4 Triangle Theorems

Four Triangle Theorems

 1/Angle Bisector Theorem PQ bisects angle P It divides the 3rd side in proportion with the other 2 sides. 2/ Altitude to the Hypotenuse Theorem The square on the altitude to the hypotenuse equals the product of the parts. 3/ Projection on the Hypotenuse Theorem c is called the projection of b on EF d is called the projection of a on EF The square on a side = the product of the hypotenuse and the projection of that side on the hypotenuse. 4/ Product of the Sides Theorem The product of the sides = the product of the altitude to the hypotenuse and the hypotenuse.

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Practice

Numerical Exercises

1/ solve for x -- justify your statements with a theorem or axiom. .

2/ solve for all indicated unknowns -- justify your statements with a theorem or axiom.

 a) find x, y and z. b) find x, y and z. .

3/ solve for all indicated unknowns -- justify your statements with a theorem.

 a) find x, and y. b) find w, x, y and z. c) find w, x, y and z. 4/ Find the area of the lot shown in the diagram. Justify your calculations .

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Solutions

1/ a) x / 14.6 = 4 /8 so x = 7.3

b) x / 15 = 7 / 14 so x = 7.5

2/ a) y 2 = 2.87(5.04) by the altitude to hypotenuse thm. so y = 3.8
x 2 = 2.87 (7.91) by the projection on the hypotenuse thm. so x = 4.76
z 2 = 5.04(7.91) by the projection on the hypotenuse thm. so z = 6.31

b) Note that AP + PB = hypotenuse = x, since 6 + x – 6 = x.

10 2 = 6x by the projection on the hypotenuse thm. so x = 50 /3
Since AP = 6 = 18 / 3,
PB = 32 / 3
y 2 = 6(32 / 3) = 64, by the altitude to hypotenuse thm. so y = 8 .

3/ a) By the Pythagorean Theorem, By the projection on the hypotenuse thm. 27 2 = 31.38x so x = 23.23
By the altitude to hypotenuse thm.y 2 = 8.15(23.23) so y = 13.76.

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b) ABC is a 30°, 60°, 90° triangle with ratios so x + w = 2(4.53) = 9.06
, and APC is a 30°, 60°, 90° triangle so z = ½ (4.53) = 2.26
By the Pythagorean Theorem, x 2 = y 2 – z 2 so x = 7.52.
Since w = 9.06 – x so w = 1.54.

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c) z = 10.2 – w
By the altitude to hypotenuse thm.1.5 2 = w(10.2 – w) so 0 = w 2 – 10.2w + 2.25
When we solve the quadratic equation, we find w = 0.23 cm
so z = 9.97 cm
By the projection on the hypotenuse thm. y 2 = 10.2(0.23) so y = 1.53 cm
By the projection on the hypotenuse thm. x 2 = 10.2(9.97) so x = 10.08 cm.

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4/ ABD is a 30°, 60°, 90° triangle with ratios Therefore, Using AB as base and AD as the height, Area ABD = 233.26 m 2 .
BD = 2(AB), so BD = 46.42 m
Now, Area BCD = 482.98 m 2 .
Total Area = 716.24 m 2 .

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