Maharashtra Board Class 10 Math Part – 2 Solution Chapter 1 Problem Set 1 – 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 |
Problem Set |
1 |
Problem set – 1
(1) Select the appropriate alternative.
(1) In ∆ABC and ∆PQR, in a one to one correspondence
AB/QR = BC/PR = CA/PQ then
(A) ∆PQR ~∆ABC
(B) ∆PQR ~∆CAB
(C) ∆CBA ~∆PQR
(D) ∆BCA ~∆PQR
Solution:
Given,
AB/QR = BC/PR = CA/PQ
Then, (B) ∆PQR ~ ∆CAB
(2) If in ∆DEF and ∆PQR, ∠D ≅∠Q, ∠R ≅∠E then which of the following statements is false?
(A) EF/PR = DF/PQ
(B) DE/PQ = EF/RP
(C) DE/QR = DF/PQ
(D) EF/RP = DE/QR
Solution:
Given,
∠D ≅∠Q, ∠R ≅∠E
∴∆PEF ~ ∆QRP by AA test
∴ (B) DE/PQ = EF/RP is a false statement.
(3) In ∆ABC and ∆DEF ∠B = ∠E, ∠F = ∠C and AB = 3DE then which of the statements regarding the two triangles is true?
(A) The triangles are not congruent and not similar
(B)The triangles are similar but not congruent.
(C)The triangles are congruent and similar.
(D) None of the statements above is true
Solution:
Given, ∆ABC & ∆DEP are equilateral triangles.
∴∆ABC ~ ∆DEF
A (∆ABC)/A(∆DEF) = 1/2 & AB = 4.
∴ 1/2 = 42/DE2 or, DE = 4√2 [Ratio of areas of similar triangles]
(5) In figure 1.71, seg XY||seg BC, then which of the following statements is true?
(A) AB/AC = AX/AY
(B) AX/XB = AY/AC
(C) AX/YC = AY/XB
(D) AB/YC = AC/XB
Solution:
Given, XY||BC
In ∆ABC & ∆AXY
∠ABC ≅∠AXY [Corresponding angles]
∠ACB ≅∠AYX [Corresponding angles]
∴∆ABC ~ ∆AXY by AA test
∴ AB/AX = AC/AY [Ratio of areas of similar triangles
(A) Or, AB/AC = AX/AY is a true statement.
(2) In ∆ABC, B – D – C and BD = 7,
BC = 20 then find following ratios.
(1) A(∆ADB)/A(∆ADC)
(2) A(∆ABD)/A(∆ABC)
(3) A(∆ADC)/A(∆ABC)
Solution:
Given, B – B – C, BD = 7, BC = 20.
∴DC = BC – BD = 20 – 7 = 13
Now, we have to construct a perpendicular AX from Vertice A to base BC.
∴In ∆ABD & ∆ADC & ∆ABC
height of three triangles = AX [∵ B – D – C]
[A is common vertice]
(1) ∴ A (∆ABD) = 1/2 × AX × BD
= 1/2 × AX × 7
= 7/2 AX
A (∆ADC) = ½ × AX × DC
= ½ × AX × 13
= 13/2 AX
∴ A(∆ABD)/A(∆ADC) = (7/2 AX)/(13/2×AX)
= 7/2 × 2/3
= 7/13
(2) A (∆ABD) = 7/2 AX
A (∆ABC) = ½ × AX × BC
= 1/2 × AX × 20
= 10AX
∴A (∆ABD)/A(∆ABC) = (7/2 AX)/10AX = 7/20
(3) A (∆ADC) = 13/2 AX
A (∆ABC) = 10AX
∴A (∆ADC)/A(∆ABC) = (13/2 AX)/10AX = 13/20
(3) Ratio of areas of two triangles with equal heights is 2:3. If base of the smaller triangle is 6 cm then what is the corresponding base of the bigger triangle?
Solution:
Given, ratio of area of two triangle is 2:3
Let, area of smaller ∆ be a1
Area of bigger ∆ be a2
Height of smaller ∆ be h
Height of bigger ∆ be h [∵ both height same]
Base of smaller ∆ is 6 — [Given]
Base of bigger ∆ be 6
∴ a1 = 1/2 × 6 × h = 3h Now, a1/a2 = 2/3 [Given]
a2 = 1/2 × b × h = bh/2 Or, 3h/(bh/2) = 2/3
Or, b = 3×3 = 9 cm.
