Geometry Ex 4.1-10th Std Maths-Book Back Question And Answer
Question 1.
Check whether the which triangles are similar and find the value of x.
Solution:
(i) In ∆ABC and ∆AED
ABAD = ACAE
83 = 1122
83 = 114 ⇒ 32 ≠ 33
∴ The two triangles are not similar.
(ii) In ∆ABC and ∆PQC
∠PQC = 70°
∠ABC = ∠PQC = 70°
∠ACB = ∠PCQ (common)
∆ABC ~ ∆PQC
5X = 63
6x = 15
x = 156 = 52
∴ x = 2.5
∆ ABC and ∆PQC are similar. The value of x = 2.5
Question 2.
A girl looks the reflection of the top of the lamp post on the mirror which is 66 m away from the foot of the lamppost. The girl whose height is 12.5 m is standing 2.5 m away from the mirror. Assuming the mirror is placed on the ground facing the sky and the girl, mirror and the lamppost are in a same line, find the height of the lamp post.
Solution:
Let the height of the tower ED be “x” m. In ∆ABC and ∆EDC.
∠ABC = ∠CED = 90° (vertical Pole)
∠ACB = ∠ECD (Laws of reflection)
∆ ABC ~ ∆DEC
ABDE = BCEC
1.5x = 0.487.6
x = 1.5×87.60.4 = 1.5×8764
= 1.5 × 219 = 328.5
The height of the Lamp Post = 328.5 m
Question 3.
A vertical stick of length 6 m casts a shadow 400 cm long on the ground and at the same time a tower casts a shadow 28 m long. Using similarity, find the height of the tower.
Solution:
In ∆ABC and ∆PQR,
∠ABC = ∠PQR = 90° (Vertical Stick)
∠ACB = ∠PRQ (Same time casts shadow)
∆BCA ~ ∆QRP
ABPQ = BCQR
6x = 428
4x = 6 × 28 ⇒ x = 6×284 = 42
Length of the lamp post = 42m
Question 4.
Two triangles QPR and QSR, right angled at P and S respectively are drawn on the same base QR and on the same side of QR. If PR and SQ intersect at T, prove that PT × TR = ST × TQ.
Solution::
In ∆PQT and ∆STR we have
∠P = ∠S = 90° (Given)
∠PTQ = ∠STR (Vertically opposite angle)
By AA similarity
∆PTQ ~ ∆STR we get
PTST = TQTR
PT × TR = ST × TQ
Hence it is proved.
Question 5.
In the adjacent figure, ∆ABC is right angled at C and DE ⊥ AB. Prove that ∆ABC ~ ∆ADE and hence find the lengths of AE and DE.
Solution:::
In ∆ABC and ∆ADE
∠ACB = ∠AED = 90°
∠A = ∠A (common)
∴ ∆ABC ~ ∆ADE (By AA similarity)
BCDE = ABAD = ACAE
12DE = 133 = 5AE
In ∆ABC, AB2 = BC2 + AC2
= 122 + 52 = 144 + 25 = 169
AB = 169−−−√ = 13
Consider, 133 = 5AE
∴ AE = 5×313 = 1513
AE = 1513 and DE = 3613
Consider, 12DE = 133
DE = 12×313 = 3613
Question 6.
In the adjacent figure, ∆ACB ~ ∆APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ.
Solution:
Given ∆ACB ~ ∆APQ
ACAP = BCPQ = ABAQ
AC2.8 = 84 = 6.5AQ
Consider AC2.8 = 84
4 AC = 8 × 2.8
AC = 8×2.84 = 5.6 cm
Consider 84 = 6.5AQ
8 AQ = 4 × 6.5
AQ = 4×6.58 = 3.25 cm
Length of AC = 5.6 cm; Length of AQ = 3.25 cm
Question 7.
If figure OPRQ is a square and ∠MLN = 90°. Prove that
(i) ∆LOP ~ ∆QMO
(ii) ∆LOP ~ ∆RPN
(iii) ∆QMO ~ ∆RPN
(iv) QR2 = MQ × RN.
Solution:
(i) In ∆LOP and ∆QMO
∠OLP = ∠OQM = 90°
∠LOP = ∠OMQ (Since OQRP is a square OP || MN)
∴ ∆LOP~ ∆QMO (By AA similarity)
(ii) In ∆LOP and ∆RPN
∠OLP = ∠PRN = 90°
∠LPO = ∠PNR (OP || MN) .
∴ ∆LOP ~ ∆RPN (By AA similarity)
(iii) In ∆QMO and ∆RPN
∠MQO = ∠NRP = 90°
∠RPN = ∠QOM (OP || MN)
∴ ∆QMO ~ ∆RPN (By AA similarity)
(iv) We have ∆QMO ~ ∆RPN
MQPR = QORN
MQQR = QRRN
QR2 = MQ × RN
Hence it is proved.
