Hii people...Please solve these questions and share thy approach. :)
@pyashraj said:@brixcelFor the 1st Its 13 n 15 ..For the 2nd Its 70??
Yes...AB= 15 and AC = 13
Both answers are right. Please share how you solved them. :)
@brixcel
(I): Here, CE= CD= 6cm..DB= FB= 8cm..Let AE=AF=x cm[Tangent Rule]
Now, Area= r*s, or, [(14+x)(x)(8)(6)]^1/2 = 4*(14+x)
Approach:
(I): Here, CE= CD= 6cm..DB= FB= 8cm..Let AE=AF=x cm[Tangent Rule]
Now, Area= r*s, or, [(14+x)(x)(8)(6)]^1/2 = 4*(14+x)
=>48x = 14*16 + 16x, or, x=7..
Thus, AC=AE + CE = 13cm..n, AB= AF + FB = 15cm..
(II): Here, TP n TQ are tangent..Hence Angle TPO= Angle OQT= 90..Considering O to be the center of the circle..
Hence, PTQ= 180-110 = 70..
@pyashraj said:@brixcelApproach:(I): Here, CE= CD= 6cm..DB= FB= 8cm..Let AE=AF=x cm[Tangent Rule]Now, Area= r*s, or, [(14+x)(x)(8)(6)]^1/2 = 4*(14+x)=>48x = 14*16 + 16x, or, x=7..Thus, AC=AE + CE = 13cm..n, AB= AF + FB = 15cm..(II): Here, TP n TQ are tangent..Hence Angle TPO= Angle OQT= 90..Considering O to be the center of the circle..Hence, PTQ= 180-110 = 70..
cool 😃 Thank you! 😃 But one doubt here... in question 2, which property have you used? I mean, for which reason PTQ+POQ = 180?
@brixcel said:cool Thank you! But one doubt here... in question 2, which property have you used? I mean, for which reason PTQ+POQ = 180?
Quadrilateral sum of opposite angles equals 180
@master.of.fate said:Quadrilateral sum of opposite angles equals 180
Thank you! I'm on the path to brush up my GEO concepts again. :|
@brixcel
Since, TPOQ is a qudrilateral..n sum of all the angles of a quadrilateral is 360..n, TPO n OQT r 90..That's why..It shld be 180-110..
@master.of.fate said:i hate geometry
I don't hate it, it's something I haven't practiced for ages, and now it all is scaring me to the hell. 
Trigo + Geo = Deadly combo. :|
Trigo + Geo = Deadly combo. :|THEn we should start with geometry,what say?
Then Here we Go: :)
A city has a circular wall surrounding it.The walls has 4 gates located at the east, west, north n south points.A peepal tree stands at a distance of 2 km to the west of the west gate outside the city. It can be seen by a person, stnding 6 km directly to the north of the east gate, but he were any closer to the east gate, the wall obstructs the view of the tree.Find the radius of the wall surrounding the city..
A.2 B.3 C.4 D.5
@pyashraj said:Then Here we Go:A city has a circular wall surrounding it.The walls has 4 gates located at the east, west, north n south points.A peepal tree stands at a distance of 2 km to the west of the west gate outside the city. It can be seen by a person, stnding 6 km directly to the north of the east gate, but he were any closer to the east gate, the wall obstructs the view of the tree.Find the radius of the wall surrounding the city.. A.2 B.3 C.4 D.5
ye to kuch ajeeb sa diagram bann raha hai. 😐
@pyashraj said:@brixcelTry Karo..u'll dfntly slve it..
Seriously having no clue....That 6kms is like a tangent ...and that 2 kms...horizontally vo bhi circle ke bahar. Radius kaise pata chalegi :(
:(my take is B 3
is it ?
i tried it solving using pythagorus triplets try solving by using 6,8,10 triplet
@brixcel
Nw have a Luk..Angle AFO shld be 90..Hence, OA^2 = OF^2 + AF^2...
Thus, AF= [(2+r)^2 - r^2]^1/2...Nw, FC= 6 km..TANGENT RULE..
Thus, AC^2 = BC^2 + AB^2..Solve accrdngly..OA is 3 km.. :)
@pyashraj said:@brixcelNw have a Luk..Angle AFO shld be 90..Hence, OA^2 = OF^2 + AF^2...Thus, AF= [(2+r)^2 - r^2]^1/2...Nw, FC= 6 km..TANGENT RULE..Thus, AC^2 = BC^2 + AB^2..Solve accrdngly..OA is 3 km..
Ahan!! Got the essence! Lot of practice needed at my side. :(
Lot of practice needed at my side. :(