Question

can someone plz ans the c part of this question​

Answer 1

Given:

Vertices of a ΔABC are A(2, 4), B(9, 4) & C(7, 11).

To find:

• To find the area, we can use the area of the triangle equation when 3 points are given.

• To find the value of BC, we can use the distance formula.

• To calculate the length of the ⊥ from A to BC, we use the area of a triangle formula. (1/2 × b × h)

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Solution for Part (i): Find the area of ΔABC.

ar(ΔABC) = ¹/₂ (x₁ (y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂))

ar(ΔABC) = ¹/₂ (2 (4 - 11) + 9(11 - 4) + 7(4 - 4))

ar(ΔABC) = ¹/₂ (2 (-7) + 9(7) + 7(0))

ar(ΔABC) = ¹/₂ (-14 + 63)

ar(ΔABC) = ¹/₂ (49)

ar(ΔABC) = 24.5 sq.units.

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Solution for Part (ii): Calculate the length of BC.

x₁ ⇒ 9

x₂ ⇒ 7

y₁ ⇒ 4

y₂ ⇒ 11

⇒ BC = √[(x₂ - x₁)² + (y₂ - y₁)²]

⇒ BC = √[(7 - 9)² + (11 - 4)²]

⇒ BC = √[(-2)² + (7)²]

⇒ BC = √[4 + 49]

⇒ BC = √53 units.

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Solution for Part (ii): Calculate the length of the perpendicular drawn from A to BC.

⇔ Area(ABC) = 1/2 × b × h

⇔ 24.5 = 1/2 × √53 × h

⇔ 24.5 × 2 = h√53

⇔ 49 = h√53

⇔ h = 49/√53

⇔ h ≈ 6.73

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Final answers:

Solution (i) - 24.5 sq.units.

Solution (ii) - √53 units.

Solution (iii) - 6.73 units.