GEOMETRY AND MENSURATION

Example1: The sides of a triangle are 13, 14, 15 cm. The in radius of the triangle is :

(1) 8 cm (2) 4 cm

(3) 3 cm (4) 2.43 cm

Example 2: Euclid had a triangle in his mind. The longest side is 20 and the other side is 10. Area of the triangle is 80. The third side is

(CAT 2001)

(1) Ö260 (2) Ö240

(3) Ö250 (4) Ö210

Example3: The radius of the biggest circle which can be fitted in an equilateral

triangle of side 6 cm is:

(1) 3Ö3 cm (2) Ö3 cm

(3) 3 cm (4) 2Ö3cm

Example4: One side of a right triangle is 3/5 times of the hypotenuse and the sum of that side and hypotenuse is 16 cm. The circumradius of the triangle is:

(1) 3 cm (2) 4 cm

(3) 5 cm (4) None of the above

Example 5: Four horses are tethered at four corners of a square plot of side 14 metres so that the adjacent horses can reach one another. There is a small circular pond of area 20 m² at the center. The area let ungrazed is (CAT 2002)

(1) 22 m² (2) 42 m²

(3) 84 m² (4) 168 m²

Example 6 : A square of side 2 cm. is cut from each corner to form a regular

octagon. What is the side of the octagon? (CAT 2001)

(1) Ö2/(Ö2 + 1) (2) Ö2/(Ö2 – 1)

(3) 2/(Ö2 + 1) (4) 2/(Ö2 – 1)

Example7: A spherical ball of radius 8 cm is cut into 4 equal parts. Total surface area of the parts is:

(1) 384 π cm² (2) 512 π cm²

(3) 256 π cm² (4) 488 π cm²

Example8: A cubic meter of silver weighting 900 grams is rolled into a square bar 16 m long. The weight of 1 cm of this bar in grams is:

(1) 5 grams (2) 5.624 gms.

(3) 0.5625 gms (4) None of the above

Example9: The base of a conical tent is 19.5 m in diameter and the height of the tent is 2.8 m. What area of canvas is required to put up such a tent?

(1) 300m² (2) 400 m²

(3) 310.7 m² (4) none of the above

Example10: The radius of a hemisphere is 7 cm. Find the altitude of a right circular cone of the same base radius and same total surface area.

(1) 10.38 cm (2) 12.12 cm

(3) 15.66 cm (4) None of the above

Example11: In a rectangular plot 160 m x 120 m, a square tank of side 40 m is dug out a depth of 8 m. The earth so dug out is evenly spread to cover the remaining area, by how much has the level of plot has been raised?

(1) 0.82 cm (2) 8.2 cm

(3) 0.82 m (4) None of the above

Example12: A cylindrical cistern whose diameter is 21 cm is party filled with water. If a rectangle block of iron measuring 40 cm wholly immersed in the water, then the rise in the water level (in cm) is:

(1) 13 cm (2) 15.24 cm

(3) 18 cm (4) 19.38 cm

Example13: The corners of an equilateral triangle of side 10cm each are cut to form a regular hexagon. The area of the hexagon is:

(1) 20 cm² (2) 28.8 cm²

(3) 24.39 cm² (4) 25.12 cm²

Example14: The surface area of hemisphere is numerically equal to its volume. Its diameter is:

(1) 4.5 units (2) 7.5 units

(3) 9 units (4) 6 units

Example15: Five cylinders of base radius 10 cm and height 1, 2, 3, 4, 5 cm

respectively are melted to form a hemisphere. The surface area of the hemisphere is:

(1) 1615 cm² (2) 1685 cm²

(3) 1729 cm² (4) 1836 cm²

Example16: A circular disc of diameter 9 cm is cut in the center and a circular hole is generated. If the weight of the disc is reduced by one third, then the diameter of the hole is:

(1) 14.3 cm (2) 13.5 cm

(3) 12.4 cm (4) None of the above

Example17: A single pipe of diameter x has to be replaced by six pipes of

diameters 10 cm each. The pipes are used to covey precious liquid in a laboratory. If the speed of the liquid is the same then the value of x is:

(1) 13cm (2) 18 cm

(3) 24.5 cm (4) 24 cm

Example18: The number of edges of a solid having 20 faces and 32 vertices is:

(1) 58 (2) 56 cm

(3) 50 (4) 59

Example19: The percentage change in the total surface area of a cylinder with volume 1000 c.c. & radius 10 cm when it’s cut into two equal parts along its cross section is:

(1) 80% (2) 10%

(3) 30% (4) 75.84%

Example 20: Instead of walking along two adjacent sides of a rectangular field, a boy took a short cut along the diagonal and saves a distance equal to half the longer side. Then the ratio of the shorter side to the longer side is (CAT 2002)

(1) ½ (2) 2/3

(3) ¼ (4) ¾

Example21: Water flows through a pipe of diameter 1/16 meter, at 6 meters/sec. The time taken to fill bath 2 metres by 3 meters by 4 meters is:

(1) 21.8 min (2) 30.33 min

(3) 15.38 min (4) None of these

Example22: A tank 72 cm long, 60 cm wide, 36 cm high contains water to a depth of 18 cm. A metal bloc 48 cm by 36 cm by 15 cm is put into the tank and totally submerged. The rise of water- level is:

(1) 5 cm (2) 5.27 cm

(3) 6.36 cm (4) 6 cm

Example23: The external diameter of a hollow metal sphere is 14 cm and its thickness is 2 cm. Find the radius of a solid sphere containing the same amount of material as the hollow sphere.

(1) 6 cm (2) 5 cm

(3) 4 cm (4) 4.5 cm

Example24: Three solid metal of radii 3 cm, 4 cm, 5 cm respectively, are melted

together. The metal is recast as a single solid sphere. The percentage reduction in the area of surface resulting from this is:

(1) 40% (2) 16%

(3) 12% (4) 28%

Example25: A bucket in the form of a frustum of a right circular cone is 48 cm deep and the radii of the top and bottom are 20 cm and 16 cm respectively. The volume of water it will hold is:

(1) 49049 c.c. (2) 30000 c.c.

(3) 31259 c.c. (4) 24076 c.c.

Example 26: The diameter of a copper sphere is 6 cm. The sphere is melted and is drawn into a long wire of uniform circular cross section. If the length of the wire is 36 cm, The radius is

(1) 1 (2) 0.5

(3) 2 (4) 0.1

Example 27 : A sphere of diameter 6 cm is dropped in a right circular cylindrical vessel, party filled with water. The diameter of the cylindrical vessel is 12cm. If the sphere is completely-submerged in water, by how much will the level of water rise in the cylindrical vessel?

(1) 2 (2) 0.5

(3) 1 (4) 0.1

Example 28: In the given figure AB and BC are 24 and 32 with included angle of 90^o. The other sides are 25 each and none of the other angle is 90^o. Find the area of the fig. (CAT 2001)

<<Geometry Unsolved Fig 1>>

(1) 684 (2) 786

(3) 880 (4) None of these.

Example 29 : Find the number of coins, 1.5 cm in diameter and 0.2 cm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.

(1) 425 (2) 450

(3) 550 (4) None of these.

Example 30: A solid iron rectangular block of dimensions 4.4 m, 2.6m and 1 m is

cast into a hollow cylindrical pipe of internal radius 30 cm and thickness 5 cm. Find the length of the pipe.

