How to use these solutions: Always draw a rough figure first and mark the given measures on it. The diagrams below show every arc you should see in your notebook (compass arcs in orange). Draw with a sharp pencil, ruler, compass and protractor, and do not erase the construction arcs.
Practice Set 1
1Draw line segments of the lengths given below and draw their perpendicular bisectors. (1) 5.3 cm (2) 6.7 cm (3) 3.8 cm
Step 1: Draw seg AB of the given length with a ruler.
Step 2: Open the compass to more than half the length of AB. With the compass point on A, draw one arc above and one arc below the segment.
Step 3: Without changing the opening, put the compass point on B and draw arcs that cut the first two arcs. Name the points of intersection C and D.
Step 4: Draw line CD. It cuts seg AB at M. Line CD is the perpendicular bisector of seg AB.
(1) l(AB) = 5.3 cm AM = MB = 2.65 cm
(2) l(AB) = 6.7 cm AM = MB = 3.35 cm
(3) l(AB) = 3.8 cm AM = MB = 1.9 cm
Verification: Measure with a protractor — m∠CMB = 90°. Measure with a divider — l(AM) = l(MB). So M is the midpoint and CD is perpendicular to AB.
Length of seg AB
l(AM) = l(MB)
(1) 5.3 cm
2.65 cm
(2) 6.7 cm
3.35 cm
(3) 3.8 cm
1.9 cm
2Draw angles of the measures given below and draw their bisectors. (1) 105° (2) 55° (3) 90°
Step 1: Draw ∠ABC of the given measure with a protractor.
Step 2: With B as centre and any convenient radius, draw an arc that cuts both arms of the angle. Name the points P (on ray BC) and Q (on ray BA).
Step 3: With P as centre and a radius more than half of PQ, draw an arc in the interior of the angle. With the same radius and Q as centre, draw another arc cutting the first arc at D.
Step 4: Draw ray BD. Ray BD is the bisector of ∠ABC.
(1) m∠ABC = 105° m∠ABD = m∠DBC = 52.5°
(2) m∠ABC = 55° m∠ABD = m∠DBC = 27.5°
(3) m∠ABC = 90° m∠ABD = m∠DBC = 45°
Verification: Measure the two parts with a protractor. Each part is half of the given angle: 105° → 52.5° each, 55° → 27.5° each, 90° → 45° each.
3Draw an obtuse-angled triangle and a right-angled triangle. Find the points of concurrence of the angle bisectors of each triangle. Where do the points of concurrence lie?
Draw any obtuse-angled triangle and any right-angled triangle. In each, bisect all three angles (use the steps of Q2). The three bisectors meet at one point, the incentre I.
Obtuse-angled ΔXYZ (∠Y is obtuse) Incentre I lies inside
Right-angled ΔLMN (∠M = 90°) Incentre I lies inside
Answer: In both triangles the point of concurrence of the angle bisectors (the incentre) lies in the interior of the triangle.
This is true for every triangle — acute, right or obtuse. The perpendiculars drawn from I to the three sides are equal, so a circle with centre I touches all three sides (the incircle, shown dotted).
4Draw a right-angled triangle. Draw the perpendicular bisectors of its sides. Where does the point of concurrence lie?
Draw right-angled ΔPQR with ∠Q = 90°. Draw the perpendicular bisectors of sides PQ, QR and PR (use the steps of Q1). They meet at one point, the circumcentre C.
Right-angled ΔPQR, ∠Q = 90°. The circumcentre C is the midpoint of hypotenuse PR.
Answer: The point of concurrence lies on the hypotenuse, at its midpoint.
Check: l(CP) = l(CQ) = l(CR) = half of the hypotenuse. A circle with centre C passes through all three vertices.
Compare: the circumcentre is inside an acute-angled triangle, on the hypotenuse of a right-angled triangle, and outside an obtuse-angled triangle.
5*Maithili, Shaila and Ajay live in three different places in the city. A toy shop is equidistant from the three houses. Which geometrical construction should be used to represent this? Explain your answer.
Construction to be used: the perpendicular bisectors of the sides of a triangle, whose point of concurrence is the circumcentre.
M — Maithili, S — Shaila, A — Ajay; T — the toy shop at the circumcentre. TM = TS = TA.
