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Chapter 7 – Triangles

Triangles

 

 

Page No 118:

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)

?CAB = ?DAB (AB bisects ?A)

AB = AB (Common)

? ?ABC ? ?ABD (By SAS congruence rule)

? BC = BD (By CPCT)

Therefore, BC and BD are of equal lengths.

Video Solution

Page No 119:

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

(iii) ?ABD = ?BAC.

Answer:

In ?ABD and ?BAC,

AD = BC (Given)

?DAB = ?CBA (Given)

AB = BA (Common)

? ?ABD ? ?BAC (By SAS congruence rule)

? BD = AC (By CPCT)

And, ?ABD = ?BAC (By CPCT)

Video Solution

Question 3:

AD and BC are equal perpendiculars to a line segment AB (See the given figure). Show that CD bisects AB.

Answer:

In ?BOC and ?AOD,

?BOC = ?AOD (Vertically opposite angles)

?CBO = ?DAO (Each 90º)

BC = AD (Given)

? ?BOC ? ?AOD (AAS congruence rule)

? BO = AO (By CPCT)

? CD bisects AB.

Video Solution

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:

In ?ABC and ?CDA,

?BAC = ?DCA (Alternate interior angles, as p || q)

AC = CA (Common)

?BCA = ?DAC (Alternate interior angles, as l || m)

? ?ABC ? ?CDA (By ASA congruence rule)

Video Solution

Question 5:

Line 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:

In ?APB and ?AQB,

?APB = ?AQB (Each 90º)

?PAB = ?QAB (l is the angle bisector of ?A)

AB = AB (Common)

? ?APB ? ?AQB (By AAS congruence rule)

? BP = BQ (By CPCT)

Or, it can be said that B is equidistant from the arms of ?A.

Video Solution

Page No 120:

Question 6:

In the given figure, AC = AE, AB = AD and ?BAD = ?EAC. Show that BC = DE.

Answer:

It is given that ?BAD = ?EAC

?BAD + ?DAC = ?EAC + ?DAC

?BAC = ?DAE

In ?BAC and ?DAE,

AB = AD (Given)

?BAC = ?DAE (Proved above)

AC = AE (Given)

? ?BAC ? ?DAE (By SAS congruence rule)

? BC = DE (By CPCT)

Video Solution

Question 7:

AB is a line segment and P is its mid-point. 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:

It is given that ?EPA = ?DPB

? ?EPA + ?DPE = ?DPB + ?DPE

? ?DPA = ?EPB

In DAP and EBP,

?DAP = ?EBP (Given)

AP = BP (P is mid-point of AB)

?DPA = ?EPB (From above)

? ?DAP ? ?EBP (ASA congruence rule)

? AD = BE (By CPCT)

Video Solution

Question 8:

In right triangle ABC, right angled at C, M is the mid-point 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 = AB

Answer:

(i) In ?AMC and ?BMD,

AM = BM (M is the mid-point of AB)

?AMC = ?BMD (Vertically opposite angles)

CM = DM (Given)

? ?AMC ? ?BMD (By SAS congruence rule)

? AC = BD (By CPCT)

And, ?ACM = ?BDM (By CPCT)

(ii) ?ACM = ?BDM

However, ?ACM and ?BDM are alternate interior angles.

Since alternate angles are equal,

It can be said that DB || AC

? ?DBC + ?ACB = 180º (Co-interior angles)

? ?DBC + 90º = 180º

? ?DBC = 90º

(iii) In ?DBC and ?ACB,

DB = AC (Already proved)

?DBC = ?ACB (Each 90)

BC = CB (Common)

? ?DBC ? ?ACB (SAS congruence rule)

(iv) ?DBC ? ?ACB

? AB = DC (By CPCT)

? AB = 2 CM

? CM =AB

Video Solution

Page No 123:

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:

(i) It is given that in triangle ABC, AB = AC

? ?ACB = ?ABC (Angles opposite to equal sides of a triangle are equal)

? ?ACB = ?ABC

? ?OCB = ?OBC

? OB = OC (Sides opposite to equal angles of a triangle are also equal)

(ii) In ?OAB and ?OAC,

AO =AO (Common)

AB = AC (Given)

OB = OC (Proved above)

Therefore, ?OAB ? ?OAC (By SSS congruence rule)

? ?BAO = ?CAO (CPCT)

? AO bisects ?A.

