PROBLEMS OF ISOSCELES TRAPEZOID

Each trapezoid is isosceles. Find the measure of each angle.

Problem 1 :

Solution :

Each pair of base angles in an isosceles trapezoid must be congruent.

So, we have

m ∠P = m ∠Q = 77^º

Because ∠S and ∠P are consecutive interior angles formed by parallel lines, they are supplementary.

So, we have

m ∠P + m ∠S = 180^º

Substitute m ∠P = 77^º.

77^º + m ∠S = 180^º

Subtract 77^º from both sides.

m ∠S = 103^º

Each pair of base angles in an isosceles trapezoid must be congruent.

So, we have

m ∠S = m∠R = 103^º

Hence,

∠1 = 77^º

∠2 = 103^º

∠3 = 103^º

Problem 2 :

Solution :

∠S + ∠R = 180

111 + ∠1 = 180

∠1 = 180 - 111

∠1 = 69

∠1 = ∠2 = 69

∠3 = 111

Problem 3 :

Solution :

∠1 = ∠P = ∠Q = 49

∠S + ∠P = 180

∠2 + ∠1 = 180

∠2 + 49 = 180

∠2 = 180 - 49

∠2 = 131

∠3 = 131

Problem 4 :

Solution :

∠X = 105, ∠Y = 105

∠W = 180 - 105

∠W = 75

∠Z = 75

Problem 5 :

Solution :

∠P = 65, ∠S = 65

∠Q + ∠P = 180

∠Q = 180 - 65

∠Q = 115

∠R = 115

Problem 6 :

Solution :

∠A = 60, ∠D = 60

∠B + ∠A = 180

∠B = 180 - 60

∠B = 120

∠C = 120

Problem 7 :

In trapezoid LMNO below, median PQ is drawn.

If LM = x + 7, ON = 3x + 11, and PQ = 25, what is the value of x?

a) 1.75    b) 3.5      c) 8     d) 17

Solution :

Given that, LM = x + 7, ON = 3x + 11

PQ = (1/2) (LM + ON)

= (1/2) (x + 7 + 3x + 11)

Applying the value of PQ, we get

25 = (1/2)(4x + 18)

25(2) = 4x + 18

50 = 4x + 18

4x = 50 - 18

4x = 32

x = 32/4

x = 8

So, option c is correct.

Problem 8 :

In the diagram below, LATE is an isosceles trapezoid with LE ≅ AT, LA = 24, ET = 40, and AT =10. Altitudes LF and AG are drawn

What is the length of LF?

a) 6      b) 8     c) 3      d) 4

Solution :

ET = 40

EF + FG + GT = 40

EF = GT

2EF + 24 = 40

2EF = 40 - 24

2EF = 16

EF = 8

In triangle AGT,

AT² = AG² + GT²

10² = AG² + 8²

AG² = 100 - 64

AG² = 36

AG = 6

LF = 6

Option a is correct.

Problem 9 :

In the diagram below of isosceles trapezoid ABCD, AB = CD = 25, AD = 26, and BC = 12.

What is the length of an altitude of the trapezoid?

a) 7      b) 14     c) 19      d) 24

Solution :

AD = AE + EF + FD

EF = 12

AE = FD

26 = 2AE + 12

2AE = 26 - 12

2AE = 14

AE = 14/2

AE = 7 = FD

In triangle FCD,

CD² = CF² + FD²

25² = CF² + 7²

625 - 49 = CF²

CF² = 576

CF = 24

So, the length of altitude is 24, option d is correct.

Problem 10 :

In isosceles trapezoid ABCD, AB ≅ CD. If BC = 20, AD = 36, and AB = 17, what is the length of the altitude of the trapezoid?

a) 10      b) 12     c) 15     d) 16

Solution :

AD = AE + EF + FD

AE = FD

EF = 20

36 = 2AE + 20

36 - 20 = 2AE

2AE = 16

AE = 16/2

AE = 8

In triangle ABE,

AB² = AE² + BE²

17² = 8² + BE²

289 - 64 = BE²

BE² = 225

BE = 15

Problem 11 :

In trapezoid ABCD below, AB || CD

If AE = 5.2, AC = 11.7, and CD = 10.5, what is the length of AB, to the nearest tenth?

a) 4.7     b) 6.5     c) 8.4      d) 13.1

Solution :

In triangles AEB and DEC,

∠AEB = ∠DEC

∠EBA = ∠EDC

Using AA, triangles AEB and DEC are similar.

EC/AE = DC/AB ----(1)

AC = AE + EC

11.7 = 5.2 + EC

EC = 11.7 - 5.2

= 6.5

By applying these values in (1), we get

6.5 / 5.2 = 10.5 / AB

6.5AB = 10.5(5.2)

AB = 10.5(5.2)/6.5

= 8.4

So, the length of AB is 8.4, option c is correct.

Problem 12 :

The diagram below shows isosceles trapezoid ABCD with AB || DC and AD ≅ BC. If m∠BAD = 2x and m∠BCD = 3x + 5, find m∠BAD.

Solution :

∠A + ∠B + ∠C + ∠D = 360

2(2x) + 2(3x + 5) = 360

2x + 3x + 5 = 360/2

5x + 5 = 180

5x = 180 - 5

5x = 175

x = 175/5

x = 35

∠BAD = 2(35)

= 70 degree