Explanation:
Show the three houses by points M, S and A and join them to form ΔMSA.
Every point on the perpendicular bisector of a segment is equidistant from its two endpoints. So the shop, being equally far from M and S, must lie on the perpendicular bisector of seg MS; being equally far from S and A, it must lie on the perpendicular bisector of seg SA.
The perpendicular bisectors of the sides of a triangle are concurrent. Their common point T, the circumcentre of ΔMSA, is the only point equidistant from all three houses. So the toy shop is at T, and l(TM) = l(TS) = l(TA).
A circle with centre T passes through all three houses. (This works only when the three houses are not in one straight line.)
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Practice Set 2
1Draw triangles with the measures given below.
(a) In ΔABC, l(AB) = 5.5 cm, l(BC) = 4.2 cm, l(AC) = 3.5 cm
Step 1: Draw a rough figure and mark the given lengths on it.
Step 2: Draw seg AB of length 5.5 cm as the base.
Step 3: Open the compass to 3.5 cm. With the compass point on A, draw an arc on one side of seg AB.
Step 4: Open the compass to 4.2 cm. With the compass point on B, draw an arc that cuts the first arc. Name the point of intersection C.
Step 5: Draw seg AC and seg BC. ΔABC is the required triangle.
ΔABC: AB = 5.5 cm, AC = 3.5 cm, BC = 4.2 cm
(b) In ΔSTU, l(ST) = 7 cm, l(TU) = 4 cm, l(SU) = 5 cm
Step 1: Draw a rough figure and mark the given lengths on it.
Step 2: Draw seg ST of length 7 cm as the base.
Step 3: Open the compass to 5 cm. With the compass point on S, draw an arc on one side of seg ST.
Step 4: Open the compass to 4 cm. With the compass point on T, draw an arc that cuts the first arc. Name the point of intersection U.
Step 5: Draw seg SU and seg TU. ΔSTU is the required triangle.
ΔSTU: ST = 7 cm, SU = 5 cm, TU = 4 cm
(c) In ΔPQR, l(PQ) = 6 cm, l(QR) = 3.8 cm, l(PR) = 4.5 cm
Step 1: Draw a rough figure and mark the given lengths on it.
Step 2: Draw seg PQ of length 6 cm as the base.
Step 3: Open the compass to 4.5 cm. With the compass point on P, draw an arc on one side of seg PQ.
Step 4: Open the compass to 3.8 cm. With the compass point on Q, draw an arc that cuts the first arc. Name the point of intersection R.
Step 5: Draw seg PR and seg QR. ΔPQR is the required triangle.
ΔPQR: PQ = 6 cm, PR = 4.5 cm, QR = 3.8 cm
Taking the longest side as the base gives the neatest figure. The triangle can also be drawn on the other side of the base.
2Draw an isosceles triangle with base 5 cm and the other sides 3.5 cm each.
Let the triangle be ΔABC with base l(BC) = 5 cm and l(AB) = l(AC) = 3.5 cm.
Step 1: Draw seg BC of length 5 cm.
Step 2: Open the compass to 3.5 cm. With the point on B, draw an arc above seg BC.
Step 3: Keeping the same opening (3.5 cm), put the point on C and draw an arc cutting the first arc at A.
Step 4: Draw seg AB and seg AC. ΔABC is the required isosceles triangle.
Isosceles ΔABC: BC = 5 cm, AB = AC = 3.5 cm
Because the two equal sides need the same compass opening, the compass is set only once for both arcs.
3Draw an equilateral triangle with side 6.5 cm.
Let the triangle be ΔPQR with l(PQ) = l(QR) = l(PR) = 6.5 cm.
Step 1: Draw seg PQ of length 6.5 cm.
Step 2: Open the compass to 6.5 cm. With the point on P, draw an arc above seg PQ.
Step 3: With the same opening and the point on Q, draw an arc cutting the first arc at R.
Step 4: Draw seg PR and seg QR. ΔPQR is the required equilateral triangle.
Equilateral ΔPQR, each side 6.5 cm
Check: each angle of the triangle measures 60°.
4Choose the lengths of the sides yourself and draw one equilateral, one isosceles and one scalene triangle.
