Math Practice - Practice problems. 1. A mast at the top of...

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Practice problems. 1. A mast at the top of a building casts a shadow whose tip is 48 feet from the base of the building. If the building is 12 feet high and its shadow is 32 feet long, what is the length of the mast? (NOTE: If the length of the mast is x, then the height of the mast above the ground is x + 12.) 2. Figure 19-6 represents an L-shaped building with dimensions as shown. On the line of sight from A to D, a stake is driven at C, a point 8 feet from the building and 10 feet from A. If ABC is a right angle, find the length of AB and the length of AD. Notice that AE is 18 feet and ED is 24 feet. Figure 19-6 - Using similar triangles. Answers: 1. 6 feet 2. AB = 6 feet AD = 30 feet
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In triangle ABC, AB is congruent to AC and the angle CAB measures 90 degrees. Take a point P on side AB. Then set Q on the line AC so that AP = AQ. Prove that triangles CAP and BAQ are congruent. Triangles ABP, BCQ, CAR are equilateral triangles. Prove that ABC and PBQ are congruent. Prove that ARQP is a parallelogram. We will use the SAS (side-angle-side) method of proving that triangles ABC and PBQ are congruent. First, since triangle ABP is equilateral, sides AB and PB are congruent.(S) Next, name the three pieces of angle PBC x, y and z, reading counterclockwise. (That is, let x be the measure of angle ABC, y be the measure of angle QBR, and z be the measure of angle PBQ.) Since x + y = 60 and z + y = 60, it follows that x = z. (A) Finally, since triangle QBC is equilateral, sides QB and CB are congruent. (S) Thus the two triangles ABC and PBQ are congruent. By Problem 1, triangles ABC and PBQ are congruent. Therefore, in particular, PQ @ AC, and AC in turn is congruent to AR (because they are two sides of an equilateral triangle). Thus PQ @ AR. Identical arguments to the ones showing that triangles ABC and PBQ are congruent show that triangle ABC is also congruent to triangle RQC. Therefore triangles PBQ and RQC are congruent to each other. In particular, QR @ PB. Since PB @ PA (because they are two sides of an equilateral triangle), QR @ PA. Thus quadrilateral ARQP has both pairs of opposite sides congruent and is therefore a parallelogram.
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The angle XOY is less than 90°. The point A is on the ray OX and the point P is on the ray OY. OABC and APDE are squares. Prove that: 1. Triangles ABP and AOE are congruent. 2. The segments BP and OE cross at a right angle. Let Q be the intersection point of BP and OE, and let R be the intersection point of BP and OA. Since the sum of the measures of the angles of a triangle is 180°, we can show that angle BQO is a right angle by showing that the measures of angles AOE and ORP add up to 90°. Since BAR is a right triangle, ÐABR+ ÐARB = 90°. By part (1), ÐABR @ ÐAOE; angles ARB and ORP are congruent because they are vertical angles. Thus ÐAOE + ÐORP = 90°, and so ÐBQO is a right angle. Two poles AA'and BB' are standing vertically. AA'and BB' are 6m long and 3m long respectively.
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This note was uploaded on 05/12/2011 for the course TED 455 taught by Professor Espisito during the Spring '11 term at CSU Dominguez Hills.

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Math Practice - Practice problems. 1. A mast at the top of...

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