Answer:
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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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Answer:
∠N = 31
Step-by-step explanation:
MN = MP
∠N = ∠P = x+3
∠M + ∠N + ∠P = 180
4x+6 + x+3 + x+3 = 6x + 12 = 180
x = 28
∠N = 28 + 3 = 31
Answer:
H or (3.5,-3)
number two:
B and (3.5,5)
Step-by-step explanation:
Answer:
R = 6
Step-by-step explanation:
Remember... Circle area is:
A = pi×R^2
Where R is the circle radius.
Then,
Pi×R^2=36×Pi
R = square root of 36
R = 6
