Problem Set 1 Similarity Class 10th Mathematics Part 2 MHB Solution

Problem Set 1

  1. In Δ ABC and ΔPQR, in a one to one correspondence ab/qr = bc/pr = ca/pq then Select…
  2. If in ΔDEF and ΔPQR, ∠D ≅ ∠Q, ∠R ≅ ∠E then which of the following statements is false?…
  3. In Δ and ΔDEF ∠B = ∠E, ∠F = ∠C and AB = 3DE then which of the statements regarding the…
  4. ΔABC and ΔDEF are equilateral triangles, A(ΔABC) : A(ΔDEF) = 1 : 2. If AB = 4 then…
  5. In figure 1.71, seg XY || seg BC, then which of the following statements is true? x x…
  6. In Δ ABC, B - D - C and BD = 7, BC = 20 then find following ratios. (1) a (deltaabd)/a…
  7. Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller…
  8. In figure 1.73, ∠ABC = ∠DCB = 90° AB = 6, DC = 8 then a (deltaabc)/a (deltadcb) = ?…
  9. In figure 1.74, PM = 10 cm A(Δ PQS) = 100 sq.cm A (ΔQRS) = 110 sq.cm then find NR.…
  10. ΔMNT ~ ΔQRS. Length of altitude drawn from point T is 5 and length of altitude drawn…
  11. In figure 1.75, A - D - C and B - E - C seg DE || side AB If AD = 5, DC = 3, BC = 6.4…
  12. In the figure 1.76, seg PA, seg QB, seg RC and seg SD are perpendicular to line AD. AB…
  13. In ΔPQR seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side…
  14. In fig 1.78, bisectors of ∠B and ∠C of ΔABC intersect each other in point X. Line AX…
  15. In □ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point…
  16. In fig 1.80, XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the…
  17. In figure 1.81, the vertices of square DEFG are on the sides of ΔABC, ∠A = 90°. Then…

Problem Set 1
Question 1.

Select the appropriate alternative.

In Δ ABC and ΔPQR, in a one to one correspondence  then


A. Δ PQR ~ Δ ABC

B. Δ PQR ~ Δ CAB

C. Δ CBA ~ Δ PQR

D. Δ BCA ~ Δ PQR


Answer:

∵ 


⇒ Δ CAB~Δ PQR


(A) doesn’t match the solution.


(C) doesn’t match the solution.


(D) doesn’t match the solution.


Question 2.

Select the appropriate alternative.

If in ΔDEF and ΔPQR, ∠D ≅ ∠Q, ∠R ≅ ∠E then which of the following statements is false?


A. 

B. 

C. 

D. 


Answer:

In Δ DEF & Δ PQR


∠ D≅ ∠ Q and ∠ R≅ ∠ E (Given)


⇒ Δ DEF ~ Δ PQR


⇒  (corresponding sides are proportional)


(A) is matching the solution, hence can’t be false.


(C) is matching the solution, hence can’t be false.


(D) is matching the solution, hence can’t be false.


Question 3.

Select the appropriate alternative.

In Δ and ΔDEF ∠B = ∠E, ∠F = ∠C and AB = 3DE then which of the statements regarding the two triangles is true?


A. The triangles are not congruent and not similar

B. The triangles are similar but not congruent.

C. The triangles are congruent and similar.

D. None of the statements above is true.


Answer:

In Δ ABC & Δ DEF


∠ B≅ ∠ E and ∠ C≅ ∠ F (Given)


⇒ Δ ABC ~ Δ DEF (By AA Test)


⇒ The triangles are similar.


And, Δ ABC ≅ Δ DEF, if AB = DE.


But, given that - AB = 3DE.


⇒ The triangles are not congruent.


(A) doesn’t match the solution.


(C) doesn’t match the solution.


(D) doesn’t match the solution.


Question 4.

Select the appropriate alternative.

ΔABC and ΔDEF are equilateral triangles, A(ΔABC) : A(ΔDEF) = 1 : 2. If AB = 4 then what is length of DE?


A. 

B. 4

C. 8

D. 


Answer:

Solution: We know that, all the angles of an equilateral triangles are equal, i.e., 60°.


⇒ Δ ABC~Δ DEF ……(By AAA Similarity Test)


⇒ 


And, (Given)


⇒ 


⇒ DE2 = 2× 42 (∵ AB = 4)


⇒ DE = √32


DE = 4√2


(A) doesn’t match the solution.


(B) doesn’t match the solution.


(C) doesn’t match the solution.


Question 5.

Select the appropriate alternative.

In figure 1.71, seg XY || seg BC, then which of the following statements is true?


A. 

B. 

C. 

D. 


Answer:

∵ segXY||segBC


⇒∠ AXY≅ ∠ ABC


And, ∠ XAY≅ ∠ BAC (Common)


⇒ Δ AXY~ Δ ABC (By AA Test)


⇒  (corresponding sides are proportional)


⇒ 


(B) doesn’t match the solution.


(C) doesn’t match the solution.


(D) doesn’t match the solution.


Question 6.

