Acme Mathematics 8 201Thus, x = 100°, y = 40°and a = 80° Example 2 : Find the value of x, ∠B and y in the adjoining figure. Solution : Here, AB = BC, ∠ACB = 70° Now, (i) ∠BAC = ∠ACB → base angle. y = 70° → ∠ACB = 70° given (ii) ∠A + ∠B + ∠ACB = 180° → Sum of interior angles of ∆ABC or, y + ∠B + 70° = 180° or, 70° + ∠B + 70° = 180° or, ∠B + 140° = 180° or, ∠B = 180 – 140° or, ∠B = 40° (iii) ∠DCB = ∠ABC → Alternate angles and AB||ED or, x = 40° → ∠B = 40°Classwork1. Match the following.Triangle is a 180°The six parts of triangle are Greater than third side. On the basis of sides triangle are Closed figure.Sum of all angles of a triangle 3 angles and 3 sidesThe angles opposite to equal sides scalene, equilateral and isoscelesSum of any two sides is are equal2. Write 'True or 'False' for the following statements.(a) A triangle can have 2 acute angles.(b) A triangle can have 2 obtuse angles.(c) All the three angles of a triangle can be less than 60°.(d) A triangle can have angles 50°, 60° and 80°.(e) A triangle can have sides 6 cm, 6 cm and 7 cm.A By70°x E C D
202 Acme Mathematics 8Exercise 5.31. Find the value of x, y, z, a, b and c in each of the following triangles.(a)B C68°63°Ax(b)30°A60° x yB C D(c)B Cx 2x3xA(d)B C Dy 60°90°xFA30°(e)B Cy°50°Ax°(f)B C3x 110°2xAD(g)E B C D130° yxA120°(h)BEF C2xyxAD3x(i)aBDyxA60°89°CE(j)B CEDA80° xy20°(k)B C Ex yA D (l)B C2xxyA50°(m)DECAB80° a70°c b(n)cba40°(o)60°2a axFAABBDDECCAD
Acme Mathematics 8 2032. Each of the two equal angles of an isosceles triangle are two times its third angle. Find the measure of all the angles of the triangle. 3. In the given figure, if ∠D = 35°, ∠DEA = 120° and ∠A = 20°. Find the measure of ∠B, and ∠BCA. 4. Verify experimentally that the sum of the length of any two sides of a triangle is greater than the length of the remaining side.5. Verify experimentally that, if two angles are equal then the sides opposite to them are equal.6. Draw three triangles of different size where AC>BC. Now complete the table and write your conclusion.Fig. AC BC ∠ABC ∠BCA Result(i)(ii)(iii)7. (a) Find the value of x from given figure.(b) Find the value of y? 8. Study the given figure then write the answer of the following questions:(a) What is the sum of ∠ABC, ∠BAC and ∠BCA? Write reason. (180°)(b) What is the degree of ∠ABC? Find out. 9. Study the given figure of triangle and answer the following questions.(a) Find the value of x.(b) What types of triangle is that ∆ABC?B C DEAAB120° DC x yxA x 2x3xCBAB C3x3x+10°4x–30°
204 Acme Mathematics 810 .(a) What is the measurement of the angle of an equilateral triangle?(i) 90° (ii) 60° (iii) 180° (iv) 45°(b) Find the value of 'x' from the given figure. x2x 3xRP Q11. (a) When two lines are parallel, then co-interior angles are .......(i) Equal (ii) Supplementary(iii) Complimentary (iv) Complete turn(b) Find the value of unknown angle.y z50°120°x E ADBC(c) In the figure along side AB|| CD, then show that ∠EGF = 90°.AGE BDy yxxF C
Acme Mathematics 8 205B. Quadrilaterala. RhombusThe rhombus is a quadrilateral. It's all sides are equal and opposite sides are parallel. In the figure ABCD is a rhombus. Its all sides are equal. i.e. AB = BC = CD = DA Its opposite sides are parallel. i.e. AB||DC and BC||AD. Its opposite angles are equal i.e. ∠A = ∠C and ∠B = ∠D Its diagonals are not equal. i.e. Diagonals AC ≠ diagonal BD. Its diagonals bisect each other at 90°. i.e. AO = OC and BO = OD. AC⊥BD. Diagonals of a rhombus bisect the angle at the vertex.i.e. ∠BAC = ∠DACExperimental verification of properties of Rhombus(i) Diagonals of a rhombus are not equal.Draw three rhombus ABCD of different sizes by using set-square.(i) ADBCA (ii)BCD(iii) ADBCIn the above rhombus AC and BD are diagonals. Now, measure the length by using scale and angles by protractor and complete the table.Fig. AC BD Result(i) 3.2 cm 2.7 cm Diagonal AC ≠ Diagonal BD(ii) 2.3 cm 2.7 cm Diagonal AC ≠ Diagonal BD(iii) 2.4 cm 3.0 cm Diagonal AC ≠ Diagonal BDHence, Diagonals of a rhombus are not equal.BACDBOACD
206 Acme Mathematics 8(ii) Diagonals of a rhombus bisect each other at right angle.Draw three rhombus ABCD of different sizes by using set-square.(i) ADBCOA (ii)BCDO(iii) ADBCOIn the above rhombus AC and BD are diagonals. O is the point of intersection of diagonals AC and BD. ∠AOB, ∠BOC, ∠COD and ∠DOA are angles made by the diagonals. AO, OC, BO and OD are the parts of diagonals AC and BD. Now, measure the length by using scale and angles by protractor and complete the table.Fig. AO OC BO OD ∠AOB ∠BOC ∠COD ∠DOA Result(i) 1.6 cm 1.6 cm 1.3 cm 1.3 cm 90° 90° 90° 90° AO = OC, BO = OD and ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°(ii) 1.1 cm 1.1 cm 1.3 cm 1.3 cm 90° 90° 90° 90°(iii) 1.2 cm 1.2 cm 1.4 cm 1.4 cm 90° 90° 90° 90°Hence, diagonals of a rhombus bisect each other at right angle. Note: Verify the result in the figures (ii) and (iii) yourself(iii) Diagonals of a rhombus bisect the angle at the vertex.(i) ADBCOA (ii)BCDO(iii) ADBCOIn the above rhombus AC and BD are diagonals. ∠ADB and ∠CDB are the parts of vertex ∠ADC. ∠DCA and ∠BCA are the parts of vertex ∠DCB. ∠CBD and ∠ABD are the parts of vertex ∠ABC. ∠DAC and ∠BAC are the parts of vertex ∠BAD. Now, measure the angles by protractor and complete the table.
Acme Mathematics 8 207Figure Vertex Angle Parts of the vertex angle Result(i)∠ADC = 100° ∠ADB = 50° ∠CDB = 50° ∠ADB = ∠CDB∠DCB = 80° ∠DCA = 40° ∠BCA = 40° ∠DCA = ∠BCA∠ABC = 100° ∠CBD = 50° ∠ABD = 50° ∠CBD = ∠ABD∠BAD = 80° ∠DAC = 40° ∠BAC = 40° ∠DAC = ∠BACHence, The diagonals of a rhombus bisect the angle at the vertex.Note: Verify the result in the figures (ii) and (iii) yourself.b. TrapeziumThe trapezium is a quadrilateral with one pair of parallel opposite sides. The parallel sides of a trapezium are called bases and the non-parallel sides of a trapezium are called legs. Sometimes the parallelogram is also called a trapezium with two parallel sides.From the above figure, we can see, sides AB and CD are parallel to each other whereasAC and BD are non-parallel sides. Also, ‘h’ is the distance between the two parallel sides which demonstrates the height of the trapezium.Shape of TrapeziumA trapezium is a closed shape or a polygon, that has four sides, four corners, four vertices and four angles. Only one pair of opposite sides of a trapezium are parallel to each other.There are many real-life examples of trapezium shape that we can see around us. For example, a table whose surface is shaped like a trapezium.Types of TrapeziumThe trapezium is categorized into three types: Isosceles Trapezium Scalene Trapezium Right TrapeziumIsosceles trapeziumIf the legs or the non-parallel sides of the trapezium are of equal length, then it is called an isosceles trapezium. In the figure, ABCD is a isosceles trapezium. Where AD=BC.Scalene TrapeziumA trapezium with all the sides and angles of different measures is called a scalene trapezium. In the figure ABCDis a scalence trapezium.CAE b2b1FhDBAD CBAD CB
208 Acme Mathematics 8Right TrapeziumA right trapezium is a trapezium that has at least two right angles, adjacent to each other. In the given figure ABCD isright trapezium, where angles A and D are right angles.Irregular TrapeziumWe know, a trapezium has exactly one pair of parallel sides and the other two sides are non-parallel. Now a regular trapezium will have the other two non-parallel sides equal, whereas an irregular trapezium will have the other two non-parallel opposite sides, unequal. In the given figure ABCD is a irregular trapezium.Properties of TrapeziumSome important properties of a trapezium are as follows: In trapezium, exactly one pair of opposite sides are parallel The diagonals intersect each other The non-parallel sides in the trapezium are unequal except in isosceles trapezium The line that joins the mid-points of the nonparallel sides is always parallel to the bases or parallel sides which is equal to half of the sum of parallel sides i.e mid-segment = AB + CD 2Where AB and CD are the parallel sides or bases In isosceles trapezium, the legs or non-parallel sides are equal The sum of the interior angles of the trapezium is equal to 360 degrees. i.e., ∠A + ∠B + ∠C + ∠D = 360° The sum of the two adjacent angles is equal to 180°. This means that the two adjacent angles are supplementary.Experimental Verification of Properties of Trapezium(ii) The sum of the interior angles of the trapezium is equal to 360 degrees.Draw three trapezium ABCD.Fig. (i) Fig. (ii) Fig. (iii)AAAB C B C B CDDDAB CDAB CDMid segment
Acme Mathematics 8 209In the given figures four angles are ∠A, ∠B, ∠C and ∠D.Now, measures the ∠A, ∠B, ∠C and ∠D and complete the table.Figure ∠A ∠B ∠C ∠D Result(i) 100° 80° 62° 118° ∠A + ∠B + ∠C + ∠D = 360° (ii) 90° 90° 60° 120° ∠A + ∠B + ∠C + ∠D = 360°(iii) 68 112° 113° 67° ∠A + ∠B + ∠C + ∠D = 360°Hence, the sum of the interior angles of the trapezium is equal to 360 degrees.(ii) The sum of the two adjacent angles made by nonparallel sides is equal to 180°Draw three trapezium ABCD.Fig. (i) Fig. (ii) Fig. (iii)AAAB C B C B CDDDIn the given figures four angles are ∠A, ∠B, ∠C and ∠D.Now, measures the ∠A, ∠B, ∠C and ∠D and complete the table.Figure ∠A ∠B ∠C ∠D Result(i) 100° 80° 62° 118° ∠A + ∠B = 180°∠C + ∠D = 180° (ii) 90° 90° 120° 60° ∠A + ∠B = 180°∠C + ∠D = 180°(iii) 68 112° 113° 67° ∠A + ∠B = 180°∠C + ∠D = 180°Hence, the sum of the two adjacent angles made by nonparallel sides is equal to 180°C. KiteA kite is a quadrilateral having closed, flat geometric shape and whose pairs of adjacentsides are equal.Properties of a Kite1. Two pairs of adjacent sides are equal.In the kite ABCD, AB = AD and BC = DC 2. The two opposite angles where the adjacent unequal sides meet are equal.o
210 Acme Mathematics 8In the kite ABCD, AB, BC and AD, DC are two pairs of adjacent unequal sides so ∠ABC = ∠ADC. 3. The two diagonals are perpendicular to each other with the longer diagonal bisecting the shorter one.In the kite ABCD, AC is longer diagonal and BD is shorter diagonalso AC ⊥ BD and BO = OD . 4. Vertex angle are bisected by diagonal.In the kite ABCD, ∠BAD and ∠BCD are vertex angle. AC is diagonal so ∠BAC = ∠DAC and ∠BCA = ∠DCA.A kite can be a rhombus with four equal sides or a square having four equal sides and each angle measuring 90°.Experimental Verification of Properties of Kites(i) The two opposite angles where the adjacent unequal sides meet are equalDraw three kites ABCD of different sizes.Fig. (i) Fig. (ii) Fig. (iii)AAACCCBBBDDDIn the kite ABCD side AB, BC and AD, DC are adjacent unequal sides. ∠ABC and ∠ADC are two opposite angles.Now, measures the ∠ABC and ∠ADC and complete the table.Figure ∠ABC ∠ADC Result(i) 99° 99° ∠ABC = ∠ADC(ii) 109° 109° ∠ABC = ∠ADC(iii) 111° 111° ∠ABC = ∠ADCHence, The two opposite angles where the adjacent unequal sides meet are equal
Acme Mathematics 8 211(ii) The two diagonals are perpendicular to each other. The longer diagonal bisect the shorter diagonal.Draw three kites ABCD of different sizes.Fig. (i) Fig. (ii) Fig. (iii)AA A OOOCC CBBBDDDAC and BD are two diagonals. AC is longer than BD. O is intersection of AC and BD.Now, measures the ∠AOB and ∠AOD. Measure BO and OD and complete the table.Figure ∠AOB ∠AOD BO OD Result(i) 90° 90° 1.1 cm 1.1 cm ∠AOB = ∠AOD = 90° so diagonal AC and BD are perpendicular. BO = OD(ii) 90° 90° 1 cm 1 cm ∠AOB = ∠AOD = 90° so diagonal AC and BD are perpendicular. BO = OD(iii) 90° 90° 1.6 cm 1.6 cm ∠AOB = ∠AOD = 90° so diagonal AC and BD are perpendicular. BO = ODHence, the two diagonals are perpendicular to each other with the longer diagonal bisecting the shorter one.(iii) The longer diagonal bisects the vertex angle.Draw three kites ABCD of different sizes.Fig. (i) Fig. (ii) Fig. (iii)AA A OOOCC CBBBDDD
212 Acme Mathematics 8AC and BD are two diagonals. AC is longer than BD. O is intersection of AC and BD. ∠BAD and ∠BCD are vertex angle.Now, measures the ∠OAB, ∠OAD and ∠OCB, ∠OCD and complete the table.Figure Vertex Angle Parts of the vertex angle Result(i)∠BAD = 102° ∠OAB = 51° ∠OAD = 51° ∠OAB = ∠OAD∠BCD = 60° ∠OCB = 30° ∠OCD = 30° ∠OCB = ∠OCD(ii) ∠BAD = 80° ∠OAB = 40° ∠OAD = 40° ∠OAB = ∠OAD∠BCD = 64° ∠OCB = 32° ∠OCD = 32° ∠OCB = ∠OCD(iii) ∠BAD = 104° ∠OAB = 52° ∠OAD = 52° ∠OAB = ∠OAD∠BCD = 66° ∠OCB = 33° ∠OCD = 33° ∠OCB = ∠OCDHence, the longer diagonal bisects the vertex angle.RememberProperties ParallelogramRectangle Square Rhombus1. Sides• Opposite sides are equal. • Adjacent sides are equal. × × • Opposite sides are parallel. 2. Angles • Opposite angles are equal. • Adjacent angles are equal. × × • All angles are right angles. × ×3. Diagonals • Diagonals are equal.× × • Diagonals bisect each other. • Diagonals bisect each other at right angle. × ×
Acme Mathematics 8 213Solved ExamplesExample 1 : Two adjacent angles of a parallelogram are in the ratio 2:4. Find all the angles of the parallelogram. Solution : Let ABCD is a parallelogram. ∠A and ∠B are its adjacent angles. Suppose ∠A = 2x° and ∠B = 4x° Then,∠C = ∠A = 2x° ∠D = ∠B = 4x° → Opposite angles of a parallelogram ABCDNow, ∠A + ∠B + ∠C + ∠D = 360° → Sum of interior angles of a quadrilateral.or, 2x° + 4x° + 2x° + 4x° = 360° or, 12x° = 360° or, x = 360°12or, x = 30° Hence, angles are, ∠A = 2x° = 2 × 30° = 60°∠B = 4x° = 4 × 30° = 120° ∠C = ∠A = 60°∠D = ∠B = 120° Example 2 : Calculate the value of x, y, a and b from the given parallelogram ABCD.Solution : Here, AB||DC Now, (i) 30° + y + 20° = 180° → Sum of interior angles of ΔABC or, y + 50° = 180° or, y = 180° – 50° or, y = 130° (ii) b = 30° → Alternate angle, AB||DC. (iii) a = 20° → Alternate angle AD||DC. (iv) x = y → Opposite angle of parallelogram ABCD. or, x = 130° A BD C2x2x4x4xA30°ba yx20°BD C
214 Acme Mathematics 8Example 3 : Calculate the length of CD from the given parallelogram ABCD. Solution : In the parallelogram ABCD. AB = CD or 4x – 1 = 2x + 5 or, 4x – 2x = 5 + 1 or, 2x = 6or, x = 62or, x = 3 Now, length of CD = (2x + 5) cm = (2 × 3 + 5) cm = 11 cmClasswork1. State whether the following statements are 'true' or 'false'.(a) All the angles of rhombus are tight angles.(b) Opposite sides of a trapezium are equal.(c) Diagonals of a rhombus bisect each other at right angle.(d) Diagonals of a square are equal and bisect each other at right angles.(e) If the diagonals of a rhombus are equal then it is a square.(f) The lengths of the diagonals of a rectangle can be 3 cm and 5 cm.(g) If adjacent sides of a parallelogram are equal, then it is a kite.2. Fill in the blanks.(a) Sum of all angles of a triangle is .....................(b) A quadrilateral has ..................... diagonals.(c) Legs of trapezium are .............................. .(d) In a rectangle, .................... sides are equal.(e) In a rhombus, ..................... sides are equal.(f) If each angle of a rhombus is right angle, then rhombus is a ............(g) If diagonals of a quadrilateral bisect each other at right angles then it is a........A BD C(4x – 1) cm(2x + 5) cm
Acme Mathematics 8 215Exercise 5.41. Find the value of x, y and z in the following parallelograms.(a) (b) (c )(d) (e) (f )(g) (h) (i)2. Find the value of unknown quantities x and y in the following parallelograms.(a) (b) (c)3. Find the value of x and y in the following parallelograms.(a) (b) (c)yzx100°x + 80°yz120° – xyz x60°70°y xzz2x3y50°60°y zx40°40°y3x – 45x – 12zx zy + 10 2y – 1040°yxz20 cm10 cm5y2xADBC(2x – 3) cm(x + 1) cm5 cmy cmQ RP SN OM P(7x + 3) cm(2y – 5) cm17 cm11 cmD COA By cm7 cm5 cm(11 – x) cmS ROP Q(3x – 1) cm11 cm(y + 3) cm12 cmN MK LO(x + 5) cm4.5 cm4.5 cm(2x + 1) cm
216 Acme Mathematics 84. Study the given parallelogram and complete the table.A BD CStatement Reasons1. ∠A = ∠C 1. 2. ∠A + ∠D = 180° 2. 3. ∠B + ∠C = 180° 3. 4. ∠D + ∠C = 180° 4. 5. Verify experimentally that:(a) Co-interior angles of parallelogram are supplementary.(b) Opposite sides of rectangle are equal.(c) Opposite sides of rectangles are parallel.(d) Diagonals of rectangle bisect to each other.(e) All angles of square are equal and 90°.6. List the same properties of square and parallelogram.7. Study the given rhombus.(a) Diagonal of a rhombus ............... the vertical angles.(b) In the given figure find the value of 'x' and 'y'.8. (a) Draw two rhombus of different sizes as shown below: B CA BEBDA E C(5x – 10) cm110°2y + 10°MP ON(3x + 20) cm
Acme Mathematics 8 217(b) Measure all the segments of the diagonals and angle between them and fill in the table as shown below. Also, write down the conclusion. Fig. EA EB EC ED ∠AEB Result1.2.Conclusion: .....................(c) What is a difference between a trapezium and parallelogram? 9. Study the given kites and calculate the value of x, y, BC and AD.Project WorkCollect the object having square and rectangular shape. Draw their nets on the copy and verify their properties. Discuss to your friends about verification of their properties.D13 cmBA5 cm5 cmxC(b)DyBA3 cm5 cmx8 cmC(a)y
218 Acme Mathematics 8C. Construction of Quadrilaterala. Construction of quadrilateral Construction of quadrilateral when four sides and one angle is given.Example 1 : Draw a quadrilateral ABCD where AB = 5 cm BC = CD = 4 cm, AD = 6 cm and ∠A = 600. Solution: We will first draw a rough sketch of quadrilateral ABCD. It help us to construct the quadrilateral. Step of construction: (i) Draw a rough sketch and write given measurements. (ii) Draw a line segment AB = 5 cm (iii) Draw an angle XAB = 60°. (Use Protroctor) (iv) With A as the centre and radius 6 cm (AD = 6 cm), draw an arc which cut the line AX at D.(v) With D as the centre and radius 4 cm (DC = 4 cm), draw an arc towards C. Again with B as the centre and radius 4 cm (BC = 4 cm), draw an arc towards C which cut the first arc at C. (vi) Join D,C and B,C. ABCD is the required quadrilateral. Rough sketchA B60°D 4 cm C5 cm6 cm 4 cmA 5 cm BAXB60°5 cmAXDB60°5 cm6 cmAXDCB60°5 cm6 cmAXDCB60°5 cm4 cm4 cm6 cm
Acme Mathematics 8 219b. Construction of Rectangle(i) Construction of rectangle in which two adjacent sides are given. Example 1 :Construct a rectangle in which two adjacent sides are 4 cm and 5.5 cm. Solution: Steps of construction: (i) Draw a rough sketch. ABCD is a rectangle. Where, AD = BC = 4 cm, AB = DC = 5.5 cm and ∠A = 90° (ii) Draw a line segment AB = 5.5 cm (iii) Draw a line AX, such that ∠XAB = 90° .(Use Protroctor)(iv) With A as centre and radius 4 cm, cut the line. AX at D. (v) With D and B as centres and 5.5 cm (DC = 5.5 cm) and 4 cm (BC = 4 cm) respectively, draw two arcs such that they intersect at C.(vi) Join D,C and B,C.ABCD is the required rectangle. A90°D4 cm5.5 cm BCRough sketchA 5.5 cm BA90°X5.5 cm BAD90°X5.5 cm4 cmBAD C90°X5.5 cm4 cmBAD C90°X5.5 cm4 cmB
220 Acme Mathematics 8C. Construction of ParallelogramTHINGS TO REMEMBERTo construct a parallelogram, we should have the knowledge of any of the following cases: (i) Two adjacent sides and the angle included between them are given (ii) One side (the base), one diagonal and the angle that the diagonal makes with the base are given (iii) Two adjacent sides and the diagonal connecting the end-point of the adjacent sides are given(i) Construction of a parallelogram when adjacent sides and angle included them are given Example 1 : Construct a parallelogram ABCD, in which AB = 5 cm, AD = 4 cm and ∠A = 70° Solution: Step of constructions.(i) Draw a rough sketch of parallelogram ABCD. Where, AB = DC = 5 cm, AD = BC = 4 cm, ∠A = 70° (ii) Draw a line segment. AB = 5 cm (iii) Draw a line AX such that ∠XAB = 70° (iv) With A as the centre and radius 4 cm (AD = 4 cm), draw an arc that cut AX at D. (v) With D as the centre and radius 5 cm (DC = 5 cm) and with B as the centre and radius 4 cm (BC = 4 cm) draw two arcs such that they intersect at C. Rough sketchA70°DB5 cm C5 cm4 cm4 cmA 5 cm BAX70°5 cm BAXD70°5 cm4 cmBAXD C70°5 cm4 cmB
Acme Mathematics 8 221(vi) Join D, C and B, C. ABCD is a required parallelogram.Example 1 : Construct a parallelogram ABCD, in which AB = 5 cm, BC = 3.5 cm and ∠ABC = 60°Solution: A BCPD 5 cm5 cm60°3.5 cm3.5 cmSteps of Construction of a parallelogram(i) Draw a line segment AB of length 5cm (ii) With the centre B, draw an angle ∠ABP = 60° (iii) With the centre B and the radius BC = 3.5cm, draw an arc to cut BP at C (iv) With the centre C and the radius CD = 5cm, draw an arc (v) With the centre A and the radius AD = 3.5cm, draw an arc. These arcs intersect at point D (vi) Join A, D and C, D Hence ABCD is the required parallelogram.AXD C70°5 cm5 cm4 cm4 cmBADB5 cm C5 cm60° 3.5 cm3.5 cmRough sketch
222 Acme Mathematics 8(ii) Construction of a parallelogram with one side, one diagonal and the angle that the diagonal makes with the base are givenExample 1 : In the parallelogram ABCD, AB = 5cm, diagonal BD = 6.5cm and ∠ABD = 45°.Solution: 5 cm45°6.5 cmA BDPCA 5 cm45°6.5 cmBD CRough sketchSteps of Construction of a parallelogram(i) Draw a line segment AB of length 5cm (ii) With the centre B, draw an angle ∠ABP = 45° (iii) With the centre B and the radius BD = 6.5cm, draw an arc to cut BP at D (iv) Join AD. (v) With the centre B and the radius BC = AD, draw an arc on the same side of BD opposite to A (vi) With centre D and the radius DC = AB, draw an arc to cut the arc in the step (v) at C (vii) Join B, C and D, C Thus, ABCD is the required parallelogram.(iii) Construction of a parallelogram when two adjacent sides and the diagonal connecting the end point of the adjacent sides are givenExample 1 : Construct a parallelogram ABCD, in which AB = 6.5 cm, AD = 3.5 cm and diagonal BD = 5.7 cm.Solution:A 6.5 cm3.5 cm5.7 cmBD CRough sketch6.5 cm5.7 cmA BDPC3.5 cm
Acme Mathematics 8 223Steps of Construction of a parallelogram(i) Draw a line segment AB of length 6.5 cm (ii) With B as the centre and the radius 5.7 cm draw an arc (iii) With A as the centre and the radius 3.5 cm draw an arc (iv) These two arcs intersect each other at point D (v) With D as the centre and the radius DC = AB = 6.5 cm, draw an arc (vi) With B as the centre and the radius BC = AD = 3.5 cm, draw an arc. These arcs meet at point C (vii) Join B, C and D, C Thus, ABCD is the required parallelogramd. Construction of SquareTHINGS TO REMEMBERTo construct a square, we should have the knowledge of any of the following cases:(i) Any one side is given(ii) The length of the diagonal is given(i) Construction of square in which one sides is given. Example 1 :Construct a square in which each sides are 4 cm. Solution: Steps of construction: (i) Draw a rough sketch. ABCD is a square. So, AD = BC = AB = DC = 4 cm and ∠A = 90° (ii) Draw a line segment AB = 4 cm A90°D4 cm4 cm BCRough sketchA 4 cm B
224 Acme Mathematics 8(iii) Draw a line AX, such that ∠XAB = 90° (iv) With A as centre and radius 4 cm, cut the line. AX at D. (v) With D and B as centres and 4 cm (DC = 4 cm) and 4 cm (BC = 4 cm) respectively, draw two arcs such that they intersect at C.(vi) Join D,C and B,C. ABCD is a required square. Exercise 5.51. Construct a quadrilateral with the following measurements. (a) Quadrilateral ABCD with AB = 5 cm, BC = 3 cm, CD = 4.5 cm, DA = 5 cm and ∠A = 50°. (b) Quadrilateral PQRS, with PQ = 4 cm, QR = 5 cm, RS = 3.3 cm, PS = 6 cm and ∠Q = 40° (c) Quadrilateral MNOP, with MN = 3.5 cm, NO = 5 cm, OP = 6 cm, MP = 2.5 cm and ∠P = 30°. 2. Construct the following rectangles. (a) ABCD, BC = 8 cm and CD = 6.5 cm (b) PQRS, PQ = 7. 8 cm and QR = 8.7 cm(c) MNOP, NO = 5.2 cm and OP = 6.5 cm3. Construct the square with the following information. (a) ABCD, with each side 3cm.(b) ABCD, with AB = 4cm A90°X4 cm BAD90°X4 cm4 cmBAD90°X4 cm4 cmBCAD90°X4 cm4 cmBC
Acme Mathematics 8 225(c) ABCD, with AD = 4.5cm (d) PQRS, with QR = 5cm4. Construct parallelograms in which the following information are given. (a) Parallelogram ABCD, where, AB = 4.5 cm, AD = 6 cm and ∠A = 30° (b) Parallelogram PQRS where, PQ = 5 cm, QR = 3 cm and ∠Q = 60°. (c) Parallelogram MNOP where, OP = 7 cm, PM = 3 cm and ∠P = 40°. 5. Construct a parallelogram PQRS from the following data: (a) PQ = 6.3 cm, QR = 4.1 cm and ∠PQR = 120° (b) PQ = 6 cm, PS = 4.5 cm and ∠SPQ = 105° 6. Construct a parallelogram ABCD from the following data: (a) AB = 5.7 cm, diagonal BD = 6.9 cm and ∠ABD = 30° (b) AB = 5.2 cm, diagonal AC = 6.2 cm and ∠BAC = 45° 7. Construct a parallelogram MNOP from the following data: (a) MN = 4.5 cm, NO = 6.2 cm and diagonal OM = 7.2 cm (b) OP = 5.2 cm, NO = 4.5 cm and diagonal NP = 7.4 cm
226 Acme Mathematics 8D. Polygons A polygon is a closed figure made by three or more line segments. Some examples ofpolygons are:Triangle Rectangle Pentagon HeptagonThis shape and are not polygona. Name of PolygonsName of Polygon Number of sidesTriangle 3Quadrilateral 4Pentagon 5Hexagon 6Heptagon 7Octagon 8Nonagon 9Decagon 10Hendecagon 11Dodecagon 12b. Regular and Irregular polygons(i) Regular and Irregular Polygon: A polygon in which all the sides and all the angles are equal is called a regular polygon. Examples of regular polygons are:Equilateral triangle Square Hexagon Pentagon
Acme Mathematics 8 227A polygon which are not regular is called irregular polygons. Examples of irregular polygons are.Isosceles triangle Rectangle Pentagon Hexagon(ii) Exterior and Interior Angles of PolygonsInterior angles: The angles x, y, z are the interior angles of the triangle. The angles a, b, c and d are the interior angles of the quadrilateral. The angles a, b, c, d and e are the interior angles of the pentagon.Exterior angles: The angles x, y and z are the exterior angles of the triangle. The angles a, b, c and d are the exterior angles of the quadrilateral. The angles a, b, c, d and e are the exterior angles of the pentagon.(iii) Sum of interior and exterior angles of the polygons: We use (n – 2) × 180° to find the sum of interior angle of polygons. Where 'n' is the number of sides.(a) Sum of interior angle of 'n' side polygon = (n – 2) × 180° (b) One interior angle of 'n' side regular polygon = n – 2 n ×180° (c) Sum of Exterior angle of the polygon = 360° x zydacba edb czyxdcabdca be
228 Acme Mathematics 8(d) One exterior angle of the regular polygon = 360°nFor example: The sum of angles x + y + z = 360°The sum of angles a + b + c + d = 360°zyxdcabNow, study the given table carefully.Polygons Number of sidesSum of all interior angles (n – 2) × 180°Triangle 3 (3 – 2) × 180° = 1 × 180° = 180°Quadrilateral 4 (4 – 2) × 180° = 2 × 180° = 360°Pentagon 5 (5 – 2) × 180° = 3 × 180° = 540°Hexagon 6 (6 – 2) × 180° = 4 × 180° = 720°Heptagon 7 (7 – 2) × 180° = 5 × 180° = 900°Octagon 8 (8 – 2) × 180° = 6 × 180° = 1080°Nonagon 9 (9 – 2) × 180° = 7 × 180° = 1260°Decagon 10 (10 – 2) × 180° = 8 × 180° = 1440°c. Number of Triangles and Number of Diagonals of the polygon Study the following polygons and their diagonals.B CAIt is a triangle. It has no diagonalA B DC It is a quadrilateral. It has two diagonals, AC and BDEither diagonal AC or diagonal BD divides the quadrilateral ABCD into two triangles. Similarly, Pentagon ABCDE has three triangles by diagonals AC and AD.Now, Look at the following table.Angles x, y and z are exterior angleABCD E
Acme Mathematics 8 229Polygons Number of sides (n) Number of Triangle Number of diagonals3 (3 – 2) = 1 –4 (4 – 2) = 2 25 (5 – 2) = 3 56 (6 – 2) = 4 97 (7 – 2) = 5 148 (8 – 2) = 6 20N - gon n n – 2 n(n – 3)2Hence, diagonal of the polygons divides the polygons into (n – 2) triangles. Note: We can calculate the number of diagonals by n(n – 3)2 , when n > 3. Solved ExamplesExample 1 : The angles of a quadrilateral are in the ratio 1: 4: 6: 7. Find the measure of each angles.Solution : Let the angles of a quadrilateral be x, 4x, 6x and 7x respectively. Then, x + 4x + 6x + 7x = 360° → Sum of interior angles of aquadrilateral. or, 18x = 360° or, x = 360°18or, x = 20°
230 Acme Mathematics 8 Now, 4x = 4 × 20° = 80° 6x = 6 × 20° = 120° 7x = 7 × 20° = 140° Hence, the angles are 20°, 80°, 120° and 140° Example 2 : Calculate the measure of each angles of regular pentagon. (a) Interior angle. (b) Exterior angle. Solution : Here, No. of sides (n) = 5 Let 'x' be the interior angle and 'y' be the exterior angle. (a) By using formula,x = n – 2n × 180°= 5 – 25 × 180°= 35 × 180° = 108°(b) By using formula,y = 360°n or, y = 360°5 or, y = 72° Hence, each interior angle is 108° and each exterior angle is 72°.Classwork1. How many diagonals can be drawn in the following polygons ?(a) Quadrilateral (b) Pentagon (c) Nonagon(d) Octagon (e) Heptagon (f) Hexagon2. Calculate each interior angle of the following regular polygons. (a) Square (b) Pentagon (c) Hexagon (d) Heptagon 3. Calculate the each exterior angle of the following regular polygons.(a) Octagon (b) Nonagon (c) Decagon (d) Dodecagon Exercise 5.61. (a) The angles of a quadrilateral are in the ratio 1: 2: 3: 4. Find the measure of each angles. (b) The angles of a quadrilateral are in the ratio 2: 4: 5: 7. Find the measure of each angles. Exterior angle6x4xx7x
Acme Mathematics 8 2312. Calculate the value of 'x'in the following figure.(a) (b) (c)(d) (e) (f)(g) (h) (i)3. Calculate the value of x and y in the following regular polygons.(a) (b) (c)(d) (e) (f)4. On the basis of the given polygon, answer the following questions.(a) Write the number of sides in the given polygon. (b) Find the measurement of the interior angles of a polygon. (c) Find the measurement of the exterior angles of a polygon.70° x90°80°110°3x 3x2x2x80°80°xx BCA Dx 80°95° 100°B4x5x3xCA20°xCB DA50°20°95°105°x120°110°D CBAEx°135°140°70°100°75°CB EAFD160°100°105°140°DCBAFEx 130°BACx yxyy xx xyxy xy
232 Acme Mathematics 8Project WorkObjective : To verify that sum of exterior angles of a polygon is 360°.Materials required :(i) Sketch pens (ii) Thick coloured paper(iii) A pair of scissors (iv) Geometry boxActivity : (a) Draw a triangle, a quadrilateral and a pentagon on the coloured paper using sketch pens. Draw their exterior angles and name them as shown in the figure.CBAHIJKLDEFG(b) Cut out these angles as shown by dotted lines in the figure.CBAHIJKLDEF G(c) Arrange cut outs of angles A, B and C together, angles D, E, F and G together, angles H, I, J, K and L together.A B EFGD JIHC K LObservation(a) Angles A, B and C form a complete angle i.e. ∠A + ∠B + ∠C = 360°(b) Angles D, E, F and G form a complete angle i.e. ∠D + ∠E + ∠F + ∠G = 360°(c) Angles H, I, J, K, L form complete angle i.e. ∠H + ∠I + ∠J + ∠K + ∠L = 360°Conclusion: Sum of exterior angles of a polygon is 360°.
