Q.451. Challenging area problem of geometry.

Question 451.
Equilateral triangle PQR is on square ABCD, as at the picture:
For given areas of right triangles (22 and 23) find unknown trapezoidal area, exact value.

Answer 451.
Let ∠ DRP = x
=> ∠ CRQ = 120° - x
Let a = length of the side of the equilateral triangle
=> (1/2) acosx * asinx = 22
=> a^2/4 sin2x = 22 ... ( 1 )
Similarly,
a^2/4 sin(240° - 2x) = 23 ... ( 2 )

Taking ratio ( 2 ) to ( 1 ),
sin(240° - 2x) / sin2x = 23/22
=> 22 * [- (√3/2) cos2x + (1/2) sin2x] = 23 * sin2x
=> - 11√3 cos2x = 12sin2x
=> tan2x = - (11√3)/12
=> sin2x = (11√3)/13√3 = 11/13
and cos2x = sin2x/tan2x = - 12/(13√3)

Plugging sin2x in ( 1 ),
a^2 = 88 / (11/13) = 104

Area of triangle = (√3/4) a^2 = 26√3

Length of side of the square
= acosx + acos(120° - x)
= acosx - (a/2)cosx + (a√3/2) sinx
= (a/2) cosx + (a√3/2) sinx

Area of the square
= [(a/2) cosx + (a√3/2) sinx]^2
= a^2 [(1/4) cos^2 x + (3/4) sin^2 x + √3 sinx cosx]
= (104) * [(1/8)(1 + cos2x) + (3/8)(1 - cos2x) + (√3/2) sin2x]
= (104) * [(1/2) - (1/4) cos2x + (√3/2) sin2x]
= (104) * [(1/2) + (1/4) (12/13√3) + (√3/2) * (11/13)]
= 52 + 30√3

=> area of the trapezium
= area of the square - area of equilateral triangle - areas of the two right triangles
= 52 + 30√3 - 22 - 23 - 26√3
= 7 + 4√3.

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Q.423. Application of Apollonius theorem in geometry.

Question 423.
The lengths of sides a, b, c in a triangle are related as a² +b² =5c². Prove that medians drawn to sides "a" and "b" are mutually perpendicular.

Answer 423.
Let AM be the length of the median on side a
and BN be the length of the median on side b
Let G be the centroid where the medians AM and BN intersect.

By Apollonius theorem,
b^2 + c^2 = 2AM^2 + 2(a/2)^2
=> AM^2 = (1/2) (b^2 + c^2 - a^2/2)
=> AG^2 = (2/3)^2 * (1/2) (b^2 + c^2 - a^2/2)
=> AG^2 = (2/9) (b^2 + c^2 - a^2/2)

Similarly, BG^2 = (2/9) (c^2 + a^2 - b^2/2)

=> AG^2 + BG^2
= (2/9) (b^2 + c^2 - a^2/2 + c^2 + a^2 - b^2/2)
= (2/9) (a^2/2 + b^2/2 + 2c^2)
= (2/9) [(1/2) (5c^2) + 2c^2] ... (plugging a^2 + b^2 = 5c^2 given)
= c^2.

=> triangle AGB is a right triangle and AG is perpendicular to BG
=> medians drawn to sides a and b are mutually perpendicular.

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Q.335. Common area of overlapping circles under given conditions.

Question 335.
Two circles of radii 5cm and 12cm are drawn partly overlapping. Centres 13cm apart. Find common area.

Answer 335.
5^2 + 12^2 = 13^2
=> The centers of the circle and their point of intersection form a right triangle.

Angle subtended by the arc of smaller circle at its center
= 2 arccos (5/13)
=> area of the sector formed by the arc subtending this angle at the center
= (1/2) * [2 arccos (5/13)] * 5^2
= 29.40 sq. units

Angle subtended by the arc of larger circle at its center
= 2 arccos (12/13)
=> area of the sector formed by the arc subtending this angle at the center
= (1/2) * [2 arccos (12/13)] * (12)^2
= 56.85 sq. units

Area of the kite formed by the centers and the intersecting points
= 5 * 12
= 60 sq. units

=> common area
= 29.40 + 56.85 - 60
= 26.25 sq. units.

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Q.332. Plane geometry

Question 332.
A square with a side length "x" is inscribed in an equilateral triangle with a side length "t". If t = kx.
find k to the nearest thousandths place.

Answer 332.
Refer to the figure in the link as under.

ABCD is a square inscribed in triangle PQR.
PM is perpendicular to the base QR.