Any suitable lengths may be chosen, as long as the sum of any two sides is greater than the third side. Sample answers:
Equilateral ΔLMN LM = MN = LN = 5 cm
Isosceles ΔDEF EF = 4 cm, DE = DF = 5.5 cm
Scalene ΔXYZ XY = 6 cm, XZ = 5 cm, YZ = 4.5 cm
Each triangle is drawn by the same method as Q1: draw the base, cut arcs from its two ends with the other two lengths, and join the point of intersection to the ends of the base.
Practice Set 3
Draw triangles with the measures given below. In each question, two sides and the angle included between them are given.
1In ΔMAT, l(MA) = 5.2 cm, m∠A = 80°, l(AT) = 6 cm
Step 1: Draw a rough figure and show the given measures in it.
Step 2: Draw seg AT of length 6 cm as the base.
Step 3: At point A, draw ray AG such that m∠GAT = 80° (with a protractor, or with a compass for 90°).
Step 4: Open the compass to 5.2 cm. With the point on A, draw an arc cutting ray AG at M.
Step 5: Join T and M. ΔMAT is the required triangle.
ΔMAT: AT = 6 cm, m∠A = 80°, AM = 5.2 cm
Check: on measuring, l(TM) ≈ 7.2 cm.
2In ΔNTS, m∠T = 40°, l(NT) = l(TS) = 5 cm
Step 1: Draw a rough figure and show the given measures in it.
Step 2: Draw seg TS of length 5 cm as the base.
Step 3: At point T, draw ray TG such that m∠GTS = 40° (with a protractor, or with a compass for 90°).
Step 4: Open the compass to 5 cm. With the point on T, draw an arc cutting ray TG at N.
Step 5: Join S and N. ΔNTS is the required triangle.
ΔNTS: TS = 5 cm, m∠T = 40°, TN = 5 cm
Check: on measuring, l(SN) ≈ 3.4 cm.
Since NT = TS, ΔNTS is isosceles; the angles at N and S are equal, 70° each.
3In ΔFUN, l(FU) = 5 cm, l(UN) = 4.6 cm, m∠U = 110°
Step 1: Draw a rough figure and show the given measures in it.
Step 2: Draw seg UF of length 5 cm as the base.
Step 3: At point U, draw ray UG such that m∠GUF = 110° (with a protractor, or with a compass for 90°).
Step 4: Open the compass to 4.6 cm. With the point on U, draw an arc cutting ray UG at N.
Step 5: Join F and N. ΔFUN is the required triangle.
ΔFUN: UF = 5 cm, m∠U = 110°, UN = 4.6 cm
Check: on measuring, l(FN) ≈ 7.9 cm.
∠U is obtuse, so in the rough figure too, show it as greater than a right angle.
4In ΔPRS, l(RS) = 5.5 cm, l(RP) = 4.2 cm, m∠R = 90°
Step 1: Draw a rough figure and show the given measures in it.
Step 2: Draw seg RS of length 5.5 cm as the base.
Step 3: At point R, draw ray RG such that m∠GRS = 90° (with a protractor, or with a compass for 90°).
Step 4: Open the compass to 4.2 cm. With the point on R, draw an arc cutting ray RG at P.
Step 5: Join S and P. ΔPRS is the required triangle.
ΔPRS: RS = 5.5 cm, m∠R = 90°, RP = 4.2 cm
Check: on measuring, l(SP) ≈ 6.9 cm.
Here ∠R = 90°, so ΔPRS is a right-angled triangle and PS is its hypotenuse.
Practice Set 4
Construct triangles of the measures given below. Here two angles and the side included between them are needed.
1In ΔSAT, l(AT) = 6.4 cm, m∠A = 45°, m∠T = 105°
Step 1: Draw a rough figure and show the given measures in it.
Step 2: Draw seg AT of length 6.4 cm as the base.
Step 3: At point A, draw ray AR such that m∠RAT = 45°.
Step 4: On the same side of seg AT, draw ray TD at point T such that m∠DTA = 105°.
Step 5: Name the point where the two rays intersect S. ΔSAT is the required triangle.
ΔSAT: AT = 6.4 cm, m∠A = 45°, m∠T = 105°
Check: m∠S = 180° − (45° + 105°) = 30°; on measuring, l(AS) ≈ 12.4 cm and l(TS) ≈ 9.1 cm.
2In ΔMNP, l(NP) = 5.2 cm, m∠N = 70°, m∠P = 40°
Step 1: Draw a rough figure and show the given measures in it.