Acme Mathematics 8 2335.3 Congruency and SimilarityA. Congruent Objects Study the following figures.Figure (i) Figure (ii) Figure (iii)In the figures (i), (ii) and (iii),The objects are same and size are also same. Such objects are called congruent objects. The figures are said to be congruent if they have the same shape and size.In the figure, ∆ABC and ∆DEF are congruent triangles. Where, ∠A = ∠D, ∠B = ∠E and ∠C = ∠F. Sides AB = DE, BC = EF and AC = DF. We use the symbol (≅) to denote congruency. Thus ∆ABC ≅ ∆DEF.Since ∠A = ∠D, ∠A and ∠D are called corresponding angles. In the same away ∠B and ∠E, ∠C and ∠F are corresponding angles. Sides AB and DE are corresponding sides. Similarly, BC and EF are corresponding sides. AC and DF are corresponding sides. B. Congruency of Triangles Two triangles are congruent if they covers each other completely when put together. Here, ∆ABC ≅ ∆PQR; when put together A falls on P, AB falls on PQ B falls on Q, BC falls on QR C falls on R AC falls on PR This relationship between the parts of two triangles is known as correspondence. Correspondence chart for ∆ABC and ∆PQRB CAE FDQ RPB CA
234 Acme Mathematics 8Corresponding verticesCorresponding sidesCorresponding anglesCorresponding triangleA ↔ P AB ↔ PQ ∠A ↔ ∠P ∆ABC ≅ ∆PQRB ↔ Q BC ↔ QR ∠ B ↔ ∠Q ∆BAC ≅ ∆QPRC ↔ R CA ↔ RP ∠C ↔ ∠R ∆CAB ≅ ∆RPQC. Conditions for congruency of triangles. The following are the conditions for the congruency of triangles. (a) S.S.S. condition (b) S.A.S .condition (c) A.S.A. condition(d) R.H.S .condition (e) A.A.S. condition(a) SSS (Side-Side-Side) conditionActivityConstruct a triangle ABC as shown in the figure. AB CDE FConstruct a triangle DEF such that its three sides, AB = DE, BC= EF and AC=DF.By using the protractor measure the angles and complete the given table. In triangle ABC In triangle DEF ResultAB =......cm DE =......cm AB = DEBC =......cm EF =......cm BC = EFAC =......cm DF =......cm AC = DF∠BAC = .... ∠EDF = .... ........∠ABC = .... ∠DEF = .... ........∠BCA = .... ∠EFD = .... ........Conclusion: Here, all sides of ∆ABC are separately equal to all the sides of ∆DEF. Thecorrespondingangles are alsoequal.So,the∆ABCand∆PQRare congruenttriangles.If the three sides of one triangle are individually equal to the three sides of another triangle, then we can say that the triangles are congruent. It is called SSS condition.
Acme Mathematics 8 235(b) SAS (Side-Angle-Side) conditionActivity Construct a triangle ABC as shown in the figure. Construct a triangle DEF such that, AB = DE, ∠ABC= ∠DEF and BC= EF.AB CDE FNow, by using the protractor and scale measure the angles and side and complete the given table.In triangle ABC In triangle DEF ResultAB =......cm DE =......cm AB = DE∠ABC =...... ∠DEF =...... ∠ABC = ∠DEFBC =......cm EF =......cm BC = EF∠BAC = .... ∠EDF = .... ........∠BCA = .... ∠EFD = .... ........AC = .... DF = .... ........Conclusion: Here, two sides and the included angle of ∆ABC are equal to the two sides and theincluded angle of ∆DEF. The other corresponding sides and angles are also equal. So, the ∆ABC and ∆DEF are congruent triangles.If two sides and the angle between them of one triangle are equal to the corresponding two sides and the angle between them of another triangle are equal, then the triangles are congruent. It is called SAS condition.(c) ASA (Angle-Side-Angle) conditionActivity Construct a triangle ABC as shown in the figure. Construct a triangle DEF such that ∠DEF = ∠ABC and ∠EFD = ∠BCA and side EF = BC. AB CDE FNow, complete the given table with the measurements of the remaining angles and sides.
236 Acme Mathematics 8(d) RHS (Right angle-Hypotenuse-Side ) conditionActivity Construct a right angled triangle ABC, right angled at B, BC as base and AC as its hypotenuse as shown in the figure. Construct a triangle DEF such that ∠DEF = ∠ABC , base EF = BC and hypotenuse DF = AC. B CAE FDNow, complete the given table with the measurements of the remaining angles and sides.By using the protractor and scale measure the angles and sides and complete the given table. In triangle ABC In triangle DEF Result∠ABC = .... ∠DEF = .... ∠ABC =∠DEFBC = ....cm EF = ....cm BC = EFAC =......cm DF =......cm AC = DF∠BCA = .... ∠EFD = .... .............∠BAC = ...... ∠EDF = .... ...........AB =......cm DE =......cm ............Conclusion: Here, In right angled ∆ABC and ∆DEF, right angle, hypotenuse and a side of ∆ABCare equal to the right angle, hypotenuse and a side of ∆DEF. The other correspondingsides and angles are also equal. So, the ∆ABC and ∆PQR are congruent triangles.By using the protractor and scale measure the angles and sides and complete the given table. In triangle ABC In triangle DEF Result∠ABC = .... ∠DEF = .... ∠ABC =∠DEFBC = .... EF = .... BC=EF∠BCA = .... ∠EFD = .... ∠BCA =∠EFDAC =......cm DF =......cm ............AB =......cm DE =......cm ..........∠BAC = .......... ∠EDF = .... ...........Conclusion: Here, two angles and the included side of ∆ABC are equal to the two angles and theincluded side of ∆DEF. The other corresponding sides and angles are also equals. So, the ∆ABC and ∆DEF are congruent triangles.If two angles and a side between them of one triangle are equal to the two angles and a side between them of another triangle, then two triangles are congruent. It is called ASA condition.
Acme Mathematics 8 237Solved ExamplesExample 1 : In the given figure, ∆ABC and ∆PQR are congruent. Find the value of 'x'.R P60° 50°4 cmQB60° C50°(3x – 2) cmAIf the hypotenuse and one side of a right angled triangle are equal to the hypotenuse and one side of another right angled triangle, then the triangles are congruent. It is called RHS condition. (e) AAS (Angle- Angle-Side) conditionActivity Construct a triangle ABC.Construct a triangle DEF with ∠DEF= ∠ABC, ∠EFD=∠BCA, and a side DF = AC. AB CDE FNow, complete the given table with the measurements of the remaining angles and sides.By using the protractor measure the angles and complete the given table. In triangle ABC In triangle DEF Result∠ABC = .... ∠DEF = .... ∠ABC =∠DEF∠BCA = .... ∠EFD = .... ∠BCA =∠EFDAC =......cm DF =......cm AC = DF∠BAC = .... ∠EDF = .... .............BC = ......cm EF = ....cm ...........AB =......cm DE =......cm ............Conclusion: Here, two angles and a non-included side of ∆ABC are equal to the correspondingtwo angles and a non-included side of ∆DEF. The corresponding sides and anglesare also equal. So, the ∆ABC and ∆DEF are congruent triangles.If a side of one triangle along with an adjacent angle, is equal to a side of another triangle, along with its adjacent angle, then the triangles are considered congruent when their corresponding parts are equal in order. It is called AAS condition. The condition is also called axiom.
238 Acme Mathematics 8Solution : Here, ∆ABC ≅ ∆PQRAB and PQ are corresponding sides. So, AB = PQ or, 3x – 2 = 4 or, 3x = 4 + 2 or, 3x = 6 or, x = 2 Hence, the value of x is 2.Example 2 : In the adjoining figure, AD = DC and AB = BC. Is ∆ABD ≅ ∆CBD ?Solution : In ∆ABD and ∆CBD (i) AB = CB → given (S) (ii) AD = CD → given (S) (iii) DB = DB → common side. (S)(iv) ∴ ∆ABD ≅ ∆CBD by SSS conditions.Exercise 5.71. In the given figures, each pair of triangles have equal parts marked on them. Name the congruence condition.(a) (b)(c) (d)BCDAFDE ABCQ RPB CAQ RY PZXMO NA BC
Acme Mathematics 8 239(e) (f) (g)2. From the pair of congruent triangles write the corresponding sides and angles.(a) (b)(c) (d)3. Find the value of 'x' in the following pairs of congruent triangles. Write the measure of corresponding unknown angles and sides also.(a) (b)4. OP is the bisector of ∠MOL, PL ⊥ OL are PM ⊥ OM. If ∆OPL ≅ ∆OMP and PL = 3 cm, find the length of PM. 5. In the figures ∆ABC and ∆DEF and congruent, AB = DE, ∠A = ∠D and ∠B = ∠E. Find the measurements of ∠B, ∠E, ∠C and sides AB and EF. A D CB B DA C B C DAB CAQ RPB CAF EDE FDRQPC BAQRPQ65° 55°4x cm R7 cmPB65° 55°12 cm C6 cmA A BC42°8 cm10 cmZYX42° (4x – 2) cm12 cmMP OLB CA60°6 cm E FD50°4 cm
240 Acme Mathematics 86. In the figure AD = BC and AC = BD. If ∠ADC = 90°, find the measure of ∠BCD. 7. In the figure ∠B = 40°, find ∠D.8. Prove that given triangles are congruent.D. Similar ObjectsStudy the given figure.Figure (i) Figure (ii) Figure (iii)In the figures (i), (ii) and (iii), the objects are same but the size are different such objectsare called similar objects. The figure having the same looks but different size are similarfigures.In the figure ∆ABC and ∆DEF are similar triangles,where, ∠A = ∠D, ∠B = ∠E and ∠C = ∠F but the sides of ∆ABC and ∆DEF and not equal in length.We use '~' to denote similarity. Thus, ∆ABC ~ ∆DEF.E. Conditions for similarity of triangles. Activity Take a cardboard paper and draw two line segments BC and EF with different measurementNow, as shown in the figure, draw the angles of same measurement at points B and E.A BD CA DOB EE FDB CAProject WorkCollect congruent object around us. Make their list and display in your classroom.B CAQ RP
Acme Mathematics 8 241Similarly, draw same angle at points C and F. Name the points of intersection of the respective sides as A and D.What shape is formed by this construction? Are the two shapes looks alike? DiscussE FDB CANow, cut out the two triangles and place them on over to other. Discuss the following questions based on the overlapping triangles.(a) Are the shape of the two tangles the same? (b) Are the angles of triangle ABC are equal to the angles of triangle DEF ?(c) Are AB = DE, BC =EF and AC= DF ? Now, Measure all the sides of both triangles. (d) Calculate ABDE , BCEF and ACDF . ABDE is the ratio of the sides opposite to the equal angles ∠C and ∠F.(e) Are the ratios equal ?The given ∆ABC and ∆DEF and similar triangle. We write it as ∆ABC ~ ∆DEF. Where, ∠BAC= ∠EDF, ∠ABC= ∠DEF, ∠BCA = ∠EFD and ABDE = BCEF = ACDF . AB and DE are corresponding sides and ∠C and ∠F are corresponding angles.BC and EF are corresponding sides and ∠A and ∠D are corresponding angles.AC and DF are corresponding sides and ∠B and ∠E are corresponding angles.Classwork1. In the given figures, triangles are similar. (i) Write the corresponding sides.(ii) Measure the sides and calculate the ratio of the corresponding sides.(a) (b)B CAE FDB CD EA
242 Acme Mathematics 82. Measure the angles and sides of the following triangles and find which of the pairs are similar.(a) (b)(c) (d)Exercise 5.81. Calculate the value of unknown quantity of the following similar triangles.(a) (b)(c) (d)(e) (f)E FDB CAE FDFP QBA CX ZYX YZQ RPBA6 cm C10 cm 8 cmEDFy cm x cm12 cm B CA3 cmyxE FD6 cm8 cm53°R QP10 cmx 12 cm 7.5 cmTM A80°y cm 12.5 cm10 cm4 cm3 cm3 cm x yABD ECB D CA3 cm 4 cmx cmBDA CO x cm10 cm9 cm5 cmy cm5 cm
Acme Mathematics 8 2431. In the adjoining figure, CD is parallel to AB. EA and FG are two straight lines.(a) Write the corresponding angle of DCE. (b) Find the value of x from the figure. (c) For what value of ∠FGD, the straight line segments FG and EA will became parallel? 2. In the adjoining figure, EF intersects straight lines AB and CD at point P and R respectively. Observe the figure and answer the following questions.(a) Write a pair of co-interior angles in the figure.(b) Find the value of x.(c) At what value ∠APR, the given line segments AB and CD will becomes parallel?3. (a) Construct a parallelogram ABCD in the following condition.AB = 5 cm, AD = 4 cm and ∠A = 30°(b) In the given figure, if MN||OP, prove that ∆MNQ ~ ∆OPQ.4. (a) Define alternative angle.(b) Find the value of x. (c) Find the value of x.(3x + 13)°(2x – 23)°HC D GF EB5x 4x6xDQCBAE P R FMONQP40°x30° ADBECADBC(5x – 15)°(3x + 15)°AMixed Exercise
244 Acme Mathematics 85. (a) Write the formula to calculate sum of the interior angle of regular polygon.(b) Write any two condition for congruency of triangle.(c) Verify experimentally that the exterior angle of a triangle is equal to the sum of its opposite interior angle.6. (a) Construct a square ABCD in which AB = 5 cm.(b) From the given figure, find the value of x and y.130° 120°xyM B CAN7. Study the figure and answer the following questions.(a) Write the relation between ∠AOD and ∠BOC(b) If ∠BOD = 110°, find the measure of ∠AOC.(c) If ∠BOC = x and ∠AOC = 2x, find the of value of x.8. In the given figure, AB//CD, ∠AGH = (3x – 5)°, ∠GHD = (x + 45)°, ∠EGB = y and ∠FHD = z. Study the figure and answer the following questions.(a) Which is the alternate angle of ∠AGH?(b) Find the value of x. (c) By how many more is the value of z than the value of y? 9. Study the given figure and answer the following questions.(a) State the type of pair of angles ∠BAC and ∠ACD.(b) If ∠ABC = 70° then find the measure of ∠BAC and ∠DEC.A DOC BACBEFH DG(x + 45)°(3x – 5)°yzB C EA D70°A
