right angle. ABCD is a rectangle by Theorem 6-5-1.

Below are some conditions you can use to determine whether a parallelogram is a rhombus.

In order to apply Theorems 6-5-1 through 6-5-5, the quadrilateral must be a parallelogram. Caution To prove that a given quadrilateral is a square, it is sufficient to show that the figure is both a rectangle and a rhombus. You will explain why this is true in Exercise 43.

Example 2B: Applying Conditions for Special Parallelograms Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid. Given: Conclusion: EFGH is a square. Step 1 Determine if EFGH is a parallelogram. Given Quad. with diags. bisecting each other  EFGH is a parallelogram.

Example 2B Continued Step 2 Determine if EFGH is a rectangle. Given. EFGH is a rectangle. with diags.   rect. Step 3 Determine if EFGH is a rhombus. with one pair of cons. sides   rhombus EFGH is a rhombus.

Example 2B Continued Step 4 Determine is EFGH is a square. Since EFGH is a rectangle and a rhombus, it has four right angles and four congruent sides. So EFGH is a square by definition. The conclusion is valid.

Check It Out! Example 2 Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid. Given: ABC is a right angle. Conclusion: ABCD is a rectangle. The conclusion is not valid. By Theorem 6-5-1, if one angle of a parallelogram is a right angle, then the parallelogram is a rectangle. To apply this theorem, you need to know that ABCD is a parallelogram .

Lesson Quiz: Part I 1. Given that AB = BC = CD = DA, what additional information is needed to conclude that ABCD is a square?

Lesson Quiz: Part II 2. Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid. Given: PQRS and PQNM are parallelograms. Conclusion: MNRS is a rhombus. valid