Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
secondary 4 | A Maths
One Answer Below
Anyone can contribute an answer, even non-tutors.
help
Date Posted: 3 years ago
Views: 253
See 1 Answer
A) PA = PB (given)
angle DAQ = angle CBR = 90 (angles of rectangle)
And Angle PAQ = angle PBR (isosceles triangle PAB)
Hence angle PAD = angle PBC
AD = BC (sides of rectangle)
hence by SAS, they are congruent //
B) PAQ congruent to PBR
C) to show a triangle is Equilateral:
Show it's isosceles, and the common angle is 60* (means the last angle must be 60 as well)
from A, PD = PC (congruency)
Now we will show that PDC is 60 deg
PDC = QDC = angle AQD (alternate angle) ---(1)
We find angle AQD:
since BR = 0.5 QD
AQ = 0.5 QD (BR = AQ)
using trigonometry:
Cos AQD = AQ / QD
AQD = cos inverse ( 1/2 )
= 60 deg
:. By (1), PDC = 60 deg
:. We shown that PDC 60 and PDC is isosceles:
PDC must be equilateral triangle (shown)
angle DAQ = angle CBR = 90 (angles of rectangle)
And Angle PAQ = angle PBR (isosceles triangle PAB)
Hence angle PAD = angle PBC
AD = BC (sides of rectangle)
hence by SAS, they are congruent //
B) PAQ congruent to PBR
C) to show a triangle is Equilateral:
Show it's isosceles, and the common angle is 60* (means the last angle must be 60 as well)
from A, PD = PC (congruency)
Now we will show that PDC is 60 deg
PDC = QDC = angle AQD (alternate angle) ---(1)
We find angle AQD:
since BR = 0.5 QD
AQ = 0.5 QD (BR = AQ)
using trigonometry:
Cos AQD = AQ / QD
AQD = cos inverse ( 1/2 )
= 60 deg
:. By (1), PDC = 60 deg
:. We shown that PDC 60 and PDC is isosceles:
PDC must be equilateral triangle (shown)
360 Tutoring Program