(4) In figure 1.73, ∠ABC = ∠DCB = 90°
AB = 6, DC = 8
Then, A(∆ABC)/A(∆DCB) =?
Solution:
Gives, ∠ABC = ∠DCB = 90
AB = 6, DC = 8
A (∆ABC) = ½ × b × h
= 1/2 × BC × AB
= 1/2 × BC × 6
= 3BC
A (∆DBC) = 1/2 × BC × DC
= 1/2 × BC × 8
= 4BC
∴A(∆ABC)/A(∆DBC) = 3BC/4BC
= 3/4
(5) In figure 1.74, PM = 10cm
A (∆PQS) = 100 sq. cm
A (∆QRS) = 110 sq. cm
Then find NR.
Solution:
Given, PM = 10cm
∠PMS = ∠RNQ = 90°
A (∆PQS) = 100 cm2
A (∆QRS) = 110 cm2
∴ A (∆PQS) = 1/2 × PM × QS
Or, 100 = 1/2 × 10 × QS
Or, 100 = 5QS
Or, QS = 20 cm
A (∆QRS) = 1/2 × QS × NR
Or, 110 = 1/2 × 20 × NR
Or, NR = 11 cm
(6) ∆MNT ~∆QRS. Length of altitude drawn from point T is 5 and length of altitude drawn from point S is 9. Find the ratio A(∆MNT)/A(∆QRS)
Solution:
Given, ∆MNT ~ ∆QRS, Right of ∆MNT with MN as base is 5
∴A(∆MNT)/(A(∆QRS) = MN2/QR2 Right of ∆QRS with as base is 9. [Ratio of areas of similar triangles] —- (i)
Also, A (∆MNT) = ½ × MN × 5
= 5/2 × MN
A (∆QRS) = ½ × QR × 9
= 9/2 × QR
∴ A (∆MNT)/A(∆QRS) = (5/2×MN)/(9/2×QR)
Or, MN2/QR2 = 5/2 × 2/9 × MN/QR
Or, MN/QR = 5/9 [From (i)]
Or, MN2/QR2 = 25/81 — (ii)
A(∆MNT)/A(∆QRS)
= MN2/QR2 = 25/81 – [From (ii)]
(7) In figure 1.75, A – D – C and B – E – C seg DE||side AB if AD = 5, DOC = 3, BC = 6.4 then find BE.
Solution:
Given, A – D – C, B – E – C,
DE||AB, AD = 5, DC = 3, BC = 6.4
In ∆ABC
DE||AB
DC/AD = CE/BE [Basic proportionality theorem]
Or, 3/5 = 6.4-x/x
Or, 3x = 32 – 5x
Or, 8x = 32
Or, x = 4
∴ BE = 4
(8) In the figure 1.76, seg PA, seg QB, seg RC and seg SD are perpendicular to line AD.
AB = 60, BC = 70, CD = 80, PS = 280 then find PQ, QR and RS.
Solution:
In the adjacent fig. it is given. PA, RC, QB, so ⊥ AD line.
AB = 60, BC = 70, CD = 80, PS = 280 considering seg so, RC & QB.
SD||RC||QB [∵ All ⊥ to AD]
∴ RS/QR = CD/BC [Property of three parallel lines and their transversal]
Or, RS/QR = 80/70 or, RS = 80QR/70 — (i)
Now, considering seg RC, QB, PA.
RC||QB||PA [∵all ⊥ to AD]
∴ QR/PQ = BC/AB [property of three parallel lines and their transversal]
Or, QR = 70/60 × PQ — (ii)
Now, since P – Q – R – S
∴ PS = PS + QR + RS
Now, Putting the value of RS QR from equation (i) & (ii)
Or, 280 = PQ + 70PQ/60 + 80QR/70
Or, 280 = PQ + 70PQ/60 + 80/70 × 70PQ/60
Or, 280 = PQ + 70PQ/60 + 4PQ/3
Or, 280 = 60PQ+70PQ+80PQ/60
Or, 280×60 = 210PQ
Or, PQ = 280×60/210
Or, PQ = 80
∴From equation (ii)
QR = 70PQ/60
= 70/60 × 80= 280/3
From equation (i)
RS = 80QR/70
= 80/70 × 280/3 = 320/3
∴ RS = 320/3, QR = 280/3, PQ = 80
(9) In ∆PQR seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY||QR.
Complete the proof by filling in the boxes.
In ∆PMQ, ray MX is bisector of ∠PMQ.
∴ —/— = —–/—- ……. (1) theorem of angle bisector.
