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K I t e
Definition
β€’KITE – is a quadrilateral with two sets of distinct
adjacent congruent sides, but opposite sides are
not congruent.
IN ANCIENT TIME
KITE were widely considered to be
useful for ensuring a good harvest or
scaring away evil spirits.
IN MODERN TIME
KITE became more widely known as
children's toys and came to be used
primarily as a leisure activity
β€’ From the definition, a kite is the only quadrilateral that we
have discussed that could be concave or non convex.
Concave or non convex kite is a kite whose diagonal do not
intersect. If a kite is concave or non convex, it is called
a dart .
β€’CONVEX KITE-
D
C
A
B
the diagonals of a kite
intersect.
β€’ The angles between the congruent sides are called vertex
angles . The other angles are called non-vertex angles . If we
draw the diagonal through the vertex angles, we would have
two congruent triangles.
B
A C
D
THEOREM 1: The non-vertex angles of a kite are congruent
and the diagonal through the vertex angle is the angle
bisector for both angles.
PROOF:
GIVEN: KITE WITH 𝐾𝐸≅𝑇𝐸 AND 𝐾𝐼≅𝑇𝐼
STATEMENTS REASONS
1. 1.
2. 2.
3. 3.
4. 4.
5. 5.
KITE WITH 𝐾𝐸≅𝑇𝐸 AND 𝐾𝐼≅𝑇𝐼 GIVEN
𝐼𝐸 β‰… 𝐼𝐸 REFLEXIVE PROPERTY
βˆ†πΎπΌπΈ β‰… βˆ†π‘‡πΌπΈ SSS CONGRUENCE POSTULATE
∠𝐾 β‰… βˆ π‘‡ CPCTC
βˆ π‘‡πΌπΈ β‰… ∠𝐾𝐼𝐸 AND ∠𝐾𝐸𝐼 β‰… βˆ π‘‡EI CPCTC
PROVE: ∠𝐾 β‰… βˆ π‘‡,
βˆ π‘‡πΌπΈ β‰… ∠𝐾𝐼𝐸 AND ∠𝐾𝐸𝐼 β‰… βˆ π‘‡EI
THEOREM 2: The diagonals of a kite are perpendicular to
each other.
PROOF:
GIVEN: Kite BCDA
STATEMENTS REASONS
1. 1.
2. 2.
3. 3.
4. 4.
D
C
A
B
Kite BCDA GIVEN
𝐡𝐢≅𝐡𝐴 AND 𝐢𝐷≅𝐴𝐷 Definition of kite
Definition of congruent segments𝐡𝐢 = 𝐡𝐴 AND 𝐢𝐷 = 𝐴𝐷
𝐢𝐴βŠ₯𝐡𝐷 If a line contains two points each of
which is equidistant from the
endpoints of a segment, then the
line is perpendicular bisector of the
segment.
PROVE: π‘ͺ𝑨βŠ₯𝑩𝑫
Theorem 3: The area of a kite is half the product of the
lengths of the diagonals.
w
PROOF:
GIVEN: Kite BCDA
STATEMENTS REASONS
1. 1.
2. 2.
3. 3.
4. 4.
5. 5.
6. 6.
7. 7.
8. 8.
Kite BCDA GIVEN
𝐢𝐴βŠ₯𝐡𝐷
The diagonals of a kite are perpendicular
to each other.
