Question

4. On the segment CD we draw the semicircle (A, a) which is tangent to the semicircle (B, b). We also draw the semicircle (0, OC), and the perpendicular EF to CD. The segments FC,FD intersect the semicircles at points H, J. The circles (M.). (L. 2) are tangent to the semicircle (O), the segment FE, and the semicircles (A). (B), respectively. a) Show that OC = a + b. b) Show that FE is a common tangent to the semicircles (A) and (B). c) Show that HEJF is a rectangle. d) Show that H) is a common external tangent to the semicircles (A) and (B). e) Show that FE? = CE.DE f) Show that the area highlighted in red equals the area of the circle (K, KE). ab g) Show that i = r2 = a+b h) Show that A MOL is isosceles. F M с m. А O E BD

Answer 1

solution :- M sox O لا اور c. EBB ginen that Aca BE b, ME - 8 L circe have rodius 92 we draw line AH and here oc - P OC is radius show that oca atb let sor LOA of semicirue OE So, OA = P-a NOW - P-a+p=2b hare AE 12 a sou p-atpa2b = a P - 2b O ATOE - AL

tangent that touches the So, 2P-2b-2a So, 2p = 20+2b P- atb so, oc : atb proved b. show that it is a common Hangent to the semicicules (A) and (0) As we know that 18 circles at one point, which is known as tongeneng at the tangonang tangent to a cirue is always perbendicular to the radius. So, ft is beobendicular to AE because AE is a radius of semicioue of a given that REFLCD 50, It also to EFIAE Er is the tangent of semicirue A Again Ef is also perpendicular is SO, EB Ef

Page No. Date: 1 perpendiculas to CD as given in question, So, if is also tangent of semicirue Ed hence the tangent of both semicirue N and B so. Er is the common tongendente semicirue A A and B c. show that HE JF is rectangle As iwe know that brine bale of rectangle 18 that curve in which every angle here CHE as <CEN - 90° because the angle insenbed in a semicirue is always a night angle because it is in semicirue А so, LEHF-90° LEND = 90° {becaus it is in semicirue B LCHE 90° so, LEJF = 90° 30, LHES = 90° as chei - 360- CEHF-44f j-LEJA

So, Enery angle is goo Hence HESF is rectangle of square me as and we every Square is rectangle So, HEJF is ree if rectangle know proved * Α. as w is a common external tangent to the semicirue and fi for this, we prove LAHj 290° and cBjH 90° we let LCHA = x So, LAHE 90-x as CCHE-90" here & AHA AE 28 so LAEH E90-3 C AEF - 90² as given that so, ckr Hax CKRH ne know that digonal Of rectangl is cavab intersected so, sor

HK Ks = fk ako So, LKEHALKHE = x Page No. as, KEECH (Date: 11 LAHRS CANE & LENK 90 x tt so C ALS2 90° L Now, let LD JB - y so, LAJE - Joy LAL - 90ansas BlaBtab SORES so, " { as KERS So, 2 KJB I goty ty SO, ekJB - 90° <BJH Hence LAHJ = LBJH = 90° Hence Hj is a common external semicarue cal and (B] tangent to the proved