14. Chapter 14. Similarity of Triangles NIOS Class 10 Mathematics Textbook Solutions

Chapter 14. Similarity of Triangles NIOS Class 10 Mathematics Textbook Solutions for Exam Preparation

Chapter 14. SIMILARITY OF TRIANGLES

CHECK YOUR PROGRESS 14.1

1. In Fig. 14.7 (i) and (ii), PQ || BC. Find the value of x in each case.

Photo

Solution: (i) Here, AP = 3 cm , PB = 4 cm , AQ = 4.5 cm  and QC = x

In ∆ABC and PQ || BC , we have

APPB=AQQC

⇒ 34 =4.5x

⇒3x=4×4.5

⇒x=4×4.53=4×1.5=6.0

⇒x=6 cm

(ii) Here, AP = 3.5 cm , PB = 4 cm , AQ = 5.25 cm  and QC = x

In ∆ABC and PQ || BC , we have

APPB=AQQC

⇒ 3.54 =5.25x

⇒3.5x=4×5.25

⇒x=4×5.253.5=4×1.5=6.0

⇒x=6 cm

2. In Fig. 14.8 [(i)], find whether DE || BC is parallel to BC or not? Give reasons for your answer.

Photo

Solution: Here, AD = 3 cm , DB = 4.5 cm , AE = 4.5 cm and 5.75 cm

 ADDB=34.5=3045=23 

 AEEC=4.55.75=450575=1823

 ADDB≠AEEC 

So, DE||BC is not parallel to BC .

CHECK YOUR PROGRESS 14.2

1. In Fig. 14.12, AD is the bisector of ∠A, meeting BC in D. If AB = 4.5 cm, BD = 3 cm, DC = 5 cm, find x.

Solution:  Here, AB = 4.5 cm , BD = 3 cm , CD = 5 cm , AC = x

Since, AD is the bisector of ∠A .

Using angle bisector theorem ,    

⇒35=4.5x

⇒3x=5×4.5

⇒x=5×4.5 3

⇒x=5×1.5=7.5

Therefore, the value of x is 7.5 cm .

2. In Fig. 14.13, PS is the bisector of ∠P of ΔPQR. The dimensions of some of the sides are given in Fig. 14.13. Find x.

Solution: Here, PQ = 4 cm , PR = 2(x – 1) , QS = x , SR = x+2

Since, PS is the bisector of ∠P of ΔPQR .

Using angle bisector theorem ,

 QSSR=PQPR 

⇒xx+2=42x-1

⇒xx+2=2x-1

⇒x2-x=2x+4

⇒x2-3x-4=0

⇒x²-4x+x-4=0

⇒xx-4+1x-4=0

⇒x-4x+1=0

x-4=0⇒x=4  or x+1=0 ⇒x=-1 (impossible)

Therefore, the value of x is 4 cm .

3. In Fig. 14.14, RS is the bisector of ∠R of ΔPQR. For the given dimensions, express p, the length of QS in terms of x, y and z.

Solution: Here, PR = x , QR = z , PS = y and QS = p

Since, RS is the bisector of ∠R of ΔPQR.  

Using angle bisector theorem ,

 QSPS=QRPR

 ⇒py=zx

⇒p=yzx

CHECK YOUR PROGRESS 14.3

1. Find values of x and y of ΔABC ~ ΔPQR in the following figures:

Photo

Fig. 14.21, Fig. 14.22,   Fig. 14.23,

Solution: Here, AB = 3 cm , BC = 4 cm , AC = y , PQ = x , QR = 6 cm and 5.25 cm

Since,  ΔABC ~ ΔPQR

  

  

Now,

 3x=46 

⇒4x=18⇒x=184⇒x=92

⇒x=4.5 cm

And

 46=y5.25

 ⇒6y=4×5.25

⇒y=4×5.256=2×1.75

⇒y=3.50⇒y=3.5  cm

(ii) Here, ∠A=60° ,∠B=x° , ∠C=50° , ∠P=60° ,∠Q=70° , ∠R=y°

Since, ΔABC ~ ΔPQR

∠A=∠P=60°

∠B=∠Q

⇒x=70° 

And ∠C=∠R 

⇒50°=y⇒y=50°

∴   x=70°  and y=50°

(iii) Here, AB = 3.5 cm ,  BC = x ,  AC = 3 cm , PQ = 6 cm , QR = 4 cm and  PR = 7 cm

Since,  ΔABC ~ ΔPQR  

 ABPR=BCQR=ACPQ

3.57=x4=36 

  3570=x4=12 

 12=x4=12

Now,

⇒x4=12⇒x=42⇒x=2

Therefore, the value of x is 2 cm.