(1) 118 (2) 116

(3) 110 (4) 112

Example31: How many sides have regular polygon whose exterior angle is 1/11 of its interior angle?

(1) 22 (2) 24

(3) 20 (4) 26

Example32: The angles of a pentagon are x^o, x + 20^o, x + 40^o, x + 60^o and

x + 80^o. The smallest angle of the pentagon is:

(1) 65^o (2) 70^o

(3) 68^o (4) 60^o

Example33: In a triangle ABC, the altitudes BD and CE are equal and ÐA = 36^o. What is the value of the angle B?

(1) 65^o (2) 60^o

(3) 75^o (4) 72^o

Example 34: In the following figure

<<Geometry Unsolved Fig 2>>

AE = EF = FB. The ratio of area of triangle CEF and rectangle ABCD is (CAT 2001)

(1) 1/6 (2) 1/8

(3) 1/9 (4) None of these.

Example35: The interior angle of a regular polygon exceeds the exterior angle by 140^o. The sides of the polygon are:

(1) 18 (2) 22

(3) 25 (4) 15

Example36: Two regular polygons have the number of their sides as 3 : 2 and the interior angles as 10 : 9. Find the number of sides of the polygon.

(1) 10, 8 (2) 14, 8

(3) 12, 8 (4) 18, 8

Example 37: In the given figure BC = AC, angle AFD = 40^o and CE = CD. The value of angle BCE = (CAT 2001)

<<Geometry Unsolved Fig 3>>

(1) 100^o (2) 50^o

(3) 60^o (4) None of these.

Example38: In a triangle PQS, R is any point on PS, such that PR = QR and QS = RS. If angle RSQ = 120^o, what is the measure of angle QPR?

(1) 15^o (2) 25^o

(3) 16.5^o (4) 30^o

Example39: O is the centre of the circumcircle of triangle ABC. If ÐA= a and ÐOCB= b. what is the measure of a + b?

(1) a + b = 65^o (2) a + b = 60^o

(3) b + a = 80^o (4) a + b =90^o

Example40: AB and CD are parallel straight lines of lengths 5 cm and 4 cm

respectively. AD and BC intersect at a point O such that AO = 10 cm. Then OD equals to:

(1) 10 cm (2) 8 cm

(3) 6 cm (4) 4 cm

Example41: PSR is a triangle right angled at S and D is the mid point of SP. If the bisector of ÐPSR and perpendicular bisector of SR meet at O, then triangle OSD is:

(1) scalene (2) equilateral

(3) isosceles right angled (4) acute angled

Example42: PR is the diameter of the circle. Find PS where PQ = 6 cm QR = 7 cm and RS = 2 cm?

(1) 7 cm (2) 5 cm

(3) 11 cm (4) 9 cm

Example43: A circle has two parallel chords of lengths 6 cm and 8 cm. If the chord are 1 cm apart and the chord is on the same side of the centre, then the diameter of the circle is of length:

(1) 5 cm (2) 6 cm

(3) 10 cm (4) 12 cm

Example 44: In the given figure, ACB is a right angled triangle. CD is the altitude. Circles are inscribed within the triangles ACD, BCD. P and Q are the centers of the circles. The distance PQ is (CAT 2002)

<<Geometry Unsolved Fig 4>>

(A) 5 (B) Ö50

(C) 7 (D) 8

Example45: In a right angled triangle ABC; ÐA = 90^o and AM is the median of BC. If AB = 6 cm and AC = 8 cm. The length of AM is:

(1) 5 cm (2) 7 cm

(3) 6.5 cm (4) 10 cm

Example46: ACB is a tangent to a circle at C, CD and CE chords such that ÐACE > ÐACD. If ÐACD = ÐBCE = 50^o, then

(1) CD = CE

(2) ED is not parallel to AB

(3) ED passes through the centre of the circle

(4) ÐCDE is a right angled triangle

Example47: A circle with centre O has a chord AB produced so that it meets the line through O in C, such that OB = BC. If angle BCO is 20^o, what is the measure of angle AOD where D lies on circle when CO is produced?

(1) 65^o (2) 60^o

(3) 75^o (4) 72^o

Example48: ABCD is a square E and F are the mid-point of BC and CD. What is the ratio of the area of DAEF to that of the square ABCD?

(1) 3: 8 (2) 5: 8

(3) 8: 3 (4) None of the above

Example49: AB is a chord of a circle whose centre is O. P is a point on the circle such that OP ^ AB and OP intersect AB at the point M. If AB = 8 cm and MP = 2 cm, then the radius of the circle is:

(1) 10 cm (2) 6 cm

(3) 5 cm (4) 4 cm

Example50: The side AB, BC, CD of a regular polygon are such that the measure of angle BAC is 15^o. The number of sides of the polygon are:

(1) 15 (2) 12

(3) 18 (4) 20

Example51: In DABC, D is the mid-point of BC which is 10 cm, AD = 5 cm and AC = 6 cm. The length of side AB is:

(1) 5 cm (2) 7 cm

(3) 6.5 cm (4) 8 cm

Example 52: In the figure given below, ABCD is a rectangle. The area of the isosceles right triangle ABE = 7 cm²; EC = 3 (BE). The area of ABCD (in cm²) is (CAT 2002)

<<Geometry Unsolved Fig5>>

(1) 21 (2) 28

(3) 42 (4) 56

Example53: The sides of the triangle are 5 cm, 4 cm and 3 cm. Find the greatest side of another similar triangle whose area is 16 times greater that the area of the given triangle.

(1) 25 cm (2) 20 cm

(3) 18 cm (4) 15 cm

Example54: P is a point on the base of the equilateral DABC such that

BP=1/3BC.Then AP²/AB²=

(1) 7/9 (2) 5/9

(3) 8/9 (4) 4/9

Example 55: The length of the common chord of two circles of radii 15 cm and 20 cm, whose centers are 25 cm apart, is (in cm) (CAT 2002)

(1) 24 (2) 25

(3) 15 (4) 20

Example 56: In the given figure, BC is the diameter of the circle with the centre O and PAT is the tangent at A. If ÐABC = 38^o, find ÐBAT.

(1) 52^o. (2) 48^o.

(3) 62^o. (4) None of these

<<Geometry Unsolved Fig 6>>

Example 57: In the given figure, O is the centre of the circle and PAQ is the tangent to the circle at A. If ÐPAB = 58^o, find ÐABQ and ÐAQB.

(1) 32^o, 26^o (2) 48^o, 24^o

(3) 42^o, 18^o (4) None of these

<<Geometry Unsolved Fig 7>>

Example 58: In the given figure, DE || BC. If AD = (4x - 3)cm, AE = (8x - 7)cm and BD = (3x - 1)cm and CE = (5x - 3), find the value of x.

(1) 2 (2) 4.

(3) 1 (4) None of these

<<Geometry Unsolved Fig 8>>

Example 59: The perimeters of two similar triangles are 30 cm and 20 cm

respectively. If one side of the first triangle is 15 cm, then find out the

proportional side of the other triangle.

(1) 20 (2) 25

(3) 15 (4) None of these

Example 60 :A vertical stick 12 cm long casts a shadow 8 cm long on the ground. At the same time tower casts a shadow 40 m long on the ground. Determine the height of the tower.

(1) 50 (2) 60

(3) 40 (4) 75

Q61. Find the area of an equilateral triangle whose vertices lie on a circle with radius 2 cm.