In Δ ABC, B - D – C and BD = 7, BC = 20 then find following ratios.



(1) 

(2) 

(3) 


Answer:

Theorem: When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides.

(1) 





(2) 




(3) 






Question 7.

Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller triangle is 6 cm then what is the corresponding base of the bigger triangle?


Answer:

(PROPERTY:Areas of triangles with equal heights are proportional to their corresponding bases.)

⇒ 


⇒ 


⇒ base(bigger triangle) = 9 cm



Question 8.

In figure 1.73, ∠ABC = ∠DCB = 90° AB = 6, DC = 8 then 



Answer:

We know that, Area of triangle = × base× height

⇒ 






Question 9.

In figure 1.74, PM = 10 cm A(Δ PQS) = 100 sq.cm A (ΔQRS) = 110 sq.cm then find NR.



Answer:

We know that, Area of triangle = × base× height

⇒ 


⇒ 


⇒ 


⇒ NR = 11 cm



Question 10.

ΔMNT ~ ΔQRS. Length of altitude drawn from point T is 5 and length of altitude drawn from point S is 9. Find the ratio 


Answer:





Question 11.

In figure 1.75, A – D – C and B – E – C seg DE || side AB If AD = 5, DC = 3, BC = 6.4 then find BE.



Answer:

By Basic Proportionality Theorem-

⇒ 


⇒ 


⇒ 3x = 32-5x


⇒ 8x = 32


⇒ x = 4 = BE



Question 12.

In the figure 1.76, seg PA, seg QB, seg RC and seg SD are perpendicular to line AD.

AB = 60, BC = 70, CD = 80, PS = 280 then find PQ, QR and RS.



Answer:

(PROPERTY:If line AX || line BY || line CZ and line l and line m are their transversals then )

⇒ 


⇒ 


⇒ 


 [1]


And 


⇒ 


⇒ 

 [2]


And,PS = 280


⇒ PQ + QR + RS = 280 ……(3)


From [1] and [2], we have


From [1],

From [2]


Question 13.

In ΔPQR seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR.



Complete the proof by filling in the boxes. In ΔPMQ, ray MX is bisector of ∠PMQ.

 .......... (I) theorem of angle bisector.

In ΔPMR, ray MY is bisector of ∠PMR.

 .......... (II) theorem of angle bisector.

But  .......... M is the midpoint QR, hence MQ = MR.



∴ XY || QR .......... converse of basic proportionality theorem.


Answer:

.......... (I) theorem of angle bisector.


AND


∴ .......... (II) theorem of angle bisector.



Question 14.

In fig 1.78, bisectors of ∠B and ∠C of ΔABC intersect each other in point X. Line AX intersects side BC in point Y. AB = 5, AC = 4, BC = 6 then find 



Answer:

By Bisector Theorem-

⇒  ……(1)


⇒ And, ……(2)


Equating (1) & (2), we get-


⇒ 


⇒ 


⇒ 



⇒ 


⇒ CY = 


⇒ CY = 


Now,


⇒ 


⇒ 



Question 15.

In □ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that 



Answer:

In Δ APD and ΔCPB

⇒ ∠ APD≅ ∠ CPB (opposite angles)


⇒ ∠ ADP≅ ∠ PBC (Alternate angles ∵ AD||BC)


⇒ Δ APD ~ Δ CPB (By AA Test)


⇒  (corresponding sides are proportional)


⇒ 



Question 16.

In fig 1.80, XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC.



Activity : 

 .......... by componendo.

.......... (I)

 ..........  test of similarity.

 .......... corresponding sides of similar triangles.

 ...from (I)


Answer:

ACTIVITY: 2AX = 3BX ∴ 

……(By Componendo)


…….(I)


Δ BCA~ ΔBYX…… (AA test of similarity).


∴ ………. (corresponding sides of similar triangles).


 ∴ AC = 15……from (I)



Question 17.

In figure 1.81, the vertices of square DEFG are on the sides of ΔABC, ∠A = 90°. Then prove that DE2 = BD × EC

(Hint : Show that ΔGBD is similar to ΔDFE. Use GD = FE = DE.)



Answer:

Proof:In □ DEFG is a square


⇒ GF||DE


⇒ GF||BC


Now,In Δ AGF and Δ DBG


⇒ ∠ AGF≅ ∠ DBG (corresponding angles)


⇒ ∠ GDB≅ ∠ FAG (Both are 90° )


⇒ Δ AGF~Δ DBG ……(1) (AA similarity)


Now,In Δ AGF and Δ EFC


⇒ ∠ AFG≅ ∠ ECF (corresponding angles)


⇒ ∠ GAF≅ ∠ FEC (Both are 90° )


⇒ Δ AGF~Δ EFC ……(2) (AA similarity)


From (1) & (2), we have-


⇒ Δ EFC~Δ DBG


⇒ 


⇒ EF× DG = BD× EC


Now,∵ DEFG is a square


⇒ DE = EF = DG


⇒ DE× DE = BD× EC


⇒ DE2 = BD× EC


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