Telangana SCERT Class 10 Maths Solution Chapter 8 Similar Triangles Exercise 8.2
(Q1) In the given figure, Ð =Ð ADE CBA
(i) Show that ∆ABC ~ ∆ADE (ii) If AD = 3.8 cm, AE = 3.6cm,
BE = 2.1 cm and BC = 4.2 cm,
find DE.
(i) Show that ∆ABC ~ ∆ADE
(ii) AD = 3.8 cm
AE = 3.6 cm
BE = 2.1 cm
BC = 4.2 cm
find DE
(I) Given : In ∆ABC,
∠ADE = ∠B
To Prove : ∆ABC ~ ∆DE |
In ∆ABC and ∆ADE
∠ ABC = ∠ ADE
∠ A = ∠ A ( Common)
By A similarity
∆ABC ~ ∆ADE
Hence Proved
(ii) Given : AD = 3.8 cm, AE = 3.6 cm, BE= 2.1 cm and BC = 4.2 cm
(Q2) The perimeters of two similar triangles are 30 cm and 20 cm respectively. If one side of the first triangle is 12 cm, determine the corresponding side of the second triangle
Ans:
The ratio of the Corresponding Sides
30/20=12/x
X = 12 x 20/30
X = 8 cm
therefore, The corresponding side of the second triangle is 8 cm
(Q3) In the given figure, AB || CD || EF.
given that AB=7.5 cm, DC= y cm
EF = 4.5 cm and BC = x cm, find
the values of x and y
(Q4) A girl height 90cm is walking away from the base of a lamp post at a speed of 1.2m/sec. if the lamp post is 3.6cm above the ground, find the length of her shadow after 4 seconds
=> Given: Height of the girl,
CD = 90cm = 0.9
Height of the lamp post AB = 3.6m
Speed of the girl = 1.2m 1sec
Distance walked by her after 4 sec
AC = 4X1.2 = 4.8m
The length of the shadow of the girl EC = x
In △ECD and △EAB
∠ECD = ∠EAB = 900
∠E = ∠E (∵ Common)
By A – A similarity,
△ECD ~ △EAB
∴ EC/EA = CD/AB
x/x+4.8 = 0.9/3.6
x/x+4.8 = 1/4
4x = x+4.8
4x – x = 4.8
3x = 4.8
X = 4.8/3
X = 1.6m
The length of the shadow and the girl = 1.6m
(Q5) Given that △ABC ~ △PQR, CM and RN are respectively the medians of △ABC prove that,
(i) △AMC ~ △PNR
(ii) CM/RN = AB/PQ
(iii) △CMB ~ △RNQ
Photo
=> Given: △ABC ~ △PQR
CM is the median of △ABC & RN is median of △PQR.
(i) Given △ABC and △PQR
= AC/PR = AB/PQ
AC/PQ = (1/2)AB/(1/2)PQ
AC/PR = AM/PN
=> ∠A = ∠P
By S.A.S. Similarity,
△AMC ~ △PNR
(ii) From (i)
In △AMK ~ △PNR
CM/RN = AM/PN
CM/RN = 2AM/2PN
= CM/RN = AB/PQ [∵ 2AM =AB, 2PN =PQ]
(iii) Given △ABC ~ △PQR
AB/PQ = BC/QR
(1/2)AB/(1/2)PQ = BC/QR
MB/NQ – BC/QR
=> ∠B = ∠Q
By S.A.S. Similarity
△CMB ~ △RNQ
(Q6) Diagonals AC and BC of a trapezium ABCD with AB II DC interest each other at the point ‘O’ using the criterion of similarity for two triangles show that OA/OC = OB/OD
=> Given in trapezium
ABCD
AB II DC
Diagonals interest at ‘O’
To prove: OA/OC = OB/OB
Proof: Consider △AOB & △COD
∠AOB = ∠COD (Vertically opposite angles)
∠OAB = ∠OCD (Alternate interior angles)
∠OBA = ∠ODC (Alternate interior angles)
By A.A.A similarity
△AOB ~ △COD
OS/OC = OB/OD
(Q7) AB, CD, PQ are perpendicular to BD, If AB PQ = Z
Prove that, 1/x + 1/y = 1/z
=>
Given: AB, CD, PQ are perpendicular to BD.
AB = x, CD = y & PQ = Z
To Prove: 1/x + 1/y = 1/z
Proof: In △ABD and △PQD
∠ABD = ∠PQD = 900
∠ADB = ∠PDQ (∵ Common)
By A.A Similarity
△ABD ~ △PQD
=> AB/PQ = BD/QD
= x/z = BD/QD
=> z/x = QD/BD ——- (i)
In △BDC and △BQP
(Q8) 8. A flag pole 4m tall casts a 6 m shadow. At the same time, a nearby building casts a shadow of 24m. How tall is the building ?
Here is your solution of Telangana SCERT Class 10 Math Chapter 8 Similar Triangles Exercise 8.2
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