(a) 3π cm2 (b) 3√3 cm2

(c) √3 π cm2 (d) 6 cm2

(e) None of these

Q62. The following statements describe a race between: Pat, Chris, Jo, and Sam.

(i) Pat is 12 seconds behind the next runner.

(ii) The leader is 20 seconds ahead of the last person.

(iii) Chris is 1 second ahead of somebody.

(iv) Sam is 19 seconds ahead of Jo.

Which of the following gives the order of the runners. (Runners listed from first to last.)

(a) Sam, Chris, Pat, and Jo

(b) Chris, Sam, Pat, and Jo

(c) Sam, Pat, Chris, and Jo

(d) Chris, Sam, Jo, and Pat

(e) Sam, Chris, Jo, and Pat

Q63. If a boat travels North for 5 miles then East for 12, then Southeast for 6, approximately how far is it from its starting point?

(a) 13 miles (b 19.01 miles

(c) 23.21 miles (d) 16.26 miles

(e) None of these

Q64. Given the following two circles, find the algebraic equation of the chord they share in common.

Circle I : (x – 1)2 + (y – 2)2 = 9

Circle II: (x + 3)2 + (y – 1)2 = 16

(a) 4x – 3y = 1 (b) 3x + 4y = 1

(c) 2x – y = 10 (d) x – 3y = 5

(e) None of these

Q65. The area of a 300' by 400' rectangle is doubled by adding a strip of width "w" around the perimeter. Approximately how wide is that strip?

<<Geometry Unsolved Fig 9>>

(a) 87.50’ (b) 83.41’

(c) 72.84’ (d) 71.22’

(e) None of these

Q66. Find the length of the perimeter of a right triangle whose area is 30 cm2 and whose hypotenuse is 13 cm.

(a) 23 + √69 cm (b) 16 + 2√24cm

(c) 30 cm (d) 7 + 3√13 cm

(e) None of these

Q67. Find the area of a triangle whose vertices are (0,0), (12, 9) and (14, 6).

(a) 27 sq units (b) 54 sq units

(c) 84 sq units (d) 168 sq units

(e) None of these

Q68. Given that it is 3 o'clock, exactly how long will it take for the minute hand to catch up with the hour hand?

(a) 15 min (b) 16 min

(c) 161/3min (d) 161/2 min

(e) None of these

Q69. If the size of a rectangle's area is twice as large as the size of its perimeter, and the length of one of its sides is 4.5, what is the length of the other side?

(a) 36 (b) 3.6

(c) 18 (d) 16

(e) None of these

Q70. A given circular cylindrical can is made up of a square piece of metal and two circular disks each with diameter c inches. What is the volume of this cylindrical can?

<<Geometry Unsolved Fig 10>>

(a) 2π – c3 in3 (b) π3c3/4 in3

(c) π3c3/6 in3 (d) 2.467c3 in3

(e) None of these

Q71. Two chords AB and DC intersect each other so that AO = 1.2, OB = 7.5, and OC = DO. How long is DC?

(a) 6 units (b) √8.7 units

(c) √35 units (d) 8.7 units

(e) None of these

Q72. Find the area of parallelogram ABCD, given that AB = BE = ED = 1, and ÐABE = 90°.

<<Geometry Unsolved Fig 11>>

(a) √2 (b) 1.707

(c) (2 – √2)/2 (d) (2 + √2)/2

(e) None of these

Q73. Find the solution set for y given that z y x, are natural numbers and the following are true:

(i) If y > 3 then x < 3

(ii) z + y < 7

(iii) x + z > 10

(a) y ε {1, 2} (b) y ε {1, 2,…,5}

(c) y ε {3, 4, 5, 6} (d) y ε {3, 4, 5, 6}

(e) y ε {1,2,…,6}

Q74. A rectangle is inscribed in a triangle such that its upper right vertex bisects the triangle's side. What is the ratio of the area of the shaded region to the area of the unshaded region?

<<Geometry Unsolved Fig 12>>

(a) 1:2 (b) 4:3

(c) 5:4 (d) 2:3

(e) None of these

Q75. Two 5-12-13 triangles are combined to form a parallelogram. Which of the following statements must be true?

(a) The parallelogram is a rectangle.

(b) The perimeter of the parallelogram is 34 units long.

(c) The area of the parallelogram is 60 square units.

(d) One of the sides of the parallelogram is 12 units long.

(e) All of these statements must be true.

Q76. To construct a circle that circumscribes a triangle one finds its center by locating the intersection of which two lines.

(a) The perpendicular bisectors of two sides.

(b) The bisectors of two of the angles.

(c) Two medians.

(d) Two altitudes.

(e) It is not possible to construct a circle that circumscribes a triangle.

Q77. Form a triangle by connecting the centers of three circles. Each of these circles is tangent to the other two and their radii are 1, 2, and 3 units long. How large is area of the triangle?

<<Geometry Unsolved Fig 13>>

(a) 12 (b) √5.π

(c) 24 (d) √17

(e) 6

Q78. Given the graph below, how many different paths are there from “S” to “E” if one never visits the same point twice?

<<Geometry Unsolved Fig 14>>

(a) Less than 11 (b) 11

(c) 12 (d) 13

(e) More than 13

Q79. A pentagon is inscribed in a circle of radius 7. How long is the circular arc that connects two neighboring vertices?

(a) 1.4π (b) 2.1π

(c) 2.8 π (d) 7π

(e) None of these

Q80. If a rectangle has a diagonal of length c and a perimeter of length p then the expression for its area is?

(a) pc (b) (p² – 4pc²)/8

(c) √(p² – 2c²)/2 (d) (p² + c²)/2

(e) None of these

Q81. A rectangular solid with a square base has dimensions 5 × 5 × 8. If its volume is quadrupled by doubling the lengths of sides of the base, by what factor is its surface area increased?

(a) 4 (b) 2

(c) <<Geometry Unsolved Fig 15>> (d) <<Geometry Unsolved Fig 16>>

(e) None of these

Q82. How long is the room shown in the figure on the right, given that it has a 7.5 foot ceiling, and an infra red beam that starts at one end of the room is bounced off the ceiling then the floor and finally hits the opposite side of the room 5.4 feet above the floor? The point at which it bounces off the ceiling is two feet from the wall.

<<Geometry Unsolved Fig 17>>

(a) 8.6 ft (b) 10.6 ft

(c) (2 + √72 )ft (d) 12 ft

(e) None of these

Q83. Find the height of a square pyramid formed by four equilateral triangles whose sides all have length 2.

(a) 1 (b) √6/2

(c) √2 (d) √3

(e) None of these

Q84. If a rectangle whose length is 9 times its width is modified so that its area is doubled but its perimeter is kept constant, what is the ratio of length to width for the new rectangle?

(a) 5 + √5 : 5 - √7 (b) 2 : 1

(c) 7 + √5 : 7 - √5 (d) 1 : 1

(e) None of these

Q85. One of the five statements below is false and the other four are true. Who told the falsehood? Ann said, “If the car was not locked then the wallet was stolen.” Bo said, “If the tickets to the game are lost then Ann's statement is false.”

Chris said, “Ann's statement is true”

Dan said, “The tickets to the game are not lost.”

Ed said. “The wallet was stolen but not the tickets to the game.”

