Which shape has opposite sides that are parallel and 4 sides that are always the same length? A. parallelogram B. rhombus C. tra
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Task 1: criteria for congruent triangles
a. (SSA) is not a valid mean for establishing triangle congruence. In this case we know <span>the measure of two adjacent sides and the angle opposite to one of them. Since we don't know anything about the measure of the third side, the second side of the triangle can intercept the third side in more than one way, so the third side can has more than one length; therefore, the triangles may or may not be congruent. In our example (picture 1) we have a triangle with tow congruent adjacent sides: AC is congruent to DF and CB is congruent to FE, and a congruent adjacent angle: </span>∠CAB is congruent to <span>∠FDE, yet triangles ABC and DEF are not congruent.
b. </span><span>(AAA) is not a valid mean for establishing triangle congruence. In this case we know the measures of the three interior sides of the triangles. Since the measure of the angles don't affect the lengths of the sides, we can have tow triangles with 3 congruent angles and three different sides. In our example (picture 2) the three angles of triangle ABC and triangle DEF are congruent, yet the length of their sides are different.
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c. <span>(SAA) is a valid means for establishing triangle congruence. In this case we know </span>the measure of a side, an adjacent angle, and the angle opposite to the side; in other words we have the measures of two angles and the measure of the non-included side, which is the AAS postulate. Remember that the AAS postulate states that if two angles and the non-included side of one triangle are congruent to two angles and the non-included side of another triangle, then these two triangles are congruent. Since SAA = AAS, we can conclude that SAA is a valid mean for establishing triangle congruence.
Task 2: geometric constructions
a. Step 1. Take a point A and point B, so AB is the radius of the circle; draw a circle at center A and radius AB.
Step 2. Draw another circle with radius AB but this time with center at B.
Step 3. Mark the two points, C and D, of intersection of both circles.
Step 4. Use the points C and D to mark a point E in the circle with center at A.
Step 5. Join the points C, D, and E to create the equilateral triangle CDE inscribed in the circle with center at A (picture 3).
b. Step 1. take a point A and point B, so AB is the radius of the circle; draw a circle at center A and radius AB.
Step 2. The point B is the first vertex of the inscribed square.
Step 3. Draw a diameter from point B to point C.
Step 4. Set a radius form point B to point D passing trough A, and draw a circle.
Step 5. Use the same radius form point C to point E using the same measure of the radius BD from the previous step.
Step 6. Draw a line FG trough were the two circles cross passing trough point A.
Step 7. Join the points B, F, C, and G, to create the inscribed square BFCG (picture 4).
c. Step 1. take a point A and point B, so AB is the radius of the circle; draw a circle at center A and radius AB.
Step 2. Draw the diameter of the circle BC.
Step 3. Use radius AB to create another circle with center at C.
Step 4. Use radius AB to create another circle with center at B.
Step 5. Mark the points D, E, F, and G where two circles cross.
Step 6. Join the points C, D, E, B, F, and G to create the inscribed regular hexagon (picture 5).
a. (SSA) is not a valid mean for establishing triangle congruence. In this case we know <span>the measure of two adjacent sides and the angle opposite to one of them. Since we don't know anything about the measure of the third side, the second side of the triangle can intercept the third side in more than one way, so the third side can has more than one length; therefore, the triangles may or may not be congruent. In our example (picture 1) we have a triangle with tow congruent adjacent sides: AC is congruent to DF and CB is congruent to FE, and a congruent adjacent angle: </span>∠CAB is congruent to <span>∠FDE, yet triangles ABC and DEF are not congruent.
b. </span><span>(AAA) is not a valid mean for establishing triangle congruence. In this case we know the measures of the three interior sides of the triangles. Since the measure of the angles don't affect the lengths of the sides, we can have tow triangles with 3 congruent angles and three different sides. In our example (picture 2) the three angles of triangle ABC and triangle DEF are congruent, yet the length of their sides are different.
</span>
c. <span>(SAA) is a valid means for establishing triangle congruence. In this case we know </span>the measure of a side, an adjacent angle, and the angle opposite to the side; in other words we have the measures of two angles and the measure of the non-included side, which is the AAS postulate. Remember that the AAS postulate states that if two angles and the non-included side of one triangle are congruent to two angles and the non-included side of another triangle, then these two triangles are congruent. Since SAA = AAS, we can conclude that SAA is a valid mean for establishing triangle congruence.
Task 2: geometric constructions
a. Step 1. Take a point A and point B, so AB is the radius of the circle; draw a circle at center A and radius AB.
Step 2. Draw another circle with radius AB but this time with center at B.
Step 3. Mark the two points, C and D, of intersection of both circles.
Step 4. Use the points C and D to mark a point E in the circle with center at A.
Step 5. Join the points C, D, and E to create the equilateral triangle CDE inscribed in the circle with center at A (picture 3).
b. Step 1. take a point A and point B, so AB is the radius of the circle; draw a circle at center A and radius AB.
Step 2. The point B is the first vertex of the inscribed square.
Step 3. Draw a diameter from point B to point C.
Step 4. Set a radius form point B to point D passing trough A, and draw a circle.
Step 5. Use the same radius form point C to point E using the same measure of the radius BD from the previous step.
Step 6. Draw a line FG trough were the two circles cross passing trough point A.
Step 7. Join the points B, F, C, and G, to create the inscribed square BFCG (picture 4).
c. Step 1. take a point A and point B, so AB is the radius of the circle; draw a circle at center A and radius AB.
Step 2. Draw the diameter of the circle BC.
Step 3. Use radius AB to create another circle with center at C.
Step 4. Use radius AB to create another circle with center at B.
Step 5. Mark the points D, E, F, and G where two circles cross.
Step 6. Join the points C, D, E, B, F, and G to create the inscribed regular hexagon (picture 5).
Answer: Plan A will cost less than Plan B for phone use that is less than or equal to 100 minutes
Step-by-step explanation:
Let m be the number of minutes of phone use.
We can convert $35 and $36 to 3500 cents and 3600 cents respectively.
Then, in Plan A, it would cost 3500 + 6m and in Plan B, it would cost 3600 + 5m.
Since we want Plan A to cost less, we can set up an inequality saying:
We can subtract 3500 + 5m from both sides:
Therefore, Plan A will cost less than Plan B for phone use that is less than or equal to 100 minutes.