AD = x and
QD = QM - MD = t/2 - x/2

tan ∠PQM = AD/QD
=> tan60° = x / (t/2 - x/2)
=> √3 = 2x / (t - x)
=> √3 (t - x) = 2x
=> √3 t = (2 + √3) x
=> k = t/x = (2 + √3) / √3 ≈ 2.155.
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Q.327. Plane geometry - Triangle area

Question 327.
The fig. shows a triangle ABC of area S with a line DE. If S₁ is area of triangle BDE , prove that  S₁/ S = BD * BE / AB * BC.

 
Answer 327.
Refer to the figure in the following figure.

Let CM and EN be perpendiculars from C and E on AB.

Right Δs BCM and BEN are similar
=> CM/EN = BC/BE ... (1)
S = Area of Δ ABC = (1/2) AB * CM
S₁ = Area of Δ BDE = (1/2) BD * EN

=> S₁/S
= [(1/2) BD * EN] / [(1/2) AB * CM]
= (BD/AB) * (EN/CM)
= (BD/AB) * (BE/BC) ... [Plugging (EN/CM) = (BE/BC) from (1).
= BD * BE / AB * BC.

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Q.326. Plane Geometry - Circle and tangents

Question 326.
 Let AB be the diameter of a semicircle, and let the tangents to it at B and at any other point D on it meet in C. If now DE be drawn perpendicular to AB, and if AC, DE meet in F, then what is the ratio of DF and EF ?

Answer 326.
Let the equation of the circle be x^2 + y^2 = r^2
Let D = (rcosθ, rsinθ)
 Eqn. of tangent at B is x = r ... (1)
Eqn. of tangent at D is xcosθ + ysinθ = r ... (2)
Solving eqns. (1) and (2),
C = [r, r(1-cosθ)/sinθ] = [r, rtan(θ/2)]

Δs ABC and AEF are similar
=> FE
= BC * (AE/AB)
= rtan(θ/2) * (r + rcosθ)/2r
=(1/2) r tan(θ/2) (1 + cosθ)
= (1/2) r tan(θ/2) * 2cos^2 (θ/2)
= (1/2) r sinθ ... (3)

DF
= DE - FE
= rsinθ - (1/2) rsinθ
= (1/2) rsinθ ... (4)

From eqns. (3) and (4),
DF = FE
=> DE : EF = 1.

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Q.320. Geometry

Question 320.
Find AG in the figure as under.

Answer 320.
As the answer is independent of the radius of the circle, the best choice of the radius will be such that DC passes through the center of the circle. As one vertical line is at a distance 4 from A and the other at a distance 9 from A, the center must be at a distance (1/2) * (4 + 9) = 6.5 from A
=> radius of the circle can be taken as 6.5.
 Refer to the figure drawn.

Δs AOC and BOD are isosceles and congruent.
[OA = OB = radius and OC = OD = radius and vertically opposite angles O are equal.]
=> x
= DF
= √[(6.5)^2 - (2.5)^2]
= 6.

Though  my  above  answer  was  selected   as  the  best, it  involves  an  unproved  assumption  of the  answer being  independent  of  the  radius  of  the  circle.

One  of  my  contacts,  Rakesh  Dubey,  who  is  very  good  at  solving  challenging  problems of  geometry  besides those  of  other  topics  of  maths,  provided  a  superior  solution  in  which  he  has  solved  the  problem  without  any  assumption  as  above.  He  had  posted  his  solution  in  comments  section  after the question  was  closed  for  answering.  His  solution  is  as  under.

Refer  to  the  following  figure.

 

Join  A  and  B  to  C  and  D.
In  right  triangle  ADB,  it  can  be  proved  with  simple  geometry  that
DF^2  =  AF * FB   ...   ...   ...   ...   ...   ( 1 )

In  right  Δs  CAG  and  BDF,
angle  ACG  =  angle  DBF
=> Δs  CAG  and  BDF  are  similar
=> AC / BD = AG / DF = x / DF
=> x = (AC * DF) / BD
=> x^2 = (AC^2 * DF^2) / BD^2   ...   ( 2 )

In  right  Δs  CAE  and  BAC,
angle A is common
=> Δs  CAE  and  BAC are similar
=> AC/BA = AE/AC
=> AC^2 = AE * BA   ...   ...   ...   ...   ( 3 )

In  right  Δs  BDF  and  BAD,
angle B is common
=> Δs  BDF  and  BAD are similar
=> BD^2 = AB * BF   ...   ...   ...   ...  ( 4 )

Plugging DF^2, AC^2 and BD^2 from ( 1 ), ( 3) and ( 4 ) in eqn. ( 2 ),
x^2 = [AE * BA * AF * FB) / (AB * BF)
=> x^2 = AE * AF = 4 * 9 = 36
=> x = AG = 6.

