Find the area of an equivalent triangle of side 16cm
64√3cm2
72√3cm2
96cm2
128√3cm2
128cm2
Correct answer is A
Area = 1/2 x 16 x 16sin60o = 64√3cm2
In the diagram above, O is the center of the circle, |SQ| = |QR| and ∠PQR = 68°. Calculate ∠PRS
34o
45o
56o
62o
68o
Correct answer is A
From the figure, < PQR = 68°
\(\therefore\) < QRS = < QSR = \(\frac{180 - 68}{2}\) (base angles of an isos. triangle)
= 56°
\(\therefore\) < PRS = 90° - 56° = 34° (angles in a semi-circle)
40
5
6
8
12
Correct answer is D
Let x be the exterior angle; interior = 3x; but x + 3x = 180o
4x = 180o; x = 45
= \(\frac{360}{45}\) = 8
In the diagram above, ∠PTQ = ∠URP = 25° and XPU = 4URP. Calculate ∠USQ.
100o
120o
125o
130o
150o
Correct answer is D
Since < URP = 25°, then < XPU = 4 x 25° = 100°
\(\therefore\) < TPQ = 180° - 100° = 80°
\(\therefore\) < PQT = 180° - (80° + 25°) = 75°
< SQR = 75° - 25° = 50° (exterior angle = 2 opp interior angles)
\(\therefore\) < USQ = 180° - 50° = 130°
The angles of a pentagon are x°, 2x°, (x + 60)°, (x + 10)°, (x -10)°. Find the value of x.
40
60
75
80
90
Correct answer is D
Sum of ∠s in a pentagon = (n - 2)180 = 540°
x° + 2x° + x° + 60° + x° + 10° + x° - 10° = 540°
6x° + 60° = 540°; x = 80°
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