NCERT Solutions Class 10 for Maths Chapter 6 Triangles : In this post, we will share with you all the detailed NCERT Solutions of Class 10 Maths Chapter 6 Triangles. This will contain both in-text and back-exercise questions for Science and Social Science, and all exercise questions for Mathematics. For all school and board level examinations, doing all the NCERT Questions is a must.

Why are NCERT Questions Important?

NCERT Questions and Answers not only help you get hold of concepts firmly and enhance your understanding, but also form the base of all types of questions asked in exams. Questions asked in exam are more or less the same type as mentioned in NCERT. Moreover, sometimes the questions in NCERT are directly asked in exams, as it is, without any changes.

Hence, it’s very important to understand NCERT Questions and Answers.

• Get NCERT Solutions of all Science Chapters of Class 10
• Get NCERT Solutions of all Mathematics Chapters of Class 10
• Get NCERT Solutions of all Social Science Chapters of Class 10
• Get Notes, PYQs, Free E-Books, and other material for Class 10
• Join our Telegram Channel for Latest Updates here

[adinserter block=”3″]

Class 10 Maths Chapter 6 – Triangles

[twy_show_add_bookmark]

Exercise 6.1 (Page 122)

Question 1:
Fill in the blanks by using the correct word given in brackets.
(i) All circles are ……………. . (congruent/similar)
(ii) All squares are …………… . (similar/congruent)
(iii) All …………….. triangles are similar. (isosceles/equilateral)
(iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are …………… and
(b) their corresponding sides are …………… (equal/proportional)

Answer 1:
(i) Similar
(ii) Similar
(iii) Equilateral
(iv) a. Equal, b. Proportional

Question 2:
Give two different examples of pair of
(i) Similar figures
(ii) Non-similar figures

Answer 2:
(i) Two twenty-rupee notes, Two two rupees coins.
(ii) One rupee coin and five rupees coin, One rupee not and ten rupees note.

Question 3:
State whether the following quadrilaterals are similar or not:

Answer 3:
From the given two figures, we can see their corresponding angles are different or unequal. Therefore they are not similar.

[adinserter block=”3″]

Exercise 6.2 (Page 128)

Question 1:
In figure. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Answer 1:
(i) Given, in △ ABC, DE∥BC

∴ AD/DB = AE/EC [Using Basic proportionality theorem]
⇒1.5/3 = 1/EC
⇒EC = 3/1.5
EC = 3×10/15 = 2 cm

Hence, EC = 2 cm.

(ii) Given, in △ ABC, DE∥BC

∴ AD/DB = AE/EC [Using Basic proportionality theorem]
⇒ AD/7.2 = 1.8 / 5.4
⇒ AD = 1.8 ×7.2/5.4 = (18/10)×(72/10)×(10/54) = 24/10
⇒ AD = 2.4

Hence, AD = 2.4 cm.

Question 2:
E and F are points on the sides PQ and PR respectively of a ΔPQR. For each of the following cases, state whether EF || QR.
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm

Answer 2:

In ΔPQR, E and F are two points on side PQ and PR respectively.
(i) PE = 3.9 cm,
EQ = 3 cm (Given)
PF = 3.6 cm, FR = 2,4 cm (Given)
∴ PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3 [By using Basic proportionality theorem]
And, PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5
So, PE/EQ ≠ PF/FR
Hence, EF is not parallel to QR.

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8cm, RF = 9cm∴ PE/QE = 4/4.5 = 40/45 = 8/9 [By using Basic proportionality theorem]And, PF/RF = 8/9So, PE/QE = PF/RFHence, EF is parallel to QR.

(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm (Given)
Here, EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm
And, FR = PR – PF = 2.56 – 0.36 = 2.20 cmSo, PE/EQ = 0.18/1.10 = 18/110 = 9/55 … (i)
And, PE/FR = 0.36/2.20 = 36/220 = 9/55 … (ii)
∴ PE/EQ = PF/FR.Hence, EF is parallel to QR

Question 3:
In the figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD

Answer 3:
In the given figure, we can see, LM || CB,
By using basic proportionality theorem, we get,

AM/AB = AL/AC……………………..(i)

Similarly, given, LN || CD and using basic proportionality theorem,
∴AN/AD = AL/AC……………………………(ii)

From equation (i) and (ii), we get,
AM/AB = AN/AD

Hence, proved.

Question 4:
In the figure, DE||AC and DF||AE. Prove that BF/FE = BE/EC

Answer 4:
In ΔABC, given as, DE || AC
Thus, by using Basic Proportionality Theorem, we get,

∴BD/DA = BE/EC ………………………………………………(i)

In ΔABC, given as, DF || AE
Thus, by using Basic Proportionality Theorem, we get,

∴BD/DA = BF/FE ………………………………………………(ii)

From equation (i) and (ii), we get
BE/EC = BF/FE

Hence, proved.

Question 5:
In the figure, DE||OQ and DF||OR, show that EF||QR.

Answer 5:
Given,
In ΔPQO, DE || OQ

So by using Basic Proportionality Theorem,
PD/DO = PE/EQ……………… ..(i)

Again given, in ΔPQO, DE || OQ ,

So by using Basic Proportionality Theorem,
PD/DO = PF/FR………………… (ii)

From equation (i) and (ii), we get,
PE/EQ = PF/FR

Therefore, by converse of Basic Proportionality Theorem,
EF || QR, in ΔPQR.

[adinserter block=”3″]

Question 6:
In the figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Answer 6:
In ΔOPQ, AB || PQ (Given)
∴ OA/AP = OB/BQ …(i) [By using Basic Proportionality Theorem]
In ΔOPR, AC || PR (Given)
∴ OA/AP = OC/CR …(ii) [By using Basic Proportionality Theorem]

From equation (i) and (ii), we get

OB/BQ = OC/CR
In ΔOQR, BC || QR. [By converse of Basic Proportionality Theorem]

Question 7:
Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Answer 7:
Given, in ΔABC, D is the midpoint of AB such that AD=DB.
A line parallel to BC intersects AC at E as shown in above figure such that DE || BC.

We have to prove that E is the mid point of AC.

Since, D is the mid-point of AB.