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Q.319. Geometry

Question 319.
The figure shows a triangle ABC. The incircle O is tangent to AC at D. M is the midpoint of AC. Line AE is perpendicular to BO extended. Prove that the measure of DM is equal to half the measure of CE.

Answer 319.

In right Δs ABL and EBL,

side BL is common and

∠ABL = ∠EBL
=> Δs ABL and EBL are congruent
=> BE = AB = c
=> CE = BC - BE = a - c ... (1)
 DM
= AM - AD
= b/2 - rcot(A/2)
= b/2 - (s - a)
= b/2 - [(a+b+c)/2 - a]
= - a/2 - c/2 + a
= (a - c)/2 ...   ...   ...   ...    (2)

From (1) and (2),
DM = (1/2) CE.

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Q.317. Geometry

Question 317.
Refer to the following geometric figure and prove that
(A1 + A2) - (A3 + A4) = 8A5.

Answer 317.

Refer to the following figure.

Let P be a point on GC such that GP = GE.

Draw a horizontal line through P forming a chord of the circle.
FQ is a vertical line through F meeting this chord and SR is another vertical line such that FS = EF
Thus, PRSE is a rectangle containg four equal rectangles each of area 2 * A5
=> area PRSE = 8 * A5 and is a part of A1.

Also, area above the horizontal chord throgh P and on right of CP = A4 and is also part of A1
CP = ED => area above the horizontal chord through P and on left of CP = A2
AE = SB => area between the horizontal chords on left of PE = raea on right of RS = A3 - A2 which is also part of A1.

=> A1 is made up of 8 * A5 + A4 + (A3 - A2)
=> (A1 + A2) - (A3 + A4) = 8 * A5.

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Q.315. Geometry

Question 315.
Find the angle α in the following figure.

Answer 315.

Refer to the figure (link below)

OD is parallel to BC

=> ∠ DOF = ∠ BCO = C/2 [corresponding angles] ... (1)

AD and AE are tangents to the incircle from A

=> AD = AE

=> Δ ADE is isosceles

and ∠ ADE = ∠ AED = (1/2) (π - A) = π/2 - A/2

=> ∠ ODF = π/2 - (π/2 - A/2) = A/2 ... (2)

α is an external angle of Δ DOF

=> α

= sum of the opposite internal angles

= ∠ DOF + ∠ ODF

= C/2 + A/2 ... [From (1) and (2)]

= π/2 - B/2

= 45°.

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Q.305. Geometry - application of Pythagorus theorem.

Question 305.
Two circles of diameters 10 cm and 15 cm are placed inside a square touching each other such that one circle touches a pair of adjacent sides and the other touches the remaining pair of adjacent sides of the square. Find the perimeter of the square.

Answer 305.
Let x = diagonal of the square
Refer to the figure as under.

x
= [R + r + √2 R + √2 r]
= [5 + 7.5 + √2 (5 + 7.5)]
= 12.5 (√2 + 1)

=> length of the side
= x / √2
and perimeter
= 4x / √2
= 50 (√2 + 1) / √2
= 85.4 cm.

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Q.286. Property of a regular tetrahedron.

Question 286.
Let O he any point within a regular tetrahedron ABCD. Lines joining the vertices A, B, C, D to O when produced meet the opposite faces in P, Q, R and S respectively. Prove that OP/AP + OQ/BQ +OR/CR+ OS/DS = 1.

Answer 286.
Join O to A, B, C and D.

It divides the tetrahedron ABCD into four smaller tetrahedrons, OBCD, OCDA, ODAB and OABC.

Let x = area of traingular base of tetrahedron and h = altitude from any vertex to the opposite face.
Let P', Q', R' and S' be the feet of perpendiculars from O on faces opposite to A, B, C and D.
=> OP/AP = OP'/h, OQ/BQ = OQ'/h, OR/CR = OR'/h and OS/DS = OS'/h

Sum of volumes of the four smaller tetrahedrons, OBCD, OCDA, ODAB and OABC
= volume of the larger tetrahedron ABCD.
=> (1/6) [OP' *x + OQ' * x + OR' * x + OS' * x) = (1/6) h * x
=> x [h * OP/AP + h * OQ/BQ + h * OR/CR + h * OS/DS] = x * h
=> OP/AP + OQ/BQ +OR/CR+ OS/DS = 1.

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Q.265. Minimum area of a triangle given its in-radius and minimum area of the circumcircle given the perimeter of the triangle

Question 265.
1 ) What is the minimum area of a triangle for a given in-radius ?
2 ) What is the minimum area of a circumcircle given the perimeter of the triangle ?

