Geometry -

Question

Please provide detailed step by step answer. I am struggling with my geometry there fore some details would be helpful to understand the basics. Thanks

alrussell · Answer

For the first one I get D

This is as Angle ABC = CAD (on the basis that ABC + BAC = 90 = BAC + CAD) with the same logic for angle BAC = ADC

If so – then the two triangles are similar so their sides are in the same proportion. On that basis line AC is the equivalent of BC in which case the ratio is 3:4 – so 4 x 4/3 = 16/3

tarek99 · Answer

for the first 1, i got 3. here's how i did it:

the moment you see a line drawn from the 90 degree angle straight down to the triangle's hypotunese at 90 degree, automatically you know that the 2 smaller triangles are similar, bisecting angle A at 45 degrees. we also know that triangle abc has the formation of 3-4-5, therefore segment AB is 5. the ratio of AC to AB is similar to the ratio of AC to AD, therefore AD must be 5. again, having 3 and 5 for triangle ACD, we know that CD must be 3.

what's the OA?

alrussell · Answer

I don’t think that it can bisect angle A at 45% - if that was the case then ABC would also be 45, in which case the triangle would be isosceles – but we know it cant be as we have lengths of 3 and 4 not 3 and 3 or 4 and 4. As such, we know that the angle is not bisected into two equal angles… 

I think… 

OA pls?

tarek99 · Answer

for the second drawing, here is how i did it:

those 2 lines are perpendicular to each other, therefore the product of their slopes must be -1:

Slope of line P:
(0-1)/(0 - -sqrt 3) = -1/sqrt 3

in order to get a product of -1, the slope of Q must be reciprical of P:

slope Q must be sqrt 3/1

therefore: sqrt 3 / 1 = (t-0)/(s-0)

therfore: t=sqrt 3 while s= 1

hope this helps!

tarek99 · Answer

alrussell, actually you're right! sorry for that, but at least i know that when such a line is drawn from the 90 degree angle, then automatically we've created similar triangles.

Amardeep Sharma · Answer

1. Answer D = 16/ 3 

Solution: Applying pythagors them in triangle ABC, we get AB = 5

Now applying pythagoras them in triangle ACD, 
we have 16 + CD^2 = AD^2.

Consider triangle ABD, we have AB^2 + AD^2 = BD^2

or 25 + AD^2 = (BC + CD)^2

or 25 + 16 + CD^2 = 9 + CD^2 + 6CD

0r CD = 16/3.... ans

2. B= 1

Solution:

Draw a straight line PQ.

Here OP = OQ = radius of the circle, where center is at origin

using the formula distance between two points we have sqrt ((y1-y2)^2 + (x1-x2)^2) we have the radius of the circle is = sqrt((sqrt3)^2 +(1)^2)=2

similarly we will have s^2 + t^2 = 4...equation of the circle

Now using pythogoras thereom we have PQ^2 = PO^2 + OQ^2

or PQ^2 = 8 ie PQ = 2*sqrt2

but PQ^2 = s^2 + t^2 + 3 + 1 - 2*sqrt3*s - 2t = 8

but we know that s^2 + t^2 = 4

so.. -sqrt 3 *s = t

putting this in the equation of the circle we have

s^2 + 3s^2 = 4 or S =1...Ans

PS: I have tried to solve step-by-step for your understanding.

Amar

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