Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.5 | Set 2

Chapter 4 of the Class 10 RD Sharma Mathematics textbook, "Triangles," covers the properties and theorems related to triangles, including similarity, congruence, and the Pythagorean theorem. Exercise 4.5 focuses on applying these properties to solve problems related to triangles.

RD Sharma Solutions for Class 10 - Mathematics - Chapter 4 Triangles - Exercise 4.5 | Set 2

This section provides detailed solutions for Exercise 4.5 from Chapter 4 of the Class 10 RD Sharma Mathematics textbook. The exercise involves problems that require students to apply various properties and theorems of triangles, such as criteria for similarity and congruence, to find unknown lengths and angles in given triangles. Solutions are presented step-by-step to aid understanding and mastery of the concepts.

Question 13. The perimeters of two similar triangles are 25 cm and 15 cm, respect. If one side of the first triangle is 9 cm, what is the corresponding side of the other triangle?

Solution:

👁 Image

Given: perimeter of two similar triangles are 25 cm, 15 cm and one side 9 cm

To find : the other side.

Let the two triangles be ABC & PQR.

Let one of its side is (AB) = 9 cm and the other side of other triangle be PQ.

∆ABC ∼ ∆PQR.      [Since the ratio of corresponding sides of similar triangles is same as the ratio of their perimeters.]

AB/PQ = BC/QR = AC/PR = 25/15

9/PQ = 25/15

25 PQ = 15 × 9

PQ = (15×9)/25

PQ =(3× 9 )/5

PQ = 27/5

PQ =  5.4 cm

Hence, the corresponding side of other ∆ is 5.4 cm.

Question 14. In ΔABC and ΔDEF, it is being given that AB = 5 cm, BC = 4 cm, CA = 4.2 cm, DE = 10 cm, EF = 8 cm, and FD = 8.4 cm. If AL ⊥ BC, DM ⊥ EF, find AL: Dm.

Solution:

👁 Image

Given,

AB = 5 cm, BC = 4 cm, CA = 4.2 cm, DE = 10 cm, EF = 8 cm, and FD = 8.4 cm

AL ⊥ BC, DM ⊥ EF

To find: AL: DM

Now In ∆ABC and ∆DEF,

AB/DE=BC/EF=AC/DF=1/2   [both triangles are similar]

[In similar triangle the ratio of corresponding altitude is same as the ratio of the corresponding sides.]

   AL: DM = 1: 2

Question 15. D and E are the points on the sides AB and AC respectively, of a ΔABC such that AD = 8 cm, DB = 12 cm, AE = 6 cm, and CE = 9 cm. Prove that BC = 5/2 DE.

Solution:

👁 Image

AE/EC=6/9=2/3.......(1)

AD/DB=8/12=2/3......(2)

from (1) and (2) we know that 

AD/DB=AE/EC

 ED∥CB                                                

Now In△AED and △ACB,

 ED∥CB  

∠AED=∠ACB

∠ADE=∠ABC

△AED∼△ACB    [by AAA criteria]

 AE/AC=ED//CB

AE/EC+AE= ED/CB [AC=AE+EC]

6/6+9= ED/CB

6/15=2/5=ED/CB

CB=5/2ED  

 Hence proved

Question 16. D is the midpoint of side BC of a ΔABC. AD is bisected at the point E and BE produced cuts AC at the point X. Prove that BE: EX = 3 : 1

Solution:

👁 Image

Given: In ∆ABC, D is the midpoint of BC, E is the midpoint of AD. BE produced meets AC at X.

To prove :  BE: EX = 3 : 1

Now In Δ BCX  and  ΔDCY,

∠CBX  = ΔCBY   (corresponding angles)

∠CXB  = ΔCYD   (corresponding angles)

ΔBCX∼ΔDCY   (AA similarity)

BC/DC = BX/ DY = CX/CY    [ corresponding sides of two similar triangles are proportional]

BX/DY = BC/DC

BX/DY = 2DC/DC   [As D is the midpoint of BC]

BX/DY = 2/1………...(i)

Now In ΔAEX and ΔADY,

∠AEX  = ΔADY   (corresponding angles)

∠AXE = ΔAYD   (corresponding angles)

ΔAEX ∼ ΔADY   (AA similarity)

AE/AD = EX/DY = AX/AY  [ corresponding sides of two similar triangles are proportional]

EX/DY = AE/AD

EX/DY = AE/2AE                 [ D is the midpoint of BC]

EX/DY = ½…………..…(ii)

On Dividing eqn. (i) by eqn. (ii) we get,

BX/EX = 4/1

BX = 4EX

BE + EX = 4EX

BE = 4EX - EX

BE = 3EX

BE /EX = 3/1

BE : EX = 3:1

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 by AB and BC.

