# RD Sharma Class 10 Solutions Triangles

## RD Sharma Class 10 Solutions Triangles

### RD Sharma Class 10 Solutions Triangles Exercise 4.1

Question 1.
Fill in the blanks 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, equilaterals) :
(iv) Two triangles are similar, if their corres-ponding angles are ……… (proportional, equal)
(v) Two triangles are similar, if their corres-ponding sides are ……………. (proportional, equal)
(vi) Two polygons of the same number of sides are similar, if (a) their corresponding angles are …………. and (b) their corres-ponding sides are …………….. (equal, proportional).
Solution:
(i) All circles are similar.
(ii) All squares are similar.
(iii) All equilaterals triangles are similar.
(iv) Two triangles are similar, if their corresponding angles are equal.
(v) Two triangles are similar, if their corresponding sides are proportional.
(vi) Two polygons of the same number of sides are similar, if (a) their corresponding angles are equal and (b) their corresponding sides are proportional.

Question 2.
Write the truth value (T/F) of each of the following statements :
(i) Any two similar figures are congruent.
(ii) Any two congruent figures are similar.
(iii) Two polygons are similar, if their corresponding sides are proportional.
(iv) Two polygons are similar if their corresponding angles are proportional.
(v) Two triangles are similar if their corresponding sides are proportional.
(vi) Two triangles are similar if their corresponding angles are proportional.
Solution:
(i) False : In some cases, the similar polygons can be congruent.
(ii) True.
(iii) False : Its corresponding angles must be equal also.
(iv) False : Angle are equal not proportional.
(v) True.
(vi) False : Sides should be proportional and corresponding angles should be equal.

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## Triangles RD Sharma Class 10 Solutions Exercise 4.2

### RD Sharma Class 10 Solutions Triangles Exercise 4.2

Question 1.
In a ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC.
(i) If AD = 6 cm, DB = 9 cm and AE = 8 cm, find AC. (C.B.S.E. 1995)
(ii) If $\frac { AD }{ DB }$= $\frac { 3 }{ 4 }$and AC = 15 cm, find AE. (C.B.S.E. 1994)
(iii) If $\frac { AD }{ DB }$= $\frac { 2 }{ 3 }$and AC = 18 cm, find AE. (C.B.S.E. 1994C)
(iv) If AD = 4, AE = 8, DB = x – 4, and EC = 3x – 19, find x. (C.B.S.E. 1992C)
(v) If AD = 8 cm, AB = 12 cm and AE = 12 cm, find CE. (C.B.S.E. 1992C)
(vi) If AD = 4 cm, DB = 4.5 cm and AE = 8 cm, find AC. (C.B.S.E. 1992C)
(vii) If AD = 1 cm, AB = 6 cm and AC = 9 cm, findAE. (C.B.S.E. 1992C)
(viii) If $\frac { AD }{ DB }$= $\frac { 4 }{ 5 }$and EC = 2.5 cm, find AE.
(ix) If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, find the value of x. (C.B.S.E. 1993C)
(x) If AD = 8x – 7, DB = 5x – 3, AE = 4x – 3 and EC = (3x – 1), find the value of x.
(xi) If AD = 4x – 3, AE = 8x – 7, BD = 3x – 1 and CE = 5x – 3, find the value of x. (C.B.S.E. 2002)
(xii) If AD = 2.5 cm, BD = 3.0 cm and AE = 3.75 cm, find the length of AC. (C.B.S.E. 2006C)
Solution:

Question 2.
In a ∆ABC, D and E are points on the sides AB and AC respectively. For each of the following cases show that DE || BC:
(i) AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm. (C.B.S.E. 1991)
(ii) AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm. (C.B.S.E. 1990)
(ii) AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm and AE = 2.8 cm.
(iv) AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and EC = 5.5 cm.
Solution:
In ∆ABC, D and E are points on the sides AB and AC respectively
(i) AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm
DB = AB – AD = 12 – 8 = 4 cm and EC = AC – AE = 18 – 12 = 6 cm

Question 3.
In a ∆ABC, P and Q are points on sides AB and AC respectively, such that PQ || BC. If AP = 2.4 cm, AQ = 2 cm, QC = 3 cm and BC = 6 cm, find AB and PQ.
Solution:
In ∆ABC,
P and Q are points on AB and AC respectively such that PQ || BC
AP = 2.4 cm, AQ = 2 cm, QC = 3 cm and BC = 6 cm

Question 4.
In a ∆ABC, D and E are points on AB and AC respectively such that DE || BC. If AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm, find BD and CE. (C.B.S.E. 2001C)
Solution:
In the ∆ABC, DE || BC
AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm

Question 5.
In the figure, state if PQ || EF.

