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Given that ABRQ is a rectangle, ABR is a right-angled triangle with the right angle at B.

Mathematics

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2 parts
a

214 cm2\sqrt{14} \text{ cm}

b

15.5∘15.5^\circ

Step-by-step solution

4 steps

Part a· 2 steps

To find the length of AS, we first need to find the length of AR.

Given that ABRQ is a rectangle, △ABR\triangle ABR is a right-angled triangle with the right angle at B.

  1. Find AR using the Pythagorean theorem in △ABR\triangle ABR.

    AR2=AB2+BR2AR^2 = AB^2 + BR^2

    Given AB=6 cmAB = 6 \text{ cm} and BR=4 cmBR = 4 \text{ cm}.

    AR2=(6 cm)2+(4 cm)2AR^2 = (6 \text{ cm})^2 + (4 \text{ cm})^2 AR2=36 cm2+16 cm2AR^2 = 36 \text{ cm}^2 + 16 \text{ cm}^2 AR2=52 cm2AR^2 = 52 \text{ cm}^2 AR=52 cm=213 cmAR = \sqrt{52} \text{ cm} = 2\sqrt{13} \text{ cm}

  2. Find AS using the Pythagorean theorem in △ASR\triangle ASR.

    The problem states that plane PQRS is perpendicular to plane ABRQ. Since SR is a side of the rectangle PQRS and SR⊥QRSR \perp QR (where QR is the line of intersection of the two planes), SR is perpendicular to the plane ABRQ.
    Therefore, △ASR\triangle ASR is a right-angled triangle with the right angle at R.

    AS2=AR2+SR2AS^2 = AR^2 + SR^2

    We found AR2=52 cm2AR^2 = 52 \text{ cm}^2 and given RS=2 cmRS = 2 \text{ cm}.

    AS2=52 cm2+(2 cm)2AS^2 = 52 \text{ cm}^2 + (2 \text{ cm})^2 AS2=52 cm2+4 cm2AS^2 = 52 \text{ cm}^2 + 4 \text{ cm}^2 AS2=56 cm2AS^2 = 56 \text{ cm}^2 AS=56 cm=4×14 cmAS = \sqrt{56} \text{ cm} = \sqrt{4 \times 14} \text{ cm} AS=214 cmAS = 2\sqrt{14} \text{ cm}

    As a decimal, AS≈7.48 cmAS \approx 7.48 \text{ cm} (to 2 decimal places).

    The length of AS is 214 cm\boxed{2\sqrt{14} \text{ cm}}

Part b

15.5∘15.5^\circ

The angle that AS makes with plane ABRQ is the angle between AS and its projection onto the plane ABRQ.

Since SR is perpendicular to plane ABRQ, the projection of S onto the plane ABRQ is R. The projection of A onto the plane ABRQ is A itself.
Therefore, the projection of the line AS onto the plane ABRQ is the line AR.
The angle AS makes with plane ABRQ is ∠SAR\angle SAR.

  1. Identify the right-angled triangle.

    From part (a), we established that △ASR\triangle ASR is a right-angled triangle at R.

  2. Use trigonometry to find ∠SAR\angle SAR.

    In △ASR\triangle ASR:
    Opposite side to ∠SAR\angle SAR is SR=2 cmSR = 2 \text{ cm}.
    Adjacent side to ∠SAR\angle SAR is AR=52 cmAR = \sqrt{52} \text{ cm}.

    tan⁡(∠SAR)=OppositeAdjacent=SRAR\tan(\angle SAR) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{SR}{AR} tan⁡(∠SAR)=252\tan(\angle SAR) = \frac{2}{\sqrt{52}} tan⁡(∠SAR)=2213\tan(\angle SAR) = \frac{2}{2\sqrt{13}} tan⁡(∠SAR)=113\tan(\angle SAR) = \frac{1}{\sqrt{13}} ∠SAR=arctan⁡(113)\angle SAR = \arctan\left(\frac{1}{\sqrt{13}}\right) ∠SAR≈15.499∘\angle SAR \approx 15.499^\circ

    Rounding to one decimal place:

    ∠SAR≈15.5∘\angle SAR \approx 15.5^\circ

    The angle that AS makes with plane ABRQ is 15.5∘\boxed{15.5^\circ}

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