(a) Ann (b) Bo

(c) Chris (d) Dan

(e) Ed

Q86. A Quadrilateral ABCD is inscribed in a circle. If the size of the angle at vertex A is 36° then the angle at vertex C is:

(a) 72° (b) 54°

(c) 126° (d) 144°

(e) None of these

Q87. The following facts are given: a gallon of paint covers 400 square feet of wall space, the room to be painted has an 8 foot ceiling, its dimensions are 20 by 14 feet, it has two doors (36 by 84 inches) and four large windows (72 by 60 inches), and you need two coats of paint. How much paint do you need to paint the walls?

(a) 1.47 gal. (b) 2.06 gal.

(c) 2.72 gal (d) 1.03 gal.

(e) None of these

Q88. For the points (0, 2), (6, 6) and (10, 0) which of the following statements are true?

I. The points form the vertices of a right triangle.

II. The points form the vertices of an isosceles triangle.

III. The largest angle is at the vertex located on (6,6).

(a) only III (b) II and III

(c) I and III (d) all are true

(e) None are true

Q89. Given you have 4 sticks, two of length 5 and two of length 8, with which you are to form a quadrilateral. If at least one of the angles is a right angle, how many different no congruent quadrilaterals could you form.

(a) 1 (b) 2

(c) 3 (d) 4

(e) More than 4

Q90. A triangle with sides 6, 8, and 10 has its shortest side doubled in length while the other two sides remain the same. What is the area of the new triangle?

(a) 30 (b) 40

(c) 15√7 (d) 8√30

(e) None of these

Q91. What is the size of an angle between two adjacent sides of a regular 12 sided polygon?

(a) 144° (b) 120°

(c) 1080^o/7 (d) 30°

(e) None of these

Q92. As shown in the figure on the right six similar triangles are each sharing one side with the next triangle and all are sharing one vertex. All angles at that vertex measure 60°. If the side of the last (smallest) triangle that is adjoining the first triangles 1/6 as large as the longest side of that first triangle, how many times larger is the area of the largest triangle as compared to the smallest?

<<Geometry Unsolved Fig 18>>

(a) 6

(b) <<Geometry Unsolved Fig 19>>

(c) <<Geometry Unsolved Fig 20>>

(d) 36

(e) None of these

Q93. Three views of the same block are shown on the right. What letter is on the side parallel to the side with the letter A?

<<Geometry Unsolved Fig 21>>

(a) E (b) O

(c) S (d) H

(e) G

Q94. An arbelos is the region formed by three mutually tangent circles whose centers are collinear, as noted in the image. If the diameters of the two smaller circles are a and b, what is the area of the arbelos.

<<Geometry Unsolved Fig 22>>

(a) abπ. (b) √abπ/4

(c) abπ/4 (d) [(√a2 + b2)π]/4

(e) None of these

Q95. A bicycle has a 70 cm diameter wheel. If you ride in a 120 km race, approximately how many revolutions does the wheel have to make to complete the race?

(a) 54,600 (b) 171,400

(c) 538,600 (d) 732,100

(e) None of these

Q96. If the volume of a tetrahedron is doubled without changing its shape, by what factor is the surface area increased?

(a) <<Geometry Unsolved Fig 23>>

(b) <<Geometry Unsolved Fig 24>>

(c) 2

(d) √8

(e) 4

Q97. A semicircle with diameter 12 inches is used to form a conical cup by bending the semicircle so that its two corners are connected and the circle's center forms the point of the cone. How large is the volume?

(a) 9π√3 (b) 27π√3

(c) 18π√6 (d) 2π√15

(e) None of these

Q98. Chord <<Geometry Unsolved Fig 25>> is parallel to the line tangent to the circle at point C. If the distance between the chord and the tangent line is 18, and the radius of the circle is 13, how long is the chord <<Geometry Unsolved Fig 26>> ?

<<Geometry Unsolved Fig 27>>

(a) 12 (b) 2√105

(c) 2√155 (d) 24

(e) None of these

Q99. A rectangle with dimensions 11 by 13 had its diagonal increased by 50% without lengthening the shorter side. Approximately how big is the area of the new rectangle?

(a) 214.5 (b) 321.75

(c) 285.852 (d) 175.139

(e) 253.597

Q100. Nikki sees that the top of a 15 foot lamp, which is 250 feet away, lines up perfectly with the peak of a distant mountain. Nikki knows that the mountain is 15 miles away so she uses the lamp to determine the mountain's height. If Nikki's eyes are 5 feet above the ground, what is the best estimate of the mountain's height relative to Nikki?

(a) 900 feet (b) 3,178 feet

(c) 4,752 feet (d) 5,625 feet

(e) 6,648 feet

Solution1: Area = √[ s(s-a) (s-b) (s-c)]

Here s = 21 cm. so, area = √(21 x 8 x 7 x 6) = 84 cm²

ð r = A/s = 84/21 = 4cm.

Ans. (2)

Solution2: Suppose the three sides are a, b, c,

i.e., a = 20, b = 10

so that s = (a + b + c)/2 = (20 + 10 + c)/2 = (30 + c)/2

By hypothesis,

80 = √[s(s – a) (s – b) (s – c) ]

= √[(30 + c)/2 ((30 + c)/2 – 20) ((30 + c)/2 – 10) x ((30 + c)/2 – c) ]

or 80 = √[ (225 – c²/4)(c²/4 – 25) ]

Write c²/4 = y, so that 80 = (225 – y)(y – 45) or y = 65

Hence, c = √(4 x 65 )= √ 260

Ans=(1)

Solution3: r = A/s, Here A = Ö3/4 x 6² and s = 9 => r = Ö3 cm.

Ans.(2)

Solution4: Forming the equations we have

a = 3/5 h a + h = 16

so 3/5 h + h = 16 or = 5/8 x 16 = 10

For right triangle the circum radius = Hypotenuse/2 = 5 cm

Ans. (3)

Solution 5: 14mtr

<<Geometry Unsolved Fig 28>>

Area of the square field = 14 x 14 = 196 m²

Now, as the horses can reach each other, the rope length is 7 mtrs

Area one horse can graze=90/360 x p7x7 (remember area of sector of a circle)

Area grazed by 4 horses = 4 x p/4r² = 22/7 x 7 x 7 = 154m².

Area of circular pond in the centre = 20m² (Given)

\Remaining area = 196 – 154 – 20 = 22m².

Ans=(1)

Solution6: Let PA = x and B = y. From rt. Ð ed DHPA,

y² = x² + x² = 2x²

or x = y/Ö2 ….(i) <<Geometry Unsolved Fig 29>>

Also x + y + x = 2

or x = (2 – y)/2

Putting it in (i),

(2 – y)/2 = y/Ö2 or Ö2(2 – y) = y

(2 – y) = Ö2y or 2 = (1 + Ö2)y

\ y = 2/(Ö2 + 1)

Ans=(3)

Solution7: 1/4^th part of sphere contains, lateral surface area + 2 semicircle of radius 8. ¼(4pr²) + 2 (1/2 pr²) = 2pr². So the total surface area = 4 x 2pr² = 512p cm².

Ans.(2)

Solution8: Volume of the cube = 100 x 100 x 100

= 10⁶cm³. Volume of the square bar = X² x 1600cm²

=>1000000 = X² x 1600 => X = 25cm

=> Volume of 1 cm length of bar = 1 x 25 x 25 = 625 cm³

=> Required weight = (900 x 625) / (1000000)

=> 0.5625 grams.

Ans.(3)

Solution9: Curved surface area of the tent = prL

since L² = r² + h² so curved surface

= p x 9.75 xÖ(2.8)² + 9.75)² = 310.7 m².