Solution:
In ∆DEF
PQ intersects DE and DF at P and Q respectively
Such that DP = 3.9 cm, PE = 3 cm DQ = 3.6 cm, QF = 2.4 cm

Question 6.
M and N are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether MN || QR.
(i) PM = 4 cm, QM = 4.5 cm, PN = 4 cm, NR = 4.5 cm
(ii) PQ = 1.28 cm, PR = 2.56 cm, PM = 0.16 cm, PN = 0.32 cm
Solution:
(i) In the ∆PQR
M and N are points on PQ and PR respectively
PM = 4 cm, QM = 4.5 cm, PN = 4 cm, RN = 4.5 cm

Question 7.
In three line segments OA, OB, and OC, points L, M, N respectively are so chosen that LM || AB and MN || BC but neither of L, M, N nor of A, B, C are collinear. Show that LN || AC.
Solution:
Given : On OA, OB and OC, points are L, M, and N respectively
Such that LM || AB, MN || BC

Question 8.
If D and E are points on sides AB and AC respectively of a ∆ABC such that DE || BC and BD = CE. Prove that ∆ABC is isosceles. (C.B.S.E. 2007)
Solution:
Given : In ∆ABC, D and E are points on the sides AB and AC such that BD = CD

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## Triangles RD Sharma Class 10 Solutions Exercise 4.3

### RD Sharma Class 10 Solutions Triangles Exercise 4.3

Question 1.
In a ∆ABC, AD is the bisector of ∠A, meeting side BC at D.
(i) If BD = 2.5 cm, AB = 5 cm and AC = 4.2 cm, find DC. (C.B.S.E. 1996)
(ii) If BD = 2 cm, AB = 5 cm and DC = 3 cm, find AC. (C.B.S.E. 1992)
(iii) If AB = 3.5 cm, AC = 4.2 cm and DC = 2.8 cm, find BD. (C.B.S.E. 1992)
(iv) If AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC.
(v) If AC = 4.2 cm, DC = 6 cm and BC = 10 cm, find AB. (C.B.S.E. 1997C)
(vi) If AB = 5.6 cm, AC = 6 cm and DC = 3 cm, find BC. (C.B.S.E. 2001C)
(vii) If AD = 5.6 cm, BC = 6 cm and BD = 3.2 cm, find AC. (C.B.S.E. 2001C)
(viii) If AB = 10 cm, AC = 6 cm and BC = 12 cm, find BD and DC. (C.B.S.E. 2001)
Solution:
In ∆ABC, AD is the angle bisector of ∠A which meet BC at D
(i) BD = 2.5 cm, AB = 5 cm and AC = 4.2 cm

⇒ 6x = 10 (12 – x) = 120 – 10x
⇒ 6x + 10x = 120
⇒ 16x = 120
x = 7.5
BD = 7.5 cm and DC = 12 – 7.5 = 4.5 cm

Question 2.
In the figure, AE is the bisector of the exterior ∠CAD meeting BC produced in E. If AB = 10 cm, AC = 6 cm and BC = 12 cm, find CE.

Solution:
In ∆ABC, AE is the bisector of exterior ∠A which meets BC produced at E.
AB = 10 cm, AC = 6 cm, BC = 12 cm Let CE = x, then BE = BC + CE = (12 + x)

Question 3.
In the figure, ∆ABC is a triangle such that $\frac { AB }{ AC }$= $\frac { BD }{ DC }$, ∠B = 70°, ∠C = 50°. Find ∠BAD.

Solution:

Question 4.
In the figure, check whether AD is the bisector of ∠A of ∆ABC in each of the following :

(i) AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm
(ii) AB = 4 cm, AC = 6 cm, BD = 1.6 cm and CD = 2.4 cm
(iii) AB = 8 cm, AC = 24 cm, BD = 6 cm and BC = 24 cm
(iv) AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 2 cm
(v) AB = 5 cm, AC = 12 cm, BD = 2.5 cm and BC = 9 cm
Solution:
(i) AB = 5 cm, AC = 10 cm, BD = 1.5 cm, CD = 3.5 cm

Question 5.
In figure, AD bisects ∠A, AB = 12 cm AC = 20 cm, and BD = 5 cm. Determine CD.

Solution:

Question 6.
In the figure, In ∆ABC, if ∠1 = ∠2, prove that $\frac { AB }{ AC }$= $\frac { BD }{ DC }$.

Solution:
Given : In ∆ABC,
AD is a line drawn from A meeting BC in D Such that ∠1 = ∠2

Question 7.
D, E and F are the points on sides BC, CA and AB respectively of ∆ABC such that AD bisects ∠A, BE bisects ∠B and CF bisects ∠C. If AB = 5 cm, BC = 8 cm and CA = 4 cm, determine AF, CE and BD.
Solution:
In ∆ABC, AD, BE and CF are the bisector of ∠A, ∠B and ∠C respectively
AB = 5 cm, BC = 8 cm and CA = 4 cm

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## Triangles RD Sharma Class 10 Solutions Exercise 4.4

### RD Sharma Class 10 Solutions Triangles Exercise 4.4

Question 1.
(i) In figure, if AB || CD, find the value of x.

(ii) In figure, if AB || CD, find the value of x.

(iii) In figure, AB || CD. If OA = 3x – 19, OB = x – 4, OC = x – 3 and OD = 4, find x. (C.B.S.E. 2000C)

Solution:
(i) In the figure, AB || CD
The diagonals of a trapezium divides each other proportionally

⇒ (x – 8)(x – 11) = 0
Either x – 8 = 0, then x = 8 or x – 11 =0, then x = 11
x = 8 or 11

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## Triangles RD Sharma Class 10 Solutions Exercise 4.5

### RD Sharma Class 10 Solutions Triangles Exercise 4.5

Question 1.
In the figure, ∆ACB ~ ∆APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ. (C.B.S.E. 1991)
Solution:
In the figure,
∆ACB ~ ∆APQ
BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm

Question 2.
In the figure, AB || QR. Find the length of PB. (C.B.S.E. 1994)

Solution:
In the figure,
In ∆PQR, AB || QR
AB = 3 cm, QR =9 cm, PR = 6 cm
In ∆PAB and ∆PQR
∠P = ∠P (common)
∠PAB = ∠PQR (corresponding angles)
∠PBA = ∠PRQ (corresponding angles)
∠PAB = ∠PQR (AAA axiom)

Question 3.
In the figure, XY || BC. Find the length of XY. (C.B.S.E. 1994C)
Solution:
In the figure
In ∆ABC XY || BC
AX = 1 cm, BC = 6 cm

Question 4.
In a right angled triangle with sides a and b and hypotenuse c, the altitude drawn on hypotenuse is x. Prove that ab = cx.
Solution:
Given : In right ∆ABC, ∠B is right angle BD ⊥ AC
Now AB = a, BC = b, AC = c and BD = x

Question 5.
In the figure, ∠ABC = 90° and BD ⊥ AC. If BD = 8 cm and AD = 4 cm, find CD.
Solution:

Question 6.
In the figure, ∠ABC = 90° and BD ⊥ AC. If AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, find BC.
Solution:
In right ∆ABC, ∠B = 90°
BD ⊥ AC
AB = 5.7 cm, BD = 3.8 and CD = 5.4 cm

Question 7.
In the figure, DE || BC such that AE = $\frac { 1 }{ 4 }$AC. If AB = 6 cm, find AD.
Solution:
In the figure, in ∆ABC, DE || BC
AE = $\frac { 1 }{ 4 }$AC, AB = 6 cm

Question 8.
In the figure, if AB ⊥ BC, DC ⊥ BC and DE ⊥ AC, prove that ∆CED ~ ∆ABC.

Solution:
Given : In the figure AB ⊥ BC, DC ⊥ BC and DE ⊥ AC
To prove : ∆CED ~ ∆ABC.
Proof: AB ⊥ BC
∠B = 90°
and ∠A + ∠ACB = 90° ….(i)
DC ⊥ BC
∠DCB = 90°
=> ∠ACB + ∠DCA = 90° ….(ii)
From (i) and (ii)
∠A = ∠DCA
Now in ∆CED and ∆ABC,
∠E = ∠B (each 90°)
∠DEA or ∠DCE = ∠A (proved)
∆CED ~ ∆ABC (AA axiom)
Hence proved.

Question 9.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using similarity criterion
for two triangles, show that $\frac { OA }{ OC }$= $\frac { OB }{ OD }$
Solution:
Given : ABCD is a trapezium in which AB || DC and diagonals AC and BD intersect each other at O
To Prove : $\frac { OA }{ OC }$= $\frac { OB }{ OD }$

Question 10.
In ∆ABC and ∆AMP are two right triangles, right angled at B and M respectively such that ∠MAP = ∠BAC. Prove that
(i) ∆ABC ~ ∆AMP
(ii) $\frac { CA }{ PA }$= $\frac { BC }{ MP }$.
Solution:
Given : In ∆ABC and ∆AMP,
∠B = ∠M = 90°
∠MAP = ∠BAC

Question 11.
A vertical stick 10 cm long casts a shadow of 8 cm long. At the same time a tower casts a shadow 30 m long. Determine the height of the tower. (CB.S.E. 1991)
Solution:
The shadows are casted by a vertical stick and a tower at the same time
Their angles will be equal

Question 12.
In the figure, A = CED, prove that ∠CAB ~ ∠CED. Also find the value of x.

Solution:
Given : In ∆ABC,
∠CED = ∠A
AB = 9 cm, BE = 2 cm, EC = 10 cm, AD = 7 cm and DC = 8 cm
To prove :
(i) ∆CAB ~ ∆CED
(ii) Find the value of x
Proof: BC = BE + EC = 2 + 10 = 12 cm
AC = AD + DC = 7 + 8 = 15 cm
(i) Now in ∆CAB and ∆CED,
∠A = ∠CED (given)
∠C = ∠C (common)
∆CAB ~ ∆CED (AA axiom)

Question 13.
The perimeters of two similar triangles are 25 cm and 15 cm respectively. If one side of first triangle is 9 cm, what is the corresponding side of the other triangle? (C.B.S.E. 2002C)
Solution:
Let perimeter of ∆ABC = 25 cm
and perimeter of ∆DEF = 15 cm
and side BC of ∆ABC = 9 cm
Now we have to find the side EF of ∆DEF
∆ABC ~ ∆DEF (given)

Question 14.
In ∆ABC and ∆DEF, it is being given that: AB = 5 cm, BC = 4 cm and CA = 4.2 cm; DE = 10 cm, EF = 8 cm and FD = 8.4 cm. If AL ⊥ BC and DM ⊥ EF, find AL : DM.
Solution:
In ∆ABC and ∆DEF,
AB = 5 cm, BC = 4 cm, CA = 4.2 cm, DE = 10 cm, EF = 8 cm and FD = 8.4 cm
AL ⊥ BC and DM ⊥ EF

Question 15.
D and E are the points on the sides AB and AC respectively of a ∆ABC such that: AD = 8 cm, DB = 12 cm, AE = 6 cm and CE = 9 cm. Prove that BC = $\frac { 5 }{ 2 }$DE.
Solution:
Given : In ∆ABC, points D and E are on the sides AB and AC respectively
and AD = 8 cm, DB = 12 cm, AE = 6 cm and CE = 9 cm

Hence proved

Question 16.
D is the mid-point of side BC of a ∆ABC. AD is bisected at the point E and BE produced cuts AC at the point X. Prove that BE : EX = 3 : 1.
Solution:
Given : In ∆ABC, D is mid point of BC, and E is mid point of AD
BE is joined and produced to meet AC at X
To prove : BE : EX = 3 : 1
Construction : From D, draw DY || BX meeting AC at Y

Hence proved.

Question 17.
ABCD is a parallelogram and APQ is a straight line meeting BC at P and DC produced at Q. Prove that the rectangle obtained by BP and DQ is equal to the rectangle contained AB and BC.
Solution:
Given : ABCD is a parallelogram.
APQ is a straight line which meets BC at P and DC on producing at Q

Question 18.
In ∆ABC, AL and CM are the perpendiculars from the vertices A and C to BC and AB respectively. If AL and CM intersect at O, prove that :
(i) ∆OMA ~ ∆OLC
(ii) $\frac { OA }{ OC }$= $\frac { OM }{ OL }$
Solution:
Given : In ∆ABC, AL ⊥ BC, CM ⊥ AB which intersect each other at O

(ii) $\frac { OA }{ OC }$= $\frac { OM }{ OL }$
Hence proved.

Question 19.
ABCD is a quadrilateral in which AD = BC. If P, Q, R, S be the mid-points of AB, AC, CD and BD respectively, show that PQRS is a rhombus.
Solution:
Given : In quadrilateral ABCD, AD = BC
P, Q, R and S are the mid points of AB, AC, CD and AD respectively
PQ, QR, RS, SP are joined

Question 20.
In an isosceles ∆ABC, the base AB is produced both the ways to P and Q such that AP x BQ = AC². Prove that ∆APC ~ ∆BCQ.
Solution:
Given : In ∆ABC, AC = BC
Base AB is produced to both sides and points P and Q are taken in such a way that
AP x BQ = AC²
CP and CQ are joined
To prove : ∆APC ~ ∆BCQ

Question 21.
A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/sec. If the lamp is 3.6 m above ground, find the length of her shadow after 4 seconds.
Solution:
Let AB be the lamp post and CD be the girl and AB = 3.6 m, CD = $\frac { 90 }{ 100 }$= 9 m
Distance covered in 4 seconds = 1.2 m x 4 = 4.8 m
BD = 4.8 m

x = 1.6
Length of her shadow = 1.6 m

Question 22.
A vertical stick of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Solution:
Let AB be stick and DE be tower.
A stick 6 m long casts a shadow of 4 m i.e., AB = 6 m and BC = 4 m
Let DE casts shadow at the same time which is EF = 28 m
Let height of tower DE = x
Now in ∆ABC and ∆DEF,
∠B = ∠E (each 90°)
∠C = ∠F (shadows at the same time)
∆ABC ~ ∆DEF (AA criterion)

Question 23.
In the figure, ∆ABC is right angled at C and DE ⊥ AB. Prove that ∆ABC ~ ∆ADE and hence find the lengths of AE and DE.

Solution:
Given: In the figure, ∆ABC is a right angled triangle right angle at C.
DE ⊥ AB
To prove:
(i) ∆ABC ~ ∆ADE
(ii) Find the length of AE and DE
Proof: In ∆ABC and ∆ADE,
∠ACB = ∠AED (each 90°)
∠BAC = ∠DAE (common)
∆ABC ~ ∆ADE (AA axiom)

Question 24.
In the figure, PA, QB and RC are each perpendicular to AC. Prove that

Solution:
Given : In the figure, PA, QB and RC are perpendicular on AC and PA = x, QB = z and RC = y

Hence proved.

Question 25.
In the figure, we have AB || CD || EF. If AB = 6 cm, CD = x cm, EF = 10 cm, BD = 4 cm and DE = y cm, calculate the values of x and y.