𝑨𝒓𝒆𝒂 𝒐𝒇 π’Œπ’Šπ’•π’† 𝑩π‘ͺ𝑫𝑨
= 𝑨𝒓𝒆𝒂 π’π’‡βˆ†π‘©π‘ͺ𝑨 + 𝑨𝒓𝒆𝒂 𝒐𝒇 βˆ†π‘ͺ𝑫𝑨
Area addition postulate
𝑨𝒓𝒆𝒂 π’π’‡βˆ†π‘©π‘ͺ𝑨 =
𝟏
𝟐
(π‘ͺ𝑨)(𝑩𝑾)
𝑨𝒓𝒆𝒂 π’π’‡βˆ†π‘ͺ𝑫𝑨 =
𝟏
𝟐
(π‘ͺ𝑨)(𝑫𝑾)
Area formula for triangles
𝐀𝒓𝒆𝒂 𝒐𝒇 π’Œπ’Šπ’•π’† 𝑩π‘ͺ𝑫𝑨
=
𝟏
𝟐
π‘ͺ𝑨 𝑩𝑾 +
𝟏
𝟐
(π‘ͺ𝑨)(𝑫𝑾)
Substitution
𝐀𝒓𝒆𝒂 𝒐𝒇 π’Œπ’Šπ’•π’† 𝑩π‘ͺ𝑫𝑨
=
𝟏
𝟐
π‘ͺ𝑨 𝑩𝑾 + 𝑫𝑾
Associative Property
π΅π‘Š + π·π‘Š = 𝐡𝐷 Segment Addition Postulate
𝐀𝒓𝒆𝒂 𝒐𝒇 π’Œπ’Šπ’•π’† 𝑩π‘ͺ𝑫𝑨 =
𝟏
𝟐
π‘ͺ𝑨 (𝑩𝑫) Substitution
PROVE:
𝑨𝒓𝒆𝒂 𝒐𝒇 π’Œπ’Šπ’•π’† 𝑩π‘ͺ𝑫𝑨 =
𝟏
𝟐
(π‘ͺ𝑨)(𝑩𝑫)
Example 1
β€’Find the area of the kite WXYZ.
20
12
12
12
U
W
Z
Y
X
Example1 Continued
20
12
12
12
U
W
Z
Y
X
We can now use the formula in
finding the area of the kite.
Area of kite WXYZ=
1
2
𝑑1𝑑2
Area of kite WXYZ=
1
2
(𝑋𝑍)(π‘Šπ‘Œ)
Area of kite WXYZ=
1
2
(24)(32)
Area of kite WXYZ=384 π‘ π‘žπ‘’π‘Žπ‘Ÿπ‘’ 𝑒𝑛𝑖𝑑𝑠
EXAMPLE 2: Given kite WXYZ
20
12
12
12
U
W
Z
Y
X
9
9
What is the length of segment XY?
EXAMPLE 2: Given kite WXYZ
20
12
12
12
U
W
Z
Y
X
9
9
π‘‹π‘Œ2
= π‘ˆπ‘‹2
+ π‘ˆπ‘Œ2
π‘‹π‘Œ2
= 92
+ 122
π‘‹π‘Œ2
= 81 + 144
π‘‹π‘Œ2
= 225
XY= 15
Example 3
β€’ Find mG and mJ.
60ο‚°132ο‚°
J
G
H
K
Since GHJK is a kite G  J
So 2(mG) + 132ο‚° + 60ο‚° = 360ο‚°
2(mG) =168ο‚°
mG = 84ο‚° and mJ = 84ο‚°
Try This!
β€’ RSTU is a kite. Find mR, mS and mT.
xο‚°
125ο‚°
x+30ο‚°
S
U
R T
x +30 + 125 + 125 + x = 360
2x + 280 = 360
2x = 80
x = 40
So mR = 70ο‚°, mT = 40ο‚° and mS = 125ο‚°
QUIZ
β€’ Given kite BCDA and point P be the point of
intersection of the diagonals , consider the given
information below and answer the question that
follows.
1. 𝐢𝐴 = 20π‘š 2.𝐢𝐴 = 14π‘š
𝐡𝐷 = 24π‘š 𝐡𝐢 = 25π‘š
What is the area of kite BCDA? 𝐡𝑃 =?
3. 𝐢𝑃 = 5
𝐢𝐷 = 13
𝐡𝐢 = 74
𝐢𝐴 =?
𝐡𝐷 =?