Ans. (2)

Solution10: Surface Area of the hemisphere = 3pR² = 462 cm²

= Total Surface Area of cone = pR² + pR Ö(H² + R²)

ð Putting the value of R = 7 cm, we get H = 12.12 cm.

Ans.(2)

Solution11: Volume of the earth dug out = 40 x 40 x 8 = 12800m3

The area over which this is spread = (160 x 120) – (40 x 40) = 17600 m²

ð The level increases by h = 12800/17600 = 0.82 m

= 82cm.

Ans.(3)

Solution12: Cross sectional area of the cistern = pR²= 346.5 cm².

Volume of the iron block= 40 x 11 x 12 = 5280cm³

ð Increase in height = 5280/346.5 = 15.24 cm.

Ans. (2)

Solution13: Since a side of the triangle is 10cm, when we cut it the side of the hexagon becomes 10/3 = 3.33 cm

= Area of the hexagon = 6 x area of an equilateral triangle

= 6 x Ö3/4 x 3.33² = 28.8 sqcm.

Ans. (2)

Solution14: We have 3pR² = 4/3 pR³ => R = 9/4 or Diameter = 9/2 = 4.5 units.

Ans. (1)

Solution15: Total volume of 5 cylinders = p x 10² x (1 + 2 + 3 +4 +5) = 4710m³.

Volume of hemisphere = 2/3 pR³ = 4710

\ R = 13.1 cm Surface Area = 3pR²

= 1616.5cm² (approx).

Ans.(1)

Solution16: Weight of the disc cut off = 1/3 of the weight of the original disc. Let D be the density => Volume = Weight/Density

r is the radius of the hole and R is radius if original disc.

ð pr²hD = 1/3 x pR²hD => r = 6.75 cm

diameter of hole= 6.75 cm x 2 = 13.5 cm.

Ans.(2)

Solution17: Volume discharge by 1 pipe = Volume discharged by 6 pipes

=> Volume per sec = Area x Speed

ð Area of bigger pipe = Total area of 6 smaller pipes

ð pR² = 6 x p x 5² = R = 12.25 cm =>D = 24.5 cm

Ans.(3)

Solution18: For this question, we require Euler’s formula:

Number of Faces + Number of Vertices = Number of edges +2

ð Number of edges = 50.

Ans.(3)

Solution19: Volume = pR²H => Volume = 1000, R = 10

=> H = 3.18 cm => Curved surface area = 200sq.cm.

Now Old Total Surface Area = 828 cm;

New Total Surface Area = 4pR² + 2pRH = 1456sq.cm.

ð Required percent charged = (1455 – 828)/828 x 100% = 75.84%.

Ans.(4)

Solution20: Let the length and breadth of the rectangular field be a and b respectively.

According to the question,

Diagonal + a/2 = a + b, As diagonal of a rectangle= √(a² + b² )

=> √( a² + b²) + a/2 = a + b => a + b – a/2 = √(a² + b² )

=> a/2 + b = √(a² + b²)

On squaring both sides, we get

a²/4 + b² + 2 x a/2 x b = a² + b²

ð a²/4 + ab = a² + b² – b²

ð => a² – a²/4 = ab

ð => 3a²/4 = ab

ð => 3a/4 = b

ð a/b = 4/3

ð => b/a = 3/4

Ans=(4)

Solution21: We have volume of bath = 2 x 3 x 4 = 24 m³

Rate of flow = Area x speed = p/4 x (1/16)² x 6 m x 60 = 1.104m³/min.

Time = 24/1.104 = 21.8 minutes.

Ans. (1)

Solution22: The rise can be found as –

Vol. increased/area of tank = (48 x 36 x 15)/(72 x 60) = 6cm Ans.(4)

Solution23: Volume of hollow sphere = 4/3 p(7³ – 5³)

volume of solid sphere= 4/3 p R³ = 4/3 p 218 = 6.018 cm.

Ans.(1)

Solution24: Let R be the radius of the recast sphere. 4/3pR³

= 4/3 p(3³ + 4³ + 5³) =>R³ = 216 => R = 6.

Total surface area of 3 spheres

= 4p(3² + 4² + 5²) = 4p x 50 = 200p cm²

Surface area of the recast sphere = 4p x 6² = 144p cm2

% Reduction in area = (200p - 144p)/200p x 100% = 28%

Ans.(4)

Solution25: Let h be the removed part of the cone. Then 16/20 = h/(h + 48)

=>h = 192 cm.

The required volume = 1/3 x p(20)² (48 + 192) – 1/3 x p(16)²(192)

= 1/3p(400 x 240 – 256 x 192) = 49049 c.c.

Ans.(1)

Solution26: Radius of the sphere = 6/2 cm = 3cm.

Volume of the sphere = [4/3 p x (3)³] cm³ = (36p)cm³.

Let the radius of the circular end of the wire be r cm.

Length of the wire = 36 cm.

Volume of the wire = (p² x 36)cm³.

Now, volume of the wire = volume of the sphere

=> 36pr² = 36p => r² = 1 => r = ±1.

But, radius cannot be negative.

Hence, the radius of the wire is 1 cm.

Ans=(1)

Solution 27: Radius of the sphere = 3cm.

Volume of the sphere [4/3 p x (3)³] cm³ = (36p)cm³.

Let the rise in the water level be h cm.

Increase in the volume when the sphere is submerged

= (p x 6 x 6 x h)cm³ = (36ph)cm³.

This volume must be equal to the volume of the sphere

\ 36ph = 36p => h = 1 cm.

Hence, rise in the water level is 1 cm.

Ans=(3)

Solution 28: From DABC,

AC² = 24² + 32² = 1600

\ AC = 40 cm

so that AE = EC = 20cm <<Geometry Unsolved Fig 30>>

\ DE = √(25² - 20²)

= √(625 – 400) = √225 = 15

Hence, required area of figure

= Area of DABC + Area of DDAC

= ½ x 32 x 24 + ½ x 40 x 15 = 384 + 300 = 684 cm².

Ans=(1)

Solution 29: Clearly, each coin is a cylinder in which

r = 0.75cm = 75/100cm = 3/4cm,

and h = 0.2 cm = 2/10cm = 1/5cm.

Volume of each coin = pr²h

= [p x (3/4)² x 1/5]cm³ = (9p/80)cm³.

For the required cylinder, we have

Radius of the base (R) = (4.5/2)cm = 2.25cm,

and height (H) = 10cm.

Volume of the cylinder formed

= [p x (2.25)² x 10]cm³ = (405p/8)cm³.

Required number of coins = volume of the cylinder formed/volume

of 1 coin

= (405p/8 x 80/9p) = 450

Ans=(2)

Solution 30: Volume of iron = (440 x 260 x 100)cm³.

Internal radius of the pipe = 30cm.

External radius of the pipe = (30 + 5)cm = 35cm

Let the length of the pipe be h cm. Then, volume of iron in the pipe

= (external volume) – (internal volume)

= [p(35)²h - p(30)²h] cm³ = ph[(35)² – (30)²]cm³.

= (65 x 5) ph cm³ = (325ph)cm³.

\ 325ph = 440 x 260 x 100

ð length = h cm = (440 x 260 x 100 x 7/325 x 22) cm

ð 11200 cm = 112m.

Hence, the length of the pipe is 112m.

