Question

mathématiques 4ème secondaire cst - la trigonométrie

1. à laide de la formule de héron, détermine laire des triangles ci - dessous.

a) triangle lzg with sides 35 dm, 10 dm, 39 dm
b) triangle svo with sides 3.4 km, 6.1 km, 7.5 km
c) triangle rkd with sides 1.7 m, 4 m, 2.8 m
réponse :
réponse :
réponse :

1. dans chaque cas, détermine laire du triangle en utilisant la formule appropriée.

a) triangle mno with sides 37 mm, 20 mm, 19 mm
b) triangle csx with base 6.4 dm and height 3.2 dm
c) triangle vfj with sides 2.2 km, 4.9 km and included angle 83.1°

Explanation:

Problem 2: Using Heron's Formula

Heron's formula for the area $A$ of a triangle with sides $a, b, c$ is:
$$A = \sqrt{s(s-a)(s-b)(s-c)}$$
where $s = \frac{a+b+c}{2}$ (semi-perimeter).

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2a) Triangle LZG: sides 35 dm, 10 dm, 39 dm

Step1: Calculate semi-perimeter $s$

$s = \frac{35 + 10 + 39}{2} = \frac{84}{2} = 42$ dm

Step2: Compute area via Heron's formula

$A = \sqrt{42(42-35)(42-10)(42-39)} = \sqrt{42 \times 7 \times 32 \times 3}$
$= \sqrt{42 \times 7 \times 96} = \sqrt{28224} = 168$ dm²

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2b) Triangle SVO: sides 3.4 km, 6.1 km, 7.5 km

Step1: Calculate semi-perimeter $s$

$s = \frac{3.4 + 6.1 + 7.5}{2} = \frac{17}{2} = 8.5$ km

Step2: Compute area via Heron's formula

$A = \sqrt{8.5(8.5-3.4)(8.5-6.1)(8.5-7.5)}$
$= \sqrt{8.5 \times 5.1 \times 2.4 \times 1.0} = \sqrt{104.04} = 10.2$ km²

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2c) Triangle RKD: sides 1.7 m, 2.8 m, 4 m

Step1: Calculate semi-perimeter $s$

$s = \frac{1.7 + 2.8 + 4}{2} = \frac{8.5}{2} = 4.25$ m

Step2: Compute area via Heron's formula

$A = \sqrt{4.25(4.25-1.7)(4.25-2.8)(4.25-4)}$
$= \sqrt{4.25 \times 2.55 \times 1.45 \times 0.25} \approx \sqrt{3.90} \approx 1.97$ m²

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Problem 3: Using appropriate area formulas

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3a) Triangle MON: sides 37 mm, 19 mm, 20 mm

Use Heron's formula

Step1: Calculate semi-perimeter $s$

$s = \frac{37 + 19 + 20}{2} = \frac{76}{2} = 38$ mm

Step2: Compute area via Heron's formula

$A = \sqrt{38(38-37)(38-19)(38-20)} = \sqrt{38 \times 1 \times 19 \times 18}$
$= \sqrt{38 \times 342} = \sqrt{12996} = 114$ mm²

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3b) Triangle CSX: base 6.4 dm, height 3.2 dm

Use area formula for triangle: $A = \frac{1}{2} \times \text{base} \times \text{height}$

Step1: Apply base-height area formula

$A = \frac{1}{2} \times 6.4 \times 3.2$

Step2: Compute final value

$A = 3.2 \times 3.2 = 10.24$ dm²

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3c) Triangle VFJ: sides 2.2 km, 4.9 km, included angle $83.1^\circ$

Use SAS area formula: $A = \frac{1}{2}ab\sin(C)$

Step1: Apply SAS area formula

$A = \frac{1}{2} \times 2.2 \times 4.9 \times \sin(83.1^\circ)$

Step2: Compute final value

$\sin(83.1^\circ) \approx 0.9926$, so $A \approx \frac{1}{2} \times 2.2 \times 4.9 \times 0.9926 \approx 5.34$ km²

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Answer:

2a) $\boldsymbol{168}$ dm²
2b) $\boldsymbol{10.2}$ km²
2c) $\boldsymbol{\approx 1.97}$ m²
3a) $\boldsymbol{114}$ mm²
3b) $\boldsymbol{10.24}$ dm²
3c) $\boldsymbol{\approx 5.34}$ km²