Solution:

👁 Image

Given : ABCD is a parallelogram and APQ is a straight line meeting BC at P and DC produced at Q.

To prove : BP x DQ = AB x BC

Now In ∆ABP and ∆QDA,

∠B = ∠D [Opposite angles of parallelogram]

∠BAP = ∠AQD [Alternate interior angles]

 ∆ABP ~ ∆QDA [By AA similarity]

AB/QD = BP/DA   [the corresponding sides of similar triangles are proportional]

DA = BC  [opposite sides of a parallelogram are equal]

AB/QD = BP/BC

AB × BC = QD × BP

Hence proved

Question 18. In ΔABC, AL and CM are the perpendiculars from the vertices A and C to BC and AB respect. If Al and CM intersect at O, prove that:

(i) ΔOMA ∼ ΔOLC

(ii)OA/OC=OM/OL

Solution:

👁 Image

Given,

AL ⊥ BC and CM ⊥ AB

To prove:

(i)  ΔOMA ∼ ΔOLC

(ii)OA/OC=OM/OL

Now In Δ OMA and ΔOLC,

∠MOA = ∠LOC                [Vertically opposite angles]

∠AMO = ∠CLO                [Each 90°]

ΔOMA ~ ΔOLC       [By AA similarity]

OA/OC=OM/OL  [Corresponding parts of similar Δ are proportional]

Hence proved

Question 19. ABCD is a quadrilateral in which AD = BC. If P, Q, R, S be the midpoints of AB, AC, CD, and BD respect. Show that PQRS is a rhombus.

Solution:

👁 Image

Given, 

ABCD is a quadrilateral in which AD = BC and P, Q, R, S are the mid points of AB, AC, CD, BD, respectively.

To prove: PQRS is a rhombus

Now,

In ΔABC, P and Q are the mid points of the sides B and AC respectively

PQ ∥ BC and PQ = 1/2 BC.             [By the midpoint theorem]

In ΔADC, Q and R are the mid points of the sides AC and DC respectively

By the mid-point theorem, we get,

QR ∥ AD  QR = 1/2 AD    [By the mid-point theorem]

QR = 1/2 BC              [AD = BC]

Now In ΔBCD,

RS∥BC and RS = 1/2 AD  [By the mid-point theorem]

RS= 1/2 BC    [AD = BC]

From above eqns.

PQ = QR = RS

 PQRS is a rhombus.

Hence proved

Question 20. In an isosceles ΔABC, the base AB is produced both ways to P and Q such that AP x BQ = AC2. Prove that Δ APC ∼ Δ BCQ

Solution:

👁 Image

Given : ΔABC is isosceles ∆ and AP x BQ = AC²

To prove : ΔAPC∼ΔBCQ.

we know that, 

ΔABC is an isosceles triangle AC = BC.

AP x BQ = AC²  (given)

AP x BQ = AC x AC

AP x BQ = AC x BC

AP/BC = ABQ……….(1).

so, ∠CAB = ∠CBA    [as  AC = BC]

180° – ∠CAP = 180° – ∠CBQ [angles opposite to equal sides are EQUAL]

∠CAP = ∠CBQ ………..(2)

In ∆APC and ΔBCQ

AP/BC = AC/BQ [From equation 1]

∠CAP = ∠CBQ [From equation 2]

ΔAPC∼ΔBCQ  [By SAS similarity criterion]

Hence proved

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 3.6m above the ground, find the length of her shadow after 4 seconds.

Solution:

👁 Image

AD=1.2 m/sec ×4 sec =4.8 m =480 cm

Suppose the length of the shadow of the girl be x cm when she at position D. Therefore,

BD = x cm

In ∆BDE and ∆BAC,

∠BDE=∠BAC  [each90°]

∠DBE=∠ABC   [Common angle ]

 ∆BDE∼∆BAC   [ by AA similarity]

BE/BC=DE/AC=BD/AB  [Corresponding sides are proportional]

90/360=x/480+x

1/4= x/480+x

4x=480+x

4x−x=480

3x=480

x=480/3=160

Hence, the length of her shadow after 4 seconds is 160 cm.