Step 2: Draw seg NP of length 5.2 cm as the base.
Step 3: At point N, draw ray NR such that m∠RNP = 70°.
Step 4: On the same side of seg NP, draw ray PD at point P such that m∠DPN = 40°.
Step 5: Name the point where the two rays intersect M. ΔMNP is the required triangle.
ΔMNP: NP = 5.2 cm, m∠N = 70°, m∠P = 40°
Check: m∠M = 180° − (70° + 40°) = 70°; on measuring, l(NM) ≈ 3.6 cm and l(PM) ≈ 5.2 cm.
3In ΔEFG, l(EG) = 6 cm, m∠F = 65°, m∠G = 45°
The side EG is given, but the given angles are ∠F and ∠G. ∠F is not at an end of seg EG, so first find ∠E using the angle sum property of a triangle.
Now the angles at both ends of seg EG are known: m∠E = 70° and m∠G = 45°.
Step 1: Draw a rough figure and show the given measures in it.
Step 2: Draw seg EG of length 6 cm as the base.
Step 3: At point E, draw ray ER such that m∠REG = 70°.
Step 4: On the same side of seg EG, draw ray GD at point G such that m∠DGE = 45°.
Step 5: Name the point where the two rays intersect F. ΔEFG is the required triangle.
ΔEFG: EG = 6 cm, m∠E = 70°, m∠G = 45°
Check: m∠F = 180° − (70° + 45°) = 65°; on measuring, l(EF) ≈ 4.7 cm and l(GF) ≈ 6.2 cm.
4In ΔXYZ, l(XY) = 7.3 cm, m∠X = 34°, m∠Y = 95°
Step 1: Draw a rough figure and show the given measures in it.
Step 2: Draw seg XY of length 7.3 cm as the base.
Step 3: At point X, draw ray XR such that m∠RXY = 34°.
Step 4: On the same side of seg XY, draw ray YD at point Y such that m∠DYX = 95°.
Step 5: Name the point where the two rays intersect Z. ΔXYZ is the required triangle.
ΔXYZ: XY = 7.3 cm, m∠X = 34°, m∠Y = 95°
Check: m∠Z = 180° − (34° + 95°) = 51°; on measuring, l(XZ) ≈ 9.4 cm and l(YZ) ≈ 5.3 cm.
Practice Set 5
Construct triangles of the measures given below. Each is a right-angled triangle; the side opposite the right angle is the hypotenuse.
1In ΔMAN, m∠MAN = 90°, l(AN) = 8 cm, l(MN) = 10 cm.
Step 1: Draw a rough figure and mark the right angle at A.
Step 2: Draw seg AN of length 8 cm as the base.
Step 3: At point A, draw ray AT perpendicular to seg AN (m∠TAN = 90°).
Step 4: Open the compass to 10 cm (the hypotenuse). With the point on N, draw an arc cutting ray AT at M.
Step 5: Join N and M. ΔMAN is the required triangle.
ΔMAN: m∠A = 90°, AN = 8 cm, hypotenuse NM = 10 cm
Check: on measuring, l(AM) ≈ 6 cm.
2In the right-angled ΔSTU, hypotenuse SU = 5 cm and l(ST) = 4 cm.
SU is the hypotenuse, so the right angle is at the opposite vertex T (m∠STU = 90°).
Step 1: Draw a rough figure and mark the right angle at T.
Step 2: Draw seg TS of length 4 cm as the base.
Step 3: At point T, draw ray TX perpendicular to seg TS (m∠XTS = 90°).
Step 4: Open the compass to 5 cm (the hypotenuse). With the point on S, draw an arc cutting ray TX at U.
Step 5: Join S and U. ΔSTU is the required triangle.
ΔSTU: m∠T = 90°, TS = 4 cm, hypotenuse SU = 5 cm
Check: on measuring, l(TU) ≈ 3 cm.
3In ΔABC, l(AC) = 7.5 cm, m∠ABC = 90°, l(BC) = 5.5 cm.
∠B = 90°, so AC (opposite ∠B) is the hypotenuse.
Step 1: Draw a rough figure and mark the right angle at B.
Step 2: Draw seg BC of length 5.5 cm as the base.