∴ AD=DB
⇒AD/DB = 1 …………………………. (i)

In ΔABC, DE || BC,
By using Basic Proportionality Theorem,

Therefore, AD/DB = AE/EC
From equation (i), we can write,

⇒ 1 = AE/EC
∴ AE = EC

Hence, proved, E is the midpoint of AC.

Question 8:
Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

Answer 8:
Given, in ΔABC, D and E are the mid points of AB and AC respectively, such that,

AD=BD and AE=EC.

We have to prove that: DE || BC.
Since, D is the midpoint of AB

∴ AD=DB
⇒AD/BD = 1……………………………….. (i)

Also given, E is the mid-point of AC.

∴ AE=EC
⇒ AE/EC = 1

From equation (i) and (ii), we get,
AD/BD = AE/EC

By converse of Basic Proportionality Theorem,
DE || BC

Hence, proved.

Question 9:
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.

Answer 9:
Given, ABCD is a trapezium where AB || DC and diagonals AC and BD intersect each other at O.

We have to prove, AO/BO = CO/DO
From the point O, draw a line EO touching AD at E, in such a way that,
EO || DC || AB

In ΔADC, we have OE || DC
Therefore, By using Basic Proportionality Theorem
AE/ED = AO/CO ……………..(i)

Now, In ΔABD, OE || AB
Therefore, By using Basic Proportionality Theorem
DE/EA = DO/BO…………….(ii)

From equation (i) and (ii), we get,
AO/CO = BO/DO
⇒AO/BO = CO/DO

Hence, proved.

Question 10:
The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium.

Answer 10:
Given, Quadrilateral ABCD where AC and BD intersects each other at O such that,

AO/BO = CO/DO

We have to prove here, ABCD is a trapezium
From the point O, draw a line EO touching AD at E, in such a way that,
EO || DC || AB

In ΔDAB, EO || AB
Therefore, By using Basic Proportionality Theorem
DE/EA = DO/OB ……………………(i)

Also, given,
AO/BO = CO/DO
⇒ AO/CO = BO/DO
⇒ CO/AO = DO/BO
⇒DO/OB = CO/AO …………………………..(ii)

From equation (i) and (ii), we get
DE/EA = CO/AO

Therefore, By using converse of Basic Proportionality Theorem,
EO || DC also EO || AB
⇒ AB || DC.

Hence, quadrilateral ABCD is a trapezium with AB || CD.

[adinserter block=”3″]

Exercise 6.3 (Page 138)

Question 1:
State which pairs of triangles in Figure, are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:

Answer 1:
(i) In  ΔABC and ΔPQR, we have
∠A = ∠P = 60° (Given)
∠B = ∠Q = 80° (Given)
∠C = ∠R = 40° (Given)
∴ ΔABC ~ ΔPQR (AAA similarity criterion)

(ii) In  ΔABC and ΔPQR, we have
AB/QR = BC/RP = CA/PQ
∴  ΔABC ~ ΔQRP (SSS similarity criterion)

(iii) In ΔLMP and ΔDEF, we have
LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6
MP/DE = 2/4 = 1/2
PL/DF = 3/6 = 1/2
LM/EF= 2.7/5 = 27/50
Here, MP/DE = PL/DF ≠ LM/EF
Hence, ΔLMP and ΔDEF are not similar.

(iv) In ΔMNL and ΔQPR, we have
MN/QP = ML/QR = 1/2
∠M = ∠Q = 70°
∴ ΔMNL ~ ΔQPR (SAS similarity criterion)

(v) In ΔABC and ΔDEF, we have
AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80°
Here, AB/DF = 2.5/5 = 1/2
And, BC/EF = 3/6 = 1/2
⇒ ∠B ≠ ∠F
Hence, ΔABC and ΔDEF are not similar.

(vi) In ΔDEF,we have
∠D + ∠E + ∠F = 180° (sum of angles of a triangle)
⇒ 70° + 80° + ∠F = 180°
⇒ ∠F = 180° – 70° – 80°
⇒ ∠F = 30°
In PQR, we have
∠P + ∠Q + ∠R = 180 (Sum of angles of Δ)
⇒ ∠P + 80° + 30° = 180°
⇒ ∠P = 180° – 80° -30°
⇒ ∠P = 70°
In ΔDEF and ΔPQR, we have
∠D = ∠P = 70°
∠F = ∠Q = 80°
∠F = ∠R = 30°
Hence, ΔDEF ~ ΔPQR (AAA similarity criterion)

Question 2:
In the figure, ΔODC ∝ ¼ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.

Answer 2:
DOB is a straight line.
Therefore, ∠DOC + ∠ COB = 180°
⇒ ∠DOC = 180° – 125°
= 55°

In ΔDOC,
∠DCO + ∠ CDO + ∠ DOC = 180°
(Sum of the measures of the angles of a triangle is 180º.)
⇒ ∠DCO + 70º + 55º = 180°
⇒ ∠DCO = 55°
It is given that ΔODC ~ ΔOBA.∴ ∠OAB = ∠OCD [Corresponding angles are equal in similar triangles.]
⇒ ∠ OAB = 55°
∴ ∠OAB = ∠OCD [Corresponding angles are equal in similar triangles.]
⇒ ∠OAB = 55°

Question 3:
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that AO/OC = OB/OD

Answer 3:

In ΔDOC and ΔBOA,
AB || CD, thus alternate interior angles will be equal,
∴∠CDO = ∠ABO

Similarly,
∠DCO = ∠BAO

Also, for the two triangles ΔDOC and ΔBOA, vertically opposite angles will be equal;
∴∠DOC = ∠BOA

Hence, by AAA similarity criterion,
ΔDOC ~ ΔBOA

Thus, the corresponding sides are proportional.
DO/BO = OC/OA
⇒OA/OC = OB/OD

Hence, proved.

Question 4:
In the fig.6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.

Answer 4:
In ΔPQR,
∠PQR = ∠PRQ
∴ PQ = PR ………………………(i)

Given,
QR/QS = QT/PRUsing equation (i), we get
QR/QS = QT/QP……………….(ii)

In ΔPQS and ΔTQR, by equation (ii),
QR/QS = QT/QP
∠Q = ∠Q
∴ ΔPQS ~ ΔTQR [By SAS similarity criterion]

Question 5:
S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.

