Math Question and Answers


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 70429 by TawaTawa last updated on 04/Oct/19

Commented by TawaTawa last updated on 04/Oct/19
If ABCD is a square, AH is the tangent of the sector BEF at G, AG:GH = 7, and DH = 2. Find the area of the square
Commented by MJS last updated on 04/Oct/19

$A G : G H = 7 ? ? ?$ $is it 1 : 7 or 7 : 1 ?$
Commented by TawaTawa last updated on 04/Oct/19

$7 : 1 sir$
Commented by MJS last updated on 04/Oct/19

$ok but then the picture is wrong . . . it looks$ $like 1 : 7$
Commented by TawaTawa last updated on 04/Oct/19

$Help me do it like 1 : 7 sir$ $God bless you . Thanks for your time sir$
Commented by MJS last updated on 04/Oct/19

$will try with 7 : 1 later$
	Answered by MJS last updated on 04/Oct/19

	$triangles H D A and A G B are similar$ ${A G}^{2} + {G B}^{2} = {B A}^{2}$ ${H D}^{2} + {D A}^{2} = {A H}^{2}$ $A G = x; G H = 7 x; H D = 2; B A = D A = s; G B = r$ $x^{2} + r^{2} = s^{2}$ $4 + s^{2} = 64 x^{2}$ $\Rightarrow r^{2} = \frac{63}{64} s^{2} - \frac{1}{16}$ $\frac{A G}{H D} = \frac{G B}{D A} = \frac{B A}{A H}$ $\frac{x}{2} = \frac{r}{s} = \frac{s}{8 x}$ $\Rightarrow r = \frac{\sqrt{s^{3}}}{4}$ $\Rightarrow \frac{s^{3}}{16} = \frac{63}{64} s^{2} - \frac{1}{16}$ $s^{3} - \frac{63}{4} s^{2} + 1 = 0$ $\Rightarrow s = - \frac{1}{4} \lor s = 8 \pm 2 \sqrt{15}$ $but s ⩾ 2 \Rightarrow s = 8 + 2 \sqrt{15}$ $r = \frac{7 \sqrt{5}}{2} + \frac{9 \sqrt{3}}{2}$ $x = \frac{\sqrt{5}}{2} + \frac{\sqrt{3}}{2}$ $a r e a = s^{2} = 124 + 32 \sqrt{15}$
	Commented by TawaTawa last updated on 04/Oct/19

	$God bless you sir$
	Commented by MJS last updated on 04/Oct/19

	$if we start with 7 : 1 we get s = - \frac{7}{4} and r \notin R$
	Commented by TawaTawa last updated on 04/Oct/19

	$Thanks for your time sir$
	Commented by ajfour last updated on 04/Oct/19

	$\Rightarrow r^{2} = \frac{63}{64} s^{2} - \frac{7}{64}$ $i t h i n k r^{2} = \frac{63}{64} s^{2} - \frac{4}{64}$ $S i r, t h i s m a y a c c o u n t f o r t h e$ $d i f f e r n c e .$
	Commented by MJS last updated on 04/Oct/19

	${you}^{'} re right, {it}^{'} s a typo . . . thank you$
	Answered by ajfour last updated on 04/Oct/19

	$l e t e q . o f c i r c l e b e x^{2} + y^{2} = r^{2}$ $l e t s q u a r e s i d e b e s = 2 \cot α$ $l e t ∠ D A H = α$ $e q . o f t a n g e n t$ $r x \sin α + r y \cos α = r^{2}$ $y - i n t e r c e p t = \frac{r}{\cos α} = s = 2 \cot α$ $A s s - r \sin α = 7 r \sin α$ $\Rightarrow s = 8 r \sin α$ $\Rightarrow 8 \sin α \cos α = 1$ $\Rightarrow \sin 2 α = \frac{1}{4} \Rightarrow \cos 2 α = \sqrt{1 - \frac{1}{16}}$ $\cos 2 α = \frac{\sqrt{15}}{4}$ $\tan 2 α = \frac{1}{\sqrt{15}}, \frac{2 \tan α}{1 - \tan^{2} α} = \frac{1}{\sqrt{15}}$ $s a y \tan α = t$ $\Rightarrow t^{2} + 2 \sqrt{15} t - 1 = 0$ $\Rightarrow t = - \sqrt{15} + \sqrt{16}$ $\frac{1}{t} = 4 + \sqrt{15}$ $A r e a o f s q u a r e, A = 4 \cot^{2} α$ $A = \frac{4}{t^{2}} = 4 (16 + 15 + 8 \sqrt{15})$ $A = 124 + 32 \sqrt{15} .$
	Commented by TawaTawa last updated on 04/Oct/19

	$God bless you sir$
	Commented by MJS last updated on 04/Oct/19

	$\sqrt{124 + 32 \sqrt{15}} = 8 + 2 \sqrt{15} \approx 15.74596 . . .$ $why do we get different solutions ?$
	Answered by Henri Boucatchou last updated on 05/Oct/19

	${W h a t}^{'} s h a p p e n i n g h e r e, p e o p l e t e x t p i c t u r e s w i t h o u t a n y q u e s t i o n ? ?$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum