Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.2
Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.2
Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.2
Miscellaneous Practice Problems
Question 1.
In the given figure, find PT given that l1 || l2
Solution:
Given that l1 || l2
∴ In ∆PQS and ∆PRT
∠P is common
∠Q = ∠R [∵ PR is the transversal for l1 and l2 corresponding angles]
∠S = ∠T [∵ corresponding angles]
∴ ∆PQS ~ ∆PRT [∵ By AAA congruency]
In similar triangles, corresponding angles are proportional.
Question 2.
From the diagram, prove that ∆SUN ~ ∆RAY
Proof:
From the ∆SUN and ∆RAY
SU = 10;
UN = 12;
SN = 14;
RA = 5,
AY = 6;
RY = 7
We have
Question 3.
The height of a tower is measured by a mirror on the ground by which the top of the tower’s reflection is seen. Find the height of the tower.
Solution:
The image and its reflection make similar shapes
Question 4.
In the figure, given that ∠1 = ∠2 and ∠3 ≡ ∠4 Prove that ∆MUG ≡ ∆ TUB.
Proof:
Question 5.
If ∆WAR ≡ ∆MOB, name the additional pair of corresponding parts. Name the criterion used by you.
Solution:
Given ∆WAR ≡ ∆MOB
∠RWA ≡ ∠BMO [∵ sum of three angles of a triangle are 180°]
∴ Criteria used here is angle sum property of triangles.
Question 6.
In the figure, ∠TMA ≡ ∠IAM and ∠TAM ≡ ∠IMA. P is the midpoint of MI and N is the midpoint of AI. Prove that ∆ PIN ~ ∆ ATM.
Solution:
Question 7.
In the figure, if ∠FEG = ∠1 then, prove that DG2 = DE.DF.
Proof:
Question 8.
In the figure, ∠TEN ≡ ∠TON = 90° and TO = TE. Prove that ∠ORN ≡ ∠ERN
Solution:
Question 9.
In the figure, PQ ≡ TS, Q is the midpoint of PR, S is the midpoint TR and ∠POU ≡ ∠TSU. Prove that QU ≡ SU.
Proof:
Question 10.
In the figure ∆TOP ≡ ∆ARM . Explain why?
Solution:
In ∆TOP and ∆ARM
OP = RM given
∠TOP = ∠ARM = 90°
given ∠OTP = ∠RAM
given ∠OPT = ∠RMA Remaining angle, by angle sum property.
∴ By ASA criteria we can say that ∆TOP ≡ ∆ARM
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