StudentAidUnit 9: Similar Figures — Exercise 9 4
9th Class Mathematics · Unit 9: Similar Figures
9.4.1.(i) What is the sum of the interior angles of a decagon (10-sided polygon)? (ii) Calculate the measure of each interior angle of a regular hexagon. (iii) What is each exterior angle of a regular pentagon? (iv) If the sum of the interior angles of a polygon is 1260°, how many sides does the polygon have?
(i)
(n-2) × 180^circ=(10-2) × 180^circ=8 × 180^circ=1440^circ
(ii)
(n-2) × 180^circn=(6-2) × 180^circ6=120^circ
(iii)
360^circn=360^circ5=72^circ
(iv)
1260^circ=(n-2) × 180^circ ⇒ 1260^circ180^circ=n-2 ⇒ 7=n-2 ⇒ n=9
(i) 1440^circ (ii) 120^circ (iii) 72^circ (iv) n=9
9.4.2.In a parallelogram ABCD, mAB = 10 cm, mAD = 6 cm and m∠BAD = 45°. Calculate the area of ABCD.
Formula
Area=overline{AB} × overline{AD}sinθ
Working
Area=10 × 6sin45^circ
Result
Area=42.43 cm2
9.4.3.In a parallelogram ABCD if m∠DAB = 70°, find the measures of all other angles in the parallelogram.
Given
mangle DAB=70^circ
mangle DAB=mangle BCD=70^circquad(becauseopposite angles of |gram)
mangle DAB+mangle ABC=180^circquad(because ADparallel BC) ⇒ mangle ABC=110^circ
Result
mangle CDA=110^circquad(because mangle ABC=mangle CDA)
9.4.4.A shape is created by cutting a square in half diagonally and then attaching a right-angled triangle to the hypotenuse of each half. Explain why this shape can tessellate and calculate the interior angle of the new shape.
Given
Sum of interior angles of a hexagon=(6-2) × 180^circ=720^circ
Shape is composed of two identical parts ⇒ sum per part=720^circdiv2=360^circ
Result
Interior angle=360^circdiv6=60^circ (equilateral part), 120^circ (isosceles part)
9.4.5.A tessellation is created by repeatedly reflecting a basic shape. The basic shape is a right-angled triangle with sides of length 3, 4, and 5 units. Find: The minimum number of reflections needed to create a tessellation that covers a square with an area of 3600 square units.
Formula
Area of triangle=frac12 × Base × Height
Working
Area of triangle=frac12 × 3 × 4=6 units2
Number of triangles=frac{area of square{area of triangle=36006=600
Result
600 reflections needed to cover the square.
9.4.6.A tessellation is created using regular hexagons. Each hexagon has a side length of 5 cm. Find the total area of the tessellation if it consists of 25 hexagons and total perimeter of the outer edge of the tessellation, assuming it's a perfect hexagon.
Formula
Area of hexagon
3sqrt32a2=3sqrt32 × 52=3sqrt32 × 25=64.95 cm2
Area of 25 hexagons
64.95 × 25=1623.75 cm2
Perimeter
6 × (5 × 5)=6 × 25=150 cm
Given
Note
Length of 1 side = 5 × 5 since each side consists of 5 hexagons
Total area=1623.75 cm2, Perimeter=150 cm
9.4.7.A rectangular floor is 12 m by 15 m. How many square tiles, each 1 m by 1 m, are needed to cover the floor?
Formula
Number of tiles=frac{area of floor{area of one tile
Working
=12 × 151 × 1=frac{180 m2}{1 m2}
Result
=180 tiles
9.4.8.A rectangular wall is 10 m tall and 12 m wide. How many gallons of paint are needed to cover the wall, if one gallon covers 35 m²?
9.4.9.A rectangular wall has a length of 10 m and a width of 4 meters. If 1 litre of paint covers 7 m², how many liters of paint are needed to cover the wall?
Formula
Paint needed=frac{area of the wall{paint per liter
Working
=10 × 47=frac{40 m2}{7 m2}
Result
=5.71 ≈ 6 liters
9.4.10.A window has a trapezoidal shape with parallel sides of 3 m and 1.5 m and a height of 2 m. Find the area of the window.
Formula
Area of trapezoidal=frac{sum of parallel sides{2} × height
Working
=3+1.52 × 2
Result
=4.5 m2