Right isosceles triangle abc has a hypotenuse of length h. line segment de is a midsegment with length 4x. what is the perimeter of triangle abc?

 If it is parallel to the hypotenuse, then the hypotenuse is twice as long. So h = 8x 
Then the legs will be (8x)/sqr(2). Since the triangle is a 45-45-90 

So the perimeter = 8x + (8x)/sqr(2)+ (8x)/sqr(2) 
=8x+ 16x/sqr(2) 
=8x+ 8xsqr2 

2)if the midsegment Is parallel to a leg, then each leg = 8x and the hypotenuse = 8x sqr2 

Then the perimeter is 8x + 8x + 8xsqr2 
= 16x+ 8xsqr2 

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