LENGHT OF THE RECTANGLE IS 7 LESS THAN THE TWICE OF THE BREADTH THEN FIND ITS DIMENSIONS IF THE PERIMETER IS 40 CM

Step-by-step explanation:

Length of rectangle = 7 less than the twice the breadthPerimeter = 40 cm

Let us say the breadth as b and length as l.

→ Length = 7 less than the twice the breadth

→ l = 2b - 7 ...(1)

Now, according to the question,

→ Perimeter of rectangle = 2(l + b)

→ 40 = 2(2b - 7 + b)

→ 40 = 2(3b - 7)

→ 40 = 6b - 14

→ 40 + 14 = 6b

→ 54 = 6b

→ 54 ÷ 6 = b

→ 9 m = Breadth

Now, finding length.

→ l = 2b - 7

→ l = 2(9) - 7 m

→ l = 18 - 7 m

→ Length = 11 m

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12

Distance=slopeslope is -2/4which equals-1/2the distance between these two points is -1/2