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Here is a prism ABCDSPQR - Edexcel - GCSE Maths - Question 20 - 2022 - Paper 3

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Here is a prism ABCDSPQR. The base ABCD of the prism is a square of side 14 cm. T is the point on BC such that BT : TC = 4 : 3. The cross section of the prism is ... show full transcript

Worked Solution & Example Answer:Here is a prism ABCDSPQR - Edexcel - GCSE Maths - Question 20 - 2022 - Paper 3

Step 1

Find TC and BT

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Answer

Given the ratio BT : TC = 4 : 3, let BT = 4x and TC = 3x. The total length BC is given by:

extBC=14extcm ext{BC} = 14 ext{ cm}

Thus,

4x+3x=14extcm4x + 3x = 14 ext{ cm}

This simplifies to:

7x=14extcm7x = 14 ext{ cm}

So,

x=2extcmx = 2 ext{ cm}

Now, substituting back, we have:

BT=4x=8extcmBT = 4x = 8 ext{ cm}

TC=3x=6extcmTC = 3x = 6 ext{ cm}

Step 2

Find CD and area of trapezium

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Answer

To find CD, we use the trapezium area formula:

extArea=12×(a+b)×h ext{Area} = \frac{1}{2} \times (a + b) \times h

Where:

  • a = AB = 14 cm
  • b = CD
  • h = (ST = CR) = 12 cm

Setting up as follows, we know:

147=12×(14+CD)×12147 = \frac{1}{2} \times (14 + CD) \times 12

Solving for CD leads to:

147=6(14+CD)147 = 6(14 + CD)

Dividing both sides by 6 gives:

24.5=14+CD24.5 = 14 + CD

Thus:

CD=24.514=10.5extcmCD = 24.5 - 14 = 10.5 ext{ cm}

Step 3

Find length ST

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Answer

Using Pythagorean theorem in triangle STC:

ST2=(TC)2+(SC)2ST^2 = (TC)^2 + (SC)^2

Where:

  • TC = 6 cm
  • SC = CD - BT = 10.5 - 8 = 2.5 cm

Thus:

ST2=62+2.52=36+6.25=42.25ST^2 = 6^2 + 2.5^2 = 36 + 6.25 = 42.25

So,

ST=42.256.5extcmST = \sqrt{42.25} ≈ 6.5 ext{ cm}

Step 4

Find angle between line ST and base ABCD

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Answer

The angle can be found using:

tan(θ)=oppositeadjacent=STBT=6.58\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{ST}{BT} = \frac{6.5}{8}

Thus:

θ=tan1(0.8125)39.8exto\theta = \tan^{-1}(0.8125) ≈ 39.8^ ext{o}

Rounding to one decimal place gives:

θ39.8exto\theta ≈ 39.8^ ext{o}

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