a) Prove that AABC/HAADE,b) Hence find the length of DB if AD 10mm,AE-6mm and EC 12mm.Solutions to Exercise 2 number 3 are on page 335.4) In the diagram below there are 2 similar triangles:a) State, with reason, the triangles that are similar.Remember that the order the triangles arewritten in is important.b) Given that AB = 10m and AC = 12m, determinethe length of AD.Solutions to Exercise 2 number 4 are on page 335CongruencyExercise 3: (Theory on pages 144-146)1) Prove that AABC = ADCB:3) Prove that AABD = ACBE:agled orA.丰CD.C2) Given 0 is the circle's centre, prove that 4) Prove that AMBE = AMCD:AABO = APQO:BMEQ.IcSolutions to Exercise 3 mumbers I and 2 are on page 335 and number 3 and 4 are on page 336QuadrilateralsExercise 4: (Theory on pages 147-150)1) Prove that quadrilateral ABCD below is a 3) ABCD below is a'parallelogram. PQRS is aparallelogram:R.quadrilateral formed by joining the

We will solve the following. As per the guidelines we can solve 3 questions only

Congruency Exercise 3: (Theory on pages 144-146) 1) Preve that AABC = ADCB: 3) Prove that AABD = ACBE: 丰 D. 2) Given O is the circle's centre, prove that AABO = APQO: 4) Prove that AMBE = AMCD: Q. Solutions to Exercise 3 numbers I and 2 are on page 335 and number 3 and 4 are on page 336 Quadrilaterals Exercise 4: (Theory on pages 147-150) 1) Prove that quadrilateral ABCD below is a 3) ABCD below is a'parallelogram. PQRS is a parallelogram: quadrilateral formed by joining the

Congruency Exercise 3: (Theory on pages 144-146) 1) Preve that AABC = ADCB: 3) Prove that AABD = ACBE: 丰 D. 2) Given O is the circle's centre, prove that AABO = APQO: 4) Prove that AMBE = AMCD: Q. Solutions to Exercise 3 numbers I and 2 are on page 335 and number 3 and 4 are on page 336 Quadrilaterals Exercise 4: (Theory on pages 147-150) 1) Prove that quadrilateral ABCD below is a 3) ABCD below is a'parallelogram. PQRS is a parallelogram: quadrilateral formed by joining the

Elementary Geometry For College Students, 7e

7th Edition

ISBN:9781337614085

Author:Alexander, Daniel C.; Koeberlein, Geralyn M.

Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.

ChapterP: Preliminary Concepts

SectionP.CT: Test

Problem 1CT

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Concept explainers

Coordinate Geometry

Coordinate geometry, also known as analytic geometry or Cartesian geometry in classical mathematics, is a type of geometry that is studied using a coordinate system. Coordinate geometry is a branch of mathematics that describes the position of points on a plane using an ordered pair of numbers called coordinates.

Question

a) Prove that AABC/HAADE,
b) Hence find the length of DB if AD 10mm,
AE-6mm and EC 12mm.
Solutions to Exercise 2 number 3 are on page 335.
4) In the diagram below there are 2 similar triangles:
a) State, with reason, the triangles that are similar.
Remember that the order the triangles are
written in is important.
b) Given that AB = 10m and AC = 12m, determine
the length of AD.
Solutions to Exercise 2 number 4 are on page 335
Congruency
Exercise 3: (Theory on pages 144-146)
1) Prove that AABC = ADCB:
3) Prove that AABD = ACBE:
agled or
A.
丰
C
D.
C
2) Given 0 is the circle's centre, prove that 4) Prove that AMBE = AMCD:
AABO = APQO:
B
M
E
Q.
Ic
Solutions to Exercise 3 mumbers I and 2 are on page 335 and number 3 and 4 are on page 336
Quadrilaterals
Exercise 4: (Theory on pages 147-150)
1) Prove that quadrilateral ABCD below is a 3) ABCD below is a'parallelogram. PQRS is a
parallelogram:
R.
quadrilateral formed by joining the

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