Answer 5:
Given, S and T are point on sides PR and QR of ΔPQR
And ∠P = ∠RTS.

In ΔRPQ and ΔRTS,
∠RTS = ∠QPS (Given)
∠R = ∠R (Common angle)

∴ ΔRPQ ~ ΔRTS (AA similarity criterion)

Question 6:
In the figure, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.

Answer 6:
It is given that ΔABE ≅ ΔACD.
∴ AB = AC [By cpct] …(i)
And, AD = AE [By cpct] …(ii)
In ΔADE and ΔABC,
AD/AB = AE/AC [Dividing equation (ii) by (i)]∠A = ∠A [Common angle]
∴ ΔADE ~ ΔABC [By SAS similarity criterion]

[adinserter block=”3″]

Question 7:
In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:

(i) ΔAEP ~ ΔCDP
(ii) ΔABD ~ ΔCBE
(iii) ΔAEP ~ ΔADB
(iv) ΔPDC ~ ΔBEC

Answer 7:
Given, altitudes AD and CE of ΔABC intersect each other at the point P.
(i) In ΔAEP and ΔCDP,

∠AEP = ∠CDP (90° each)
∠APE = ∠CPD (Vertically opposite angles)

Hence, by AA similarity criterion,
ΔAEP ~ ΔCDP

(ii) In ΔABD and ΔCBE,
∠ADB = ∠CEB ( 90° each)
∠ABD = ∠CBE (Common Angles)

Hence, by AA similarity criterion,
ΔABD ~ ΔCBE

(iii) In ΔAEP and ΔADB,
∠AEP = ∠ADB (90° each)
∠PAE = ∠DAB (Common Angles)

Hence, by AA similarity criterion,
ΔAEP ~ ΔADB

(iv) In ΔPDC and ΔBEC,
∠PDC = ∠BEC (90° each)
∠PCD = ∠BCE (Common angles)

Hence, by AA similarity criterion,
ΔPDC ~ ΔBEC

Question 8:
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.

Answer 8:
Given, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Consider the figure below,

In ΔABE and ΔCFB,
∠A = ∠C (Opposite angles of a parallelogram)
∠AEB = ∠CBF (Alternate interior angles as AE || BC)
∴ ΔABE ~ ΔCFB (AA similarity criterion)

Question 9:
In the figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:

(i) ΔABC ~ ΔAMP
(ii) CA/PA = BC/MP

Answer 9:
Given, ABC and AMP are two right triangles, right angled at B and M respectively.

(i) In ΔABC and ΔAMP, we have,
∠CAB = ∠MAP (common angles)
∠ABC = ∠AMP = 90° (each 90°)
∴ ΔABC ~ ΔAMP (AA similarity criterion)

(ii) As, ΔABC ~ ΔAMP (AA similarity criterion)
If two triangles are similar then the corresponding sides are always equal,
Hence, CA/PA = BC/MP

Question 10:
CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, Show that:

(i) CD/GH = AC/FG
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF

Answer 10:
Given, CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively.

(i) From the given condition,
ΔABC ~ ΔFEG.
∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE

Since, ∠ACB = ∠FGE
∴ ∠ACD = ∠FGH (Angle bisector)

And, ∠DCB = ∠HGE (Angle bisector)
In ΔACD and ΔFGH,

∠A = ∠F
∠ACD = ∠FGH
∴ ΔACD ~ ΔFGH (AA similarity criterion)
⇒CD/GH = AC/FG

(ii) In ΔDCB and ΔHGE,
∠DCB = ∠HGE (Already proved)
∠B = ∠E (Already proved)
∴ ΔDCB ~ ΔHGE (AA similarity criterion)

(iii) In ΔDCA and ΔHGF,
∠ACD = ∠FGH (Already proved)
∠A = ∠F (Already proved)
∴ ΔDCA ~ ΔHGF (AA similarity criterion)

[adinserter block=”3″]

Question 11:
In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.

Answer 11:
Given, ABC is an isosceles triangle.
∴ AB = AC
⇒ ∠ABD = ∠ECF

In ΔABD and ΔECF,
∠ADB = ∠EFC (Each 90°)
∠BAD = ∠CEF (Already proved)
∴ ΔABD ~ ΔECF (using AA similarity criterion)

Question 12:
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see Fig 6.41). Show that ΔABC ~ ΔPQR.

Answer 12:
Given, ΔABC and ΔPQR, AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR
i.e. AB/PQ = BC/QR = AD/PM

We have to prove: ΔABC ~ ΔPQR
As we know here,
AB/PQ = BC/QR = AD/PM

⇒AB/PQ = BC/QR = AD/PM (D is the midpoint of BC. M is the midpoint of QR)
⇒ ΔABD ~ ΔPQM [SSS similarity criterion]
∴ ∠ABD = ∠PQM [Corresponding angles of two similar triangles are equal]
⇒ ∠ABC = ∠PQR

In ΔABC and ΔPQR

AB/PQ = BC/QR ………………………….(i)
∠ABC = ∠PQR ……………………………(ii)

From equation (i) and (ii), we get,
ΔABC ~ ΔPQR [SAS similarity criterion]

Question 13:
D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD

Answer 13:
Given, D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC.

In ΔADC and ΔBAC,
∠ADC = ∠BAC (Already given)
∠ACD = ∠BCA (Common angles)
∴ ΔADC ~ ΔBAC (AA similarity criterion)

We know that corresponding sides of similar triangles are in proportion.
∴ CA/CB = CD/CA
⇒ CA² = CB.CD.

Hence, proved.

Question 14:
Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR.

Answer 14:
Given: Two triangles ΔABC and ΔPQR in which AD and PM are medians such that;
AB/PQ = AC/PR = AD/PM

We have to prove, ΔABC ~ ΔPQR
Let us construct first: Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN.

In ΔABD and ΔCDE, we have
AD = DE [By Construction.]
BD = DC [Since, AP is the median]

and, ∠ADB = ∠CDE [Vertically opposite angles]
∴ ΔABD ≅ ΔCDE [SAS criterion of congruence]
⇒ AB = CE [By CPCT] …………………………..(i)

Also, in ΔPQM and ΔMNR,
PM = MN [By Construction.]
QM = MR [Since, PM is the median]
and, ∠PMQ = ∠NMR [Vertically opposite angles]
∴ ΔPQM = ΔMNR [SAS criterion of congruence]
⇒ PQ = RN [CPCT] ………………………………(ii)

Now, AB/PQ = AC/PR = AD/PM
From equation (i) and (ii),

⇒CE/RN = AC/PR = AD/PM
⇒ CE/RN = AC/PR = 2AD/2PM
⇒ CE/RN = AC/PR = AE/PN [Since 2AD = AE and 2PM = PN]
∴ ΔACE ~ ΔPRN [SSS similarity criterion]

Therefore, ∠2 = ∠4
Similarly, ∠1 = ∠3

∴ ∠1 + ∠2 = ∠3 + ∠4
⇒ ∠A = ∠P …………………………………………….(iii)

Now, in ΔABC and ΔPQR, we have
AB/PQ = AC/PR (Already given)

From equation (iii),
∠A = ∠P
∴ ΔABC ~ ΔPQR [ SAS similarity criterion]

Question 15:
A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Answer 15:
Given, Length of the vertical pole = 6m
Shadow of the pole = 4 m
Let Height of tower = h m
Length of shadow of the tower = 28 m

In ΔABC and ΔDEF,
∠C = ∠E (angular elevation of sum)
∠B = ∠F = 90°
∴ ΔABC ~ ΔDEF (AA similarity criterion)
∴ AB/DF = BC/EF (If two triangles are similar corresponding sides are proportional)
∴ 6/h = 4/28
⇒h = (6×28)/4
⇒ h = 6 × 7
⇒ h = 42 m

Hence, the height of the tower is 42 m.

[adinserter block=”3″]

Question 16:
If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR prove that AB/PQ = AD/PM.

Answer 16:

It is given that ΔABC ~ ΔPQR
We know that the corresponding sides of similar triangles are in proportion.∴ AB/PQ = AC/PR = BC/QR …(i)
Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R …(ii)
Since AD and PM are medians, they will divide their opposite sides.∴ BD = BC/2 and QM = QR/2 …(iii)
From equations (i) and (iii), we get
AB/PQ = BD/QM …(iv)
In ΔABD and ΔPQM,
∠B = ∠Q [Using equation (ii)]
AB/PQ = BD/QM [Using equation (iv)]
∴ ΔABD ~ ΔPQM (By SAS similarity criterion)⇒ AB/PQ = BD/QM = AD/PM.

Exercise 6.4 (Page 143)

Question 1:
Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.

Answer 1:
Given, ΔABC ~ ΔDEF,
Area of ΔABC = 64 cm2
Area of ΔDEF = 121 cm2
EF = 15.4 cm

As we know, if two triangles are similar, ratio of their areas are equal to the square of the ratio of their corresponding sides,
= AC2/DF2 = BC2/EF2
∴ 64/121 = BC2/EF2

⇒ (8/11)2 = (BC/15.4)2
⇒ 8/11 = BC/15.4
⇒ BC = 8×15.4/11
⇒ BC = 8 × 1.4
⇒ BC = 11.2 cm

Question 2:
Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.

Answer 2:
Given, ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect each other at point O.

In ΔAOB and ΔCOD, we have
∠1 = ∠2 (Alternate angles)
∠3 = ∠4 (Alternate angles)
∠5 = ∠6 (Vertically opposite angle)
∴ ΔAOB ~ ΔCOD [AAA similarity criterion]

As we know, If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides. Therefore,

Area of (ΔAOB)/Area of (ΔCOD) = AB²/CD²
= (2CD)²/CD² [∴ AB = 2CD]
∴ Area of (ΔAOB)/Area of (ΔCOD)
= 4CD²/CD² = 4/1

Hence, the required ratio of the area of ΔAOB and ΔCOD = 4:1

Question 3:
In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area (ΔABC)/area (ΔDBC) = AO/DO.

Answer 3:

Given: ABC and DBC are triangles on the same base BC. Ad intersects BC at O.
To Prove: area (ΔABC)/area (ΔDBC) = AO/DO.
Construction: Let us draw two perpendiculars AP and DM on line BC.
Proof: We know that area of a triangle = 1/2 × Base × Height

In ΔAPO and ΔDMO,
∠APO = ∠DMO (Each equals to 90°)
∠AOP = ∠DOM (Vertically opposite angles)
∴ ΔAPO ~ ΔDMO (By AA similarity criterion)∴ AP/DM = AO/DO
⇒ area (ΔABC)/area (ΔDBC) = AO/DO.

Question 4:
If the areas of two similar triangles are equal, prove that they are congruent.

Answer 4:
Say ΔABC and ΔPQR are two similar triangles and equal in area

Now let us prove ΔABC ≅ ΔPQR.
Since, ΔABC ~ ΔPQR

∴ Area of (ΔABC)/Area of (ΔPQR) = BC²/QR²
⇒ BC²/QR² =1 [Since, Area(ΔABC) = (ΔPQR)
⇒ BC²/QR²
⇒ BC = QR

Similarly, we can prove that
AB = PQ and AC = PR
Thus, ΔABC ≅ ΔPQR [SSS criterion of congruence]

Question 5:
D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC.

Answer 5:
Given, D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC.

To Find: area(ΔDEF) and area(ΔABC)

Solution: In ΔABC, we have
F is the mid point of AB (Given)
E is the mid-point of AC (Given)
So, by the mid-point theorem, we have
FE || BC and FE = 1/2BC
⇒ FE || BC and FE || BD [BD = 1/2BC]
∴ BDEF is parallelogram [Opposite sides of parallelogram are equal and parallel]
Similarly in ΔFBD and ΔDEF, we have
FB = DE (Opposite sides of parallelogram BDEF)
FD = FD (Common)
BD = FE (Opposite sides of parallelogram BDEF)
∴ ΔFBD ≅ ΔDEF
Similarly, we can prove that
ΔAFE ≅ ΔDEF
ΔEDC ≅ ΔDEF
If triangles are congruent,then they are equal in area.
So, area(ΔFBD) = area(ΔDEF) …(i)
area(ΔAFE) = area(ΔDEF) …(ii)
and, area(ΔEDC) = area(ΔDEF) …(iii)
Now, area(ΔABC) = area(ΔFBD) + area(ΔDEF) + area(ΔAFE) + area(ΔEDC) …(iv)
area(ΔABC) = area(ΔDEF) + area(ΔDEF) + area(ΔDEF) + area(ΔDEF)
⇒ area(ΔDEF) = 1/4area(ΔABC) [From (i), (ii) and (iii)]
⇒ area(ΔDEF)/area(ΔABC) = 1/4

Hence, area(ΔDEF):area(ΔABC) = 1:4

Question 6:
Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Answer 6:

Given: AM and DN are the medians of triangles ABC and DEF respectively and ΔABC ~ ΔDEF.
To Prove: area(ΔABC)/area(ΔDEF) = AM²/DN²
Proof: ΔABC ~ ΔDEF (Given)
∴ area(ΔABC)/area(ΔDEF) = (AB²/DE²) …(i)
and, AB/DE = BC/EF = CA/FD …(ii)

In ΔABM and ΔDEN, we have
∠B = ∠E [Since ΔABC ~ ΔDEF]
AB/DE = BM/EN [Prove in (i)]
∴ ΔABC ~ ΔDEF [By SAS similarity criterion]
⇒ AB/DE = AM/DN …(iii)
∴ ΔABM ~ ΔDEN
As the areas of two similar triangles are proportional to the squares of the corresponding sides.
∴ area(ΔABC)/area(ΔDEF) = AB²/DE² = AM²/DN²

[adinserter block=”3″]

Question 7:
Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Answer 7:

Given, ABCD is a square whose one diagonal is AC. ΔAPC and ΔBQC are two equilateral triangles described on the diagonals AC and side BC of the square ABCD.

Area(ΔBQC) = ½ Area(ΔAPC)

Since, ΔAPC and ΔBQC are both equilateral triangles, as per given,

∴ ΔAPC ~ ΔBQC [AAA similarity criterion]

∴ area(ΔAPC)/area(ΔBQC) = (AC²/BC²) = AC²/BC²

Since, Diagonal = √2 side = √2 BC = AC

⇒ area(ΔAPC) = 2 × area(ΔBQC)

⇒ area(ΔBQC) = 1/2area(ΔAPC)

Hence, proved.

Question 8:
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is
(A) 2 : 1
(B) 1 : 2
(C) 4 : 1
(D) 1 : 4

Answer 8:

ΔABC and ΔBDE are two equilateral triangle.
D is the mid point of BC.
∴ BD = DC = 1/2BC
Let each side of triangle is 2a.As, ΔABC ~ ΔBDE
∴ area(ΔABC)/area(ΔBDE)
= AB²/BD² 
= (2a)²/(a)²
 = 4a²/a²
 = 4/1
= 4:1
Hence, the correct option is (C).

Question 9:
Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio
(A) 2 : 3
(B) 4 : 9
(C) 81 : 16
(D) 16 : 81

Answer 9:

Let ABC and DEF are two similarity triangles
ΔABC ~ ΔDEF (Given)
and,
 AB/DE = AC/DF
= BC/EF = 4/9 (Given)
∴ area(ΔABC)/area(ΔDEF) = AB²/DE² [the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides]∴ area(ΔABC)/area(ΔDEF)
= (4/9)²
 = 16/81
= 16:81
Hence, the correct option is (D).

[adinserter block=”3″]

Exercise 6.5 (Page 150)

Question 1:
Sides of triangles are given below. Determine which of them are right triangles? In case of a right triangle, write the length of its hypotenuse.

(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm

Answer 1:
(i) Given, sides of the triangle are 7 cm, 24 cm, and 25 cm.
Squaring the lengths of the sides of the, we will get 49, 576, and 625.

49 + 576 = 625
(7)2 + (24)2 = (25)2

Therefore, the above equation satisfies, Pythagoras theorem. Hence, it is right angled triangle.
Length of Hypotenuse = 25 cm

(ii) Given, sides of the triangle are 3 cm, 8 cm, and 6 cm.
Squaring the lengths of these sides, we will get 9, 64, and 36.

Clearly, 9 + 36 ≠ 64
Or, 32 + 62 ≠ 82

Therefore, the sum of the squares of the lengths of two sides is not equal to the square of the length of the hypotenuse.

(iii) Given, sides of triangle’s are 50 cm, 80 cm, and 100 cm.
Squaring the lengths of these sides, we will get 2500, 6400, and 10000.

However, 2500 + 6400 ≠ 10000
Or, 502 + 802 ≠ 1002
As you can see, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.

Therefore, the given triangle does not satisfies Pythagoras theorem.
Hence, it is not a right triangle.

(iv) Given, sides are 13 cm, 12 cm, and 5 cm.
Squaring the lengths of these sides, we will get 169, 144, and 25.

Thus, 144 +25 = 169
Or, 122 + 52 = 132

The sides of the given triangle are satisfying Pythagoras theorem.
Therefore, it is a right triangle.
Hence, length of the hypotenuse of this triangle is 13 cm.

Question 2:
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM × MR.

Answer 2:

Proof: In ΔPQM,
we have
PQ² = PM² + QM² [By Pythagoras theorem]
Or,
PM² = PQ² – QM² …(i)
In ΔPMR, we havePR² = PM² + MR² [By Pythagoras theorem]
Or, PM² = PR² – MR² …(ii)
Adding (i) and (ii),
we get2PM² = (PQ² + PM²) – (QM² + MR²)
= QR² – QM² – MR²       
 [∴ QR² = PQ² + PR²]
= (QM + MR)² – QM² – MR²
= 2QM × MR∴ PM² 
= QM × MR

Question 3:
 In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that
(i) AB² = BC × BD
(ii) AC² = BC × DC
(iii) AD² = BD × CD

Answer 3:
(i) In ΔADB and ΔCAB,
∠DAB = ∠ACB (Each 90°)
∠ABD = ∠CBA (Common angles)

∴ ΔADB ~ ΔCAB [AA similarity criterion]

⇒ AB/CB = BD/AB
⇒ AB² = CB × BD

(ii) Let ∠CAB = x

In ΔCBA,
∠CBA = 180° – 90° – x
∠CBA = 90° – x

Similarly, in ΔCAD
∠CAD = 90° – ∠CBA
= 90° –x
∠CDA = 180° – 90° – (90° – x)
∠CDA = x

In ΔCBA and ΔCAD, we have
∠CBA = ∠CAD
∠CAB = ∠CDA
∠ACB = ∠DCA (Each 90°)

∴ ΔCBA ~ ΔCAD [AAA similarity criterion]
⇒ AC/DC = BC/AC
⇒ AC² =  DC × BC

(iii) In ΔDCA and ΔDAB,
∠DCA = ∠DAB (Each 90°)
∠CDA = ∠ADB (common angles)

∴ ΔDCA ~ ΔDAB [AA similarity criterion]

⇒ DC/DA = DA/DA
⇒ AD² = BD × CD

Question 4:
ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC² .

