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.
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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.

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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
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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} .
( Plane Geometry MathRoom Index )
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