Answer 265.
1 )
Of all the triangles drawn for the incircle with radius, r, the equilateral triangle will have the least area.
Let a/2 = half the length of side of the equilateral triangle
=> (a/2) / r = cot30°
=> a/2 = r√3 cm
=> minimum area of the triangle
= 3 * (a/2) * r
= 3 * r√3 * r
= 3√3 r^2 sq. units.

2)
For minimum area of circumcircle, we have to find the minimum value of circumradius, R
Minimum R
= (p/3) / 2sin60°, where p = perimeter ... [Using the formula a/sinA = 2R]
= p / (3√3)
=> minimum area of a circumscribing circle for a given perimeter, p
= π(Rmin)^2
= πp^2/(27) sq. units.

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Q.254. Common area between intersecting circles

Question 254.
Find the common area of two circles of equal radii, R, each passing through the centre of the other.

Answer 254.
Let the circles be drawn in the x-y coordinate plane with their centers at (-R/2, 0) and (R/2, 0) as shown in the figure.

Their equations are
(x + R/2)^2 + y^2 = R^2 ... (1) and
(x - R/2)^2 + Y^2 = R^2 ... (2)

They intersect on the y-axis.
=> Required area
4 times the shaded area
= 4 ∫ (x=0 to R/2) √[R^2 - (x - R/2)^2] dx
= 4 [(x - R/2)/2 * √[R^2 - (x - R/2)^2] + (R^2/2) * arcsin[(x - R/2) / R] ... (x=0 to R/2)
= 4 [ - R/4 * (√3/2) R - (R^2/2) arcsin(-1/2) ]
= (2π/3) R^2 - (√3/2) R^2
= (2π/3 - √3/2) R^2.

Alternate Geometrical Approach:
Join the center of the right-hand circle to the point of intersection of the two circles on +ve y-axis.
Refer to the figure.

Base of the triangle = R/2
Height of the triangle = √[R^2 - (R/2)^2] = (√3/2) R
Area of the gtriangle = (1/2) * (R/2) * (√3/2) R = (1/4) * (√3/2) R^2
 Angle subtended by the arc of the circle at gthe center = arccos (R/2) / R = π/6
=> area of the sector of the circle = (π/6) R^2

Required area
= 4 * Shaded area
= 4 * [area of the sector of the circle - area of the right triangle.]
= 4 * [(π/6) R^2 - (1/4) * (√3/2) R^2]
= (2π/3 - √3/2) R^2.

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Q.219. Geometry.

Question 219.
Find distance x in terms of areas A, B, C in the following figure.

Answer 219.

A, B and C are the areas of parts A, B and C.

Let the area of the circular sector outside the square = m
=> B = A - (1/2) (2Rx - m)
and C = A - (2Rx - m) + x^2

=> C - 2B = x^2 - A
=> A + C - 2B = x^2
=> x = √(A + C - 2B).

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Q.182. Geometry, in-radius.

Question 182.

In a triangle ABC, in-radius r = 4 cm. Let incircle touch the sides BC, CA and AB at D, E and F respectively. It is given that BD = 8 cm and DC = 6 cm. Find the sides AB and AC.

Answer 182.

Refer to the figure.

Let AF = AE = x cm

Also note that BF = BD = 8 cm and CE = CD = 6 cm.

Area of ΔABC

= area of (ΔIBC + ΔICA + IAB)

= (1/2) * 4 * (14 + 6 + x + 8 + x)
= 4(x + 14) ... (1)
 Semi-perimeter, s = x + 14 cm
=> area of ΔABC using Heron's formula
= √[s(s-a)(s-b)(s-c)]
= √[(x+14) * x * (8) * (6)]
= 4 √[3x(x+14)] ... (2)

From (1) and (2),
4 √[3x(x+14)] = 4(x+14)
=> √(3x) = √(x+14)
=> 3x = x+14
=> x = 7 cm
 => AB = x + 8 = 7 + 8 = 15 cm
and AC = x + 6 = 7 + 6 = 13 cm.

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Q.159. Length of a helix

Question 159.
What is the length of the wire needed to make a spring of length L, diameter D and  number of turns, N.

Answer 159.
Let
L = length of the spring

D = diameter of the spring and
N = number of turns of the spring.

Now, the distance between successive turns = L/N
If the cylindrical surface of the spring is cut open along the length, it will be a rectangle with one end of the spring at one corner and the other end of one turn at the diagonally opposite corner of the rectangle of length = πD and width = L/N
=> length of one turn of the helix = √[(πD)^2 + (L/N)^2]
=> Total length of the wire needed to make the spring
= N * √[(πD)^2 + (L/N)^2]
= √[(πND)^2 + L^2].

