Due in 10!!! Fill in the blanks…

Mathematics

1 answer:

Alenkinab [10]3 years ago

5 0

sorry this is late!!

#1 - A circular grid like this one can be helpful for performing dilations

#2 - To perform a dilation, we need a (In order to perform a dilation, students will need to know the center of dilation (which can be communicated using the coordinate grid), the coordinates of the polygon that they are dilating (also communicated using the coordinate grid), and the scale factor.)

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Please i really need help

Shkiper50 [21]

Answer:

I honestly do not know im sorry something else?

Step-by-step explanation:

5 0

3 years ago

Given: △ABC, m∠A=60°, m∠C=45°, AB=9 Find: Perimeter of △ABC, Area of △ABC

stellarik [79]

Answer:

Perimeter of ΔABC: $\frac{27}{2}$ + $\frac{9}{2} * \sqrt6$ units

Area of ΔABC: $\frac{81}{8}*\sqrt3 + \frac{243}{8}$ units

Skills required: HS Geo, Special Triangles

Step-by-step explanation:

1) The best option is to break down this triangle. Let's draw an altitude from Point B down to Segment AC. The point from the altitude that intersects AC is Point D. BD is the height of our triangle, AC is the base.

2) Angle A is 60 degrees, and since Angle BDA is 90 degrees, Angle ABD is 30 degrees. We can use the 30-60-90 degree right triangle property for the triangle BDA.

This states that if the side opposite the 30 degree angle is $x$ , the side opposite the 60 degree angle is $x*\sqrt3$ , and the side opposite the 90 degree angle is $2x$ .

AB is 9 units, and it is opposite the 90 degree angle. This means that $2x=9, x = \frac{9}{2}$ ==> This then means that AD, the segment opposite the 30 degree angle in this triangle is $\frac{9}{2}$ units. Segment BD (the height) is $\frac{9}{2} * \sqrt3$ .

3) Angle C is 45 degrees, and Angle BDC is 90 degrees, which means that Angle CBD is 45 degrees. We can use the 45-45-90 degree right triangle property for the triangle BCD.

This states that if the side opposite the 45 degree angle is $x$ , the other side opposite a 45 degree angle is also $x$ , but the hypotenuse (side opposite the right (90 degree) angle) is $\sqrt{2}*x$ .

BD is $\frac{9}{2} * \sqrt3$ , which means DC is the same. BC, which is the hypotenuse is BD multiplied by square-root-2, which is $\frac{9}{2} * \sqrt6$ .

4) Area is $\frac{1}{2}*b*h$ , the base (b) is AC (which is $\frac{9}{2}+\frac{9}{2}*\sqrt3$ ), the height is BD ( $\frac{9}{2}*\sqrt3$ ). When multiple you will get $\frac{81}{4}*\sqrt3 + \frac{243}{4}$ , then this multiplied by 1/2 is