Question 22. A vertical stick of length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28m long. Find the height of the tower.

Solution:

👁 Image

Let AB be stick and DE be tower.

A 6 m long stick casts a shadow of 4 m 

So,  AB = 6 m and BC = 4 m

Let DE casts shadow at the same time, So 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]

AB/DE=BC/EF      [ Corresponding sides are proportional ]

6/x=4/28

x=42 m

Hence height of tower is 42 m

Question 23. In the fig. given, ΔABC is a right-angled triangle at C and DE ⊥ AB. Prove that ΔABC ∼ ΔADE.hence find the lengths of AEand DE.

Solution:

👁 Image

Given : ΔACB is right-angled triangle at C = 90°.

According to the figure : BC = 12 cm , AD=3 cm, DC = 2 cm.

AC = AD + DC = 3 +2= 5 cm

Now In ∆ACB,

AB² = AC² + BC² [by Pythagoras theorem]

AB² = 5² + 12²

AB² = 25 + 144 = 169

AB= √169 = 13  

AB = 13 cm

Now In  ΔABC & ΔADE,

∠BAC = ∠DAE     [common]

∠ACB = ∠AED      [each 90°]

ΔABC∼ΔADE     [by AA similarity criterion]

AB/AD = BC/DE = AC/AE    [ corresponding sides of two similar triangles are proportional]

13/3 = 12/ DE = 5/AE

13/3 = 12/DE

13 DE = 12×3  

DE = 36/13

13/3 = 5/AE

13 AE = 5×3

AE = 15/13

Hence, the length of DE= 36/13 & AE = 15/13

Question 24. In the figure, PA, QB and RC are each perpendicular to AC. Prove that 1/X+1/Z=1/Y

👁 Image

Given:

PA, QB and RC are each perpendicular to AC

To prove : 1/X+1/Z=1/Y

Now In ΔPAC and ΔQBC,

∠PCA = ∠QCB  [common angle]

∠PAC = ∠QBC [Corresponding angles]

ΔPAC ~ ΔQBC [by AA criteria]

PA/QB = AC/BC  [ Corresponding parts of similar triangles are proportional]

x/y = AB/BC  

y/x = BC/AC

Now In ΔRCA and ΔQBA,

∠RAC = ∠QAB [common angle]

∠RCA = ∠QBA [Corresponding angle]

ΔRCA ~ΔQBA  [by AA criteria]

RC/QB = AC/AB [ Corresponding parts of similar triangles are proportional]

z/y= AC/AB

y/z= AB/AC

y/z + y/z = BC + AC/ AC = 1  [by adding both eq we get]

y/z + y/z = 1

1/x + 1/z = 1/y [multiplying both sides by y]

Hence proved

Question 25. In fig. given, 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:

👁 Image

Given : AB || CD || EF , AB = 6 cm, CD = x cm, BD = 4 cm, and DE = y cm and EF = 10 cm.

Now In ∆ECD and ∆EAB,

∠CED = ∠AEB    [common]

∠ECD = ∠EAB   [corresponding angles]

∆ECD ~ ∆EAB ……....(1) [By AA similarity]

∴ EC/EA = CD/AB   [Corresponding parts of similar triangles are proportional]

EC / EA = x/6 ……………(2)

In ∆ACD and ∆AEF

∠CAD = ∠EAF [common]

∠ACD = ∠AEF [corresponding angles]

∆ACD ~ ∆AEF [By AA similarity]

∴ AC/AE = CD/EF     [Corresponding parts of similar triangles are proportional]

AC /AE = x/10 ……………..(3)

EC/EA+ AC /AE =  x/6 + x/10  [by Adding eq 2 & 3]

(EC + AC) /AE =( 5x + 3x)/30

AE / AE = 8x /30

1 = 8x/30

x = 30/8  

x = 3.75 cm

∆ECD ~ ∆EAB    [From eq  (i)]

DC/AB = ED /EB   [Corresponding parts of similar triangle are proportional]

3.75/6 = y/y+4

6y = 3.75(y+4)

6y = 3.75y + 15

6y - 3.75y = 15

2.25y = 15

y = 15/2.25

y = 6.67 cm

Hence, x = 3.75 cm and y = 6.67 cm.

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• Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.5 | Set 1
• Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.6 | Set 2