PROBLEMS ON ISOSCELES TRAPEZOID WITH DIAGONALS

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What is Trapezium ?

A trapezium is a convex quadrilateral with exactly one pair of opposite sides parallel to each other. The trapezium is a two-dimensional are called legs. It is also called a trapezoid. Sometimes the parallelogram is also called a trapezoid with two parallel sides.

isocelestrapeziumq1

Diagonals are congruent.

Find the length of the diagonal indicated for each trapezoid.

Problem 1 :

DF = 9.1, Find CE

diagonalofisotrapeq1

Solution :

CE and DF are diagonals.

DF = 9.1, so CE = 9.1

Solve for x. Each figure is a trapezoid.

Problem 2 :

DF = 6 and CE = 2x - 10

diagonalofisotrapeq2

Solution :

DF and CE are diagonals. So, they are equal.

6 = 2x - 10

2x = 6 +10

2x = 16

x = 16/2

x = 8

Problem 3 :

DF = 12 and EG = -2x + 36

diagonalofisotrapeq3

Solution :

DF and EG are diagonals, so they are equal.

DF = EG

12 = -2x + 36

12 - 36 = -2x

-24 = -2x

x = 24/2

x = 12

Find the length of the diagonal indicated for each trapezoid.

Problem 4 :

JL = 2x + 4, KM = 4x - 8. Find JL

diagonalofisotrapeq4

Solution :

JL and KM are diagonals.

JL = MK

2x + 4 = 4x - 8

2x - 4x = -8 - 4

-2x = -12

dividing by -2 on both sides.

x = 12/2

x = 6

Problem 5 :

UW = 3x + 11, TV = 8x - 9 and Find UW

diagonalofisotrapeq5

Solution :

UW and TV are diagonals.

UW = TV

3x + 11 = 8x - 9

Subtracting 8x and 11, we get

3x - 8x = -9 - 11

-5x = -20

x = 20/5

x = 4

Problem 6 :

The bases of an isosceles trapezoid ABCD measure 10 cm and 20 cm. The height (altitude) is 12 cm. How long are the legs AB and CD?

diagonalofisotrapeq6

Solution :

diagonalofisotrapeq6s

In triangle ABE,

AB2 = AE2 + BE2

AB2 = 52 + 122

AB2 = 25 + 144

AB2 = 169

AB = 13

Since is it isosceles trapezoid, CD is 13.

Problem 7 :

In isosceles trapezoid KIME, ∠K and ∠E are the base angles. If IK = 11 and ME = 3x - 1, what is the value of x?

Solution :

In KIME, KI and ME are non parallel sides.

IM and KE are parallel sides.

Non parallel sides will be equal in length.

IK = ME

11 = 3x - 1

Add 1 on both sides.

11 + 1 = 3x

3x = 12

x = 12/3

x = 4

Problem 8 :

𝐴𝐡𝐢𝐷 is an isosceles trapezoid with bases 𝐴𝐡 and 𝐷𝐢. If 𝐴𝐷 = 3x + 4 and 𝐡𝐢 = x + 12. Find the value of x and the length of 𝐴𝐷.

Solution :

AD = BC

3x + 4 = x + 12

3x - x = 12 - 4

2x = 8

x = 8/2

x = 4

Problem 9 :

Trapezoid 𝐴𝐡𝐢𝐷 is an isosceles. Find the missing angle measures.

isosceles-trapezoid-q1

Solution :

DA = BC (given)

∠A = βˆ B

∠B = 65

Since AB and DC are parallel, then co-interior angles will be supplementary.

∠A + βˆ D = 180

65 + βˆ D = 180

∠D = 180 - 65

= 115

∠C = 115

Problem 10 :

A hip roof slopes at the ends of the buildings as well as the front and back. The front of this hip roof is in the shape of an isosceles trapezoid. If one angle measures 30 degree, find the measures of the other three angles.

isosceles-trapezoid-q2

Solution :

acute angle = 30

Since it is isosceles trapezoid, the other angle is also 30 degree.

Obtuse angle = 180 - 30

= 150

So, the required angles are 30, 30, 150 and 150.

Problem 11 :

Find the length of the shorter base of a trapezoid if the length of the median is 34 meters and the length of the longer base is 49 meters.

Solution :

Length of longer base = 49 meters

Length of shorter base = x

Median = 34 meter

M = (1/2) (longer base + shorter base)

34 = (1/2) (49 + x)

34(2) = 49 + x

49 + x = 68

x = 68 - 49

x = 19

So, the length of the shorter base is 19 meters.

Problem 12 :

One base angle of an isosceles trapezoid is 45 degree. Find the measures of the other three angles.

Solution :

Acute angle = 45 degree

Obtuse angle = 180 - 45

= 135

Since it is isosceles trapezoid, the required angles are 45, 45, 135 and 135.

Problem 13 :

Show that the quadrilateral with vertices at Q(0, 3), R(0, 6), S(-6, 0) and T(-3, 0) is a trapezoid. Decide whether the trapezoid is isosceles. Then find the length of the midsegment of the trapezoid.

isosceles-trapezoid-q3

Solution :

Length of ST = βˆš(-6 + 3)+ (0 - 0)2

√(-3)2+ (0 - 0)2

= 3

Length of TQ = βˆš(-3 + 0)+ (0 - 3)2

√9 + 9

= βˆš18

= √(2 x 3 x 3)

= 3√2

Length of QR = βˆš(0 - 0)+ (3 - 6)2

√02+ (-3)2

= 3

Length of RS = βˆš(0 + 6)+ (6 - 0)2

√(36 + 36)

= βˆš72

= √(2 x 6 x 6)

= 6√2

It is not isosceles trapezium. Since it is not isosceles trapezium, we cannot use midsegment theorem.

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