Practice Set 1.2
- Given below are some triangles and lengths of line segments. Identify in which figures,…
- In Δ PQR, PM = 15, PQ = 25, PR = 20, NR = 8. State whether line NM is parallel to side…
- In Δ MNP, NQ is a bisector of ∠N. If MN = 5, PN = 7 MQ = 2.5 then find QP.…
- Measures of some angles in the figure are given. Prove that ap/pb = aq/qc…
- In trapezium ABCD, side AB || side PQ || side DC, AP = 15, PD = 12, QC = 14, find BQ.…
- Find QP using given information in the figure.
- In figure 1.41, if AB || CD || FE then find x and AE.
- In Δ LMN, ray MT bisects ∠ LMN If LM = 6, MN = 10, TN = 8, then find LT.…
- In Δ ABC, seg BD bisects ∠ ABC. If AB = x, BC = x + 5, AD = x - 2, DC = x + 2, then…
- In the figure 1.44, X is any point in the interior of triangle. Point X is joined to…
- In ΔABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB ≅ seg AC then prove…
Practice Set 1.2
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)