In ∆PMR, ray MY is bisector of ∠PMR
∴ —-/—- = —-/—– …….(ii) Theorem of angle bisector.
But MP/MQ = MP/MR …. M is the midpoint QR, hence MQ = MR.
∴ PX/XQ = PY/YR
∴ XY||QR ….. converse of basic proportionality theorem.
Solution:
In ∆PMQ, ray MX is bisector of ∠PMQ
∴PX/XQ = PM/MB [Theorem of angle bisector]
In ∆PMR, ray MY is bisector of ∠PMR
∴ PY/YR = PM/MR [Theorem of angle bisector)
But, MP/MQ = MP/MR — M is the midpoint of QR, hence MQ = MR
∴ PX/XQ = PY/YR
∴ XY||QR – converse of basic proportionality theorem.
(10) In fig 1.78, bisectors of ∠B and ∠C of ∆ABC intersect each other in point X. Line AX intersects side BC in point Y. AB = 5, AC = 4, BC = 6 then find AX/XY.
Solution:
In the adjacent fig
BX bisects ∠B
CX bisects ∠C
AB = 5, AC = 4, BC = 6
In ∆AYB
∵ BX bisects ∠B
AX/XY = AB/BY [Theorem of angle bisector] —- (i)
In ∆AYC
∵ CX bisects ∠C
AX/AY = AC/CY [Theorem of angle bisector]
∴ AB/BY = AC/CY
Or, AB×CY = AC×BY
Or, 5CY = 4BY
Or, CY = 4BY/5 — (ii)
Now, BC = 6
Or, BY + CY = 6 [∵ B – Y – C]
Or, putting value of CY from equation (ii)
Or, BY + 4BY/5 = 6
Or, 5BY + 4BY = 6×5
Or, 9BY = 30
Or, BY = 10/3
Now, in equation (i) putting value of BY = 10/3 we get.
AX/AY = AB/BY
= 5/(10/3) = 3/2 = AX/AY = 2/3
(11) In — ABCD, seg AD||seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that AP/PD = PC/BP
Solution:
In the adjacent fig it is given that, AD||BC.
In ∆APD & ∆BPC
∠APD ≅∠BPC [Vertically opposite angles]
∠PAD ≅∠PCB [Alternate angles]
∴∆APD ~∆BPC by AA test
∴ AP/PC = PD/BP [Property of similar of similar triangles]
Or, AP/PD = PC/BP proved
(12) In fig 1.80, XY||seg AC.
If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC.
Solution:
Given, XY||AC
2AX = 3BX XY = 9
Now, 2AX = 3BX
∴ AX/BX = 8/2
AX+BX/BX = 3+2/2 = 5/2 — (i)
In ∆BCA &∆BYX
∠B is common
∠BAC ≅∠BXY [Corresponding angles ∵ XY||AC]
∴∆BCA ~∆BYX by AA test
∴ BA/BX = AC/XY [Corresponding sides of similar triangles]
Or, AX+BX/BX = AC/9 [∵ A – X – B]
Or, 5/2 = AC /9 [From equation (i)]
Or, AC = 45/2
(13) In figure 1.81, the vertices of square DEFG are on the sides of ∆ABC. ∠A = 90°. Then prove that DE2 = BD×EC [Hint: Show that ∆CFE. Use GD = FE = DE.]
Solution:
In the adjacent fig it is given that. DEFG is a square.
∠A = 90°
In ∆GBD &∆ABC
∠B is common.
∠GDB = ∠GDE = 90°
[∵∠GDE is angle of square DEFG and B – D – E – C also, ∠GDE Co linear angle to ∠GDB ∴∠GDB = 180 – ∠GDE
= 180 – 90°= 90°]
∴∠GDB = ∠A = 90°
∴∆GBD ~∆ABC by AA test
In ∆CEF &∆ABC
∠C is common.
∠CEF = ∠FED = 90°
[∵ is angle of square DEGF and B – D – E – C also, ∠FED collinear angle to ∠CEF so, ∠CEF = 180° – ∠FED = 180° – 90° => 90°]
∴∆CEF ~∆GBD
∵Both triangles are similar to ∆ABC.
Or, CE × BD = EGD [Now similar ∆]
Or, DE × DE = CE × BD [∵ EF = GD — DE = FG are sides of square DEFG]
Or, DE2 = CE × BD Proved
Here is your solution of Maharashtra Board Class 10 Math Part – 2 Chapter 1 Similarity Problem Set 1
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