Solution:
In the figure, AB || CD || EF
AB = 6 cm, EF = 10 cm, BD = 4 cm, CD = x cm and DE = y cm
In ∆ABE, CE || AB
∆CED ~ ∆AEB

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### RD Sharma Class 10 Solutions Triangles Exercise 4.6

Question 1.
Triangles ABC and DEF are similar.
(i) If area (∆ABC) = 16 cm², area (∆DEF) = 25 cm² and BC = 2.3 cm, find EF. (C.B.S.E. 1992)
(ii) If area (∆ABC) = 9 cm², area (∆DEF) = 64 cm² and DE = 5.1 cm, find AB.
(iii) If AC = 19 cm and DF = 8 cm, find the ratio of the area of two triangles. (C.B.S.E. 1992C)
(iv) If area (∆ABC) = 36 cm², area (∆DEF) = 64 cm² and DE = 6.2 cm, find AB. (C.B.S.E. 1992)
(v) If AB = 1.2 cm and DE = 1.4 cm, find the ratio of the areas of ∆ABC and ∆DEF. (C.B.S.E. 1991C)
Solution:

Question 2.
In the figure, ∆ACB ~ ∆APQ. If BC = 10 cm, PQ = 5 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ. Also, find the area (∆ACB) : area (∆APQ).

Solution:

Question 3.
The areas of two similar triangles are 81 cm² and 49 cm² respectively. Find the ratio of their corresponding heights, what is the ratio of their corresponding medians ?
Solution:
Areas of two similar triangles are 81 cm² and 49 cm²
The ratio of the areas of two similar triangles are proportion to the square of their corresponding altitudes and also squares of their corresponding medians
Ratio in their altitudes = √81 : √49 = 9 : 7
Similarly, the ratio in their medians = √81 : √49 = 9 : 7

Question 4.
The areas of two similar triangles are 169 cm² and 121 cm² respectively. If the longest side of the larger triangle is 26 cm, find the longest side of the smaller triangle.
Solution:
Triangles are similar Area of larger triangle = 169 cm²
and area of the smaller triangle =121 cm²
Length of longest sides of the larger triangles = 26 cm
Let the length of longest side of the smaller triangle = x

Question 5.
The areas of two similar triangles are 25 cm² and 36 cm² respectively. If the altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other.
Solution:
Area of first triangle = 25 cm²
Area of second = 36 cm²
Altitude of the first triangle = 2.4 cm
Let altitude of the second triangle = x
The triangles are similar

Question 6.
The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas.
Solution:
Length of the corresponding altitude of two triangles are 6 cm and 9 cm
triangles are similar

Question 7.
ABC is a triangle in which ∠A = 90°, AN ⊥ BC, BC = 12 cm and AC = 5 cm. Find the ratio of the areas of the ∆ANC and ∆ABC.
Solution:
In ∆ABC, ∠A = 90°
AN ⊥ BC
BC = 12 cm, AC = 5 cm

Question 8.
In the figure, DE || BC
(i) If DE = 4 cm, BC = 6 cm and area (∆ADE) = 16 cm², find the area of ∆ABC.
(ii) If DE = 4 cm, BC = 8 cm and area of (∆ADE) = 25 cm², find the area of ∆ABC. (C.B.S.E. 1991)
(iii) If DE : BC = 3 : 5, calculate the ratio of the areas of ∆ADE and the trapezium BCED

Solution:

Question 9.
In ∆ABC, D and E are the mid-points of AB and AC respectively. Find the ratio of the areas of ∆ADE and ∆ABC.
Solution:
In ∆ABC, D and E are the mid points of AB and AC respectively

Question 10.
The areas of two similar triangles are 100 cm² and 49 cm² respectively. If the altitude of the bigger triangle is 5 cm, find the corresponding altitude of the other. (C.B.S.E. 2002)
Solution:
∆ABC ~ ∆DEF
area ∆ABC = 100 cm²
and area ∆DEF = 49 cm²

Question 11.
The areas of two similar triangles are 121 cm² and 64 cm² respectively. If the median of the first triangle is 12.1 cm, find the corresponding median of the other. (C.B.S.E. 2001)
Solution:
∆ABC ~ ∆DEF
area of ∆ABC = 121 cm² area of ∆DEF = 64 cm²
AL and DM are the medians of ∆ABC and ∆DEF respectively
AL = 12.1 cm

Question 12.
In ∆ABC ~ ∆DEF such that AB = 5 cm and (∆ABC) = 20 cm² and area (∆DEF) = 45 cm², determine DE.
Solution:
∆ABC ~ ∆DEF
area (∆ABC) = 20 cm²
area (∆DEF) = 45 cm²

Question 13.
In ∆ABC, PQ is a line segment intersecting AB at P and AC at Q such that PQ || BC and PQ divides ∆ABC into two parts equal in area. Find $\frac { BP }{ AB }$.
Solution:
In ∆ABC, PQ || BC and PQ divides ∆ABC in two parts ∆APQ and trap. BPQC of equal in area
i.e., area ∆APQ = area BPQC

Question 14.
The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, find the length of QR. (C.B.S.E. 2004)
Solution:
∆ABC ~ ∆PQR
area (∆ABC) : area (∆PQR) = 9 : 16
and BC = 4.5 cm

Question 15.
ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 cm, prove that area of ∆APQ is one sixteenth of the area of ∆ABC. (C.B.S.E. 2005)
Solution:
In ∆ABC, P and Q are two points on AB and AC respectively such that
AP = 1 cm, PB = 3 cm, AQ = 1.5 cm and QC = 4.5 cm

Question 16.
If D is a point on the side AB of ∆ABC such that AD : DB = 3 : 2 and E is a point on BC such that DE || AC. Find the ratio of areas of ∆ABC and ∆BDE. (C.B.S.E. 2006C)
Solution:
In ∆ABC, D is a point on AB such that AD : DB = 3 : 2

Question 17.
If ∆ABC and ∆BDE are equilateral triangles, where D is the mid point of BC, find the ratio of areas of ∆ABC and ∆BDE. [CBSE 2010]
Solution:
∆ABC and ∆DBE are equilateral triangles Where D is mid point of BC

Question 18.
Two isosceles triangles have equal vertical angles and their areas are in the ratio 36 : 25. Find the ratio of their corresponding heights.
Solution:
Two isosceles triangles have equal vertical angles
So their base angles will also be the equal to each other
Triangles will be similar Now, ratio in their areas = 36 : 25

Question 19.
In the figure, ∆ABC and ∆DBC are on the same base BC. If AD and BC intersect

Solution:
Given : Two ∆ABC and ∆DBC are on the same base BC as shown in the figure
AC and BD intersect eachother at O

Question 20.
ABCD is a trapezium in which AB || CD. The diagonals AC and BD intersect at O. Prove that
(i) ∆AOB ~ ∆COD
(ii) If OA = 6 cm, OC = 8 cm, find

Solution:
Given : ABCD is a trapezium in which AB || CD
Diagonals AC and BD intersect each other at O

Question 21.
In ∆ABC, P divides the side AB such that AP : PB = 1 : 2. Q is a point in AC such that PQ || BC. Find the ratio of the areas of ∆APQ and trapezium BPQC.
Solution:
In ∆ABC, P is a point on AB such that AP : PQ = 1 : 2
PQ || BC
Now we have to find the ratio between area ∆APQ and area trap BPQC

Question 22.
AD is an altitude of an equilateral triangle ABC. On AD as base, another equilateral triangle ADE is constructed. Prove that Area (∆ADE) : Area (∆ABC) = 3 : 4. [CBSE 2010]
Solution:
Given: In equilateral ∆ABC, AD ⊥ BC and with base AD, another equilateral ∆ADE is constructed

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## Triangles RD Sharma Class 10 Solutions Exercise 4.7

### RD Sharma Class 10 Solutions Triangles Exercise 4.7

Question 1.
If the sides of a triangle are 3 cm, 4 cm and 6 cm long, determine whether the triangle is a right-angled triangle. (C.B.S.E. 1992)
Solution:
We know that if the square of the hypotenuse (longest side) is equal to the sum of squares of other two sides then it is right triangle
Now the sides of a triangle are 3 cm, 4 cm and 6 cm
(Longest side)² = (6)² = 36
and sum of two smaller sides = (3)² + (4)² = 9 + 16 = 25
36 ≠ 25
It is not a right-angled triangle

Question 2.
The sides of certain triangles are given below. Determine which of them are right triangles :
(i) a = 1 cm, b = 24 cm and c = 25 cm
(ii) a = 9 cm, b = 16 cm and c = 18 cm
(iii) a = 1.6 cm, b = 3.8 cm and c = 4 cm
(iv) a = 8 cm, b = 10 cm and c = 6 cm (C.B.S.E. 1992)
Solution:
We know that if the square of hypotenuse is equal to the sum of squares of other two sides, then it is a right triangle
(i) Sides of a triangle are a = 7 cm, b = 5.24 cm and c = 25 cm
(Longest side)² = (25)² = 625
Sum of square of shorter sides = (7)² + (24)² = 49 + 576 = 625
625 = 625
This is right triangle
(ii) Sides of the triangle are a = 9 cm, b = 16 cm, c = 18 cm
(Longest side)² = (18)² = 324
and sum of squares of shorter sides = (9)² + (16)² = 81 + 256 = 337
324 ≠ 337
It is not a right-angled triangle
(iii) Sides of the triangle are a = 1.6 cm, 6 = 3.8 cm, c = 4 cm
(Longest side)² = (4)² =16
Sum of squares of shorter two sides + (1.6)² + (3.8)² = 2.56 + 14.44 = 17.00
16 ≠ 17
It is not a right triangle
(iv) Sides of the triangle are a = 8 cm, b = 10 cm, c = 6 cm
(Longest side)² = (10)² = 100
Sum of squares of shorter sides = (8)² + (6)² = 64 + 36 = 100
100 = 100
It is a right triangle

Question 3.
A man goes 15 metres due west and then 8 metres due north. How far is he from the starting point ?
Solution:
Let a man starts from O, the starting point to west 15 m at A and then from A, 8 m due north at B
Join OB

Now in right ∆OAB
OB² = OA² + AB² (Pythagoras Theorem)
OB² = (15)² + (8)² = 225 + 64 = 289 = (17)²
OB = 17
The man is 17 m away from the starting point

Question 4.
A ladder 17 m long reaches a window of a building 15 m above the ground. Find the distance of the foot of the ladder from the building.
Solution:
Length of ladder = 17 m
Height of window = 15 m

Let the distance of the foot of ladder from the building = x
Using Pythagoras Theorem
AC² = AB² + BC²
=> (17)² = (15)² + x²
=> 289 = 225 + x²
=> x² = 289 – 225
=> x² = 64 = (8)²
x = 8
Distance of the foot of the ladder from the building = 8m