Ans=(4)

Solution31: If x be the exterior angle, then according to the question

x = 1/11 (180^o – x) => 11x =180o – x => 12x = 180^o or x = 15^o

Number of sides = 360^o/15^o = 24 (Remember formulae)

Ans.(2)

Solution32: Since the sum of the interior angles of a Pentagon

= (2 x 5 – 4) x 90^o = 540^o

So, x + x + 20 + x + 40 + x + 60 + x + 80 = 540

ð 5x + 200 = 540 => 5x = 340 or 68^o

Ans. (3)

Solution33: D BEC ~= D BDC (Draw the figure yourself)

\ ÐB = ÐC = (180 – 36)/2 = 72^o each.

Ans. (4)

Solution34: Let AB = x, BC = y

so that, AE = EF = FB = x/3

Area of DCEF = Area of DBCE – Area of DBCF

= ½ x 2x/3 x y –1/2 x x/3 x y = xy/6

Also, Area of rect. ABCD = xy

Hence, reqd. ratio = (xy/6)/xy = 1/6.

Ans=(1)

Solution35: Let x be the exterior angle, so (180^o – x) is the interior angle

^ð (180^o – x) – x = 140^o => 2x = 40^o

\ x =20^o

Sum of the exterior angles of n sides polygon is 360^o

\Number of sides = 360^o/20^o = 18.

Ans.(1)

Solution36: Let the sides be 3n and 2n

Now, (6n – 4)/3n x 2n/(4n – 4) = 10/9

so 18n – 12 = 20n – 20 or 2n = 8 or n = 4

\ sides are 12, 8.

Ans.(3)

Solution37: From DAFD,

x + 40^o + y = 180^o

or x + y = 140^o …(i)

Again, from Ds BAC and ECD,

2x + t = 180^o …(ii) <<Geometry Unsolved Fig 31>>

and 2y + l = 180^o …(iii)

Adding (ii) and (iii),

2(x + y) + t + l = 360^o

or 2 x 140 + t + l = 360^o [Using (i)]

or t + l =80^o (iv)

Finally, t + z + l = 180^o or 80^o + z = 180^o [Using (iv)]

or z = 100^o

Ans=(1)

Solution38: Draw the diagram yourself

As RS = SQ (given)

\ ÐQRS = ½ (180^o – 120^o) = 30^o

\ ÐQRP = 180^o – 30^o = 150^o

Hence ÐQPS = ½*(180^o – 150^o) = 15^o (as PR = RQ).

Ans.(1)

Solution39: Draw the diagram yourself

OB = OC (As O is circumcentre), so ÐOBC = b

Also, ÐBOC = 2a (angle subtended by chord BC at centre is twice of in alternate segment)

In D BOC, we have b + 2a + b = 180^o

\ a + b =90^o.

Ans.(4)

Solution40: Draw the diagram yourself

As DAOB and DCDO are similar

\ AB/CD = AO/DO => 5/4 = 10/DO

\ DO = 8 cm.

Ans.(2)

Solution41: Draw the figure yourself

Clearly ÐOSD = ÐSOD => SD = OD

Further ÐSDO = 90^o

\ DOSD is isosceles right angled.

Ans.(3)

Solution42: Draw the figure yourself

PR2 = PQ² + QR²

PR² = 6² – 7² = 85

Again PR2 = PS² + RS²

\ 85 = PS² + 2²

ð PS² = 81 or PS = 9

Ans.(4)

Solution43: Draw the figure yourself

Let OP = x

In DOPB: 4² + x² = r². In DOQD: 3² +√ (1 + x²) = r²

\ 4² + x² = 3² + (1 + x)². i.e. x = 3. \ 2r = 2 4² – x² = 10

Ans.(3)

Solution44:

AB = √ (15² + 20² )= √(225 + 400) = 25

Area of DABC = ½ x 15 x 20 = 150 sq. units.

\1/2 x 25 x CD = 150 => CD = 12 units.

From DADC, AD = √(15² + 12² )= Ö81 = 9 units

\ BD = 16 units.

From DADC, S = AC + CD + DA/2 = 15 + 12 + 9/2 = 18

\ radius = D/S = (½ x 9 x 12)/18 = 3

From DBCD, S = BD + DC + CB/2 = 16 + 12 + 20/2 = 24

\ radius = D/S = (1/2 x 16 x 12)/24 = 96/24 = 4

\ Required distance PQ = 3 + 4 = 7.

Ans=(3)

Solution45: Draw the figure yourself

In a right triangle mid point on hypotenuse is always equidistance from each vertex i.e. BM = MC = AM

But BC = 6² + 8² = 10

\ AM = 5 cm.

Ans.(1)

Solution46: CD = CE. ((Draw the figure first)

Ans.(1)

Solution47 Draw the figure yourself

ÐBCO = 20^o and OB = BC

So ÐBOC = 20^o

ÐABO = 20^o + 20^o = 40^o

(Exterior angle ABO = Sum of two Interior opposite angles)

\ ÐOAB = 40^o (AO = BO)

Thus ÐAOB = 180^o – (40^o + 40^o) = 100^o

\ ÐAOD = 180^o – (100^o + 20^o) = 60^o

Ans.(2)

Solution48: Draw the figure yourself

If a is the side of the square then area of the square = a²…..(i)

DABE + DECF + DADF

½ x a a/2 + ½ x a/2 x a/2 x ½ x a x a/2 = 5a²/8

\ Area of DAEF = a² – 5a²/8 = 3a²/8…….(ii)

From (i) and (ii), ratio of areas of DAEF : square ABCD

3a²/8: a² = 3:8.

Ans.(1)

Solution49 Draw the figure yourself

: OM = OP – MP = r – 2

\ OA² = OM² + AM²

r² = (r – 2)² + 4² or r = 5 cm.

Ans.(3)

Solution50: Draw the figure yourself

Since ABC is an isosceles triangle, so angle ACB is also 15. Hence

ÐCBA = 150^o

and exterior ÐCBA=30^o

\number of sides = 360^o/30^o = 12.

Ans.(2)

Soluition51 Draw the figure you150^o yorrself

AB² + AC² = 2(BD² + AD²) (Remember this formulae)

ð AB² + 36 = 2(25 + 25)

ð AB² = 100 – 36 = 64

\ AB = 8.

Ans.(4)

Solution52: AB = BE (As ABE is isosceles triangle)

\ ½ x AB x BE = 7

ð 1/2AB² = 7

ð AB = Ö14

EC = 3BE (Given)

\ BC = 4BE = 4AB

Now area of ABCD = AB x BC = AB x 4AB = 4AB² = 4 x 14 = 56cm².

Ans=(4)

Solution53: Since the area is 16 times as the original, so the sides are Ö16 =times i.e. 4 times the sides of the given triangle. (Remember properties of similar triangles)

Hence the required length of the side

= 4 x 5 = 20 cm

Ans.(2)

Solution54: Draw the figure yourself

Draw AD perpendicular on BC.

AB=BC=AC (Given)

And BP=1/3BC (given)=1/3 AB

BD=DC=1/2BC (Remember properties of equilateral triangles)=1/2AB

So PD=BD-BP=1/2BC-1/3BC

=1/6BC=1/6AB

Now in DAPD,

AP²=AD²+PD² (Pythagoras theorem)

Or AP²=(AB²- BD²)+(1/6AB) ² (Pythagorus theorem in DABD)

Or AP²=(AB² -1/4AB² +1/36AB²

Or AP² =7/9AB²

So AP²/ AB²=7/9

Ans (1)

Solution55:

<<Geometry Unsolved Fig 32>>

Let O and O’ be the centers of the circles of radii 20 cm and 15 cm,

respectively and let PQ be their common chord. We have,

OP = 20cm, O’P = 15cm, OO’ = 25cm.