Video Solution

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 ?ADC and ?ADB,

AD = AD (Common)

?ADC =?ADB (Each 90º)

CD = BD (AD is the perpendicular bisector of BC)

? ?ADC ? ?ADB (By SAS congruence rule)

?AB = AC (By CPCT)

Therefore, ABC is an isosceles triangle in which AB = AC.

Video Solution

Page No 124:

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 ?AEB and ?AFC,

?AEB and ?AFC (Each 90º)

?A = ?A (Common angle)

AB = AC (Given)

? ?AEB ? ?AFC (By AAS congruence rule)

? BE = CF (By CPCT)

Video Solution

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:

(i) In ?ABE and ?ACF,

?AEB = ?AFC (Each 90º)

?A = ?A (Common angle)

BE = CF (Given)

? ?ABE ? ?ACF (By AAS congruence rule)

(ii) It has already been proved that

?ABE ? ?ACF

? AB = AC (By CPCT)

Video Solution

Question 5:

ABC and DBC are two isosceles triangles on the same base BC (see the given figure). Show that ?ABD = ?ACD.

Answer:

Let us join AD.

In ?ABD and ?ACD,

AB = AC (Given)

BD = CD (Given)

AD = AD (Common side)

? ?ABD ?ACD (By SSS congruence rule)

? ?ABD = ?ACD (By CPCT)

Video Solution

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:

In ?ABC,

AB = AC (Given)

? ?ACB = ?ABC (Angles opposite to equal sides of a triangle are also equal)

In ?ACD,

AC = AD

? ?ADC = ?ACD (Angles opposite to equal sides of a triangle are also equal)

In ?BCD,

?ABC + ?BCD + ?ADC = 180º (Angle sum property of a triangle)

? ?ACB + ?ACB +?ACD + ?ACD = 180º

? 2(?ACB + ?ACD) = 180º

? 2(?BCD) = 180º

? ?BCD = 90º

Video Solution

Question 7:

ABC is a right angled triangle in which ?A = 90º and AB = AC. Find ?B and ?C.

Answer:

It is given that

AB = AC

? ?C = ?B (Angles opposite to equal sides are also equal)

In ?ABC,

?A + ?B + ?C = 180º (Angle sum property of a triangle)

? 90º + ?B + ?C = 180º

? 90º + ?B + ?B = 180º

? 2 ?B = 90º

? ?B = 45º

? ?B = ?C = 45º

Video Solution

Question 8:

Show that the angles of an equilateral triangle are 60º each.

Answer:

Let us consider that ABC is an equilateral triangle.

Therefore, AB = BC = AC

AB = AC

? ?C = ?B (Angles opposite to equal sides of a triangle are equal)

Also,

AC = BC

? ?B = ?A (Angles opposite to equal sides of a triangle are equal)

Therefore, we obtain

?A = ?B = ?C

In ?ABC,

?A + ?B + ?C = 180°

? ?A + ?A + ?A = 180°

? 3?A = 180°

? ?A = 60°

? ?A = ?B = ?C = 60°

Hence, in an equilateral triangle, all interior angles are of measure 60º.

Video Solution

Page No 128:

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 P, 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:

(i) In ?ABD and ?ACD,

AB = AC (Given)

BD = CD (Given)

AD = AD (Common)

? ?ABD ? ?ACD (By SSS congruence rule)

? ?BAD = ?CAD (By CPCT)

? ?BAP = ?CAP …. (1)

(ii) In ?ABP and ?ACP,

AB = AC (Given)

?BAP = ?CAP [From equation (1)]

AP = AP (Common)

? ?ABP ? ?ACP (By SAS congruence rule)

? BP = CP (By CPCT) … (2)

(iii) From equation (1),

?BAP = ?CAP

Hence, AP bisects ?A.

In ?BDP and ?CDP,

BD = CD (Given)

DP = DP (Common)

BP = CP [From equation (2)]

? ?BDP ? ?CDP (By S.S.S. Congruence rule)

? ?BDP = ?CDP (By CPCT) … (3)

Hence, AP bisects ?D.

