PSEB Solutions for Class 9 Maths Chapter 7 Triangles
PSEB 9th Class Maths Solutions Chapter 7 Triangles
Ex 7.1
Question 1.
In quadrilateral ACBD. AC = AD and AB bisects ∠ A (see the given figure). Show that ∆ ABC ≅ ∆ ABD. What can you say about BC and BD?
Answer:
In ∆ ABC and ∆ ABD,
AC = AD (Given)
∠ BAC = ∠ BAD (AB bisects ∠ A)
AB = AB (Common)
∴ ∆ ABC ≅ ∆ ABD (SAS rule)
∴ BC = BD (CPCT)
Thus, BC and BD are equal.
Question 2.
ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see the given figure). Prove that (i) ∆ ABD ≅ ∆ BAC, (ii) BD = AC and (iii) ∠ ABD = ∠ BAC
Answer:
In ∆ ABD and ∆ BAC,
AD = BC (Given)
∠ DAB = ∠ CBA (Given)
AB = BA (Common)
∴ ∆ ABD ≅ ∆ BAC (SAS rule)
∴ BD = AC (CPCT)
∴ ∠ ABD = ∠ BAC (CPCT)
Question 3.
AD and BC are equal perpendiculars to a line segment AB (see the given figure). Show that CD bisects AB.
Answer:
AD and BC are equal perpendiculars to line segment AB.
∴ AD = BC and ∠ OAD = ∠ OBC = 90°.
Now, in ∆ ADO and ∆ BCO,
AD = BC
∠ OAD = ∠ OBC
∠ AOD = ∠ BOC (Vertically opposite angles)
∴ ∆ ADO ≅ ∆ BCO (AAS rule)
∴ OA = OB (CPCT)
CD intersects AB at O and OA = OB.
Hence, CD bisects AB.
Question 4.
l and m are two parallel lines intersected by another pair of parallel lines p and q (see the given figure). Show that:
∆ ABC ≅ ∆ CDA.
Answer:
l || m and AC is their transversal.
∴ ∠ BCA = ∠ DAC (Alternate angles)
p l| q and AC is their transversal.
∴ ∠ BAC = ∠ DCA (Alternate angles)
Now, in ∆ ABC and ∆ CDA,
∠ BCA = ∠ DAC
∠ BAC = ∠ DCA
AC = CA (Common)
∴ ∆ ABC ≅ ∆ CDA (ASA rule)
Question 5.
Ray l is the bisector of an angle ∠ A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠ A (see the given figure). Show that:
(i) ∆ APB ≅ ∆ AQB
(ii) BP = BQ or B is equidistant from the arms of ∠ A.
Answer:
l is the bisector of ∠ PAQ and B is any point on l.
∴ ∠ PAB = ∠ QAB
BP and BQ are perpendiculars from B to AP and AQ.
∴ ∠ BPA = ∠ BQA = 90°.
Now, in ∆ APB and ∆ AQB,
∠ PAB = ∠ QAB
∠ BPA = ∠ BQA
AB = AB (Common)
∴ ∆ APB ≅ ∆ AQB (AAS rule)
∴ BP = BQ (CPCT)
BP and BQ are perpendiculars from B to arms AP and AQ of ∠ A.
∴ BP is the distance of B from AP and BQ is the distance of B from AQ.
Thus, B is equidistant from the arms of ∠ A.
Question 6.
In the given figure, AC = AE, AB = AD and ∠ BAD = ∠ EAC. Show that BC = DE
Answer:
∠ BAD = ∠ EAC
∴ ∠ BAD + ∠ DAC = ∠ EAC + ∠ DAC
∴ ∠ BAC = ∠ DAE (Adjacent angles)
Now, in ∆ BAC and ∆ DAE,
AC = AE (Given)
AB = AD (Given)
∠ BAC = ∠ DAE
∴ ∆ BAC ≅ ∆ DAE (SAS rule)
∴ BC = DE (CPCT)
Question 7.
AB is a line segment and P is its midpoint. D and E are points on the same side of AB such that ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB (see the given figure). Show that:
(i) ∆ DAP ≅ ∆ EBP
(ii) AD = BE
Answer:
∠ BAD = ∠ ABE
∴ ∠ PAD = ∠PBE (∵ P lies on AB.)
∠ EPA = ∠ DPB
∴ ∠ EPA + ∠ EPD = ∠ DPB + ∠ EPD
∴ ∠ APD = ∠ BPE (Adjacent angles)
P is the midpoint of AB.
∴ AP = BP
Now, in ∆ DAP and ∆ EBP
∠ PAD = ∠ PBE
∠ APD = ∠ BPE
AP = BP
∴ ∆ DAP ≅ ∆ EBP (ASA rule)
∴ AD = BE (CPCT)
Question 8.
In right triangle ABC, right angled at C, M is the midpoint of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that:
(i) ∆ AMC ≅ ∆ BMD
(ii) ∠ DBC is a right angle
(iii) ∆ DBC ≅ ∆ ACB
(iv) CM = 1/2 AB
Answer:
In ∆ AMC and ∆ BMD,
AM = BM (∵ M is the midpoint of AB.)
CM = DM (Given)
∠ AMC = ∠ BMD (Vertically opposite angles)
∴ By SAS rule, ∆ AMC ≅ ∆ BMD [Result (i)]
∴ ∠ MCA = ∠ MDB (CPCT)
∠ MCA and ∠ MDB are alternate angles formed by transversal CD of lines AC and BD and they are equal.
∴ AC || BD
Now, ∠ DBC and ∠ ACB are interior angles on the same side of transversal BC of AC || BD.
∴ ∠ DBC + ∠ ACB = 180°
∴ ∠ DBC + 90° = 180° (Given : ∠ C = 90°)
∴ ∠ DBC = 90°
Thus, ∠ DBC is a right angle. [Result (ii)]
Now, ∆ AMC ≅ ∆ BMD
∴ AC = BD
In ∆ DBC and ∆ ACB,
BD = CA
∠ DBC = ∠ ACB (Right angles)
BC = CB (Common)
∴ ∆ DBC ≅ ∆ ACB [Result (iii)]
∴ DC = AB (CPCT)
DM = CM and M lies on line’ segment CD.
∴ DC = 2 CM
∴ AB = 2CM
∴ 1/2AB = CM
∴ CM = 1/2AB
Ex 7.2
Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠ B and ∠ C intersect each other at O. Join A to O. Show that:
(i) OB = OC
(ii) AO bisects ∠ A
Answer:
In ∆ ABC, AB = AC
∴ ∠ ABC = ∠ ACB (Theorem 7.2)
∴ 1/2 ∠ ABC = 1/2 ∠ ACB
∴ ∠ OBC = ∠ OCB (BO bisects ∠ ABC and CO bisects ∠ ACB)
Now, in ∆ OBC, ∠ OBC = ∠ OCB
∴ OB = OC (Theorem 7.3)
Similarly, ∠ ABC = ∠ ACB gives
∴ 1/2 ∠ ABC = 1/2 ∠ ACB
∴ ∠ ABO = ∠ ACO
Now, in ∆ ABO and ∆ ACO,
AB = AC (Given)
∠ ABO = ∠ ACO
and OB = OC
∴ ∆ ABO ≅ ∆ ACO (SAS rule)
∴ ∠ BAO = ∠ CAO (CPCT)
But, ∠ BAO + ∠ CAO = ∠ BAC (Adjacent angles)
∴ ∠ BAO = ∠ CAO = 1/2 ∠ BAC
Thus, AO bisects ∠ A.
Question 2.
In ∆ ABC, AD is the perpendicular bisector of BC (see the given figure). Show that ∆ ABC is an isosceles triangle in which AB = AC.
Answer:
In ∆ ABC, AD is the perpendicular bisector of BC.
∴ BD = CD and ∠ ADB = ∠ ADC = 90°
In ∆ ADB and ∆ ADC,
AD = AD (Common)
∠ ADB = ∠ ADC (Right angles)
and BD = CD
∴ ∆ ADB ≅ ∆ ADC (SAS rule)
∴ AB = AC (CPCT)
Now, in ∆ ABC, AB = AC.