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Q.155. Geometry - Finding sum of angles.

Question 155.

Find the sum of the missing angle measures as shown in the following diagram.

Answer 155.

Refer to the following figure.

Missing three angles of the inner hexagon are written in terms of "d" from the given information.
Adding the angles of the hexagon and equatting to 720°
=> c + d = 180° ... ( 1 )

 Missing two angles of the lower small triangle are written in terms of "a" and "d".
Adding the angles of the triangle,
a + d = 133° ... ( 2 )

 For the upper small triangle, external angle = external angle of right top triangle
=> e + 47 = d + 40°
=> d - e = 7° ... ( 3 )

 Subtracting ( 3 ) from ( 1 ),
c + e = 173° ... ( 4 )
For the left top triangle, external angle c = a + b.
Plugging c = a + b in ( 4 ),
a + b + e = 173° ... ( 5 )

 Adding ( 1 ) and ( 5 ),
a + b + c + d + e
= 180 + 173°
= 353°.

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Q.125. Geometry/Trigonometry

Question 125.

Rectangle is split, by diagonal, in two right triangles and circles are inscribed in triangles. Distance between the centers of the circles is 1 and width of the rectangle is also 1. Find length of rectangle a=?

Answer 125.

If r = in-radius of the circle, from the traingle,

then, r = area of traingle / semi-perimeter

=> r = a/[1 + a + √(1+a^2)] ... (1)

Constructing a right triangle with a line joining the

centers of the circles as hypotenuse and sides parallel

to sides of the rectangle,

(1 - 2r)^2 + (a - 2r)^2 = 1 ... (2)

Rationalizing the value of r in eqn. (1),

=> r = a[1 + a - √(1+a^2)] / [1 + a + √(1+a^2)]

* [1 + a - √(1+a^2)]

=> r = (1/2) [1 + a - √(1+a^2)]

=> 2r = 1 + a - √(1+a^2)

=> 1 -2r = √(1+a^2) - a

and a - 2r = √(1+a^2) - 1

Plugging in eqn. (2),
[√(1+a^2) - a]^2 + [√(1+a^2) - 1]^2 = 1
=> 2(1 + a^2) + a^2 - 2(a + 1)√(1+a^2) = 0
=> (3a^2 + 2) = 2(a + 1)√(1+a^2)

Squarring,
9a^4 + 12a^2 + 4 = 4(a^2 + 2a + 1)(1 + a^2)
=> 9a^4 + 12a^2 + 4 = 4(a^4 + 2a^3 + 2a^2 + 2a + 1)
=> 5a^4 - 8a^3 + 4a^2 - 8a = 0
=> 5a^3 - 8a^2 + 4a - 8 = 0 ... [because a ≠ 0]

Putting in Wolfram Alpha (Link 1) gives
a ≈ 1.68771 and its exact value as
a = (8/15) + (1/15)*[2492 - 300√(69)]^(1/3)] + (1/15)[623 + 75√(69)]^(1/3)].

Putting the equations (1) and (2) in wolfram alpha (Link 2)
a ≈ 1.68771 and r ≈ 0.362993
For exact values of a and r, click on exact forms in the link.

Wolfram Alpha Link 1
Wolfram Alpha Link 2

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Q.115. Geometry

Question 115.
Given quadrilateral ABCD where AB = BC = CD,  ∠ ABC = 70°, ∠ BCD = 170°,
find ∠ BAD without using a calculator.

The following solution selected as the Best Answer by the asker was worked out by my son, Kuntal, a networking professional with Cisco Systems despite being out of touch with the subject. I posted it in my account as he is not registered with YA!

Originally, Mr. Duke was the first to post an elegant solution to this problem of geometry. His ability to think of unique solutions to the problems of math is amazing! He drew my attention to this problem and inspired me to attempt. I too posted my solution which was followed by a stream of good solutions from Kuntal, rozeta53, Rakesh Dubey and Half-Blood Prince, who are my other valuable contacts of YA!
The three answers of Half-Blood Prince can be read in comments.

Answer 115.

ABCD is the given quadrilateral. Draw equilateral Δ BEC. E may lie inside or outside the quadrilateral ABCD or on side AD. We assume that it is inside ABCD and prove that this is not possible. With the given quadrilateral and the construction, different angles are as shown in the figure.

=> Sum of all angles of the quadrilateral
= α° + 85° + 70° + 170° + 35° + β°
= 360° + α° + β°
=> α° + β° = 0.

This can be proved even if E were outside ABCD.

=>  E is on AD so that α° + β° = 0.
=> ∠ BAD = 85°.

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