Let OL = x, \LO’ = 25 - x

Again let PQ = y \ PL = y/2

In right triangle OLP

OL = √ (OP² – LP² )= 20² – (y/2)² => x² = 400 – y²/4 …(i)

In right triangle O’LP, O’L = √(15² – y²/4)

(25 – x)² = 225 – y²/4 …(ii)

By (i) And (ii) we get

x² – 625 + 50x – x² = 400 – y²/4 – 225 + y²/4

=> 50x =625 + 175 => x = 800/50 = 16

From equation (i)

256 = 400 – y²/4 => y²/4 = 400 – 256 = 144 => y² = 144 x 4

ð y = 12 x 2 = 24cm.

Ans=(1)

Solution56: ÐBAC = 90^o [angle in a semicircle].

In DABC, we have

ÐABC + ÐACB + ÐBAC = 180^o

ð 38^o + ÐACB + 90^o = 180^o

ð ÐACB = (180^o – 128^o) = 52^o.

\ ÐBAT = ÐACB = 52^o [angles in alternate segments].

Ans=(1)

Solution57: Let BOQ intersect the circle at C.

Join CA.

ÐACB = ÐPAB = 58^o [Ðs in alternate segments.]

Now, in DACB, we have

ÐCAB + ÐABC + ÐACB = 180^o

ð 90^o + ÐABC + 58^o =180^o

ð ÐABC = (180^o – 148^o) = 32^o.

\ ÐABQ = 32^o [because ÐABQ = ÐABC].

Also, ÐPAB = ÐABQ + ÐAQB [Exterior-Angle Theorem]

ð 58^o = 32^o + ÐAQB

ð ÐAQB = (58^o - 32^o) = 26^o.

Hence, ÐABQ = 32^o and ÐAQB = 26^o.

Ans=(1)

Solution 58: Using BPT theorem, we have

AD/BD = AE/CE => (4x – 3)/(3x – 1) = (8x – 7)/(5x – 3)

ð (4x – 3)/(5x – 3) = (8x – 7)/(3x – 1)

ð 20x² – 27x + 9 = 24x² – 29x + 7

ð 4x² – 2x – 2 = 0 => 2x² – x – 1 = 0

ð (2x + 1)(x – 1) = 0

ð x = - ½ or x = 1.

But, x = -1/2 makes AD = [4 x (-1/2) - 3)] cm = -5 cm.

Since distance can never be negative, therefore x ¹ –1/2.

Hence, x = 1

Ans=(3)

Solution 59: For similar triangles,

Ratio of perimeters = Ratio of Proportional sides

ð 30/20 = 15/X => X =10cm

Ans=(4)

Solution 60 : Use similar triangle property: (Draw the figure yourself)

Height of tower : Height of stick = Shadow of tower : Shadow of

stick

X : 12 cm = 40 m : 8 cm => X = 60 m.

Ans-(2)

61(b). In an equilateral triangle, the centroid, as well as the in center and circumcenter, is located two -thirds of the distance from the vertex to the opposite side.

Then

2(x)2 = x2 + 32 => 3x2 = 9 => x = . This makes the area ½(2√3)³ = 3√3

<<Geometry Unsolved Fig 33>>

62(b). Placing the runners on a number line, we see that there is a spread of 20 seconds from first to last, that Jo was at the end, Sam in second 19 second ahead of Jo and one second behind Chris, the winner. Pat then was in third 12 seconds behind Sam.

63(b). In the figure, we see that

CE = DE = 3√2.

Also FD = BC + CE = 12 + 3√2 and AF = 5 – 3√2.

Thus the distance we want,

AD = <<Geometry Unsolved Fig 34>> = <<Geometry Unsolved Fig 35>> = 16.26

<<Geometry Unsolved Fig 36>>

64(a). Points on both circles must satisfy both equations, so both (1) (x – 1)2 + (y + 2)2 = 9 and (2) (x + 3)2 + (y – 1)2 = 16. Squaring both, and subtracting gives, (x2 + 6x + 9 + y2 – 2y + 1) – (x2 – 2x + 1 + y2 + 4y + 4) = 16 – 9, or simply 8x – 6y = 2 => 4x – 3y = 1.

65(d). The new length and width are 400 + 2w and 300 + 2w, so new area is

(400 + 2w)(300 + 2w) = 2(400)(300) =>

120000 + 1400w + 4w2 = 240000 =>

w2 + 350w – 3000 = 0 =>

w = <<Geometry Unsolved Fig 37>>

66(c). A good guess, since the hypotenuse is 13, would be the 5 – 12 – 13 right triangle. The area is indeed 30 and the perimeter is 30. If you did not guess this, you could use the equations a2 + b2 = 132 = 169 and 1/2 ab = 30 => 2ab = 120. Adding and subtracting the second equation from the first gives a2 + 2ab + b2 = 169 + 120 = 289 and a2 – 2ab + b2 = 169 – 120 = 49. Thus (a + b)2 = 289 => a + b = 17 and (a – b)2 = 49 => a – b = 7. These last two imply that a = 12 and b = 5.

67(a). The simplest way to find this is to put a rectangle around the given triangle and then subtract off the areas of the right triangles that surround it. (See Figure). The area of the entire rectangle is 126, but after subtracting the areas of the 3 right triangles, whose areas are 54, 3, and 42, we are left with an area of 27.

<<Geometry Unsolved Fig 38>>

68(e). We want the two angles, from vertical or the 12 on the clock, going clockwise, to be equal for both hands. The minute hands starts at an angle of zero and rotates at 1/60 revolutions per minute. The hour hand, which begins at 1/4th of a revolution, moves at 1/720 revolutions per minute. (It takes 12 hours times 60 minutes to make one complete revolution.) Setting these equal we have

(1/ω) t= ¼ + 1/720 t => 12t = 180 + t => 1 lt = 180 => t = <<Geometry Unsolved Fig 39>>

69(a). Let the sides have length x and 4.5.

Then we have 4.5x = 2(2x + 9), so 0.5x = 18 => x = 36.

70(b). The circumference of the can us cπ = 2πr, so the radius is r = c/2. This side of the square piece is the same as the circumference. The volume is

V = πr2h = . <<Geometry Unsolved Fig 40>>

71(a). By the power of a point formula, we know that x2 = 1.2 × 7.5 = 9 => x = 3

<<Geometry Unsolved Fig 41>>

72(d). The area of a rectangle is base times height. In the figure we see that the base is √2 + 1 and the height is √2/2 , so the area is A = Bh

= (√2 + 1)√2/2 = (2 + √2)/2

<<Geometry Unsolved Fig 42>>

73(c). z + y < 7 => 0 < z < 7 and 0 < y < 7. x + z > 10 and 0 < z < 7 => x ³ 3. x ³ 3 => y ³ 3. y ³ 3 and 0 < y < 7 => y = 3, 4, 5, 6.

74(e). In the figure we see that the area of the shaded region is x(a + b). The area of the entire triangle is
1/2(2x)(2a +2b) = 2x(a + b).