(iv) ?BDP ? ?CDP

? ?BPD = ?CPD (By CPCT) …. (4)

?BPD + ?CPD = 180 (Linear pair angles)

?BPD + ?BPD = 180

2?BPD = 180 [From equation (4)]

?BPD = 90 … (5)

From equations (2) and (5), it can be said that AP is the perpendicular bisector of BC.

Video Solution

Question 2:

AD is an altitude of an isosceles triangles ABC in which AB = AC. Show that

(i) AD bisects BC (ii) AD bisects ?A.

Answer:

(i) In ?BAD and ?CAD,

?ADB = ?ADC (Each 90º as AD is an altitude)

AB = AC (Given)

AD = AD (Common)

??BAD ? ?CAD (By RHS Congruence rule)

? BD = CD (By CPCT)

Hence, AD bisects BC.

(ii) Also, by CPCT,

?BAD = ?CAD

Hence, AD bisects ?A.

Video Solution

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:

(i) In ?ABC, AM is the median to BC.

? BM = BC

In ?PQR, PN is the median to QR.

? QN = QR

However, BC = QR

? BC = QR

? BM = QN … (1)

In ?ABM and ?PQN,

AB = PQ (Given)

BM = QN [From equation (1)]

AM = PN (Given)

? ?ABM ? ?PQN (SSS congruence rule)

?ABM = ?PQN (By CPCT)

?ABC = ?PQR … (2)

(ii) In ?ABC and ?PQR,

AB = PQ (Given)

?ABC = ?PQR [From equation (2)]

BC = QR (Given)

? ?ABC ? ?PQR (By SAS congruence rule)

Video Solution

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 ?BEC and ?CFB,

?BEC = ?CFB (Each 90°)

BC = CB (Common)

BE = CF (Given)

? ?BEC ? ?CFB (By RHS congruency)

? ?BCE = ?CBF (By CPCT)

? AB = AC (Sides opposite to equal angles of a triangle are equal)

Hence, ?ABC is isosceles.

Video Solution

Question 5:

ABC is an isosceles triangle with AB = AC. Drawn AP ? BC to show that ?B = ?C.

Answer:

In ?APB and ?APC,

?APB = ?APC (Each 90º)

AB =AC (Given)

AP = AP (Common)

? ?APB ? ?APC (Using RHS congruence rule)

? ?B = ?C (By using CPCT)

Video Solution

Page No 132:

Question 1:

Show that in a right angled triangle, the hypotenuse is the longest side.

Answer:

Let us consider a right-angled triangle ABC, right-angled at B.

In ?ABC,

?A + ?B + ?C = 180° (Angle sum property of a triangle)

?A + 90º + ?C = 180°

?A + ?C = 90°

Hence, the other two angles have to be acute (i.e., less than 90º).

? ?B is the largest angle in ?ABC.

? ?B > ?A and ?B > ?C

? AC > BC and AC > AB

[In any triangle, the side opposite to the larger (greater) angle is longer.]

Therefore, AC is the largest side in ?ABC.

However, AC is the hypotenuse of ?ABC. Therefore, hypotenuse is the longest side in a right-angled triangle.

Video Solution

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:

In the given figure,

?ABC + ?PBC = 180° (Linear pair)

? ?ABC = 180° – ?PBC … (1)

Also,

?ACB + ?QCB = 180°

?ACB = 180° – ?QCB … (2)

As ?PBC < ?QCB,

? 180º – ?PBC > 180º – ?QCB

? ?ABC > ?ACB [From equations (1) and (2)]

? AC > AB (Side opposite to the larger angle is larger.)

Video Solution

Question 3:

In the given figure, ?B < ?A and ?C < ?D. Show that AD < BC.

Answer:

In ?AOB,

?B < ?A

? AO < BO (Side opposite to smaller angle is smaller) … (1)

In ?COD,

?C < ?D

? OD < OC (Side opposite to smaller angle is smaller) … (2)

On adding equations (1) and (2), we obtain

AO + OD < BO + OC

AD < BC

Video Solution

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:

Let us join AC.

