Maharashtra Board Class 10 Math Part – 2 Solution Chapter 1 Practice Set 1.3 – 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.
|Maharashtra Class 10|
Math Part 2 Solution
Practice set 1.3
(1) In figure 1.55, ∠ABC = 75°,
∠EDC = 75° state which two triangles are similar and by which test? Also write the similarity of these two triangles by a proper one to one correspondence.
In the adjacent fig, it is given that ∠ABC = ∠CDE = 75°
∴In ∆ABC and ∆CDE
∠DCE is common.
∴ ∆ABC ~ ∆CDE by AA test
(2) Are the triangles in figure 1.56 similar? If yes, by which test?
In the adjacent fig, PQ = 6, QR = 8, PR = 10, LM = 3, MN = 4, LN = 5
In ∆PQR & ∆LMN
∴ PQ/LM = 6/3 = 2/1
QR/MN = 8/4 = 2/1, PR/LN = 10/5 = 2/1
∴ PQ/LM = QR/MN = PR/LN
∴∆PQR ~ ∆LMN by SSS test
(3) As shown in figure 1.57, two poles of height 8 m and 4 m are perpendicular to the ground. If the length of shadow of smaller pole due to sunlight is 6 m then how long will be the shadow of the bigger pole at the same time?
In the adjacent triangles PR⊥ to BQ AC⊥BC
PR = 4, QR = 6, AC = 8, BC = x
In ∆PQR & ∆ABC
∠PRQ = ∠ACB = 90°
∠QPR = ∠BAC [∵ Angles made by sunlight on top are congruent]
∴∆PQR ~ ∆ABC by AA test
By property of similar triangles we can say that.
∴ PR/AC = QR/BC or, 4/8 = 6/x or, x = 12
(4) In ∆ABC, AP⊥BC, BQ⊥AC
B – P – C, A – Q – C then prove that,
If AP = 7, BQ = 8, BC = 12 then find AC.
In the adjacent triangles if it is given that,
B – P – C, A – Q – C
In ∆CPA & ∆CQB
∠APC = ∠CQB = 90°
∠BCQ = ∠ACP [Common angle of two triangle]
∴∆CPA ~ ∆CQB by AA test
Now, AP = 7, AB = 8, BC = 12 given
∴By property of similar triangles are can say that AP/BQ = AC/BC
Or, 7/8 = AC/12
Or, AC = 21/2 = 10.5
(5) Given: In trapezium PQRS, side PQ||side SR, AR = 5AP, AS = 5AQ then prove that, SR = 5PQ
In the adjacent trap PQRS
PQ||SR, AR = 5AP —- (i)
Now, in ∆ASR & ∆APQ
∠SAR = ∠PAQ [Opposite vertically angle]
∠ASR = ∠AQP [Alternate angle ∵ PQ||SR]
∴∆ASR ~ ∆APQ by AA test
∴By property of similar triangle.
AR/AP = AS/AQ = SR/PQ also, AR/AP = SR/PQ
Or, 5AP/AP = AS/AQ – From equation (i) or, 5AP/AP = SR/PQ – from equation (ii)
Or, AS = 5AQ Prove/ Or, SR = 5PQ proved
(6) In trapezium ABCD, (Figure 1.60) side AB||Side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6, OB = 15 then find OD.
In the adjacent trap ABCD
AB||DC, AB = 20, DC = 6, OB = 15
∆AOB & ∆DOC
∠AOB = ∠DOC [Vertically opposite angles]
∠ABO = ∠CDO [Alternate angles ∵AB||DC]
∴ ∆ADB ~ ∆DOC by AA test
∴ By property of similar triangles we can say that,
AB/DC = OB/OD = OA/OC
∴ AB/DC = OB/OD
Or, 20/6 = 15/OD
Or, OD = 15×6/20 = 9/2 = 4.5
(7) ABCD is a parallelogram point E is on side BC. Line DE intersects ray AB in point T. Prove that DE×BE = CE×TE
In the adjacent ABCD is a parallelogram.
∵ ABCD is a parallelogram.
∴ AD || BC & AB||DC
BT is an extention of seg AB as mentioned in question.
∴In ∆DEC & ∆BET
∠BET = ∠DEC [Vertically opposite angles]
∠TBE = ∠DCE [Alternate angles ∵ AT||DC]
∴∆DEC ~ ∆BET by AA test
∴by property of similar triangles we can say that,
DE/TE = CE/BE = DC/TB
Or, DE/TE = CE/BE
Or, DE × BE = CE×TE Proved
(8) In the figure, seg AC and seg BD intersect each other in point P and AP/CP = BP/DP. Prove that,
AP/CP = BP/DP. Prove that,
In the adjacent fig,
AP/CP = BP/DP [Given]
∴In∆APB & ∆CPD
AP/CP = BP/DC [Given]
∠APB = ∠CPD [Vertically opposite angles]
∴∆APB ~ ∆CPD by SAS test
(9) In the figure, in ∆ABC, point D on side BC such that,
∠BAC = ∠ADC
Prove that, CA2 = CB×CD
In the adjacent fig it is given that,
∠BAC = ∠ADC
∴In ∆ABC & ∆ADC
∠BAC = ∠ADC [Given]
∠ACB = ∠ACD [Common angle of two triangle]
∴∆ABC ~ ∆ADC by AA test.
∴by the property similar we can say that
CA/CB = AD/AB = CD/CA
∴ CA/CB = CD/CA
Or, CA2 = CB × CA proved
Here is your solution of Maharashtra Board Class 10 Math Part – 2 Chapter 1 Similarity Practice Set 1.3
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