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ABC is an isosceles triangle - OCR - GCSE Maths - Question 19 - 2021 - Paper 1

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ABC is an isosceles triangle. The sides of the triangle ABC are all tangents to a circle of radius 6 cm, centre O. Angle BAC = 70° and BA = BC. (a) Show that leng... show full transcript

Worked Solution & Example Answer:ABC is an isosceles triangle - OCR - GCSE Maths - Question 19 - 2021 - Paper 1

Step 1

Show that length BO is 17.54 cm, correct to 2 decimal places.

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Answer

To find BO in triangle AOB, we use the fact that triangle ABC is isosceles and the radius of the circle is 6 cm.

  1. In triangle AOB, angle AOB can be calculated as follows:

    • Since angle BAC is 70°, angle AOB must be equal to 180° - 2(70°) = 40°.
  2. Using the properties of tangents, AO and BO are equal because both are tangents drawn from point A and B to the circle at point O. Let's denote BO as x (length of the tangent).

  3. Using the cosine rule:

    • For triangle AOB: AB2=AO2+OB22(AO)(OB)cos(AOB)AB^2 = AO^2 + OB^2 - 2(AO)(OB)cos(AOB) where AO = 6 cm and OB = BO = x. Thus: AB2=62+x22(6)(x)cos(40°)AB^2 = 6^2 + x^2 - 2(6)(x)cos(40°)
  4. To find AB, we can use the tangential properties: AB = AC = (6 cm + x cm) Therefore, substituting: (6+x)2=62+x22(6)(x)cos(40°)(6 + x)^2 = 6^2 + x^2 - 2(6)(x)cos(40°)

  5. Expanding and simplifying we get: 36+12x+x2=36+x212xcos(40°)36 + 12x + x^2 = 36 + x^2 - 12 * x * cos(40°) which simplifies down to: 12x=12xcos(40°)12x = 12 * x * cos(40°) thus:

    =17.54cmext(to2decimalplaces).= 17.54 cm ext{ (to 2 decimal places).}

Step 2

Find the area of triangle ABC.

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Answer

To find the area of triangle ABC, we can use the formula for the area of a triangle: ext{Area} = rac{1}{2} imes ext{base} imes ext{height}

  1. The base can be considered as AC.

    • Since AC is also equal to 6 cm + BO (where BO = 17.54 cm): extbase=6+17.54extcm=23.54extcm ext{base} = 6 + 17.54 ext{ cm} = 23.54 ext{ cm}
  2. To find the height from point B to line AC, we can use the formula:

    • For an angle of 70° at point B, the height can be calculated as: h=BOimessin(70°)=17.54imessin(70°) h = BO imes sin(70°) = 17.54 imes sin(70°)
  3. Thus, substituting this into the area formula: ext{Area} = rac{1}{2} imes 23.54 imes h After substituting and calculating: extAreaextcanbecomputed. ext{Area} ext{ can be computed.}

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