Question 5.
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops. (C.B.S.E. 1996C, 2002C)
Solution:
Two poles AB and CD which are 6 m and 11 m long respectively are standing oh the ground 12 m apart

Draw AE || BD so that AE = BD = 12 m and ED = AB = 6 m
Then CE = CD – ED = 11 – 6 = 5 m
Now in right ∆ACE
Using Pythagoras Theorem,
AC² = AE² + EC² = (12)² + (5)² = 144 + 25 = 169 = (13)²
AC = 13
Distance between their tops = 13 m

Question 6.
In an isosceles triangle ABC, AB = AC = 25 cm, BC = 14 cm. Calculate the altitude from A on BC. (C.B.S.E. 1994)
Solution:
∆ABC is an isosceles triangle in which AB = AC = 25 cm .
AD ⊥ BC BC = 14 cm

Perpendicular AD bisects the base i.e . BD = DC = 7 cm
Let perpendicular AD = x
In right ∆ABD,
AB² = AD² + BD² (Pythagoras Theorem)
=> (25)² = AD² + (7)²
=> 625 = AD² + 49
=> AD² = 625 – 49
=> AD² = 576 = (24)²
=> AD = 24
Perpendicular AD = 24 cm

Question 7.
The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its tip reach ?
Solution:
In first case,
The foot of the ladder are 6 m away from the wall and its top reaches window 8 m high
Let AC be ladder and BC = 6 m, AB = 8 m

Now in right ∆ABC,
Using Pythagoras Theorem
AC² = BC² + AB² = (6)² + (8)² = 36 + 64 = 100 = (10)²
AC = 10 m
In second case,
ED = AC = 10 m
BD = 8 m, let ED = x
ED² = BD² + EB²
=> (10)² = (8)² + x²
=> 100 = 64 + x²
=> x² = 100 – 64 = 36 = (6)²
x = 6
Height of the ladder on the wall = 6 m

Question 8.
Two poles of height 9 m and 14 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.
Solution:
Let CD and AB be two poles which are 12 m apart
AB = 14 m, CD = 9 m and BD = 12 m
From C, draw CE || DB
CB = DB = 12 m
EB = CD = 9 m
and AE = 14 – 9 = 5 m

Now in right ∆ACE,
AC² = AE² + CE² (Pythagoras Theorem)
= (5)² + (12)²
= 25 + 144 = 169 = (13)²
AC = 13
Distance between their tops = 13 m

Question 9.
Using Pythagoras theorem, determine the length of AD in terms of b and c shown in the figure. (C.B.S.E. 1997C)

Solution:
In right ∆ABC, ∠A = 90°
AB = c, AC = b

Question 10.
A triangle has sides 5 cm, 12 cm and 13 cm. Find the length to one decimal place, of the perpendicular from the opposite vertex to the side whose length is 13 cm. (C.B.S.E. 1992C)
Solution:
A triangle has sides 5 cm, 12 cm and 13 cm
(Longest side)² = (13)² = 169
Sum of squares of shorter sides = (5)² + (12)² = 25 + 144= 169
169 = 169
It is a right triangle whose hypotenuse is 13 cm

BD = 4.6 cm

Question 11.
ABCD is a square, F is the mid point of AB. BE is one third of BC. If the area of ∆FBE = 108 cm² find the length of AC. (C.B.S.E. 1995)
Solution:
In square ABCD, F is mid point of AB i.e.,

Question 12.
In an isosceles triangle ABC, if AB = AC = 13 cm and the altitude from A on BC is 5 cm, find BC. (C.B.S.E. 2000)
Solution:
In ∆ABC, AB = AC
AB = AC = 13 cm, AD = 5 cm
AD bisects BC at D
BD = $\frac { 1 }{ 2 }$BC
=> BC = 2BD

Question 13.
In a ∆ABC, AB = BC = CA = 2a and AD ⊥ BC. Prove that
(i) AD = a √3
(ii) area (∆ABC) = √3 a² (C.B.S.E. 1991)
Solution:
In ∆ABC, AB = BC = AC = 2a
AD bisects BC at D
BD = DC = $\frac { 1 }{ 2 }$BC = a

Question 14.
The lengths of the diagonals of a rhombus are 24 cm and 10 cm. Find each side of the rhombus. (C.B.S.E. 1993C)
Solution:
ABCD is a rhombus whose diagonals AC = 24 cm and BD = 10 cm
The diagonals of a rhombus bisect eachother at right angles

AO = OC = $\frac { 24 }{ 2 }$= 12 cm
and BO = OD = $\frac { 10 }{ 2 }$= 5 cm
Now in right ∆AOB,
AB² = AO² + BO² (Pythagoras Theorem)
(12)² + (5)² = 144 + 25 = 169 = (13)²
AB = 13
Each side of rhombus = 13 cm

Question 15.
Each side of a rhombus is 10 cm. If one of its diagonals is 16 cm, find the length of the other diagonal.
Solution:
In rhombus ABCD, diagonals AC and BD bisect eachother at O at right angles
Each side = 10 cm and one diagonal AC = 16 cm

