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ABC and EDC are straight lines - Edexcel - GCSE Maths - Question 5 - 2017 - Paper 2

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ABC and EDC are straight lines. EA is parallel to DB. EC = 8.1 cm. DC = 5.4 cm. DB = 2.6 cm. (a) Work out the length of AE. AC = 6.15 cm. (b) Work out the length ... show full transcript

Worked Solution & Example Answer:ABC and EDC are straight lines - Edexcel - GCSE Maths - Question 5 - 2017 - Paper 2

Step 1

(a) Work out the length of AE.

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Answer

To find the length of AE, we can use the properties of similar triangles since EA is parallel to DB.

The ratio of the lengths of segments on line EC to line DC is given by:

ECDC=8.15.4=1.5\frac{EC}{DC} = \frac{8.1}{5.4} = 1.5

Let AE be denoted as xx. Using the same ratio, we can write:

AEDB=x2.6=8.15.4\frac{AE}{DB} = \frac{x}{2.6} = \frac{8.1}{5.4}

Cross-multiply:

5.4x=8.1×2.65.4x = 8.1 \times 2.6

Calculating the right side:

5.4x=21.065.4x = 21.06

Now solve for xx:

x=21.065.43.9x = \frac{21.06}{5.4} \approx 3.9

Thus, the length of AE is approximately 3.9 cm.

Step 2

(b) Work out the length of AB.

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Answer

To find the length of AB, we can use the segments given.

First, we note that:

AC=AE+EBAC = AE + EB

Since we know AC and AE, we can express EB as:

EB=ACAEEB = AC - AE

Substituting the known values:

EB=6.153.9=2.25EB = 6.15 - 3.9 = 2.25

Now, we relate it back to the total segment AB through the similar triangles property:

ABAC=DBEB\frac{AB}{AC} = \frac{DB}{EB}

where DB = 2.6 cm and EB = 2.25 cm. Thus, we can express AB as:

AB=AC×DBEBAB = AC \times \frac{DB}{EB}

Substituting the values:

AB=6.15×2.62.256.15×1.155567.1AB = 6.15 \times \frac{2.6}{2.25} \approx 6.15 \times 1.15556 \approx 7.1

Therefore, the length of AB is approximately 7.1 cm.

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