Answer:
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Class 9
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>>In Fig. 6.43, if PQ PS, PQ∥ SR, SQR = 2
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In Fig. 6.43, if PQ⊥PS,PQ∥SR,∠SQR=28
0
and ∠QRT=65
0
, then find the values of x and y.
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Solution
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Given, PQ⊥PS,PQ∥SR,∠SQR=28
∘
,∠QRT=65
∘
According to the question,
x+∠SQR=∠QRT (Alternate angles as QR is transversal.)
⇒x+28
∘
=65
∘
⇒x=37
∘
Also ∠QSR=x
⇒∠QSR=37
∘
Also ∠QRS+∠QRT=180
∘
(Linear pair)
⇒∠QRS+65
∘
=180
∘
⇒∠QRS=115
∘
Now, ∠P+∠Q+∠R+∠S=360
∘
(Sum of the angles in a quadrilateral.)
⇒90
∘
+65
∘
+115
∘
+∠S=360
∘
⇒270
∘
+y+∠QSR=360
∘
⇒270
∘
+y+37
∘
=360
∘
⇒307
∘
+y=360
∘
⇒y=53
∘
Step-by-step explanation:
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Slope = (y2 - y1)/(x2 - x1)
Slope = (6 -0)/(3 - 0)
Slope = 6/3 = 2
Answer
2
Answer:
Area = 14.8
Step-by-step explanation:
Area = 5/2 + a^2*sqrt(5 + 2 sqrt(5) )
This formula given the side length, calculates the area.
a = 2
Area = 2.5 + 2^2 * sqrt(5 + 2*sqrt(5) )
Area = 2.5 + 4 * sqrt(5 + 2*2.236)
Area = 2.5 + 4* sqrt(5+ 4.4721)
Area = 2.5 + 4*sqrt(9.4721)
Area = 2.5 + 4*3.0776
Area = 2.5 + 12.3107
Area = 14.8107
Answer:
36
Step-by-step explanation:
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