Right Triangles. Draw and label a figure to illustrate each situation. 1) SL is an altitude of...

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CHAPTER 5 SECTION 2 Right Triangles

Transcript of Right Triangles. Draw and label a figure to illustrate each situation. 1) SL is an altitude of...

Page 1: Right Triangles. Draw and label a figure to illustrate each situation. 1) SL is an altitude of triangle RST and a perpendicular bisector of side RT. 2)

CHAPTER 5SECTION 2

Right Triangles

Page 2: Right Triangles. Draw and label a figure to illustrate each situation. 1) SL is an altitude of triangle RST and a perpendicular bisector of side RT. 2)

WARM-UPDraw and label a figure to illustrate each situation.

1) SL is an altitude of triangle RST and a perpendicular bisector of side RT.

2) Triangle HIJ is an isosceles triangle with vertex angle at J. JN is an altitude of triangle HIJ.

State whether each sentence is always, sometimes, or never true.

3) The perpendicular bisectors of a triangle intersect at more than on point.

4) The altitude of a triangle contains the midpoint of the opposite side.

5) The medians of a triangle contain the midpoints of the opposite sides.

Page 3: Right Triangles. Draw and label a figure to illustrate each situation. 1) SL is an altitude of triangle RST and a perpendicular bisector of side RT. 2)

WARM-UPDraw and label a figure to illustrate each situation.

1) SL is an altitude of triangle RST and a perpendicular bisector of side RT.

2) Triangle HIJ is an isosceles triangle with vertex angle at J. JN is an altitude of triangle HIJ.

R

S

TL

J

NIH

Page 4: Right Triangles. Draw and label a figure to illustrate each situation. 1) SL is an altitude of triangle RST and a perpendicular bisector of side RT. 2)

WARM-UPState whether each sentence is always, sometimes, or never true.

3) The perpendicular bisectors of a triangle intersect at more than on point.Never

4) The altitude of a triangle contains the midpoint of the opposite side.Sometimes

5) The medians of a triangle contain the midpoints of the opposite sides.Always

Page 5: Right Triangles. Draw and label a figure to illustrate each situation. 1) SL is an altitude of triangle RST and a perpendicular bisector of side RT. 2)

VOCABULARYLL Theorem- If the legs of one right triangle are congruent to the corresponding legs of another right triangle, then the triangles are congruent.

HA Theorem- If the hypotenuse and an acute angle of one right triangle are congruent to the hypotenuse and an acute angle of another right triangle, then the two triangles are congruent.

Page 6: Right Triangles. Draw and label a figure to illustrate each situation. 1) SL is an altitude of triangle RST and a perpendicular bisector of side RT. 2)

VOCABULARY CONT.LA Theorem– If one leg and an acute angle of one right triangle are congruent to the corresponding leg and acute angle of another right triangle, then the triangles are congruent.

HL Postulate- If the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and corresponding leg of another right triangle, then the triangles are congruent.

Page 7: Right Triangles. Draw and label a figure to illustrate each situation. 1) SL is an altitude of triangle RST and a perpendicular bisector of side RT. 2)

Example 1) State the additional information needed to prove each pair of triangles congruent by the given theorem or postulate.

<B and <D are right angles

A) HL

Q

M

L

V

U

T

SR

D

CB

A

B) LL

ST is congruent to TUC) LA

R

P

N

LN is congruent to QR or NM is congruent to RP

Page 8: Right Triangles. Draw and label a figure to illustrate each situation. 1) SL is an altitude of triangle RST and a perpendicular bisector of side RT. 2)

Example 2) Find the values of x and y so that triangle MNO is congruent to triangle RST.

15 cm

58

RS

T

(47 – 8x) cm

M

ON

(3y – 20)Assume triangle MNO is congruent to triangle RST. Then <N is congruent to <R and MO is congruent to RT.

m<N = m<R58 = 3y – 2078 = 3y26 = y

MO = RT47 – 8x = 15-8x = -32x = 4

Page 9: Right Triangles. Draw and label a figure to illustrate each situation. 1) SL is an altitude of triangle RST and a perpendicular bisector of side RT. 2)

Example 3) Find the values of x and y so that triangle JKL is congruent to triangle MLK.

L

(2x - 15)

J

M

K

(3 – 10x)

Assume triangle JKL is congruent to triangle MLK.

m<JLK = m<LKM8y = 10y – 8-2y = -8y = 4

JK = LM3 – 10x = 2x - 153 - 12x = -15-12x = -18x = 3/2 or 1.5

(8y)(10y - 8)

Page 10: Right Triangles. Draw and label a figure to illustrate each situation. 1) SL is an altitude of triangle RST and a perpendicular bisector of side RT. 2)

Example 4) Find the value of x so that triangle ABC is congruent to triangle XYZ by the indicated theorem or postulate.

A

X

Z Y

B C

A) AB = 2x + 6, BC = 15, AC = 3x + 4, XY = 20, YZ = x + 8; LL

Let XY be congruent to AB and YZ be congruent to BC.

XY = AB20 = 2x + 614 = 2x7 = x

YZ = BCx + 8 = 15x = 7

So x is equal to 7.

Page 11: Right Triangles. Draw and label a figure to illustrate each situation. 1) SL is an altitude of triangle RST and a perpendicular bisector of side RT. 2)

Example 4) Find the value of x so that triangle ABC is congruent to triangle XYZ by the indicated theorem or postulate.

A

X

Z Y

B C

B) m<Z = 55, BC = 15x + 2, m<C = 55, AB = 24, YZ = 4x + 13; LA

Let <Z be congruent to <C and YZ be congruent to BC.

YZ = BC4x + 13 = 15x + 213 = 11x + 211 = 11x1 = x

So x is equal to 1.

Page 12: Right Triangles. Draw and label a figure to illustrate each situation. 1) SL is an altitude of triangle RST and a perpendicular bisector of side RT. 2)

Example 4) Find the value of x so that triangle ABC is congruent to triangle XYZ by the indicated theorem or postulate.

A

X

Z Y

B C

C) YX = 21, m<X = 9x + 9, AB = 21, m<A = 11x - 3; LA

Let XY be congruent to AB and <X be congruent to <A.

m<X = m<A9x + 9 = 11x - 39 = 2x – 312 = 2x6 = x

So x is equal to 6.

Page 13: Right Triangles. Draw and label a figure to illustrate each situation. 1) SL is an altitude of triangle RST and a perpendicular bisector of side RT. 2)

Example 4) Find the value of x so that triangle ABC is congruent to triangle XYZ by the indicated theorem or postulate.

A

X

Z Y

B C

D) AC = 28, AB = 7x + 4, ZX = 9x + 1, YX = 5(x + 2); HL

Let AC be congruent to ZX and AB be congruent to YX.

XY = AB5(x + 2) = 7x + 45x + 10 = 7x + 410 = 2x + 46 = 2x3 = x

AC = ZX28 = 9x + 127 = 9x3 = x

So x is equal to 3.