Chapter 1 Similarity 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 ∠QPR.

Solution:

word image 7670 2
word image 7670 3

Question 2.
In ∆PQR PM = 15, PQ = 25, PR = 20, NR = 8. State whether line NM is parallel to side RQ. Give reason.

Solution:

PN + NR = PR [P – N – R]
∴ PN + 8 = 20
∴ PN = 20 – 8 = 12
Also, PM + MQ = PQ [P – M – Q]
∴ 15 + MQ = 25
Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 3
∴ line NM || side RQ [Converse of basic proportionality theorem]

Question 3.
In ∆MNP, NQ is a bisector of ∠N. If MN = 5, PN = 7, MQ = 2.5, then find QP.

Solution:

word image 7670 7

Question 4.


Solution:

Proof
∠APQ = ∠ABC = 60° [Given]
∴ ∠APQ ≅ ∠ABC
∴ side PQ || side BC (i) [Corresponding angles test]
In ∆ABC,
sidePQ || sideBC [From (i)]
word image 7670 10

Question 5.
In trapezium ABCD, side AB || side PQ || side DC, AP = 15, PD = 12, QC = 14, find BQ.

Solution:

side AB || side PQ || side DC [Given]
word image 7670 12

Question 6.
Find QP using given information in the figure.

Solution:

In ∆MNP, seg NQ bisects ∠N. [Given]
word image 7670 14

Question 7.
In the adjoining figure, if AB || CD || FE, then find x and AE.

Solution:

line AB || line CD || line FE [Given]
word image 7670 16
∴ X = 6 units
Now, AE AC + CE [A – C – E]
= 12 + x
= 12 + 6
= 18 units
∴ x = 6 units and AE = 18 units

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

Solution:

In ∆LMN, ray MT bisects ∠LMN. [Given]
word image 7670 18

Question 9.
In ∆ABC,seg BD bisects ∠ABC. If AB = x,BC x+ 5, AD = x – 2, DC = x + 2, then find the value of x.

Solution:

In ∆ABC, seg BD bisects ∠ABC. [Given]
word image 7670 20
∴ x(x + 2) = (x – 2)(x + 5)
∴ x2 + 2x = x2 + 5x – 2x – 10
∴ 2x = 3x – 10
∴ 10 = 3x – 2x
∴ x = 10

Question 10.
In the adjoining figure, 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.

Solution:

Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 12

Question 11.
In ∆ABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB = seg AC, then prove that ED || BC.

Solution:
word image 7670 24

Question 1.
i. Draw a ∆ABC.
ii. Bisect ∠B and name the point of intersection of AC and the angle bisector as D.
iii. Measure the sides.

iv. Find ratios  and 
v. You will find that both the ratios are almost equal.
vi. Bisect remaining angles of the triangle and find the ratios as above. Verify that the ratios are equal. (Textbook pg. no. 8)
Solution:

Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 15
Note: Students should bisect the remaining angles and verify that the ratios are equal.

Question 2.
Write another proof of the above theorem (property of an angle bisector of a triangle). Use the following properties and write the proof.
i. The areas of two triangles of equal height are proportional to their bases.
ii. Every point on the bisector of an angle is equidistant from the sides of the angle. (Textbook pg. no. 9)

Given: In ∆CAB, ray AD bisects ∠A.

Construction: Draw seg DM ⊥ seg AB A – M – B and seg DN ⊥ seg AC, A – N – C.
Solution:

Proof:
In ∆ABC,
Point D is on angle bisector of ∠A. [Given]
∴DM = DN [Every point on the bisector of an angle is equidistant from the sides of the angle]
word image 7670 29

Question 3.
i. Draw three parallel lines.
ii. Label them as l, m, n.
iii. Draw transversals t1 and t2.
iv. AB and BC are intercepts on transversal t1.
v. PQ and QR are intercepts on transversal t2.
vi. Find ratios AB/BC and PQ/QR. You will find that they are almost equal. Verify that they are equal.(Textbook pg, no. 10)
Solution:

Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 17
(Students should draw figures similar to the ones given and verify the properties.)

Question 4.
In the adjoining figure, AB || CD || EF. If AC = 5.4, CE = 9, BD = 7.5, then find DF. (Textbook pg, no. 12)

Solution:

Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 19

Question 5.
In ∆ABC, ray BD bisects ∠ABC. A – D – C, side DE || side BC, A – E – B, then prove that AB/BC = AE/EB (Textbook pg, no. 13)

Solution:

Maharashtra Board Class 10 Maths Solutions Chapter 1 Similarity Practice Set 1.2 21