Answer 4:

Given that ΔABC is an isosceles triangle right angled at C.
In ΔACB, ∠C = 90°AC = BC (Given)
AB² = AC² + BC² ([By using Pythagoras theorem]
= AC² + AC² [Since, AC = BC]
AB² = 2AC²

Question 5:
ABC is an isosceles triangle with AC = BC. If AB² = 2AC², prove that ABC is a right triangle.

Answer 5:
Given, ΔABC is an isosceles triangle having AC = BC and AB² = 2AC²

In ΔACB,
AC = BC
AB² = 2AC²
AB² = AC² + AC²
= AC² + BC² [Since, AC = BC]

Hence, by Pythagoras theorem  ΔABC is right angle triangle.

[adinserter block=”3″]

Question 6:
ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Answer 6:
Given, ABC is an equilateral triangle of side 2a.

Draw, AD ⊥ BC
In ΔADB and ΔADC,

AB = AC
AD = AD
∠ADB = ∠ADC [Both are 90°]

Therefore, ΔADB ≅ ΔADC by RHS congruence.
Hence, BD = DC [by CPCT]

In right angled ΔADB,

AB² = AD² + BD²
(2a)² = AD² + a² 
⇒ AD^{2 =} 4a² – a²
⇒ AD^{2 =} 3a²
⇒ AD ⁼ √3a

Question 7:
Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.

Answer 7:

ABCD is a rhombus whose diagonals AC and BD intersect at O. [Given]
We have to prove that,
 AB² + BC² + CD² + AD² = AC² + BD²
Since, the diagonals of a rhombus bisect each other at right angles.
Therefore, AO = CO and BO = DO
In ΔAOB
,∠AOB = 90°
AB² = AO² + BO² … (i) [By Pythagoras]
Similarly,
 AD² = AO² + DO² … (ii)
DC² = DO² + CO² … (iii)

BC² = CO² + BO² … (iv)
Adding equations (i) + (ii) + (iii) + (iv) 
we get,
AB² + AD² +DC² +BC²  =  2(AO² + BO² + DO² + CO² )
= 4AO² + 4BO²
 [Since, AO = CO and BO =DO]
= (2AO)²+ (2BO)²
= AC² + BD²

Question 8:
In Fig. 6.54, O is a point in the interior of a triangle.

ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that:
(i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE² ,
(ii) AF² + BD² + CE² = AE² + CD² + BF².

Answer 8:
Given, in ΔABC, O is a point in the interior of a triangle.
And OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.
Join OA, OB and OC

(i) By Pythagoras theorem in ΔAOF, we have
OA² = OF² + AF²

Similarly, in ΔBOD
OB² = OD² + BD²

Similarly, in ΔCOE
OC² = OE² + EC²

Adding these equations,
OA² + OB² + OC² = OF² + AF² + OD² + BD² + OE² + EC²
OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE².

(ii) AF² + BD² + EC² = (OA² – OE²) + (OC² – OD²) + (OB² – OF²)
∴ AF² + BD² + CE² = AE² + CD² + BF².

Question 9:
A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

Answer 9:

Let BA be the wall and AC be the ladder,
Therefore, by Pythagoras theorem,

AC² =AB² + BC²
10² = 8² + BC²
BC² = 100 – 64
BC² = 36
BC= 6m

Therefore, the distance of the foot of the ladder from the base of the wall is 6 m.

Question 10:
A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

Answer 10:

Let AB be the pole and AC be the wire.
By Pythagoras theorem,

AC² =AB² + BC²
24² = 18² + BC²
BC² = 576 – 324
BC² = 252
BC= 6√7m

Therefore, the distance from the base is 6√7m.

Question 11:
An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after 11/2 hours.

Answer 11:
Speed of first aeroplane = 1000 km/hr
Distance covered by first aeroplane due north in  hours (OA) = 100 × 3/2 km = 1500 km
Speed of second aeroplane = 1200 km/hr
Distance covered by second aeroplane due west in hours (OB) = 1200 × 3/2 km = 1800 km
In right angle ΔAOB, we have
AB² =AO² + OB²⇒ AB² =(1500)2 + (1800)2
⇒ AB = √2250000 + 3240000= √5490000
⇒ AB = 300√61 km
Hence, the distance between two aeroplanes will be 300√61 km.

[adinserter block=”3″]

Question 12:
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.

Answer 12:

Let CD and AB be the poles of height 11 m and 6 m.
Therefore, CP = 11 – 6 = 5 m
From the figure, it can be observed that AP = 12m
Applying Pythagoras theorem for ΔAPC, we get
AP² =PC² + AC²
(12m)² + (5m)² = (AC)²
AC² = (144+25)m² = 169 m²
AC = 13m
Therefore, the distance between their tops is 13 m.

Question 13:
 D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE² + BD² = AB² + DE².

Answer 13:

By Pythagoras theorem in ΔACE, we get
AC² +CE² = AE² ………………………………………….(i)

In ΔBCD, by Pythagoras theorem, we get
BC² +CD² = BD² ………………………………..(ii)

From equations (i) and (ii), we get,
AC² +CE² + BC² +CD² = AE² + BD² …………..(iii)

In ΔCDE, by Pythagoras theorem, we get
DE² =CD² + CE²

In ΔABC, by Pythagoras theorem, we get
AB² =AC² + CB²

Putting the above two values in equation (iii), we get
DE² + AB² = AE² + BD².

Question 14:
The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Figure). Prove that 2AB² = 2AC² + BC².

