Each of the numbers from 1 to 9 is placed, one per circle, into the pattern shown. The sums along each of the four sides are equal. How many different.

Slides:



Advertisements
Similar presentations
POLYGONS.
Advertisements

Whiteboardmaths.com © 2008 All rights reserved
FUNCTIONS OF ANY ANGLE, OBLIQUE TRIANGLES
Section 8.5. Find the area of parallelograms. Base of a parallelogram Height of a parallelogram Parallelogram Rhombus.
20 Questions Chapter 10 Review. 1. Polygons The sum of the measures of the interior angles of a convex polygon is How many sides does the polygon.
EVERYTHING YOU NEED TO KNOW TO GET A GRADE C
MATHCOUNTS Countdown Round.
 A typical problem involving the area and perimeter of a rectangle gives us the area, perimeter and/r length and width of the rectangle. We may also.
Area Formulas and Parallelograms
- Four sides - Four angles - Four vertices  - The diagonals are equal in length and perpendicular and bisect each other.
The circumference of the circle is 50  feet. The area of the shaded region = ________ example 1 :
Geometry Review AREA 1. Find the measure of each interior angle of the regular polygon shown below. 2.
8-4 Angles in Polygons Problem of the Day How many different rectangles are in the figure shown? 100.
QUIZ REVIEW What to study and example problems....
Geometry--Ch. 6 Review Classify each polygon as regular/irregular, concave/convex, and name it based on its number of sides: 1)2) irregular concave decagon.
Short Answer Practice Problems
CIRCUMFERENCE AND AREA
Is there anything invariant about circles?. Bell Ringer CDCD Compare the ratio of the circumference to the diameter C = 9.42 D = 3 C =
Definition: Rectangle A rectangle is a quadrilateral with four right angles.
Sullivan Algebra and Trigonometry: Section R.3 Geometry Review Objectives of this Section Use the Pythagorean Theorem and Its Converse Know Geometry Formulas.
Area: Parallelograms, Rectangles, Squares and Trapezoids.
Basic Geometry Review-Shapes and Angles. Review Topics Squares Triangles Rectangles Polygons Obtuse Angle Acute Angle Right Angle Finished?
Triangles Sum.
Triangle Sum Properties
A parallelogram with opposite equal acute and obtuse angles and four equal sides. Diagonals 4 equal sides.
Fraction Division: A Whole Number Divided by a Fraction 1  = ? 1515 To get the answer, ask: 1  ? = 1515 How many groups of can be made from 1? 1515.
Divisibility Rules. What Are Divisibility Rules? They are a quick way to tell if one number can be completely divided into another without a remainder.
The figure is composed of a right triangle and a semi-circle. What is the area of the shaded region? The figure is not drawn to scale. 7 in 24 in.
Lesson 5.5 Use Inequalities in a Triangle. Theorem 5.10 A B C 8 5 IF AB > BC, THEN C > A The angle opposite the longest side is the largest angle; pattern.
Bellwork Most Missed Quarter III Exam (58.9%): Find the perimeter of if AB = 16 cm, and.
Two sides of an isosceles triangle measure 3 and 7. Which of the following could not be the measure of the third side? A) 9 B) 7 C) 5 d) 3.
EXAMPLE 2 Find measures in a triangle Find the measures of P, Q, and R. The diagram shows that PQR is equilateral. Therefore, by the Corollary to the Base.
Special Right Triangles. What are Special Right Triangles? There are 2 types of Right triangles that are considered special. We will talk about only one.
Perimeter and Area of rectangles, parallelograms and triangles
Practice Quiz Circles.
18.01 Area & Perimeter of Rectangles
SAT MATH Drill Lesson 5.
Find the area and circumference of each circle.
An inequality is a mathematical sentence that contains the following symbols:
Addition Subtraction Multiplication and Division Estimation Grab Bag
Date: Topic: Rhombi, Rectangles, and Squares (7.2)
Area of Triangles.
Geometry - Area Divisibility Rules
State Countdown Round MATHCOUNTS State Countdown Round.
EXAMPLE 1 Describe a visual pattern
EXPANDING DOUBLE BRACKETS
Chapter 4 – Scale Factors and Similarity Key Terms
Combine Like Terms and Distribute
9-1.
Geometric Patterns Name: ___________________________
8.1.1 Solving Simple Equations
Quadrant.
TRIANGLE INEQUALITY THEOREM
The Pythagorean Theorem and Its Converse
Area of Rhombus.
SHAPES By: Ms. Conquest.
Area of Quadrilateral.
Types of Quadrilaterals
Triangle sum property.
Parallelogram.
Parallelograms, Triangles, Rhombuses Rectangles & Trapezoids Regular
Unit 6 Quadrilaterals Section 6.3 Properties of Rectangles
Area of combined figures
EXPANDING DOUBLE BRACKETS
The Pythagorean Theorem
Find the Area of a Shaded Region
The Pythagoras Theorem c a a2 + b2 = c2 b.
The parallelogram and the isosceles triangle have the same perimeter.
This is a square. This is a square. How can you tell. How can you tell
Arc and Sector Word Problem 1
Presentation transcript:

Each of the numbers from 1 to 9 is placed, one per circle, into the pattern shown. The sums along each of the four sides are equal. How many different numbers can be placed in the middle circle to satisfy these conditions? Solution: Because the sums along each side must be equal, therefore sums of the 8 numbers must be divisible by 4. Note that 1+2+3+4+5+6+7+8+9 = 45 = 4 × 11 + 1. Therefore the numbers which have a remainder of 1 when divided by 4 are the candidates to be placed in the middle. Answer:

When Anura was 8 years old his father was 31 years old When Anura was 8 years old his father was 31 years old. Now his father is twice as old as Anura is. How old is Anura now? Solution: Now, after 𝑥 year, Anura’s father’s age is 31 + 𝑥 and Anura’s age is 8 + 𝑥 Anura’s father’s age is 2 × Anura’s age 31 + 𝑥 = 2 × (8 + 𝑥) = 16 + 2𝑥 𝑥 = 15 Anura’s age = 8 + 15 =____ Answer: 23

In rectangle 𝐴𝐵𝐶𝐷, 𝐴𝐵 = 12𝑐𝑚 and 𝐴𝐷 = 5𝑐𝑚 In rectangle 𝐴𝐵𝐶𝐷, 𝐴𝐵 = 12𝑐𝑚 and 𝐴𝐷 = 5𝑐𝑚. Point P, Q, R and S are all on diagonal AC, so that 𝐴𝑃 = 𝑃𝑄 = 𝑄𝑅 = 𝑅𝑆 = 𝑆𝐶. What is the total area of the shaded region, in cm2? Solution: The area for 𝐴𝐵𝐶𝐷 = 5 × 12 = 60 𝑐𝑚2 The area for △𝐴𝐵𝐶 and △𝐴𝐷𝐶 = 60 ÷ 2 = 30 𝑐𝑚2 Since 𝐴𝑃=𝐶𝑆= 1 5 𝐴𝐶, therefore the area for △𝐴𝑃𝐷=△𝐴𝑃𝐵=△𝐶𝑆𝐷=△𝐶𝑆𝐵= 1 5 △𝐴𝐵𝐶= 1 5 ×30=6 𝑐𝑚2 Total area of shaded region =6 +6+6+6=____ Answer: 24 𝑐𝑚2

In triangle 𝐴𝐵𝐶, 𝐴𝑃 = 𝐴𝑄 and 𝐵𝑄 = 𝐵𝑅. Determine angle 𝑃𝑄𝑅 in degrees. Solution: 𝐴𝑃 = 𝐴𝑄⇒∠𝐴𝑃𝑄 = ∠𝐴𝑄𝑃 = ∝° 𝐵𝑄 = 𝐵𝑅⇒∠𝐵𝑅𝑄 = ∠𝐵𝑄𝑅 =𝛽° ∠𝑃𝐴𝑄 = 180° − 2∝° ∠𝑅𝐵𝑄 = 180° − 2𝛽° From △𝐴𝐵𝐶, 70° + (180° − 2∝°) + (180° − 2𝛽°) = 180° 2∝° + 2𝛽° = 250° ∝° + 𝛽° = 125° ∠𝑃𝑄𝑅 = 180° − ∝° − 𝛽° = 180° − (∝° + 𝛽°) = 180° − 125° = 55 ° Answer: 55°