Practice Set 1.2 Similarity Class 10th Mathematics Part 2 MHB Solution

Practice Set 1.2

  1. Given below are some triangles and lengths of line segments. Identify in which figures,…
  2. In Δ PQR, PM = 15, PQ = 25, PR = 20, NR = 8. State whether line NM is parallel to side…
  3. In Δ MNP, NQ is a bisector of ∠N. If MN = 5, PN = 7 MQ = 2.5 then find QP.…
  4. Measures of some angles in the figure are given. Prove that ap/pb = aq/qc…
  5. In trapezium ABCD, side AB || side PQ || side DC, AP = 15, PD = 12, QC = 14, find BQ.…
  6. Find QP using given information in the figure.
  7. In figure 1.41, if AB || CD || FE then find x and AE.
  8. In Δ LMN, ray MT bisects ∠ LMN If LM = 6, MN = 10, TN = 8, then find LT.…
  9. In Δ ABC, seg BD bisects ∠ ABC. If AB = x, BC = x + 5, AD = x - 2, DC = x + 2, then…
  10. In the figure 1.44, X is any point in the interior of triangle. Point X is joined to…
  11. In ΔABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB ≅ seg AC then prove…

Practice Set 1.2
Question 1.

Given below are some triangles and lengths of line segments. Identify in which figures, ray PM is the bisector of ∠OPR.



Answer:

Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.

Therefore, we’ll find the ratio for all the triangle.Hence, for


(1) 


= 2.33


And 


= 2.33


⇒ 


⇒ In (1), ray PM is a bisector.


(2)


= 0.75


And 


= 0.7


⇒ 


⇒ In (2), ray PM is not a bisector.


(3)


= 1.1


And 


= 1.11


⇒ 


⇒ In (3), ray PM is a bisector.



Question 2.

In Δ PQR, PM = 15, PQ = 25, PR = 20, NR = 8. State whether line NM is parallel to side RQ. Give reason.



Answer:

By Converse of basic Proportionality Theorem

(Theorem: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.)


⇒ If, then line NM is parallel to side RQ.


∴ we’ll check if.


⇒ 





And,





⇒ , therefore line NM || side RQ



Question 3.

In Δ MNP, NQ is a bisector of ∠N. If MN = 5, PN = 7 MQ = 2.5 then find QP.



Answer:

Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.

⇒ 


⇒ 


⇒ QP× 5 = 2.5× 7


⇒ QP = 


⇒ QP = 3.5



Question 4.

Measures of some angles in the figure are given. Prove that 



Answer:

Here, PQ||BC (∵ ∠ APQ≅ ∠ ABC)

(PROPERTY: If a transversal intersects two lines so that corresponding angles are congruent, then the lines are parallel)


∴ By Basic Proportionality Theorem


(Theorem : If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion.)


⇒ 



Question 5.

In trapezium ABCD, side AB || side PQ || side DC, AP = 15, PD = 12, QC = 14, find BQ.



Answer:

By Basic Proportionality Theorem

(Theorem : If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion.)


⇒ 


⇒ 


⇒ BQ = 


⇒ BQ = 17.5



Question 6.

Find QP using given information in the figure.



Answer:

Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.

And ∵ NQ is angle bisector of ∠N




⇒ QP = 


⇒ QP = 22.4



Question 7.

In figure 1.41, if AB || CD || FE then find x and AE.



Answer:

Theorem: The ratio of the intercepts made on a transversal by three parallel lines is equal to the ratio of the corresponding intercepts made on any other transversal by the same parallel lines.

⇒ 


⇒ 


⇒ x = 


⇒ x = 6


Now,AE = AC + CE


= 12 + x


= 12 + 6


⇒ AE = 18



Question 8.

In Δ LMN, ray MT bisects ∠ LMN If LM = 6, MN = 10, TN = 8, then find LT.



Answer:

Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.

⇒ 


⇒ LT = 


⇒ LT = 


⇒ LT = 4.8



Question 9.

In Δ ABC, seg BD bisects ∠ ABC. If AB = x, BC = x + 5, AD = x – 2, DC = x + 2, then find the value of x.



Answer:

Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.

⇒ 


⇒ 


⇒ x(x + 2) = (x-2)(x + 5)


⇒ x2 + 2x = x2-2x + 5x-10


⇒ x2 + 2x-x2 + 2x-5x + 10 = 0


⇒ x = 10



Question 10.

In the figure 1.44, X is any point in the interior of triangle. Point X is joined to vertices of triangle. Seg PQ || seg DE, seg QR || seg EF. Fill in the blanks to prove that, seg PR || seg DF.



Proof : IN 



(Basic proportionality theorem)

In 



(converse of basic proportionality theorem)


Answer:

Proof: In ΔXDE, PQ||DE….. (Given)


∴  …..(I)


(Basic proportionality theorem)


In ΔXDE, QR||EF …….(Given)


∴  ………(II) (Basic Proportionality Theorem)


∴  ……… from (I) and (II)


∴ seg PR||Seg DE ………..


(converse of basic proportionality theorem)



Question 11.

In ΔABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB ≅ seg AC then prove that ED || BC.


Answer:

PROOF:



Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.


⇒  ……(1)


and  …..(2) (∵ , BD and CE are angle bisectors of ∠B and ∠C respectively.)


Now, ∵ seg AB ≅ seg AC


⇒ AB = AC


⇒ 


⇒ R.H.S of (1) & (2) are equal.


⇒ L.H.S of (1) & (2) will be equal.


∴ Equating L.H.S of (1) &(2), we get-


⇒ 


⇒ ED||BC (By converse basic proportionality theorem)


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