Answer 14:
Given that in ΔABC, we have
AD ⊥BC and BD = 3CD
In right angle triangles ADB and ADC, we have
AB² =AD² + BD² …(i)
AC² =AD² + DC² …(ii) [By Pythagoras theorem]

Subtracting equation (ii) from equation (i), we get
AB² – AC² = BD² – DC²= 9CD² – CD²  [∴ BD = 3CD]               
= 9CD² = 8(BC/4)²
 [Since, BC = DB + CD = 3CD + CD = 4CD]
Therefore, AB² – AC² = BC²/2
⇒ 2(AB² – AC²) = BC²
⇒ 2AB² – 2AC² = BC²
∴ 2AB² = 2AC² + BC².

Question 15:
 In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD² = 7AB².

Answer 15:

Question 16:
In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Answer 16:

Question 17:
Tick the correct answer and justify: In ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.
The angle B is:
(A) 120°
(B) 60°
(C) 90°
(D) 45°

Answer 17:
Given that, AB = 6√3 cm, AC = 12 cm, and BC = 6 cm
We can observe that
AB² = 108
AC² = 144
And, BC² = 36
AB² + BC² = AC²
The given triangle, ΔABC, is satisfying Pythagoras theorem.
Therefore, the triangle is a right triangle, right-angled at B.
∴ ∠B = 90°
Hence, the correct option is (C).

[adinserter block=”3″]

Exercise 6.6 (Page 152)

Question 1:
In Figure, PS is the bisector of ∠ QPR of ∆ PQR. Prove that QS/PQ = SR/PR

Answer 1:
Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.
Given, PS is the angle bisector of ∠QPR. Therefore,
∠QPS = ∠SPR………………………………..(i)

As per the constructed figure,
∠SPR=∠PRT(Since, PS||TR)……………(ii)
∠QPS = ∠QRT(Since, PS||TR) …………..(iii)

From the above equations, we get,
∠PRT=∠QTR

Therefore,
PT=PR

In △QTR, by basic proportionality theorem,
QS/SR = QP/PT

Since, PT=TR
Therefore,
QS/SR = PQ/PR

Hence, proved.

Question 2:
 In Fig. 6.57, D is a point on hypotenuse AC of ∆ABC, such that BD ⊥AC, DM ⊥ BC and DN ⊥ AB. Prove that: (i) DM² = DN . MC (ii) DN² = DM . AN.

Answer 2:
Given that, D is a point on hypotenuse AC of ∆ABC, DM ⊥ BC and DN ⊥ AB.
Now, join NM.
Let BD and NM intersect at O.

(i) In ∆DMC and ∆NDM,∠DMC = ∠NDM (Each equal to 90°)
∠MCD = ∠DMN
Let MCD = ∠1
Then,
∠MDC = 90° − (90°-∠1)= ∠1 (∵∠MCD + ∠MDC + ∠DMC = 180°) 
∴ ∠ODM = 90° − (90° − ∠1)= ∠1
⇒ ∠DMN = ∠1
∴ ∆DMO ~ ∆NDM  (AA similarity criterion)
∴ DM/ND = MC/DM(Corresponding sides of the similar triangles are proportional)
⇒ DM² = MC ND

(ii) In ∆DNM and ∆NAD,∠NDM = ∠AND (Each equal to 90°)
∠DNM = ∠NAD
Let ∠NAD = ∠2
Then, ∠NDA = 90° − ∠2
∵∠NDA + ∠DAN + ∠DNA = 180°
∴ ∠ODN = 90° − (90° − ∠2) = ∠2
∴ ∠DNO = ∠2
∴ ∆DNM ~ ∆NAD  (AA similarity criterion)
∴ DN/NA = DM/ND
⇒ DN/AN = DM/DN
⇒ DN² = DM×AN

Question 3:
In Figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that
AC²= AB²+ BC²+ 2 BC.BD.

Answer 3:
By applying Pythagoras Theorem in ∆ADB, we get,
AB² = AD² + DB² ……………………… (i)

Again, by applying Pythagoras Theorem in ∆ACD, we get,
AC² = AD² + DC²
AC² = AD² + (DB + BC) ²
AC² = AD² + DB² + BC² + 2DB × BC

From equation (i), we can write,
AC² = AB² + BC² + 2DB × BC

Hence, proved.

Question 4:
In Figure, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that
AC²= AB²+ BC² – 2 BC.BD.

Answer 4:
By applying Pythagoras Theorem in ∆ADB, we get,
AB² = AD² + DB²

We can write it as;
⇒ AD² = AB² − DB² ……………….. (i)

By applying Pythagoras Theorem in ∆ADC, we get,
AD² + DC² = AC²

From equation (i),
AB² − BD² + DC² = AC²
AB² − BD² + (BC − BD) ² = AC²
AC² = AB² − BD² + BC² + BD² −2BC × BD
AC² = AB² + BC² − 2BC × BD

Hence, proved.

Question 5:
In Figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that :
(i) AC² = AD² + BC.DM + 2 (BC/2) ²
(ii) AB² = AD² – BC.DM + 2 (BC/2) ²
(iii) AC² + AB² = 2 AD² + ½ BC²

Answer 5:
(i) By applying Pythagoras Theorem in ∆AMD, we get,
AM² + MD² = AD² ………………. (i)

Again, by applying Pythagoras Theorem in ∆AMC, we get,
AM² + MC² = AC²
AM² + (MD + DC) ² = AC²
(AM² + MD² ) + DC² + 2MD.DC = AC²

From equation(i), we get,
AD² + DC² + 2MD.DC = AC²

Since, DC=BC/2, thus, we get,
AD² + (BC/2) ² + 2MD.(BC/2) ² = AC²
AD² + (BC/2) ² + 2MD × BC = AC²

Hence, proved.

(ii) By applying Pythagoras Theorem in ∆ABM, we get;
AB² = AM² + MB²

= (AD² − DM²) + MB²
= (AD² − DM²) + (BD − MD) ²
= AD² − DM² + BD² + MD² − 2BD × MD
= AD² + BD² − 2BD × MD
= AD² + (BC/2)² – 2(BC/2) MD
= AD² + (BC/2)² – BC MD

Hence, proved.

