Strategy to solve complex problems Two right triangles are superimposed on one another. One of them is displaced by 9 cm, find the area of the shaded portion.

Slides:



Advertisements
Similar presentations
TWO STEP EQUATIONS 1. SOLVE FOR X 2. DO THE ADDITION STEP FIRST
Advertisements

Objective - To use basic trigonometry to solve right triangles.
Multiply Polynomials When multiplying polynomials, we always use the Distributive Property.
Law of Sines and Cosines
ACTIVITES MENTALES Collège Jean Monnet Préparez-vous !
Fractions VI Simplifying Fractions
Repeated Addition Or Adding up.
Multiplication Facts Review. 6 x 4 = 24 5 x 5 = 25.
Exponents You will have 20 seconds to complete each of the following 16 questions. A chime will sound as each slide changes. Read the instructions at.
Reducing Fractions. Factor A number that is multiplied by another number to find a product. Factors of 24 are (1,2, 3, 4, 6, 8, 12, 24).
Merrill pg. 765, feet meters feet meters meters ° 16. 1° °
0 - 0.
1  1 =.
2 pt 3 pt 4 pt 5 pt 1 pt 2 pt 3 pt 4 pt 5 pt 1 pt 2 pt 3 pt 4 pt 5 pt 1 pt 2 pt 3 pt 4 pt 5 pt 1 pt 2 pt 3 pt 4 pt 5 pt 1 pt Time Money AdditionSubtraction.
MULTIPLICATION EQUATIONS 1. SOLVE FOR X 3. WHAT EVER YOU DO TO ONE SIDE YOU HAVE TO DO TO THE OTHER 2. DIVIDE BY THE NUMBER IN FRONT OF THE VARIABLE.
MULT. INTEGERS 1. IF THE SIGNS ARE THE SAME THE ANSWER IS POSITIVE 2. IF THE SIGNS ARE DIFFERENT THE ANSWER IS NEGATIVE.
FACTORING Think Distributive property backwards Work down, Show all steps ax + ay = a(x + y)
Addition Facts
August 27 AP physics.
BALANCING 2 AIM: To solve equations with variables on both sides.
GCSE Higher Revision Starters 7
Factorise the following 10x a – 20 36m a + 27b + 9c 9y² - 12y 30ab³ + 35a²b 24x4y³ - 40x²y.
Richmond House, Liverpool (1) 26 th January 2004.
Multiplication Tables Test 3 and 5 Times Tables 3 and 5 times tables There are 10 questions. Each one will stay on the screen for 15 seconds. Write down.
7.3 Area of Complex Figures
Areas of Complex Shapes
Area of triangles.
Area in the amount of space inside an enclosed region. Area of Rectangle = base x height Base =10 Height = 6 Area = (10)(6) = 60 square units.
Mathematical Similarity
Squares and Square Root WALK. Solve each problem REVIEW:
8 2.
Mathematics Involving Shape and Space a Algebra. The 9-Dot Problem.
dd vv Fast constant negative Slow constant negative At rest Getting slower In POS direction Slow positive velocity Same velocity.
Lets play bingo!!. Calculate: MEAN Calculate: MEDIAN
UNIT 2: SOLVING EQUATIONS AND INEQUALITIES SOLVE EACH OF THE FOLLOWING EQUATIONS FOR y. # x + 5 y = x 5 y = 2 x y = 2 x y.
Past Tense Probe. Past Tense Probe Past Tense Probe – Practice 1.
Special Right Triangles
Addition 1’s to 20.
Strategy to solve complex problems
Test B, 100 Subtraction Facts
CH 8 Right Triangles. Geometric Mean of 2 #’s If you are given two numbers a and b you can find the geometric mean. a # = # b 3 x = x 27 Ex ) 3 and 27.
Main Idea/Vocabulary composite figure Find the area of composite figures.
10-7 The Quadratic Formula
Week 1.
Bottoms Up Factoring. Start with the X-box 3-9 Product Sum
Practice Skip Counting
Preview Warm Up California Standards Lesson Presentation.
Special Right Triangles
SOLVING EQUATIONS WITH ADDITION & SUBTRACTION By: Erica Wagner Let’s get started! Let’s get started!
35 cm 40 cm Area of rectangle = length × breadth Area of cardboard = 40 cm × 35 cm = 1400 cm² Area of picture = 30 cm × 25 cm = 750 cm² Area of cardboard.
Volume of Triangular Prism. Volume of a Triangular Prism Length Volume of a prism = Area x length Area of triangle = ½ x base x height.
8cm 5cm Area = 8 x 5 = 40cm 2 A parallelogram can be split up into a rectangle and 2 triangles – each with the same area. 10cm 5cm.
1 Trig. Day 3 Special Right Triangles. 2 45°-45°-90° Special Right Triangle 45° Hypotenuse X X X Leg Example: 45° 5 cm.
Perimeter and Area of rectangles, parallelograms and triangles
Which of the following statements is true for this triangle?
عناصر المثلثات المتشابهة Parts of Similar Triangles
Solve: 1. 4<
45°-45°-90° Special Right Triangle
Mathsercise-C Ready? Equations 1 Here we go!.
Special Right Triangles
Special Right Triangles
Special Right Triangles
12cm Area 64cm2 Area 100cm2 ? ? Area ?cm2 12cm Area 36cm2
12cm Area 64cm2 Area 100cm2 ? ? Area ?cm2 12cm Area 36cm2
Special Right Triangles
Similar Triangles Make a model of each triangle using Centimeters (cm)
Presentation transcript:

Strategy to solve complex problems Two right triangles are superimposed on one another. One of them is displaced by 9 cm, find the area of the shaded portion of the triangle. 15 cm 27 cm

Strategy to solve complex problems Two right triangles are superimposed on one another. One of them is displaced by 9 cm, find the area of the shaded portion of the triangle. 15 cm 27 cm

Strategy to solve complex problems Two right triangles are superimposed on one another. One of them is displaced by 9 cm, find the area of the shaded portion of the triangle. 15 cm 27 cm

Strategy to solve complex problems Two right triangles are superimposed on one another. One of them is displaced by 9 cm, find the area of the shaded portion of the triangle. 15 cm 27 cm

Strategy to solve complex problems Two right triangles are superimposed on one another. One of them is displaced by 9 cm, find the area of the shaded portion of the triangle. 15 cm 27 cm

Strategy to solve complex problems Two right triangles are superimposed on one another. One of them is displaced by 9 cm, find the area of the shaded portion of the triangle. 15 cm 27 cm

9 cm Strategy to solve complex problems Two right triangles are superimposed on one another. One of them is displaced by 9 cm, find the area of the shaded portion of the triangle. 15 cm 27 cm

Strategy to solve complex problems Two right triangles are superimposed on one another. One of them is displaced by 9 cm, find the area of the shaded portion of the triangle. 9 cm 15 cm 27 cm

Strategy to solve complex problems 9 cm 15 cm 27 cm

Strategy to solve complex problems 9 cm 15 cm 27 cm ?

Strategy to solve complex problems 9 cm 15 cm 27 cm 18:27 = ?: = ? cm ?

Strategy to solve complex problems 9 cm 15 cm 27 cm 18:27 = 10: = cm 10

Strategy to solve complex problems 9 cm 15 cm 27 cm 18 cm 10 Area: cm 2

Strategy to solve complex problems 9 cm 15 cm 27 cm 18 cm 10 Area: cm 2

Strategy to solve complex problems 9 cm 15 cm 27 cm 18 cm 10 Area: cm

Strategy to solve complex problems 9 cm 15 cm 27 cm 18 cm 10 Area: cm 2 9 ( ) 2

Strategy to solve complex problems 9 cm 15 cm 27 cm 18 cm 10 Area: cm 2 9 ( ) = cm 2