CHECK YOUR PROGRESS 14.4 

1. In Fig. 14.27, ABC is a right triangle with A = 90° and C = 30° . Show that ΔDAB ~ΔDCA ~ ΔACB .

Solution:  Given, ∠A=90° , ∠ADB=∠ADC=90° and ∠C=30°

∠B=180°-∠A+∠C=180°-90°+30°=180°-120°=60° 

∠BAD=180°-∠B+∠ADB=180°-60°+90°=180°-150°=30° 

∠CAD=180°-∠C+∠ADC=180°-30°+90°=180°-120°=60° 

In ΔDAB and ΔDCA , we have,

∠ADB=∠ADC=90°

∠BAD=∠C=30°

∠B=∠CAD=60° 

ΔDAB ~ΔDCA [AAA]

Again, ΔDCA  and ΔACB

∠ADC=BAC=90°

∠C=∠C=30°  [common angle]

∠CAD=∠B=60°

 ΔDCA ~ ΔACB  [AAA]

So,  ΔDAB ~ΔDCA ~ ΔACB

Hence proved .

2. Find the ratio of the areas of two similar triangles if two of their corresponding sides are of length 3 cm and 5 cm.

Solution: Let the two triangles be ABC and PQR

Let BC = 3 cm and QR = 5 cm

  Area(∆ABC)Area(∆PQR)=BCQR2=352=925   

3. In Fig. 14.28, ABC is a triangle in which DE||BC. If AB = 6 cm and AD = 2 cm, find the ratio of the areas of ΔADE and trapezium DBCE.

Solution: Given, AB = 6 cm and AD = 2 cm

ADAB=26=13

Area∆ADEArea∆ABC=ADAB2=132=19

Let ,x and 9x are the Area∆ADE and Area∆ABC respectively .

Area∆DBCE=Area∆ABC-Area∆ADE=9x-x=8x

4. P, Q and R are respectively the mid-points of the sides AB, BC and CA of the ΔABC. Show that the area of ΔPQR is one-fourth the area of ΔABC.

Solution: Given, P, Q and R are the mid-points of the sides AB, BC and CA of the ΔABC respectively .

To prove:

Area∆PQR=14Area∆ABC

Proof: Since P and Q are the mid-points of AB and BC, then

PQ∥AC and PQ=12AC

 Q and R are the mid-points of BC and CA, then

 QR∥AB and QR=12AB

 R and P are the mid-points of CA and AB,

 RP∥BC and RP=12BC

In ΔAPR and ΔPQR , we have

 AP = PB =12 AB (P is midpoint)

AR = RC =12AC (R is midpoint)

PR = PR  [common side ]

 ΔAPR ≅ ΔPQR [SSS]

So,  
Area∆APR=Area∆PQR

Similarly ,    Area∆PQB=Area∆PQR and Area∆RQC=Area∆PQR  
Area∆ABC=Area∆APR+Area∆PQR+Area∆PQB+Area∆RQC 

⇒Area∆ABC=Area∆PQR+Area∆PQR+Area∆PQR+Area∆PQR

⇒Area∆ABC=4 Area∆PQR

⇒Area∆PQR=14 Area∆ABC

Hence prove,

5. In two similar triangles ABC and PQR, if the corresponding altitudes AD and PS are in the ratio of 4 : 9, find the ratio of the areas of ΔABC and ΔPQR.

[ Hint: Use  ABPQ=ADPS=BCQR=CAPR ]

Solution: Since, ∆ABC~ ∆PQR  , we have

ABPQ=ADPS=BCQR=CAPR=49

Area∆ABCArea∆PQR=ADPS2=492=1681

6. If the ratio of the areas of two similar triangles is 16 : 25, find the ratio of their corresponding sides.

Solution: Let the two triangles be ABC and PQR

So,

 ABPQ=BCQR=ACPR

 Area(∆ABC)Area(∆PQR)=1625

  Area∆ABCArea∆PQR=BCQR2

⇒1625=BCQR2

⇒BCQR2=452

---

Posted 1 month ago