Download Geometry Sample Problems and more Lecture notes Geometry in PDF only on Docsity!
Geometry Sample Problems
Sample Proofs – Below are examples of some typical proofs covered in Jesuit Geometry classes. Shown first are blank proofs that can be used as sample problems, with the solutions shown second.
Proof #
Given: a triangle with m–3 = 90∞
Prove: – 1 and – 2 are complementary
Statements Reasons
- m–3 = 90∞ 1. Given
Proof # Given: PQ bisects – SPT, SP @ PT
Prove: DSPQ @ DTPQ
Statements Reasons
- PQ bisects – SPT, SP @ PT 1. Given
Proof # D Given: AB @ AC, BD @ CD
Prove: AD bisects – CAB
Statements Reasons
- AB @ AC, BD @ CD 1. Given
1 (^3 )
Proof # Given: p’gram ABCD w/ diagonals AC & BD
Prove: AO @ OC and DO @ OB
Statements Reasons
- p’gram ABCD w/ diagonals AC & BD 1. Given
Proof # Given: AE @ EC, DE @ EB
Prove: ABCD is a p’gram
Statements Reasons
- AE @ EC, DE @ EB
ABCD is a p’gram
- Given
Defn of a p’gram
Proof # Given: trapezoid ABCD AD @ BC
Prove: AC @ BD Statements Reasons
Proof # D Given: AB @ AC , BD @ CD
Prove: AD bisects – CAB
Statements Reasons
- AB @ AC , BD @ CD 1. Given
- AD @ AD 2. Reflexive prop. of congruence
- DACD @ DABD 3. SSS congruence postulate
- – CAD @ –BAD 4. CPCTC
- AD bisects – CAB 5. Defn of angle bisector
Proof #
Given: p’gram ABCD w/ diagonals AC &
BD
Prove: AO @ OC and DO @ OB
Statements Reasons
- p’gram ABCD w/ diagonals AC & BD 1. Given
- AB ║ DC 2. Defn of parallelogram
- – BAO @ –DCO and – ABO @ –CDO 3. Alt Int – ‘s @
- AB @ CD 4. Opposite sides of a p’gram^ @
- DABO @ DDCO 5. ASA congruence theorem
- AO @ OC and DO @ OB 6. CPCTC
Proof #
Given: AE @ EC and DE @ EB
Prove: ABCD is a p’gram
Statements Reasons
- AE @ EC and DE @ EB 1. Given
- – BEC @ –AED 2. Vertical – ‘s @
- DBEC @ DAED 3. SAS congruence postulate
- – CBE @ –ADE 4. CPCTC
- BC ║ AD 5. Alt int^ – ‘s^ @ Æ^ lines ║
- – AEB @ –CED 6. Vertical – ‘s @
- DAEB @ DCED 7. SAS congruence postulate
- – DCE @ –BAE 8. CPCTC
- AB ║ DC 9. Alt int^ – ‘s^ @ Æ^ lines ║
- ABCD is a p’gram 10. Defn of a parallelogram
Proof #
Given: trapezoid ABCD with AD @ BC
Prove: AC @ BD
Statements Reasons
- Trapezoid ABCD, AD @ BC 1. Given
- ABCD is isosceles Trapezoid 2. Definition of isosceles trapezoid
- DC @ DC 3. Reflexive prop of^ @
- – BCD @ –ADC 4. Base – ‘s in an isosceles trapezoid are @
- DBCD @ DADC 5. SAS
- AC @ BD 6. CPCTC
Proof # Given: p’gram ABCD
Prove: ABCD is a rhombus
Statements Reasons
- p’gram ABCD 1. Given
- – CBD @ –ABD, – BDC @ –BDA 2. Given
- BD @ BD 3. Reflexive prop of^ @
- DBCD @ DBAD 4. ASA congruence theorem
- CD @ AD 5. CPCTC
- BC @ AB 6. CPCTC
- CD @ AB , AD @ BC 7. Opposite sides of parallelogram are congruent
- AB @ BC @ AD @ CD 8. Transitive prop of^ @
- ABCD is a rhombus 9. Defn of a rhombus
Example 4:
Q
R S
SR is tangent to circle Q at R. Find RS. (QS = 20, QR = 8)
Answer : The radius, QR, is perpendicular to RS at the tangent point R. Therefore, – QRS is a right angle.
Use the Pythagorean theorem to find “leg” RS. 8 2 + ( RS )^2 = 202 ; 336 =( RS )^2 RS = 4 21
Example 5: Given: PS = 6, SR = 8, find the value of SQ.
P 6 S 8 R
Q
Answer : Since SQ is an altitude of triangle PQR, SQ is the geometric mean of segments PS and SR.
Therefore we have the ratio SR
SQ
SQ
PS
6 SQ
SQ
( SQ )^2 = 48
SQ = 4 3
Regular Geometry and XL style questions
Example 1 and 2 are the types of problems Jesuit expects both Geometry and Geometry XL students to be able to do. Example 3 is a Geometry XL type of problem. The problems show our expectations of students regarding all the steps necessary to complete a problem.
Equations of Circles
The standard equation of a circle with radius r and center (h, k) is: (x h)^2 + (y k)^2 = r^2
Example 1 : Write the standard equation of the circle whose center is ( 2, 3) and whose radius is 4.
Solution: (x h)^2 + (y k)^2 = r^2 (x ( 2))^2 + (y 3)^2 = 4^2 (x + 2)^2 + (y 3)2 =^16
Example 2 : Write the standard equation of the circle whose center is (1, 1) and passes through the point ( 1, 4).
Solution: The radius is the distance from ( 1, 4) to the center (1, 1).
r = ( 1 1) 2 + (4 1) 2 = ( 2) 2 + (3) 2 = 13
Thus, the equation is: (x 1)^2 + (y 1)^2 = ( 13 )
2 or (x 1)^2 + (y 1)^2 = 13