Question 8.
If ∆ABC ~ ∆DEF such that area of ∆ABC is 9cm2 and the area of ∆DEF is 16 cm2 and BC = 2.1 cm. Find the length of EF.
Solution:
Given ∆ABC ~ ∆DEF
916 = (2.1)2EF2
(34)2 = (2.1EF)2
34 = 2.1EF
EF = 4×2.13 = 2.8 cm
Legth of EF = 2.8 cm
Question 9.
Two vertical poles of heights 6 m and 3 m are erected above a horizontal ground AC. Find the value of y.
Solution:
In the ∆PAC and ∆BQC
∠PAC = ∠QBC = 90°
∠C is common
∆PAC ~ QBC
APBQ = ACBC
6y = ACBC
∴ BCAC = y6 …..(1)
In the ∆ACR and ∆QBC
∠ACR = ∠QBC = 90°
∠A is common
∆ACR ~ ABQ
RCQB = ACAB
3y = ACAB
ABAC = y3 ……..(2)
By adding (1) and (2)
BCAC + ABAC = y6 + y3
1 = 3y+6y18
9y = 18 ⇒ y = 189 = 2
The Value of y = 2m
Question 10.
Construct a triangle similar to a given triangle PQR with its sides equal to 23 of the corresponding sides of the triangle PQR (scale factor 23 ).
Solution:
Given ∆PQR, we are required to construct another triangle whose sides are 23 of the corresponding sides of the ∆PQR
Steps of construction:
(i) Construct a ∆PQR with any measurement.
(ii) Draw a ray QX making an acute angle with QR on the side opposite to the vertex P.
(iii) Locate 3 points Q1, Q2 and Q3 on QX.
So that QQ1 = Q1Q2 = Q2Q3
(iv) Join Q3 R and draw a line through Q2 parallel to Q3 R to intersect QR at R’.
(v) Draw a line through R’ parallel to the line RP to intersect QP at P’. Then ∆ P’QR’ is the required triangle.
Question 11.
Construct a triangle similar to a given triangle LMN with its sides equal to 45 of the corresponding sides of the triangle LMN (scale factor 45 ).
Solution:
Given a triangle LMN, we are required to construct another ∆ whose sides are 45 of the corresponding sides of the ∆LMN.
Steps of Construction:
Construct a ∆LMN with any measurement.
Draw a ray MX making an acute angle with MN on the side opposite to the vertex L.
Locate 5 Points Q1, Q2, Q3, Q4, Q5 on MX.
So that MQ1 = Q1Q2 = Q2Q3 = Q3Q4 = Q4Q5
Join Q5 N and draw a line through Q4. Parallel to Q5N to intersect MN at N’.
Draw a line through N’ parallel to the line LN to intersect ML at L’.
∴ ∆L’ MN’ is the required triangle.
Question 12.
Construct a triangle similar to a given triangle ABC with its sides equal to 65 of the corresponding sides of the triangle ABC (scale factor 64).
Solution:
Given triangle ∆ABC, we are required to construct another triangle whose sides are 65 of the corresponding sides of the ∆ABC.
Steps of construction
(i) Construct an ∆ABC with any measurement.
(ii) Draw a ray BX making an acute angle with BC.
(iii) Locate 6 points Q1, Q2, Q3, Q4, Q5, Q6 on BX such that
BQ1 = Q1Q2 = Q2Q3 = Q3Q4 = Q5Q6
(iv) Join Q5 to C and draw a line through Q6 parallel to Q5 C intersecting the extended line BC at C’.
(v) Draw a line through C’ parallel to AC intersecting the extended line segment AB at A’.
∆A’BC’ is the required triangle.
Question 13.
Construct a triangle similar to a given triangle PQR with its sides equal to 73 of the corresponding sides of the triangle PQR (scale factor 73).
Solution:
Given triangle ABC, we are required to construct another triangle whose sides are 73 of the corresponding sides of the ∆ABC.
Steps of construction
(i) Construct a ∆PQR with any measurement.
(ii) Draw a ray QX making an acute angle with QR on the side opposite to the vertex P.
(iii) Locate 7 points Q1, Q2, Q3, Q4, Q5, Q6, Q7 on QX.
So that
QQ1 = Q1Q2 = Q2Q3 = Q3Q4 = Q5Q6 = Q6Q7
(iv) Join Q3 to R and draw a line through Q7 parallel to Q3R intersecting the extended line segment QR at R’.
(v) Draw a line through parallel to RP.
Intersecting the extended line segment QP at P’.
∴ ∆P’QR’ is the required triangle.
