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melamori03 [73]
2 years ago
4

Find the area of \triangle ABC△ABCtriangle, A, B, C.

Mathematics
2 answers:
Bezzdna [24]2 years ago
7 0

Answer:

6.0square units

Step-by-step explanation:

Area of a triangle = 1/2 × base × height

The height of the given triangle is CD and the base is AB

To get the height CD, we need to use the Pythagoras theorem on the triangle ACD

AC² = CD² + AD²

CD² = 3²-1.8²

CD² = 5.76

CD = √5.76

CD = 2.4

According to ∆ABC,

AB² = AC²+BC²

Since AB = AD+DB then;

(AD+DB)² = AC²+BC²

AD² + 2(AD)(DB)+DB² = AC²+BC²

DB² = AC²+BC² - AD² - 2(AD)(DB)... 1

According to ∆BCD

BC² = CD²+DB²

DB² = BC²-CD²...2

Equating 1 and 2

AC²+BC² - AD² - 2(AD)(DB) = BC²-CD²

AC² - AD² - 2(AD)(DB) + CD² = 0

AC = 3

AD = 1.8

CD = 2.4

On substituting

3²-1.8² - 2(1.8)DB +2.4² = 0

9-3.24-3.6DB+5.76 = 0

11.52 - 3.6DB = 0

3.6DB = 11.52

DB = 11.52/3.6

DB = 3.2

The base of the triangle

AB = AD+DB

AB = 1.8+3.2

AB = 5.0

Area of the triangle = 1/2 × 2.4 × 5

= 1.2 × 5

= 6.0square units

Sergio [31]2 years ago
4 0

Answer:

= 6 square unit

Step-by-step explanation:

Area of a triangle is 1/2 × base × height

<u>first we find CD</u>

we use Pythagoras theorem

we represent CD with x

3² = 1.8² + x²

9 = 3.24 + x²

x² = 9 - 3.24

x² = 5.76

x = √5.76

x = 2.4

so

CD = 2.4

Area of a triangle = 1/2 × base × height

According to ∆ABC,

AB² = AC²+BC²

Since AB = AD+DB then;

(AD+DB)² = AC²+BC²

AD² + 2(AD)(DB)+DB² = AC²+BC²

DB² = AC²+BC² - AD² - 2(AD)(DB)... 1

According to ∆BCD

BC² = CD²+DB²

DB² = BC²-CD²...2

Equating 1 and 2

AC²+BC² - AD² - 2(AD)(DB) = BC²-CD²

AC² - AD² - 2(AD)(DB) + CD² = 0

AC = 3

AD = 1.8

CD = 2.4

On substituting

3²-1.8² - 2(1.8)DB +2.4² = 0

9-3.24-3.6DB+5.76 = 0

11.52 - 3.6DB = 0

3.6DB = 11.52

DB = 11.52/3.6

DB = 3.2

The base of the triangle

AB = AD+DB

AB = 1.8+3.2

AB = 5.0

Area of the triangle = 1/2 × 2.4 × 5

= 1.2 × 5

= 6 square unit

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