FIND MISSING ANGLES AND SIDES OF TRAPEZIUM

What is trapezium ?

A trapezium is a quadrilateral which has a pair of parallel opposite sides.

Properties :

• A trapezium has two parallel sides and two non parallel sides.
• The diagonal of regular trapezium bisect each other.
• The length of the midsegment is equal to half the sum of parallel bases.
• Two pairs of adjacent angles of a trapezium formed between the parallel sides and one of the non parallel side, add up to 180 degrees.

Find the values of the variables. Then find the lengths of the sides.

Problem 1 :

Solution :

By observing the figure,

4y + 6 = 7y – 3

Comparing like terms.

4y – 7y = -3 – 6

-3y = -9

y = 3

To find the lengths of the sides :

Substitute y = 3 in 4y + 6

= 4(3) + 6

= 12 + 6

= 18

Base of the trapezium = 5y + 1.4

5(3) + 1.4

15 + 1.4

16.4

So, the lengths of the sides are 18, 18, 16.4, 4.8.

Problem 2 :

The figure is an isosceles trapezoid. Determine the value of the variable

a) x = 3     b) x = 2      c) x = 18x + 48      d) x = 51

Solution :

17x = x + 48

Subtract x from both sides.

17x – x = x – x + 48

16x = 48

Divide both sides by 16.

16x/16 = 48/16

x = 3

So, the value is x = 3.

Problem 3 :

Find the value of x in the trapezium ABCD given below.

Solution :

Sum of co interior angles = 180

x - 20 + x + 40 = 180

2x + 20 = 180

2x = 180 - 20

2x = 160

x = 160/2

x = 80

Problem 4 :

Find the value of y, if the perimeter of the trapezium is 142.

Solution :

y + 12 + y + 12 + y + y - 12 = 164

4y + 12 = 164

4y = 164 - 12

4y = 152

y = 152/4

y = 38

Problem 5 :

Incan architecture often features trapezoidal doorways and windows. Find m∠M, m∠K, and m∠L in the doorway.

Solution :

m∠J = 85

m∠M = 85

KL and JM are parallel,

m∠J + m∠K = 180

85 + m∠K = 180

m∠K = 180 - 85

= 95

m∠L = 95

Problem 6 :

Your cousin claims there is enough information to prove that JKLM is an isosceles trapezoid. Is your cousin correct? Explain

Solution :

m∠J = m∠K

JM and KL are equal, when two sides are equal then equal sides only will create equal angles.

JM = KL

So, it is isosceles trapezoid.

Problem 7 :

The bases of a trapezoid lie on the lines y = 2x + 7 and y = 2x − 5. Write the equation of the line that contains the midsegment of the trapezoid

Solution :

y = 2x + 7 and y = 2x − 5

Equation of midsegment of the trapezoid = (1/2)(2x + 7 + 2x - 5)

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

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

= 2x + 1

So, equation of midsegment of the trapezoid is 2x + 1.

Problem 8 :

In trapezoid PQRS, PQ || RS and MN is the midsegment of PQRS. If RS = 5 ⋅ PQ, what is the ratio of MN to RS?

a) 3 : 5     b) 5 : 3    c)  1 : 2    d)  3 : 1

Solution :

RS = 5 ⋅ PQ

RS/PQ = 5/1

MN = (1/2) PQ + RS

= (1/2)(1 + 5)

= 6/2

= 3

MN : RS = 3 : 5

So, option a is correct.

Problem 9 :

If EG = FH, is trapezoid EFGH isosceles?

If m∠HEF = 70° and m∠FGH = 110°, is trapezoid EFGH isosceles? Explain

Solution :

HG || EF

m∠HEF = 70° and m∠FGH = 110°

In triangles EHG and FGH,

HG = HG (common)

EG = FH (given)

m∠HEF + m∠EHG = 180

70° + m∠EHG = 180

m∠EHG = 180 - 70

m∠EHG = 110

So, EH and FG are equal. Then trapezoid EFGH isosceles.

Problem 10 :

In trapezoid JKLM, ∠J and ∠M are right angles, and JK = 9 centimeters. The length of midsegment NP of trapezoid JKLM is 12 centimeters. Sketch trapezoid JKLM and its midsegment. Find ML. Explain your reasoning

Solution :

NP = (1/2) (JK + ML)

12 = 1/2 (9 + ML)

12(2) = 9 + ML

24 = 9 + ML

24 - 9 = ML

ML = 15 cm

So, the length of the base is 15 cm.

Problem 11 :

Give the most specific name for the quadrilateral. Explain your reasoning.

Solution :

It is trapezium, but it is not isosceles. Because diagonals are not congruent.