CBSE Class 10 Answered
question
Asked by arindeep.singh | 02 Oct, 2020, 13:20: PM
Expert Answer
Question:
Solution:
Let TR = y
Since OT is perpendicular bisector of PQ.
Therefore, PR=QR=4cm
In right triangle OTP and PTR, we have,
TP2 = TR2 + PR2
Also, OT2 = TP2 + OP2
OT2 = (TR2+PR2) + OP2
(y+3)2 = y2 + 16 + 25 ... (OR = 3, as OR2 = OP2 - PR2)
6y=32
y=
TP2=TR2+PR2
TP2=+42 = +16 =
TP= cm
Let TR = y
Since OT
is perpendicular bisector of PQ.
Therefore, PR=QR=4cm
In right triangle OTP and PTR, we have,
TP2=TR2+PR2
Also, OT2=TP2+OP2
OT2=(TR2+PR2) + OP2
(y+3)2=y2+16+25 (OR = 3, as OR2 = OP2 - PR2)
Answered by Renu Varma | 03 Oct, 2020, 16:08: PM
Application Videos
Concept Videos
CBSE 10 - Maths
Asked by yalagondanikhil | 16 Apr, 2024, 12:25: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by riazu945 | 16 Jan, 2022, 13:00: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by sinhkaran79 | 14 Mar, 2021, 12:59: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by s.saumya1011 | 05 Nov, 2020, 14:10: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by brjkishorchauhan03 | 04 Nov, 2020, 12:10: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by shivappahb308 | 21 Sep, 2020, 14:15: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by VasupradUboveja376 | 22 Jun, 2020, 00:03: AM
ANSWERED BY EXPERT