Maharashtra Board Class 10 Math Part – 2 Solution Chapter 1 Practice Set 1.2 – Similarity
Balbharati Maharashtra Board Class 10 Math Part – 2 Solution Chapter 1: Similarity. Marathi or English Medium Students of Class 10 get here Similarity full Exercise Solution.
Std |
Maharashtra Class 10 |
Subject |
Math Part 2 Solution |
Chapter |
Similarity |
Practice Set |
1.2 |
Practice set 1.2
(1) Given below are some triangles and lengths of line segments. Identify in which figures, ray PM is the bisector of ∠QPR.
Solution:
In option (i) the ray PM bisects ∠QPR, her is the proof.
3M/MR = 3.5/1.5
= 7/3
QP/RP = 7/3
∴ QM/MR = QP/RP
∴By properly of angle bisector we can by the ray PM bisects ∠QPR
(2) In ∆PQR, PM = 15, PQ = 25
PR = 20, NR = 8. State whether line NM is parallel to side RQ. Give reason.
Solution:
In ∆PQR, it is given that
PM = 15, PQ = 25, PR = 20, NR = 8
∴ PN = PR – NR
= 20 – 8
= 1
MQ = PQ – PM
= 25 – 15
= 10
Now, PN/NR = 12/8 = 3/2 & PM/MQ = 15/10 = 3/2
∴ PN/NR = PM/NQ
∴By basic proportionality theorem we can say that NM is parallel to RQ.
(3) In ∆MNP, NQ is a bisector of ∠N. If MN = 5, PN = 7 MQ = 2.5 then find QP.
Solution:
Given, In ∆MNP, NQ bisects ∠N
MN = 5, PN = 7, MQ = 2.5
In ∆MNP, by property of angle bisector,
MQ/QP = MN/PN [∵ NQ bisects ∠N]
Or, 2.5/QP = 5/7
Or, QP = 7×2.5/5 = 3.5
(4) Measures of some angles in the figure are given. Prove that
AP/PB = AQ/QC
Solution:
In the adjacent fig
∠APQ = ∠ABC = 60°
Now, AP and AB are collinear.
∴seg PQ and seg BC are parallel,
Since ∠APQ &∠ABC are corresponding angles to seg PQ and seg BC.
∴In ∆ABC
PQ||BC,
∴By basic proportionality theorem,
∴ AP/PB = AQ/QC [Proved]
(5) In trapezium ABCD, Side AB||side PQ||Side DC, AP = 15, PD = 12, QC = 14, find BQ.
Solution:
Given, In trap ABCD, AB||PQ||DC, AP = 15, PD = 12, QC = 14
In trap ABCD,
By property of three parallel lines and their transversal we can say that.
AP/PD = BQ/QC [∵ AB||PQ||DC]
Or, 15/12 = BQ/14 or, BQ = 15×14/12 = 35/3 = 11.66
(6) Find QP using given information in the figure.
Solution:
In the adjacent fig.
QN bisects ∠MNP as started.
MN = 25, NQ = 14
NP = 40
∴In ∆MNP,
By property of angle bisector,
MQ/QP = MN/NP [∵ QN bisector ∠NNP]
Or, 14/QP = 25/40
Or, QP = 40×14/25 = 112/5 = 22.4
(7) In figure 1.41, if AB||CD||FE then find x and AE.
Solution:
In the adjacent fig it is given that, AB||CD||FE
Now, by property of three parallel lines and their transversal.
We can say that.
AC/CE = BO/DF
Or, 12/x = 8/4
Or, x = 12/2 = 6
∴ AE = AC + CE
= 12+6
= 18
(8) In ∆LMN, ray MT bisects ∠LMN if LM = 6, MN = 10, TN = 8, then find LT.
Solution:
In ∆LMN it is given that, ray MT bisects ∠LMN, LM = 6, MN = 10, TN = 8,
In ∆LMN,
By property of angles bisector we can say that,
LT/TN = LM/MN [∵ ray NT bisects ∠LMN]
Or, LT/8 = 6/10
Or, LT = 48/10 = 4.8
(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, it is given that,
BD bisects ∠ABC, AB = x, BC = x+5, AD = x – 2, DC = x + 2
∴In ∆ABC,
By the property of angle bisector we can say that
AD/DC = AB/BC
Or, x-2/x+2 = x/x+5
Or, x (x + 2) = (x + 5) (x – 2)
Or, x2 + 2x = x2 – 2x + 5x – 10
Or, x2 + 2x = x2 + 3x – 10
Or, 3x – 2x = 10
Or, x = 10
(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, segPR || seg DF.
Solution:
In the adjacent fig it is given that,
PQ||DE, QR||SP,
In ∆DEX
X by basic proportionality theorem,
XP/PD = XQ/BE [∵ PQ||DE] —— (i)
Also, In ∆EFX
By basic proportionality theorem,
XQ/QE = XR/RF [∵ QR||EF] —– (ii)
Comparing Equation (i) & (ii) we get,
XP/PD = XR/RF
∴In ∆DXF
XP/PD = XR/RF
Now, By converse of basic proportionality theorem we can say that seg PR||seg DF Proved
(11) In ∆ABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB ≅ seg AC then prove that ED||BC.
Solution:
In ∆ABC, it is given that:
BD bisects ∠ABC & CE bisects ∠ACB
AB≅AC
∴In ∆ABC,
By property of angle bisector,
AD/DC = AB/BC [∵ BD bisects ∠ABC] ——- (i)
Also,
By angle property of angle bisector,
AE/EB = AC /BC [∵ CE bisects ∠ACB]
AE/EB = AB/BC [∵ AB ≅ AC] (ii)
On comparing equation (i) &(ii) we get,
AD/DC = AE/EB
∴In ∆ABC,
By converse of basic proportion theorem we can say that ED||BC proved
Here is your solution of Maharashtra Board Class 10 Math Part – 2 Chapter 1 Similarity Practice Set 1.2
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