Flashcards
Lines And Agles
Cards in this deck (9)
- 1Lines And Angles Class-9Here are some questions ad whosoever wants to answer can just raise their hands.
Let's Begin! - 2In the given figure, if PQ || RS, then find the measure of angle m.Solution:
Here, PQ || RS, PS is a transversal.
⇒ ∠PSR = ∠SPQ = 56°
Also, ∠TRS + m + ∠TSR = 180°
14° + m + 56° = 180°
⇒ m = 180° – 14 – 56 = 110°
- 3If AB || EF and EF || CD, then find the value of x.Solution:
Since EF || CD ∴ y + 150° = 180°
⇒ y = 180° – 150° = 30°
Now, ∠BCD = ∠ABC
x + y = 70°
x + 30 = 70
⇒ x = 70° – 30° = 40°
Hence, the value of x is 40°.
- 4The two complementary angles are in the ratio 1 : 5. Find the measures of the angles.Solution:
Let the two complementary angles be x and 5x.
∴ x + 5x = 90°
⇒ 6x = 90°
⇒ x = 15°
Hence, the two complementary angles are 15° and 5 × 15° i.e., 15° and 75°. - 5In the given figure, lines AB and CD intersect at O. Find the value of x.Solution:
Here, lines AB and CD intersect at O.
∴ ∠AOD and ∠BOD forming a linear pair
⇒ ∠AOD + ∠BOD = 180°
⇒ 7x + 5x = 180°
⇒ 12x = 180°
⇒ x = 15°
- 6In the given figure, if x°, y° and z° are exterior angles of ∆ABC, then find the value of x° + y° + z°.Solution:
We know that, an exterior angle of a triangle is equal to sum of two opposite interior angles.
⇒ x° = ∠1 + ∠3
⇒ y° = ∠2 + ∠1
⇒ z° = ∠3 + ∠2
Adding all these, we have
x° + y° + z° = 2(∠1 + ∠2 + ∠3)
= 2 × 180°
= 360°
- 7In the given figure, AB || CD, ∠FAE = 90°, ∠AFE = 40°, find ∠ECD.Solution:
In AFAE,
ext. ∠FEB = ∠A + F
= 90° + 40° = 130°
Since AB || CD
∴ ∠ECD = FEB = 130°
Hence, ∠ECD = 130°.
- 8In the given figure, prove that m || n.Solution:
In ∆BCD,
ext. ∠BDM = ∠C + ∠B
= 38° + 25° = 63°
Now, ∠LAD = ∠MDB = 63°
But, these are corresponding angles. Hence,
m || n
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- Flashcards on Lines And Agles, created by SHAGUN GARG on 09/08/2021.
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- SHAGUN GARG
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