(England, Wales, Non-
UK 8 (60
minutes)
In the diagram, PQRS, JQK and LRK are straight lines
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J
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Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
Question 7
Question 8
Question 9
Question 10
Question 11
M
33°
vo
270/R
L
S
Question 12
Question 13
What is the size of the angle JKL?

∠JKL = 38°

Step-by-step explanation:

PQRS, JQK and LRK are straight lines

Let's take the straight lines in the diagrams one after the other to find what they consist.

The related diagram can be found at (question ID: 18713345)

Find attached the diagram used for solving the question.

For straight line PQRS,

2x°+y°+x°+2y° = 180°

(Sum of angles on a Straight line = 180°)

Collect like terms

3x° + 3y° = 180°

Also straight line PQRS = straight line PQR + straight line SRQ

For straight line PQR,

2y + x + ∠RQM = 180° ....equation 1

For straight line SRQ,

2x + y + ∠MRQ = 180° ....equation 2

Straight line PQRS = addition of equation 1 and 2

By collecting like times

3x +3y + ∠RQM + ∠MRQ = 360°....equation 3

Given ∠QMR = 33°

∠RQM + ∠MRQ + ∠QMR = 180° (sum of angles in a triangle)

∠RQM + ∠MRQ + 33° = 180°

∠RQM + ∠MRQ = 180-33

∠RQM + ∠MRQ = 147° ...equation 4

Insert equation 4 in 3

3x° +3y° + 147° = 360°

3x +3y = 360 - 147

3x +3y = 213

3(x+y) = 3(71)

x+y = 71°

∠JQP = ∠RQK = 2y° (vertical angles are equal)

∠LRS = ∠QRK = 2x° (vertical angles are equal)

∠QRK + ∠RQK + ∠QKR = 180° (sum of angles in a triangle)

2x+2y + ∠QKR = 180

2(x+y) + ∠QKR = 180

2(71) + ∠QKR = 180

142 + ∠QKR = 180

∠QKR = 180 - 142

∠QKR = 38°

∠JKL = ∠QKR = 38°

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step-by-step explanation:

hello :

by applying the distributive property :

6 × 4.3 = 6 × ( 4 + 0.3 )

            =   (6 ×   4) + (6 ×   0.3)

            = 24 + 1.8

6 × 4.3 = 25.8