Step 3: At point B, draw ray BT perpendicular to seg BC (m∠TBC = 90°).
Step 4: Open the compass to 7.5 cm (the hypotenuse). With the point on C, draw an arc cutting ray BT at A.
Step 5: Join C and A. ΔABC is the required triangle.
ΔABC: m∠B = 90°, BC = 5.5 cm, hypotenuse CA = 7.5 cm
Check: on measuring, l(BA) ≈ 5.1 cm.
4In ΔPQR, l(PQ) = 4.5 cm, l(PR) = 11.7 cm, m∠PQR = 90°.
∠Q = 90°, so PR (opposite ∠Q) is the hypotenuse.
Step 1: Draw a rough figure and mark the right angle at Q.
Step 2: Draw seg QP of length 4.5 cm as the base.
Step 3: At point Q, draw ray QT perpendicular to seg QP (m∠TQP = 90°).
Step 4: Open the compass to 11.7 cm (the hypotenuse). With the point on P, draw an arc cutting ray QT at R.
Step 5: Join P and R. ΔPQR is the required triangle.
ΔPQR: m∠Q = 90°, QP = 4.5 cm, hypotenuse PR = 11.7 cm
Check: on measuring, l(QR) ≈ 10.8 cm.
5Students should take examples of their own and practise construction of triangles.
Practice question — choose your own measures. For example: ΔDEF with m∠E = 90°, l(EF) = 6 cm, hypotenuse DF = 7.5 cm. Remember that the hypotenuse must be longer than the other given side.
Practice Set 6
1Write the names of pairs of congruent line segments. (Use a divider to find them.)
Seven segments drawn from point G
On measuring with a divider: seg GM, seg GR, seg GN and seg GE are all of the same length; seg GB and seg GC are of the same length (longer); seg GD is the shortest and matches none of the others.
(i) seg GM ≅ seg GR
(ii) seg GN ≅ seg GE
(iii) seg GB ≅ seg GC
(iv) seg GM ≅ seg GN
Since GM, GR, GN and GE are all equal, any two of them form a congruent pair — for example seg GR ≅ seg GE is also correct. Always check with a divider on your own textbook page.
2On the line below, the distance between any two adjoining points shown on it is equal. Hence, fill in the blanks. (Points in order: Q, P, Z, Y, X, W, A, B, C)
Take the distance between two adjoining points as 1 unit. Then the positions are: Q = 0, P = 1, Z = 2, Y = 3, X = 4, W = 5, A = 6, B = 7, C = 8. Segments with the same number of units are congruent.
(i) seg AB ≅ seg ___ — seg AB ≅ seg QP Length 1 unit. Other correct answers: PZ, ZY, YX, XW, WA, BC.
(ii) seg AP ≅ seg ___ — seg AP ≅ seg BZ Length 5 units. Other correct answers: QW, CY.
(iii) seg AC ≅ seg ___ — seg AC ≅ seg QZ Length 2 units. Other correct answers: PY, ZX, YW, XA, WB.
(iv) seg ___ ≅ seg BY — seg QX ≅ seg BY Length 4 units. Other correct answers: PW, ZA, XC.
(v) seg ___ ≅ seg YQ — seg ZW ≅ seg YQ Length 3 units. Other correct answers: PX, YA, XB, WC.
(vi) seg BW ≅ seg ___ — seg BW ≅ seg QZ Length 2 units. Other correct answers: PY, ZX, YW, XA, AC.
Practice Set 7
1Some angles are given below. Using the symbol of congruence, write the names of the pairs of congruent angles in these figures.
Measures of the angles in the figures:
First figureOB bisects ∠AOC = 90°, so m∠AOB = m∠BOC = 45°. m∠DOC = 30°.
Other figuresm∠PQR = 90°, m∠RST = 45°, m∠LMN = 30°.
Angles of equal measure are congruent. So the pairs of congruent angles are:
∠AOB ≅ ∠BOC (45° each)
∠AOB ≅ ∠RST (45° each)
∠BOC ≅ ∠RST (45° each)
∠AOC ≅ ∠PQR (90° each)
∠DOC ≅ ∠LMN (30° each)
∠RST may also be written as ∠TSR, and ∠LMN as ∠NML. The remaining angles in the first figure, ∠AOD = 60° and ∠BOD = 15°, have no congruent partner here.
