Answer:
.
See the diagram attached below.
Let the chords be AB and AC with common point A.
AD is the diameter. Join B with D and C with D to form two triangles.
We need to prove that AB=AC.
\begin{gathered}In\ \triangle ABD\ and \triangle ACD;\\Given\ that\ \angle BAD=\angle CAD----(condition\ 1)\\since\ AD\ is\ diameter, \angle ABD=\angle ACD = 90^0\\So\ \angle ADB=\angle ADC--------(condition\ 2)\\AD=AD\ (common\ side)-----(condition\ 3)\\ \\So\ the\ triangles\ are\ congruent\ by\ ASA\ rule.\\Hence\ AB=AC.\end{gathered}
In △ABD and△ACD;
Given that ∠BAD=∠CAD−−−−(condition 1)
since AD is diameter,∠ABD=∠ACD=90
0
So ∠ADB=∠ADC−−−−−−−−(condition 2)
AD=AD (common side)−−−−−(condition 3)
So the triangles are congruent by ASA rule.
Hence AB=AC.
Answer:
x = 3
Step-by-step explanation:
8 - 2x = 2
- 2x = 2 - 8
- 2x = - 6
- x = - 6/2
- x = - 3
x = 3
Answer:
Exercise 1:
base [b]=8cm
perpendicular [p]=6cm
hypotenuse [h]=?
<u>By</u><u> </u><u>using</u><u> </u><u>Pythagoras</u><u> </u><u>law</u>
h²=p²+b²
h²=6²+8²
h=√100
h=10cm
<u>So</u><u> </u><u>another</u><u> </u><u>side's</u><u> </u><u>length</u><u> </u><u>is</u><u> </u><u>1</u><u>0</u><u>c</u><u>m</u>
<u>Exercise</u><u> </u><u>2</u><u>:</u>
base [b]=6m
perpendicular [p]=bm
hypotenuse [h]=8m
By using Pythagoras law
h²=p²+b²
8²=b²+6²
b²=8²-6²
b=√28=2√7 0r 5.29 or 5.3
So height of kite is√<u>28</u><u>o</u><u>r</u><u> </u><u>2√7 0r 5.29 or 5.3 m</u>
Step-by-step explanation:
[Note: thanks for translating]
Truest sense that the world of war in my opinion was gon in my own name in