What is the area of kite BCDA?
D
C
A
B

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Math reviewers-theorems-on-kite

  • 1. K I t e
  • 2. Definition β€’KITE – is a quadrilateral with two sets of distinct adjacent congruent sides, but opposite sides are not congruent.
  • 3.
  • 4. IN ANCIENT TIME KITE were widely considered to be useful for ensuring a good harvest or scaring away evil spirits.
  • 5. IN MODERN TIME KITE became more widely known as children's toys and came to be used primarily as a leisure activity
  • 6. β€’ From the definition, a kite is the only quadrilateral that we have discussed that could be concave or non convex. Concave or non convex kite is a kite whose diagonal do not intersect. If a kite is concave or non convex, it is called a dart .
  • 8. β€’ The angles between the congruent sides are called vertex angles . The other angles are called non-vertex angles . If we draw the diagonal through the vertex angles, we would have two congruent triangles. B A C D
  • 9. THEOREM 1: The non-vertex angles of a kite are congruent and the diagonal through the vertex angle is the angle bisector for both angles. PROOF: GIVEN: KITE WITH 𝐾𝐸≅𝑇𝐸 AND 𝐾𝐼≅𝑇𝐼 STATEMENTS REASONS 1. 1. 2. 2. 3. 3. 4. 4. 5. 5. KITE WITH 𝐾𝐸≅𝑇𝐸 AND 𝐾𝐼≅𝑇𝐼 GIVEN 𝐼𝐸 β‰… 𝐼𝐸 REFLEXIVE PROPERTY βˆ†πΎπΌπΈ β‰… βˆ†π‘‡πΌπΈ SSS CONGRUENCE POSTULATE ∠𝐾 β‰… βˆ π‘‡ CPCTC βˆ π‘‡πΌπΈ β‰… ∠𝐾𝐼𝐸 AND ∠𝐾𝐸𝐼 β‰… βˆ π‘‡EI CPCTC PROVE: ∠𝐾 β‰… βˆ π‘‡, βˆ π‘‡πΌπΈ β‰… ∠𝐾𝐼𝐸 AND ∠𝐾𝐸𝐼 β‰… βˆ π‘‡EI
  • 10. THEOREM 2: The diagonals of a kite are perpendicular to each other. PROOF: GIVEN: Kite BCDA STATEMENTS REASONS 1. 1. 2. 2. 3. 3. 4. 4. D C A B Kite BCDA GIVEN 𝐡𝐢≅𝐡𝐴 AND 𝐢𝐷≅𝐴𝐷 Definition of kite Definition of congruent segments𝐡𝐢 = 𝐡𝐴 AND 𝐢𝐷 = 𝐴𝐷 𝐢𝐴βŠ₯𝐡𝐷 If a line contains two points each of which is equidistant from the endpoints of a segment, then the line is perpendicular bisector of the segment. PROVE: π‘ͺ𝑨βŠ₯𝑩𝑫
