5 Minute Check on Lesson 11 1 Transparency

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5 -Minute Check on Lesson 11 -1 Transparency 11 -2 Find the area and

5 -Minute Check on Lesson 11 -1 Transparency 11 -2 Find the area and the perimeter of each parallelogram. Round to the nearest tenth if necessary. 17 ft 11 cm 1. 2. 13 cm A = 204 ft² A = 101. 1 cm² 12 ft P = 58 ft P = 48 cm 45° 3. 4. 60° A = 39. 7 m² P = 25. 2 m A = 171. 5 in² P = 58 in 11 in 18 in 6. 3 m 5. Find the height and base of this parallelogram if the area is 168 square units x+2 x h = 12 , b = 14 units 6. Find the area of a parallelogram if the height is 8 cm and the base length is 10. 2 cm. Standardized Test Practice: A 28. 4 cm² B 29. 2 cm² C 81. 6 cm² Click the mouse button or press the Space Bar to display the answers. D 104. 4 cm²

Lesson 11 -2 Areas of Triangles, Trapezoids, and Rhombi

Lesson 11 -2 Areas of Triangles, Trapezoids, and Rhombi

Objectives • Find areas of triangles – A = ½ bh • Find areas

Objectives • Find areas of triangles – A = ½ bh • Find areas of trapezoids – A = ½ (b 1 + b 2)h • Find areas of rhombi – A = ½ d 1 · d 2 (note: this is the one area formula not on SOL formula sheet)

Vocabulary • base – the “horizontal” distance of the figure (bottom side) • height

Vocabulary • base – the “horizontal” distance of the figure (bottom side) • height – the “vertical” distance of the figure • area – the amount of flat space defined by the figure (measured in square units) • perimeter – once around the figure

Area of Triangles, Trapezoids & Rhombi Triangle Area R A = ½ * b

Area of Triangles, Trapezoids & Rhombi Triangle Area R A = ½ * b * h = ½ * ST * RW h is height (altitude) b is base (┴ to h) h K Trapezoid Area A = ½* h* (b 1 + b 2) = ½ * LN * (JK + LM) h is height (altitude) b 1 and b 2 are bases (JK & LM) (bases are parallel sides) h L T W b 1 N J S b 2 M A B d 1 Rhombus Area A = ½ * d 1 * d 2 = ½ * AD * BC d 1 and d 2 are diagonals d 2 C D

Triangle Area Example R Find the area of triangle RST h A = ½

Triangle Area Example R Find the area of triangle RST h A = ½ bh = ½ 20(h) = 10 h square units S 45° 10 W (side opposite 45°) h = ½ hyp √ 2 No hypotenuse! ∆ RSW is right isosceles; so legs are equal! h = 10 So, area = 10(10) = 100 square units T 10

Trapezoids Area Example Find the area of trapezoid JKLM A = ½ (b 1

Trapezoids Area Example Find the area of trapezoid JKLM A = ½ (b 1 + b 2)h = ½ (12 + 20)(h) = 16 h square units (side opposite 60°) h = ½ hyp √ 3 J 60° 20 N 14 K h M 12 L h = ½ (14) √ 3 h = 7 √ 3 So, area = 16(7√ 3) ≈ 193. 99 square units

Rhombi Area Example A Find the area of rhombus ABCD A = ½ (d

Rhombi Area Example A Find the area of rhombus ABCD A = ½ (d 1 · d 2) 5 = ½ (2(3) · 2(4)) = ½ (48) = 24 square units What if we try to find the area by adding the 4 triangles together? A = 4 (½ bh) = 2 bh A = 2(3)(4) = 2 (12) = 24 square units!! B 3 4 D 5 C

Find the area of quadrilateral ABCD if AC = 35, BF = 18, and

Find the area of quadrilateral ABCD if AC = 35, BF = 18, and DE = 10. The area of the quadrilateral is equal to the sum of the areas of Area formula Substitution Simplify. Answer: The area of ABCD is 490 square units.

Find the area of quadrilateral HIJK if IK = 16, HL = 5 and

Find the area of quadrilateral HIJK if IK = 16, HL = 5 and JM = 9 Answer:

Rhombus RSTU has an area of 64 square inches. Find US if RT =

Rhombus RSTU has an area of 64 square inches. Find US if RT = 8 inches. Use the formula for the area of a rhombus and solve for d 2. Answer: US is 16 inches long.

Trapezoid DEFG has an area of 120 square feet. Find the height of DEFG.

Trapezoid DEFG has an area of 120 square feet. Find the height of DEFG. Use the formula for the area of a trapezoid and solve for h. Answer: The height of trapezoid DEFG is 8 feet.

a. Rhombus ABCD has an area of 81 square centimeters. Find BD if AC

a. Rhombus ABCD has an area of 81 square centimeters. Find BD if AC = 6 centimeters. Answer: 27 cm b. Trapezoid QRST has an area of 210 square yards. Find the height of QRST. Answer: 6 yd

STAINED GLASS This stained glass window is composed of 8 congruent trapezoidal shapes. The

STAINED GLASS This stained glass window is composed of 8 congruent trapezoidal shapes. The total area of the design is 72 square feet. Each trapezoid has bases of 3 and 6 feet. Find the height of each trapezoid. First, find the area of one trapezoid. From Postulate 11. 1, the area of each trapezoid is the same. So, the area of each trapezoid is 72 8 or 9 square feet. Next, use the area formula to find the height of each trapezoid.

Area of a trapezoid Substitution Add. Multiply. Divide each side by 4. 5. Answer:

Area of a trapezoid Substitution Add. Multiply. Divide each side by 4. 5. Answer: Each trapezoid has a height of 2 feet.

INTERIOR DESIGN This window hanging is composed of 12 congruent trapezoidal shapes. The total

INTERIOR DESIGN This window hanging is composed of 12 congruent trapezoidal shapes. The total area of the design is 216 square inches. Each trapezoid has bases of 4 and 8 inches. Find the height of each trapezoid. Answer: 3 in.

Summary & Homework • Summary: – The formula for the area of a triangle

Summary & Homework • Summary: – The formula for the area of a triangle can be used to find the areas of many different figures – Congruent figures have equal areas • Homework: – pg 606 -608; 13 -18, 30 -34