(iii) By applying Pythagoras Theorem in ∆ABM, we get,
AM² + MB² = AB² ………………….… (i)

By applying Pythagoras Theorem in ∆AMC, we get,
AM² + MC² = AC² …………………..… (ii)

Adding both the equations (i) and (ii), we get,

2AM² + MB² + MC² = AB² + AC²
2AM² + (BD − DM) ² + (MD + DC) ² = AB² + AC²
2AM²+BD² + DM² − 2BD.DM + MD² + DC² + 2MD.DC = AB² + AC²
2AM² + 2MD² + BD² + DC² + 2MD (− BD + DC) = AB² + AC²
2(AM²+ MD²) + (BC/2) ² + (BC/2) ² + 2MD (-BC/2 + BC/2) ² = AB² + AC²
2AD² + BC²/2 = AB² + AC²

[adinserter block=”3″]

Question 6:
Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

Answer 6:

Now, draw AM⊥DC and BN⊥D  (produced).Proof:In right ∆AMD and ∆BNC,AD = BC (Opposite sides of a parallelogram)
AM = BN (Both are altitudes of the same parallelogram to the same base) ,
△AMD ⩭ △BNC (RHS congruence criterion)
MD = NC (CPCT) —(i)

In right △BND,
∠N = 90°
BD² = BN² + DN²(By Pythagoras theorem)
= BN² + (DC + CN)² (∵ DN = DC + CN) 
= BN² + DC² + CN² + 2DC.CN [∵ (a + b)² = a² + b² + 2ab]
= (BN² + CN²) + DC² + 2DC.CN
= BC² + DC² + 2DC.CN — (ii) (∵In right △BNC with ∠N = 90°)
BN² + CN² = BC²  (By Pythagoras theorem)

In right △AMC,
∠M = 90°
AC² = AM² + MC²(∵MC = DC – DM)
= AM² + (DC – DM)²[∵ (a + b)² = a² + b² + 2ab]
= AM² + DC² + DM² – 2DC.DM
(AM² + DM²) + DC² – 2DC.DM
= AD² + DC² – 2DC.DM

[∵ In  right triangle AMD with ∠M = 90°, AD² = AM² + DM² (By Pythagoras theorem)]
= AD² + AB² = 2DC.CN  — (iii)
[∵ DC = AB, opposite sides of parallelogram and BM = CN from eq (i)]
Now, on adding Eqs. (iii) and (ii), we get
AC² + BD² = (AD² + AB²) + (BC² + DC²)
= AB² + BC² + CD² + DA²

Question 7:
In Figure, two chords AB and CD intersect each other at the point P. Prove that :

(i) ∆APC ~ ∆ DPB
(ii) AP . PB = CP . DP

Answer 7:

Firstly, let us join CB, in the given figure.
(i) In ∆APC and ∆DPB,

∠APC = ∠DPB (Vertically opposite angles)
∠CAP = ∠BDP (Angles in the same segment for chord CB)

Therefore,
∆APC ∼ ∆DPB (AA similarity criterion)

(ii) In the above, we have proved that ∆APC ∼ ∆DPB
We know that the corresponding sides of similar triangles are proportional.

∴ AP/DP = PC/PB = CA/BD
⇒AP/DP = PC/PB
∴AP. PB = PC. DP

Hence, proved.

Question 8:
In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that:

(i) ∆ PAC ~ ∆ PDB
(ii) PA . PB = PC . PD.

Answer 8:
(i) In ∆PAC and ∆PDB,
∠P = ∠P (Common Angles)

As we know, exterior angle of a cyclic quadrilateral is ∠PCA and ∠PBD is opposite interior angle, which are both equal.
∠PAC = ∠PDB

Thus, ∆PAC ∼ ∆PDB(AA similarity criterion)

(ii) We have already proved above,
∆APC ∼ ∆DPB

We know that the corresponding sides of similar triangles are proportional.
Therefore,

AP/DP = PC/PB = CA/BD
AP/DP = PC/PB
∴ AP. PB = PC. DP

Question 9:
In Figure, D is a point on side BC of ∆ ABC such that BD/CD = AB/AC. Prove that AD is the bisector of ∠ BAC.

Answer 9:
In the given figure, let us extend BA to P such that;
AP = AC.
Now join PC.

Given, BD/CD = AB/AC
⇒ BD/CD = AP/AC

By using the converse of basic proportionality theorem, we get,
AD || PC
∠BAD = ∠APC (Corresponding angles) ……………….. (i)
And, ∠DAC = ∠ACP (Alternate interior angles) …….… (ii)

By the new figure, we have;
AP = AC
⇒ ∠APC = ∠ACP ……………………. (iii)

On comparing equations (i), (ii), and (iii), we get,
∠BAD = ∠APC

Therefore, AD is the bisector of the angle BAC.
Hence, proved.

Question 10:
Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

Answer 10:

To find AC, we have to use Pythagoras theorem in ∆ABC, is such way;
AC² = AB²+ BC²
AB² = (1.8 m) ² + (2.4 m) ²
AB² = (3.24 + 5.76) m²
AB² = 9.00 m²
⟹ AB = √9 m = 3m

Thus, the length of the string out is 3 m.
As its given, she pulls the string at the rate of 5 cm per second.
Therefore, string pulled in 12 seconds = 12 × 5 = 60 cm = 0.6 m

Let us say now, the fly is at point D after 12 seconds.
Length of string out after 12 seconds is AD.

AD = AC − String pulled by Nazima in 12 seconds
= (3.00 − 0.6) m
= 2.4 m

In ∆ADB, by Pythagoras Theorem,

AB² + BD² = AD²
(1.8 m) ² + BD² = (2.4 m) ²
BD² = (5.76 − 3.24) m² = 2.52 m²
BD = 1.587 m

Horizontal distance of fly = BD + 1.2 m
= (1.587 + 1.2) m = 2.787 m
= 2.79 m

[twy_show_add_bookmark]

[adinserter block=”3″]

For More Content related to Class 10 –

• Get NCERT Solutions of all Science Chapters of Class 10
• Get NCERT Solutions of all Mathematics Chapters of Class 10
• Get NCERT Solutions of all Social Science Chapters of Class 10
• Get Notes, PYQs, Free E-Books, and other material for Class 10
• Join our Telegram Channel for Latest Updates here

[adinserter block=”3″]