Since the area of the shaded region is one -half the area of the entire triangle, the ratio of the shaded region to the unshaded region is 1:1.

<<Geometry Unsolved Fig 43>>

75(c). The three configurations are shown here. In each case the area must be the sum of the two areas, since there is no overlap. The other statements are false, depending on the arrangement.

<<Geometry Unsolved Fig 44>>

76(a). The center of the circumscribed circle must be the same distance from each vertex. The perpendicular- bisector of one side is equidistant from two vertices, so the point of intersection of two perpendicular-bisectors is equidistant from all 3 vertices.

77(a). The segments joining the centers will be formed by the radii, so the sides of the triangle are 3, 4, and 5 units long, making it a 3-4-5 right triangle with are 6.

78(a). There are 2 ways to get to point B, so we can start at B and then double our answer. A tree diagram shows there are 8 ways to get from B to S without visiting any point twice.

<<Geometry Unsolved Fig 45>>

79(c). The radius is 7, so the circumference is 14π. Divide this by 5 to get 2.8π for the arc length.

80(e). The perimeter is 2a + 2b = p => a + b = p/2.

Squaring we have (a + b)2 = (p/2)² => a2 + 2ab + b2 = (p/4)².

But the diagonal c2 = a2 + b2, so subtracting we get 2ab = (p/4)² – c² => ab = (p² – 4c²)/8 .

81(d). The volume of the original solid is 200 cubic units. When two sides are doubled, the volume goes to 800. The original surface areas was 2.52 + 4.(5.8) = 210. The new surface area will be 2.102 + 4(10.8) = 520. The ratio of the volumes is 520/210 = 52/21 =2(10/21).

82(b). In the figure was see that triangle MBN is similar to triangle ORN, so BM/BN = OR/RN => 3/2 = 7.5/RN => RN = 5. Similarly, triangle NQO is similar to triangle PDO, so NQ/QO = PD/DO => 7.5/5 = 5.4/Do => DO = 3.6. Thus the length of the rectangle is 10.6.

<<Geometry Unsolved Fig 46>>

83(c). Draw in diagonal AC and look at triangle APC. The length of AC is 2√2. so that makes triangle APC an isosceles right triangle sna the altitude is √2.

<<Geometry Unsolved Fig 47>>

84(a). The old area was 9w2 and the perimeter was 20w. The new area will be 18w2. The perimeter stays the same so the new length and width, L and W are such that L = 10w – W, and (10w – W)W = 18w2. This is quadratic in W, so w2 – 10wW – 18w2 = 0

=> W = 10w ± √(100w² – 72w²)/2. This simplifies to W = (5 ± √7) => L = <<Geometry Unsolved Fig 48>>

. Taking the length

to be the longer side, the ratio is (5 + √7)/(5 – √7).

85(e). Each of the five statements involves only three statements. We will call these C for the Car is locked, T for the Tickers are stolen, and W for the wallet is stolen. A truth table of all 8 possibilities is shown here:

Since we are told that exactly one of the statements is false and the rest true, this only happens in row 4, where Ed makes the false statement and the others make true ones.

86(d). Opposite angles in an inscribed quadrilateral must be supplementary, since the arcs they subtend form the entire circle. Thus the measure of the angle at vertex C must be 144 degrees.

87(e). The surface area to paint is 2(8.20) + 2(8.14) – 2(3.7) – 4(6.5) = 382. Since we need two coats, we need enough paint to cover 764 sq feet, or 1.91 gallons.

88(d). In the figure we see the slopes and lengths of the sides of this triangle. Since two of the slopes are negative reciprocals, and since the converse of the Pythagorean Theorem applies, we do have an isosceles right triangle.

<<Geometry Unsolved Fig 49>>

89(d). The four possible figures are shown here. Note that the problem does not say that one of the angles of the quadrilateral has to be a right angle, only that two of the sides meet at a right angle. They allows the one concave polygon.

<<Geometry Unsolved Fig 50>>

90(c). The sides of the new triangle will be 8, 10 and 12. Thus the perimeter is 30 and the semi perimeter is 15. Using Heron’s Formula, the area is

√s[(s – a)(s – b)(s – c)]= √(15 x 7 x 5 x 3) = 15√7.

91(e). Each angle of a regular n-sided polygon is (n – 2)180/n, so in this case each angle is (12 – 2)180/12 = 150.

92(c). Since each of the triangles are similar, ratios of corresponding sides leads us to conclude that x6 = 6, so x = 61/6 . The areas of the triangles are 1/2 . 1 . x . sin(60o) and 1/2x5 . x6 . sin(60o) so the ratio is of their areas is

<<Geometry Unsolved Fig 51>> = x10 = (6^1/6)¹⁰ = 6^5/3.

<<Geometry Unsolved Fig 52>>

93(e). The cube can be flattened to look like the figure, showing that G is on the opposite side of the block from A.

<<Geometry Unsolved Fig 53>>

94(c). The radius of the largest circle is (a + b)/2 while the radii of the two smaller circles are a/2 and b/2. Thus the area of the shaded region is

π/2{ ((a + b)/2)² - (a/2)² - (b)/2)²}

This simplifies to

π/2{ (a² + 2ab +b² - a² - b²)/4} = πab/4

95(a). If the diameter of the wheel is 70cm, the circumference is 70πcm. Dividing this into 120km. 1000m/km. 100cm/m = 120000000cm yields about 54567.4 revolutions. This is pretty close to choice a.

96(b). The volume of a regular solid is proportional to one edge cubed. The surface area is proportional to the side squared, so the surface area is proportional to the volume raised to the two-thirds power, SA= kV2/3 . If the volume is doubled, the new surface area will be (2V )2/3 = 22/3V2/3.

97(a). As shown in the figure, the midpoint of the diameter becomes the vertex of the cone and points A and B meet on the circumference of the base of the cone. The radius OA becomes the slant height of the cone and the half circumference of the semicircle becomes the circumference of the cone. So 6π = 2πr , where r is the radius of the cone. The radius is then 3, the slant height 6, so the altitude is (√62 – 32) = √27 = 3√3. Thus the volume of the cone is

<<Geometry Unsolved Fig 54>>

<<Geometry Unsolved Fig 55>>

98(d). Through the center of the circle O, drop the perpendicular to the tangent line. This will have length 18, and will include the radius of length 13 and segment OB of length 5. By the Pythagorean theorem, AB = 12 and the chord has length 24, since the segment perpendicular to the chord also bisects it

<<Geometry Unsolved Fig 56>>

99(e). The original diagonal AC has length √(11² + 13²) = √290. The new diagonal, AE has length 1.5√290, so the new side of the rectangle has length √[(1.5√290)² - 11²] = √531.50 = 23.054. Thus the new area is approximately 11×23.054 = 253.597.

<<Geometry Unsolved Fig 57>>

100(b). First, several simplifying assumptions need to be made. First, we need to assume that the 15 miles is to the center of the mountain, the point directly below the peak. Next we need to decide whether the mountain is 15 miles from the lamp or Nikki. (It is not clear from the problem statement.) Let’s assume it is from the lamp. Then, using similar triangles, we have 10/250 = h/15.5280 => h = 10.15.5280/250 = 3168. Choice b is the only one even close. f we assume that the mountain is 15 miles from where Nikki is standing we get 10/250 = h/15.5280 +250 => h = 10 (15 × 5280 + 250)/250 = 3178. This equals choice b.

<<Geometry Unsolved Fig 58>>