Respuesta :
Answer:
Question 42
area of a triangle = 1/2 × base × height
Given:
- area = 14 in²
- height = x + 3
- base = x
Substituting given values into the formula:
⇒ 14 = 1/2 × x × (x + 3)
⇒ 28 = (x + 3)x
⇒ 28 = x² + 3x
⇒ x² + 3x - 28 = 0
⇒ (x - 4)(x + 7) = 0
⇒ x = 4, x = -7
As length is positive, x = 4 only.
⇒ height = 4 + 3 = 7 in
⇒ base = 4 in
-----------------------------------------------------------------------------------------
Question 43
volume of a cuboid = length x width x height
Given:
- volume = 72 in³
- length = 6 in
- height = x
- width = 2x - 5
Substituting given values into the formula:
⇒ 72 = 6 × (2x - 5) × x
⇒ 12 = (2x - 5)x
⇒ 12 = 2x² - 5x
⇒ 2x² - 5x - 12 = 0
⇒ (x - 4)(2x + 3) = 0
⇒ x = 4, x = -1.5
As length is positive, x = 4 only.
⇒ height = 4 in
⇒ width = 2(4) - 5 = 3 in
-----------------------------------------------------------------------------------------
Question 44
Pythagoras' Theorem: a² + b² = c²
(where a and b are the legs, and c is the hypotenuse, of a right triangle)
Given:
- a = x - 7
- b = x
- c = 13 m
Substituting given values into the formula:
⇒ (x - 7)² + x² = 13²
⇒ x² - 14x + 49 + x² = 169
⇒ 2x² - 14x - 120 = 0
⇒ x² - 7x - 60 = 0
⇒ (x - 12)(x + 5) = 0
⇒ x = 12, x = -5
As length is positive, x = 12 only.
⇒ one leg = 12 m
⇒ other leg = 12 - 7 = 5 m