Acme Mathematics 8 24510. Solve the following problems.(a) In the given figure, ∠A = ∠P, ∠B = ∠R, AB = PR, BC = (3x – 1) cm, PQ = 8 cm and QR = (x + 7) cm then by which axiom, ∆ABCand ∆PQR are congruent? Alsofind the value of x.(b) Construct a parallelogram PQRS in which QR = 5 cm, QS = 6.5 cm and ∠SQR = 45°.11. In the figure, AD//BC. If ∠DAE = ∠EAB = x and ∠ABE = ∠EBC = y then, answer the following questions.(a) What is the value of ∠DAB + ∠ABC?(b) Find the measure of ∠AEB?(c) Which side of ∆AEB is the longest? Give reason.12. On the basis of the given figure and answer the following questions.(a) From figure, write the name of corresponding angle equal to ∠MEA.(b) If ∠CFE = 110°, what is the value of x? Find it.(c) If EF = FG, then what is the value of y? Find it.13. In a parallelogram ABCD, ∠BAD = 60°, AD = 4 cm and AB = 4.5 cm.(a) Construct a parallelogram ABCD.(b) Verify experimentally that the sum of exterior angle of a traingle is equal to it's interior angle.14. In the given figure straight lines AB and CD are intersected by a transversal MN at points E and F. By observing the figure. Answer the following questions.(a) Write a pair of alternate angle in the figure.(b) Find the value of x.(c) At what value of ∠BEG line segments AB and CD will be parallel?B P CA R(3x – 1) cm Q(x + 7) cm8 cmAB CDE110° yxC F G DNME A B60°ACBDNFEMGx2x
246 Acme Mathematics 815. Answer the following questions. (a) Construct a parallelogram ABCD with AB = 5 cm, BC = 7 cm and ∠BAC = 60°.(b) Sketch two triangles ABC and ACD from the parallelogram ABCD. Prove the ∆ABC is congruent to ∆CDA.16. (a) Construct a parallelogram ABCD with AB = 7 cm, BD = 7 cm and AD = 5.4 cm.(b) Find the area of parallelogram so formed according to the question (a) by measuring the height.17. (a) Construct a rectangle ABCD with AB = 8 cm, BC = 6 cm and ∠ABC = 90°.(b) By which axiom ∆ABC and ∆PQR are congruent? Also find the value of x?B (x + 3) cm CAQ (2x + 1) cm RP18. In the figure, ∆ABC is an isosceles triangle. Where, AC//FD, ∠CAB = 26° and ∠DBE = x° are given.(a) Write the name of alternate angle that is equal with ∠ACB.(b) What is the value of x°? Find it.(c) Experimentally verify that the base angles ∠ABC and ∠ACB of an isosceles triangle ABC are equal by making two different size of isosceles triangle.19. In the figure, A diagonal BD is drawn in the rectangle ABCD having length 8 cm and breadth 6 cm.(a) According to the given measurement, construct the given rectangle ABCD by using compass.ABCEF Dx°26°DACB8 cm6 cm
Acme Mathematics 8 247(b) By which axiom triangles ABD and CDB are congruent? Justify with facts and reason.20. (a) Draw the triangle ABC with different shape and size as shown below:BBC C(i) (ii)AA(b) Measure all the angle of both the triangle and fill in the table as shown below and write down the conclusion:Fig ∠ABC ∠BCA ∠CAB Result(i)(ii)Conclusion:(c) In ∆XYZ, XY = 6a, XZ = 5a + 2 and YZ = 2a + 3 and XY = XZ. Then find the perimeter of ∆XYZ. 21. Answer the following questions.(a) If a° and c° are two equal acute angles of right angle triangle, what is the value of a°.(b) If x°, 2x° and 30° are the angles of a triangle find x° and 2x°.(c) Verify experimentally that sum of the angles of a triangle 180°.22. In the given figure, PQ = PR and MP⊥QR(a) Prove that ∆PMQ ≅ ∆PMR (b) If QM = (3x + 1) cm and MR = 13 cm, find the value of x.(c) In the given figure ∆PQR ~ ∆PMN, MN||QR, MN = 10 cm, QR = 15 cm and PQ = 12 cm. Find the length of PM. Q M RPQ RP12 cmM N10 cm15 cm
248 Acme Mathematics 823. (a) Write any one property of isosceles triangle.(b) Experimentally verify that the longer diagonal of a kite bisect the vertex angle.. (Two different figure are necessary). (c) The pairs of given triangles are similar. Find the length of the sides marked x and y.AB y C18 cm14 cmMN 24 cm Ox 21 cm24. (a) In the figure, AB||CDand AC = CE. Prove that ∆ABC ≅ ∆CDE and BC = CD.(b) Construct a rectangle CDEF having diagonal CE = 6 cm and angle between diagonal CE and side CD = 60°.ABDEC
Acme Mathematics 8 249EvaluationTime: 72 minutes Full Marks: 301. (a) Construct a rectangle ABCD in which the length of the diagonal is 6cm each and the angle between them is 600. [3](b) Verify experimentally that the opposite angles of parallelogram are qual.Two figures of different measures are requird. [3] (c) In the given figure, ∆ABC~∆PQC then find the valuesof x and y. [2]2. (a) Construct a parallelogram ABCD with AB = 7cm, AD = 4cm and ∠DAB = 60o. [3](b) In the giving figure, if ∆PQR ~ ∆PAB. Find the measurement of QR. [2] 3. (a) Draw a parallelogram where diagonals are 6cm & 8cm and angle between then is 30o. [3](b) In the given figure BC||DE then prove ∆ABC and ∆ADE are similar. [2] 4. (a) Define square. [1](b) Construct a square having length of a side 4cm. [3] 5. Answer the following questions. (a) Construct a parallelogram ABCD with AB = 7cm, BC = 5cm and ∠ABC = 60o. [3](b) Sketch two triangles ABC and ACD from the parallelogram ABCD. Prove the ∆ABC is congruent to ∆CDA. [2] (c) In the given figure,D is the mid point of BC. AD is perpendicular to BC.Prove that ∆ABD≅∆ACD. [3]C Q BAP5cm y24cm20cm10cmx cmPBQAR4cm8cm 2cmAEBDCB CAD
250 Acme Mathematics 85.4 SolidREVISIONA. Solids Any thing which occupies space is called a solid. A solid has length, breadth and height. A geometrical box, brick, soap, ice cream cone, ball and tin of juice, etc. are the solids. Some figures of solids are given below.Some of the solids are bounded by curved surfaces. For example, ice-cream cone. Different types of solids are cube, cuboid, cylinder, cone, sphere, prism, etc.Cube Cuboid Sphere Cone Cylinder PyramidB. Net, Model and Skeleton of Some Solids (a) Cube(b) CuboidCube Skeleton of cube Net of cubeCuboid Skeleton of cuboid Net of cuboid