Hence, ∆ ABC is an isosceles triangle in which AB = AC.
Question 3.
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see the given figure). Show that these altitudes are equal.
Answer:
In ∆ ABC, AC = AB
∴ ∠ ABC = ∠ ACB
∴ ∠ FBC = ∠ ECB
Now, in ∆ FBC and ∆ ECB,
∠ FBC = ∠ ECB
∠ BFC = ∠ CEB (Right angles)
BC = CB (Common)
∴ ∆ FBC ≅ ∆ ECB (AAS rule)
∴ CF = BE (CPCT)
Thus, the altitudes CF and BE on equal sides AB and AC respectively of ∆ ABC are equal.
Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see the given figure). Show that
(i) ∆ ABE ≅ ∆ ACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.
Answer:
In ∆ ABE and ∆ ACF,
∠ AEB = ∠ AFC (Right angles)
∠ A = ∠ A (Common)
BE = CF (Given)
∴ ∆ ABE ≅ ∆ ACF (AAS rule)
∴ AB = AC (CPCT)
Thus, ∆ ABC is an isosceles triangle.
Question 5.
ABC and DBC are two isosceles triangles on the same base BC (see the given figure). Show that ∠ ABD = ∠ ACD.
Answer:
∠ ABC and ∠ DBC are adjacent angles.
∴ ∠ ABC + ∠ DBC = ∠ ABD ………… (1)
∠ ACB and ∠ DCB are adjacent angles.
∴ ∠ ACB + ∠ DCB = ∠ ACD ………….. (2)
In ∆ ABC, AB = AC.
∴ ∠ ABC = ∠ ACB (Theorem 7.2)
In ∆ DBC, DB = DC.
∴ ∠ DBC = ∠ DCB (Theorem 7.2)
∴ ∠ ABC + ∠ DBC = ∠ ACB + ∠ DCB
∴ ∠ ABD = ∠ ACD [From (1) and (2))
Question 6.
∆ ABC is an isosceles triangle in which AB = AC. side BA is produced to D such that AD = AB (see the given figure). Show that ∠ BCD is a right angle.
Answer:
AB = AC and AD = AB
∴ AC = AD
In ∆ ABC, AB = AC
∴ ∠ ACB = ∠ ABC (Theorem 7.2) ……………… (1)
In A ADC, AC = AD
∴ ∠ ACD = ∠ ADC (Theorem 7.2) ……………… (2)
Adding (1) and (2),
∠ ACB + ∠ ACD = ∠ ABC + ∠ ADC
∴ ∠ BCD = ∠ DBC + ∠ BDC
(Adjacent angles and A lies on BD)
In ∆ BCD,
∠ DBC + ∠ BDC + ∠ BCD = 180°
∴ ∠ BCD + ∠ BCD = 180° (from (3)]
∴ 2 ∠ BCD = 180°
∴ ∠ BCD = 90°
Thus, ∠ BCD is a right angle.
Question 7.
ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠ B and ∠ C.
Answer:
In ∆ ABC, AB = AC
∴ Z C = Z B (Theorem 7.2)
In ∆ ABC,
∠ A + ∠ B + ∠ C = 180°
∴ 90° + ∠ B + ∠ B = 180° (Given and ∠ C = ∠ B)
∴ 2 ∠ B = 90°
∴ ∠ B = 45°
∴ ∠ C = 45°
Question 8.
Show that the angles of an equilateral triangle are 60° each.
Answer:
∆ ABC is an equilateral triangle.
∴ AB = BC = AC
In ∆ ABC, AB = BC
∴ ∠ C = ∠ A (Theorem 7.2)
In ∆ ABC, AB = AC
∴ ∠ C = ∠ B (Theorem 7.2)
Hence, ∠ A = ∠B = ∠ C.
Now, in ∆ ABC, ∠ A + ∠ B + Z C = 180°
∴ ∠ A = ∠ B = ∠ C = 180°/3 = 60°
Thus, the angles of ah equilateral triangle are 60° each.
Ex 7.3
Question 1.