AO = OC = $\frac { 16 }{ 2 }$= 8 cm
Now in right angled triangle AOB,
AB² = AO² + OB² (Pythagoras Theorem)
(10)² = (8)² + (BO)²
=> 100 = 64 + BO²
=> BO² = 100 – 64 = 36 = (6)²
BO = 6
BD = 2BO = 2 x 6 = 12 cm

Question 16.
Calculate the height of an equilateral triangle each of whose sides measures 12 cm.
Solution:
Each side of the equilateral ∆ABC = 12 cm
AD ⊥ BC which bisects BC at D

BD = DC = $\frac { 12 }{ 2 }$= 6 cm

Question 17.
In the figure, ∠B < 90° and segment AD ⊥ BC. Show that:
(i) b² = h² + a² + x² – 2ax
(ii) b² = a² + c² – 2ax

Solution:
Given : In ∆ABC, ∠B < 90°
AD = c, BC = a, CA = b AD = h, BD = x, DC = a – x
To prove: (i) b² = h² + a² + x² – 2ax
(ii) b² = a² + c² – 2ax
Proof: (i) In right ∆ADC, AC² = AD² + DC² (Pythagoras Theorem)
=> b² = h² + (a – x)² = h² + a² + x² – 2ax
(ii) Similarly in right ∆ADB
AB² = AD² + BD²
c² = h² + x² ….(i)
b² = h² + a² + x² – 2ax = h² + x² + a² – 2ax
= c² + a² – 2ax {From (i)}
= a² + c² – 2ax
Hence proved.

Question 18.
In an equilateral ∆ABC, AD ⊥ BC, prove that AD² = 3 BD². (C.B.S.E. 2002C)
Solution:
Given : ∆ABC is an equilateral in which AB = BC = CA .

Question 19.
∆ABD is a right triangle right-angled at A and AC ⊥ BD. Show that
(i) AB² = BC.BD
(ii) AC² = BC.DC
(iii) AD² = BD.CD
(iv) $\frac { { AB }^{ 2 } }{ { AC }^{ 2 } }$= $\frac { BD }{ DC }$[NCERT]
Solution:
In ∆ABD, ∠A = 90°
AC ⊥ BD

Question 20.
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 ?
Solution:
Let AB be the vertical pole whose height = 18 m

Question 21.
Determine whether the triangle having sides (a – 1) cm, 2 √a cm and (a + 1) cm is a right angled triangle. [CBSE 2010]
Solution:
Sides of a triangle are (a – 1) cm, 2 √a cm and (a + 1) cm
Let AB = (a – 1) cm BC = (a + 1) cm
and AC = 2 √a

Question 22.
In an acute-angled triangle, express a median in terms of its sides.
Solution:
In acute angled ∆ABC,
AD is median and AL ⊥ BC
Using result of theorem, sum of the squares of any two sides is equal to the twice the square of half of the third side together with twice the square of the median which bisects the third side

Question 23.
In right-angled triangle ABC in which ∠C = 90°, if D is the mid point of BC, prove that AB² = 4AD² – 3AC².
Solution:
Given : In right ∆ABC, ∠C = 90°
D is mid point of BC

Question 24.
In the figure, D is the mid-point of side BC and AE ⊥ BC. If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that :

Solution:

Question 25.
In ∆ABC, ∠A is obtuse, PB x AC and QC x AB. Prove that:
(i) AB x AQ = AC x AP
(ii) BC² = (AC x CP + AB x BQ)
Solution:
Given : In ∆ABC, ∠A is an obtuse angle PB x AC and QC x AB on producing them

Question 26.
In a right ∆ABC right-angled at C, if D is the mid-point of BC, prove that BC² = 4 (AD² – AC²).
Solution:
Given : In ∆ABC, ∠C = 90°
D is mid-point of BC

=> 4AD² = 4AC² + BC²
=> BC² = 4AD² – 4AC²
=> BC² = 4 (AD² – AC²)
Hence proved.

Question 27.
In a quadrilateral ABCD, ∠B = 90°, AD² = AB² + BC² + CD², prove that ∠ACD = 90°.
Solution:
Given : ABCD is a quadrilateral in which ∠B = 90° and AD² = AB² + BC² + CD²
To prove : ∠ACD = 90°
Construction : Join AC

proof : In right ∆ABC, ∠B = 90°
AC² = AB² + BC² ….(i) (Pythagoras Theorem)
But AD² = AB² + BC² + CD² (given)
=> AD² = AC² + CD² {From (i)}
∆ACD is a right angle with right angle ACD
Hence proved.

Question 28.
An aeroplane leaves an airport and flies due north at a speed of 1000 km/hr. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km/ hr. How far apart will be the two planes after 1$\frac { 1 }{ 2 }$hours?
Solution:

Speed of the first plane = 1000 km/hr
Distance travelled in 1$\frac { 1 }{ 2 }$hour due north = 1000 x $\frac { 3 }{ 2 }$= 1500 km
Speed of the second plane = 1200 km/hr
Distance travelled in 1$\frac { 1 }{ 2 }$hours due west

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