  • 11. Theorem 3: The area of a kite is half the product of the lengths of the diagonals. w PROOF: GIVEN: Kite BCDA STATEMENTS REASONS 1. 1. 2. 2. 3. 3. 4. 4. 5. 5. 6. 6. 7. 7. 8. 8. Kite BCDA GIVEN 𝐢𝐴βŠ₯𝐡𝐷 The diagonals of a kite are perpendicular to each other. 𝑨𝒓𝒆𝒂 𝒐𝒇 π’Œπ’Šπ’•π’† 𝑩π‘ͺ𝑫𝑨 = 𝑨𝒓𝒆𝒂 π’π’‡βˆ†π‘©π‘ͺ𝑨 + 𝑨𝒓𝒆𝒂 𝒐𝒇 βˆ†π‘ͺ𝑫𝑨 Area addition postulate 𝑨𝒓𝒆𝒂 π’π’‡βˆ†π‘©π‘ͺ𝑨 = 𝟏 𝟐 (π‘ͺ𝑨)(𝑩𝑾) 𝑨𝒓𝒆𝒂 π’π’‡βˆ†π‘ͺ𝑫𝑨 = 𝟏 𝟐 (π‘ͺ𝑨)(𝑫𝑾) Area formula for triangles 𝐀𝒓𝒆𝒂 𝒐𝒇 π’Œπ’Šπ’•π’† 𝑩π‘ͺ𝑫𝑨 = 𝟏 𝟐 π‘ͺ𝑨 𝑩𝑾 + 𝟏 𝟐 (π‘ͺ𝑨)(𝑫𝑾) Substitution 𝐀𝒓𝒆𝒂 𝒐𝒇 π’Œπ’Šπ’•π’† 𝑩π‘ͺ𝑫𝑨 = 𝟏 𝟐 π‘ͺ𝑨 𝑩𝑾 + 𝑫𝑾 Associative Property π΅π‘Š + π·π‘Š = 𝐡𝐷 Segment Addition Postulate 𝐀𝒓𝒆𝒂 𝒐𝒇 π’Œπ’Šπ’•π’† 𝑩π‘ͺ𝑫𝑨 = 𝟏 𝟐 π‘ͺ𝑨 (𝑩𝑫) Substitution PROVE: 𝑨𝒓𝒆𝒂 𝒐𝒇 π’Œπ’Šπ’•π’† 𝑩π‘ͺ𝑫𝑨 = 𝟏 𝟐 (π‘ͺ𝑨)(𝑩𝑫)
  • 12. Example 1 β€’Find the area of the kite WXYZ. 20 12 12 12 U W Z Y X
  • 13. Example1 Continued 20 12 12 12 U W Z Y X We can now use the formula in finding the area of the kite. Area of kite WXYZ= 1 2 𝑑1𝑑2 Area of kite WXYZ= 1 2 (𝑋𝑍)(π‘Šπ‘Œ) Area of kite WXYZ= 1 2 (24)(32) Area of kite WXYZ=384 π‘ π‘žπ‘’π‘Žπ‘Ÿπ‘’ 𝑒𝑛𝑖𝑑𝑠
  • 14. EXAMPLE 2: Given kite WXYZ 20 12 12 12 U W Z Y X 9 9 What is the length of segment XY?
  • 15. EXAMPLE 2: Given kite WXYZ 20 12 12 12 U W Z Y X 9 9 π‘‹π‘Œ2 = π‘ˆπ‘‹2 + π‘ˆπ‘Œ2 π‘‹π‘Œ2 = 92 + 122 π‘‹π‘Œ2 = 81 + 144 π‘‹π‘Œ2 = 225 XY= 15
  • 16. Example 3 β€’ Find mG and mJ. 60ο‚°132ο‚° J G H K Since GHJK is a kite G  J So 2(mG) + 132ο‚° + 60ο‚° = 360ο‚° 2(mG) =168ο‚° mG = 84ο‚° and mJ = 84ο‚°
  • 17. Try This! β€’ RSTU is a kite. Find mR, mS and mT. xο‚° 125ο‚° x+30ο‚° S U R T x +30 + 125 + 125 + x = 360 2x + 280 = 360 2x = 80 x = 40 So mR = 70ο‚°, mT = 40ο‚° and mS = 125ο‚°
  • 18. QUIZ β€’ Given kite BCDA and point P be the point of intersection of the diagonals , consider the given information below and answer the question that follows. 1. 𝐢𝐴 = 20π‘š 2.𝐢𝐴 = 14π‘š 𝐡𝐷 = 24π‘š 𝐡𝐢 = 25π‘š What is the area of kite BCDA? 𝐡𝑃 =? 3. 𝐢𝑃 = 5 𝐢𝐷 = 13 𝐡𝐢 = 74 𝐢𝐴 =? 𝐡𝐷 =? What is the area of kite BCDA? D C A B