∆ ABC and ∆ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the given figure). If AD is extended to intersect BC at D show that,
(i) ∆ ABD ≅ ∆ ACD
(ii) ∆ ABP ≅ ∆ ACP
(iii) AP bisects ∠ A as well as ∠ D.
(iv) AP is the perpendicular bisector of BC.
Answer:
∆ ABC and ∆ DBC are isosceles triangles on the same base BC.
∴ In ∆ ABC, AB = AC and in ∆ DBC, DB = DC.
In ∆ ABD and ∆ ACD,
AB = AC
DB = DC
and AD = AD (Common)
∴ ∆ ABD s ∆ ACD (SSS rule) [Result (i)]
∴ ∠ BAD = ∠ CAD (CPCT)
In ∆ ABP and ∆ ACP
AB = AC
∠ BAP = ∠ CAP (∵ ∠ BAD = ∠ CAD)
and AP = AP (Common)
∴ ∆ ABP ≅ ∆ ACP (SAS rule) [Result (ii)]
∴ BP = CP (CPCT)
In ∆ DBP and ∆ DCR
DB = DC
BP = CP
and DP = DP (Common)
∴ ∆ DBP ≅ ∆ DCP (SSS rule)
From ∆ ABP ≅ ∆ ACR ∠ BAP = ∠ CAP (CPCT)
∴ AP bisects ∠A.
From ∆ DBP ≅ ∆ DCR ∠BDP = ∠ CDP (CPCT)
∴ DP bisects ∠D.
Thus, AP bisects ∠A as well as ∠D. [Result (iii)]
∆ A ABP ≅ ∆ ACP
∴ BP = CP and ∠ APB = ∠ APC (CPCT)
But, ∠ APB + ∠ APC = 180° (Linear pair)
∴ ∠ APB = ∠ APC = = 90°
Thus, BP = CP and AP ⊥ BC.
∴ AP is the perpendicular bisector of BC. [Result (iv)]
Question 2.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that:
(i) AD bisects BC
(ii) AD bisects ∠ A
Answer:
AD is an altitude of A ABC.
∴ ∠ ADB = ∠ ADC = 90°
In ∆ ADB and ∆ ADC,
hypotenuse AB = hypotenuse AC (Given)
∠ ADB = ∠ ADC (Right angles)
AD = AD (Common)
∴ ∆ ADB ≅ ∆ ADC (RHS rule)
∴ BD = CD and ∠ BAD = ∠ CAD (CPCT)
Now, BD = CD means D is the midpoint of BC.
Hence, AD bisects BC. [Result (i)]
Moreover, ∠ BAD = ∠ CAD and
∠ BAD + ∠ CAD = ∠ BAC.
Hence, AD bisects ∠A. [Result (ii)]
Question 3.
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆ PQR (see the given figure). Show that:
(i) ∆ ABM ≅ ∆ PQN
(ii) ∆ ABC ≅ ∆ PQR
Answer:
In ∆ ABC, AM is a median.
∴ BM = CM = 1/2 BC
In ∆ PQR, PN is a median.
∴ QN = RN = 1/2 QR
Now, BC = QR (Given)
∴ 1/2 BC = 1/2 QR
∴ BM = QN
In ∆ ABM and ∆ PQN,
AB = PQ (Given)
AM = PN (Given)
BM = QN (Proved)
∴ ∆ ABM ≅ ∆ PQN (SSS rule) [Result (i)]
∴ ∠ ABM = ∠ PQN (CPCT)
∴ ∠ ABC = ∠ PQR
Now, in ∆ ABC and ∆ PQR,
AB = PQ
∠ ABC = ∠ PQR
BC = QR .
∴ ∆ ABC ≅ ∆ PQR (SAS rule) [Result (ii)]
Question 4.
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Answer:
In ∆ FBC and ∆ ECB,
CF = BE (Given)
∠ CFB = ∠ BEC = 90° (Given)
BC = CB (Common)
∴ A FBC ≅ A ECB (RHS rule)
∴ ∠ FBC = ∠ ECB (CPCT)
∴ ∠ ABC = ∠ ACB
Now, in ∆ ABC, ∠ ABC = ∠ ACB
∴ AC = AB (Theorem 7.3)
Hence, ∆ ABC is an isosceles triangle.
Question 5.
ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠ B = ∠ C.
Answer:
In ∆ ABC, AP is an altitude.
∴ ∠ APB = ∠ APC = 90°
In ∆ APB and ∆ APC,
∠ APB = ∠ APC = 90°
AB = AC (Given)
AP = AP (Common)
∴ ∆ APB ≅ ∆ APC (RHS rule)
∴ ∠ ABP = ∠ AGP (CPCT)
∴ ∠ ABC = ∠ ACB
Thus, in ∆ ABC, ∠ B = ∠ C.
Ex 7.4
Question 1.
Show that in a right angled triangle, the hypotenuse is the longest side.
Answer:
In ∆ ABC, ∠ B is a right angle.
∴ ∠ B = 90° and AC is the hypotenuse.
In ∆ ABC,
∠A + ∠B + ∠C = 180°
∴ ∠ A + 90° + ∠ C = 180°
∴ ∠ A + ∠ C = 90°
Now, ∠ A and ∠ C are both positive (in degrees) and their sum is 90°.
∴ ∠ A < 90° and ∠ C < 90°
∴ ∠ A < Z B and ∠ C < ∠ B
∴ BC < AC and AB < AC (Theorem 7.7)
Hence, hypotenuse AC is greater than each of the other two sides BC and AB. Thus, the hypotenuse is the longest side in a right angled triangle.
Question 2.
In the given figure, sides AB and AC of ∆ ABC are extended to points P and Q respectively. Also, ∠PBC < ∠ QCB. Show that AC > AB.
Answer:
∠ PBC < ∠ QCB (Given)
∴ – ∠ PBC > – ∠ QCB (Multiplying an inequality by (-1), it reverses)
∴ 180° – ∠ PBC > 180° – ∠ QCB (Adding 180° on both the sides) …………. (1)
Now, ∠ ABC and ∠ PBC as well as ∠ ACB and ∠ QCB from a linear pair.
∴ ∠ ABC = 180° – ∠ PBC and
∠ ACB = 180° – ∠ QCB
Substituting these values in (1), we get
∠ ABC > ∠ ACB
Now, in ∆ ABC, ∠ ABC > ∠ ACB A
∴ AC > AB (Theorem 7.7)
Question 3.
In the given figure, ∠ B < ∠ A and ∠ C < ∠ D. Show that AD < BC.
Answer:
In ∆ OAB, ∠ B < ∠ A
∴ OA < OB (Theorem 7.7) …………….. (1)
In ∆ OCD, ∠ C < ∠ D
∴ OD < OC (Theorem 7.7) ……………… (2)
Adding (1) and (2),
OA + OD < OB + OC
∴ AD < BC (As O is the point of intersection of AD and BC, it lies on both the line segments.)
Question 4.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD ,(see the given figure). Show that ∠A > ∠C and ∠B > ∠D.
Answer:
Construction: In quadrilateral ABCD, draw diagonal AC.
AB is the smallest side of ABCD and CD is the longest side of ABCD.
∴ AB < BC and AD < CD.
In ∆ ABC, AB < BC
∴ ∠ ACB < ∠ BAC …… (1)
In ∆ CDA, AD < CD
∴ ∠ DCA < ∠ DAC ……… (2)
Adding (1) and (2),
∠ ACB + ∠ DCA < ∠ BAC + ∠ DAC
∴ ∠ BCD < ∠ BAD
∴ ∠ BAD > ∠ BCD
Thus, in quadrilateral ABCD, ∠ A > ∠ C. Similarly, after constructing diagonal BD and using the inequalities in A ABD and A CBD, it can be proved that ∠ B > ∠ D.
Question 5.
In the given figure, PR > PQ and PS bisects ∠ QPR. Prove that ∠ PSR > ∠ PSQ.
Answer:
PS is the bisector of ∠QPR.
∴ ∠QPS = ∠RPS = 1/2 ∠QPR ……………. (1)
∠ PSR is an exterior angle of A PQS and ∠ PSQ is an exterior angle of A PRS.
∴ PSR = ∠ Q + ∠QPS and
∠PSQ = ∠R + ∠RPS …………….. (2)
Now, in A PQR, PR > PQ
∴ ∠ Q > ∠ R
∴ ∠Q + 1/2 ∠ QPR > ∠R + 1/2 ∠ QPR
∴ ∠Q + ∠QPS > ∠R + ∠RPS [from (1)]
∴ ∠PSR > ∠PSQ
Question 6.
Show that of all line segments drawn from a given point not on a given line, the perpendicular line segment is the shortest.
Answer:
AB is a line and P is a point not on AB.
PM is the perpendicular line segment drawn from P to line AB.
N is any point on AB, other than M.
In ∆ PMN, ∠ M = 90°
∴ ∠ N < 90°
Thus, in ∆ PMN, ZN < ZM.
∴ PM < PN
This is true for any location of point N.
Hence, of all the line segments drawn from a point not on a given line, the, perpendicular line segment is the shortest.
Ex 7.5
Question 1.
ABC is a triangle. Locate a point in the interior of ∆ ABC which is equidistant from all the vertices of ∆ ABC.
Answer:
In ∆ ABC, draw l, the perpendicular bisector of side AB and m, the perpendicular bisector of side BC. Name the point of intersection of l and m as P.
P is a point on the perpendicular bisector of AB. Hence, P is equidistant from A and B.
∴ PA = PB
P is a point on the perpendicular bisector of BC. Hence, P is equidistant from B and C.
∴ PB = PC
Thus, PA = PB = PC
Hence, P is the required point which is equidistant from all the vertices of ∆ ABC.
Note: Since ∆ ABC given here is an acute angled triangle, point P lies in the interior of ∆ ABC. If ∆ ABC is a right angled Mangle, point P lies on the hypotenuse. Actually, in that case, point P will be the midpoint of the hypotenuse. Lastly, if ∆ ABC is an obtuse angled triangle, point P lies in the exterior of ∆ ABC. This point P is called the circumcentre of ∆ ABC.
Question 2.
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
Answer:
In ∆ ABC, draw the bisectors of ∠B and ∠C to intersect each other at point I.
I is a point on the bisector of ∠ B. Hence, I is equidistant from sides BA and BC. Similarly, I is a point on the bisector of ZC. Hence, I is equidistant from sides BC and CA.
Thus, point I is the required point which is equidistant from all the three sides AB, BC and CA of ∆ ABC.
This point I is called the incentre of ∆ ABC. It always lies in the interior of ∆ ABC irrespective of the type of ∆ ABC.
Question 3.
In a huge park, people are concentrated at three points (see the given figure):
A: where there are different slides and swings for children.
B: near which a man-made lake is situated.
C: which is near to a large parking and exit.
Where should an ice cream parlour be set up so that maximum number of persons can approach it ?
(Hint: The parlour should be equidistant from A, B and C.)
Answer:
First of all, construct ∆ ABC with the given points A, B and C as vertices. Then, as shown in example 1, draw the perpendicular s bisectors of any two sides of ∆ ABC and name their point of intersection as P.
Now, the ice cream parlour should be set up at the location given by point P as it is equidistant from all the three places (points) A, B and C.
Question 4.
Complete the hexagonal and star shaped < Rangolies [see figure (1) and (2)] by filling them with as many equilateral triangles of > side 1 cm as you can. Count the number < of triangles in each case. Which has more triangles?
Answer:
In figure (1), if we join the opposite vertices, we get three longest diagonals of hexagon ABCDEE By the intersection of these diagonals we get point O and six equilateral triangles – ∆ OAB, ∆ OBC, ∆ OCD, ∆ ODE, ∆ OEF and ∆ OFA. Each side of all these equilateral triangles will measure 5 cm. In each of these six triangles, we can fill 25 (1 + 3 + 5 + 7 + 9) equilateral triangles with side 1 cm each. Hence in the hexagonal Rangoli ABCDEF, we can fill 25 × 6 = 150 triangles.
Similarly, in figure (2), we can fill 150 triangles in the inner hexagon and 150 triangles in the six triangles on the boundary of the hexagon. Thus, in figure (2), 150 + 150 = 300 triangles can be filled.
Hence, more triangles can be filled in figure (2).
MCQ
Multiple Choice Questions and Answer
Answer each question by selecting the proper alternative from those given below each question to make the statement true:
Question 1.
In ∆ ABC, ∠A = ∠C, AC = 5 and BC = 4. Then, the perimeter of ∆ ABC is ……………. .
A. 9
B. 14
C. 13
D. 15
Answer:
C. 13
Question 2.
In ∆ PQR, PQ = PR, QR is extended to S and ∠PRS = 110°. Then, ∠PQR = ……………… .
A. 30°
B. 50°
C. 80°
D. 70°
Answer:
D. 70°
Question 3.
In ∆ ABC and ∆ DEF, AB = DE, BC = EF and ∠B = ∠E. If the perimeter of ∆ ABC is 20, then the perimeter of ∆ DEF is ……………. .
A. 10
B. 20
C. 15
D. 40
Answer:
B. 20
Question 4.
In ∆ ABC and ∆ PQR, AB = PQ, ∠A = ∠P and ∠B = ∠Q. If ∠A + ∠C = 130°, then ∠Q = …………….. .
A. 65°
B. 130°
C. 50°
D. 100°
Answer:
C. 50°
Question 5.
In ∆ PQR, ∠P = ∠Q = ∠R. If PQ = 6, then the perimeter of ∆ PQR is ………………. .
A. 12
B. 9
C. 18
D. 24
Answer:
C. 18
Question 6.
In ∆ ABC, AB < AC. Then …………… holds good.
A. ∠A < ∠B
B. ∠B < ∠C
C. ∠C < ∠A
D. ∠C < ∠B
Answer:
D. ∠C < ∠B
Question 7.
In ∆ PQR, ∠R > ∠Q. Then, ………………. holds good.
A. PQ > PR
B. QR > PQ
C. PR > PQ
D. PQ > QR
Answer:
A. PQ > PR
Question 8.
In ∆ ABC, AB > BC and BC > AC. Then, the smallest angle of ∆ ABC is …………………. .
A. ∠A
B. ∠C
C. ∠B
D. ∠A or ∠C
Answer:
C. ∠B
Question 9
………………….. cannot be the measures of the sides of a triangle.
A. 10, 12, 14
B. 2, 3, 4
C. 8, 9, 10
D. 2, 4, 10
Answer:
D. 2, 4, 10
Question 10.
In ∆ PQR, PQ = 4, QR = 6 and PR = 5. Then, …………….. is the angle with greatest measure in ∆ PQR.
A. ∠P
B. ∠Q
C. ∠R
D. ∠QRP
Answer:
A. ∠P
Question 11.
In ∆ XYZ, ∠X = 45° and ∠Z = 60°. Then, …………….. is the longest side of ∆ XYZ.
A. XY
B. YZ
C. XZ
D. XY or YZ
Answer:
C. XZ
Question 12.
In ∆ ABC, ∠B = 30° and BC is extended to D. If ∠ACD = 110°, then the longest side of ∆ ABC is ………………. .
A. AB
B. BC
C. CA
D. AB or AC
Answer:
B. BC
Question 13.
In ∆ ABC, AB = 4 and BC = 7, Then, ……………… holds good.
A. AC < 7
B. AC > 4
C. 4 < AC < 7
D. 3 < AC < 11
Answer:
D. 3 < AC < 11
Question 14.
In ∆ PQR, PQ = 3 and QR = 7. Then, …………….. holds good.
A. PR = 4
B. PR = 10
C. 10 > PR > 4
D. 7 > PR > 3
Answer:
C. 10 > PR > 4
Question 15.
In ∆ ABC, the bisectors of ∠B and ∠C intersect at I. If ∠A = 70°, then ∠BIC = ………………… .
A. 35°
B. 75°
C. 100°
D. 125°
Answer:
D. 125°