In ?ABC,

AB < BC (AB is the smallest side of quadrilateral ABCD)

? ?2 < ?1 (Angle opposite to the smaller side is smaller) … (1)

In ?ADC,

AD < CD (CD is the largest side of quadrilateral ABCD)

? ?4 < ?3 (Angle opposite to the smaller side is smaller) … (2)

On adding equations (1) and (2), we obtain

?2 + ?4 < ?1 + ?3

? ?C < ?A

? ?A > ?C

Let us join BD.

In ?ABD,

AB < AD (AB is the smallest side of quadrilateral ABCD)

? ?8 < ?5 (Angle opposite to the smaller side is smaller) … (3)

In ?BDC,

BC < CD (CD is the largest side of quadrilateral ABCD)

? ?7 < ?6 (Angle opposite to the smaller side is smaller) … (4)

On adding equations (3) and (4), we obtain

?8 + ?7 < ?5 + ?6

? ?D < ?B

? ?B > ?D

Video Solution

Question 5:

In the given figure, PR > PQ and PS bisects ?QPR. Prove that ?PSR >?PSQ.

Answer:

As PR > PQ,

? ?PQR > ?PRQ (Angle opposite to larger side is larger) … (1)

PS is the bisector of ?QPR.

??QPS = ?RPS … (2)

?PSR is the exterior angle of ?PQS.

? ?PSR = ?PQR + ?QPS … (3)

?PSQ is the exterior angle of ?PRS.

? ?PSQ = ?PRQ + ?RPS … (4)

Adding equations (1) and (2), we obtain

?PQR + ?QPS > ?PRQ + ?RPS

? ?PSR > ?PSQ [Using the values of equations (3) and (4)]

Video Solution

Page No 133:

Question 6:

Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Answer:

Let us take a line l and from point P (i.e., not on line l), draw two line segments PN and PM. Let PN be perpendicular to line l and PM is drawn at some other angle.

In ?PNM,

?N = 90º

?P + ?N + ?M = 180º (Angle sum property of a triangle)

?P + ?M = 90º

Clearly, ?M is an acute angle.

? ?M < ?N

? PN < PM (Side opposite to the smaller angle is smaller)

Similarly, by drawing different line segments from P to l, it can be proved that PN is smaller in comparison to them.

Therefore, it can be observed that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Video Solution

Question 1:

ABC is a triangle. Locate a point in the interior of ?ABC which is equidistant from all the vertices of ?ABC.

Answer:

Circumcentre of a triangle is always equidistant from all the vertices of that triangle. Circumcentre is the point where perpendicular bisectors of all the sides of the triangle meet together.

In ?ABC, we can find the circumcentre by drawing the perpendicular bisectors of sides AB, BC, and CA of this triangle. O is the point where these bisectors are meeting together. Therefore, O is the point which is equidistant from all the vertices of ?ABC.

Video Solution

Question 2:

In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

Answer:

The point which is equidistant from all the sides of a triangle is called the incentre of the triangle. Incentre of a triangle is the intersection point of the angle bisectors of the interior angles of that triangle.

Here, in ?ABC, we can find the incentre of this triangle by drawing the angle bisectors of the interior angles of this triangle. I is the point where these angle bisectors are intersecting each other. Therefore, I is the point equidistant from all the sides of ?ABC.

Video Solution

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 parlor should be equidistant from A, B and C)

Answer:

Maximum number of persons can approach the ice-cream parlour if it is equidistant from A, B and C. Now, A, B and C form a triangle. In a triangle, the circumcentre is the only point that is equidistant from its vertices. So, the ice-cream parlour should be set up at the circumcentre O of ?ABC.

In this situation, maximum number of persons can approach it. We can find circumcentre O of this triangle by drawing perpendicular bisectors of the sides of this triangle.

Video Solution

Question 4:

Complete the hexagonal and star shaped rangolies (see the given figures) 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:

It can be observed that hexagonal-shaped rangoli has 6 equilateral triangles in it.

Area of ?OAB 

Area of hexagonal-shaped rangoli

Star-shaped rangoli has 12 equilateral triangles of side 5 cm in it.

Area of star-shaped rangoli =  

Therefore, star-shaped rangoli has more equilateral triangles in it.

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