Answer:
3a) The value of x = 56
3b) The measure of ∠ H T M = 90°
3c) The radius of the circle = 53
Step-by-step explanation:
3a) ∵ A F is a tangent to the circle O at point F
∵ Secant AH intersects circle O at point T
∴ (A F)² = (A T)(A H)
∴ 7( x + 7) = (21)² ⇒ ÷ 7
∴ x + 7 = 63
∴ x = 63 - 7 = 56
3b) ∵ HM is a diameter
∴ The measure of the arc HM = 180° ⇒ semi-circle
∵ ∠ H T M is inscribed angle subtended by the arc HM
∴ m ∠ H T M = half the measure of arc HM
∴ m ∠ H T M = 180° ÷ 2 = 90°
3c) ∵ Δ H T M is a right angle triangle at T
∴ (H M)² = (M T)² + (H T)² ⇒ Pythagorean theorem
∴ (H M)² = (90)² + (56)²
∴ (H M)² = 11236
∴ HM =
= 106
∴ OM = 106 ÷ 2 = 53
∵ OM is the radius of the circle O
∴ The radius = 53
Let the width be x.
Length is 8 feet more than width. Length = x + 8
Area = x(x + 8)
width increased by 4, that is, (x + 4)
Length decreased by 5, (x + 8 - 5) = (x + 3)
Area = (x + 4)(x +3)
Area remains the same
x(x + 8) = (x+4)(x +3)
x² + 8x = x(x +3) + 4(x +3)
x² + 8x = x² +3x + 4x +12
x² + 8x = x² +7x +12 Eliminate x² from both sides
8x = 7x + 12
8x - 7x = 12
x = 12
Dimensions of original rectangle : x, x + 8
12, 12 +8 = 12, 20
Original rectangle is 20 feet by 12 feet
Answer:
It should be 5x-6y=30
Step-by-step explanation: