In the parallelogram PQRS, PE is perpendicular to QR. Find the area of the parallelogram.
60cm2
65cm2
72cm2
132cm2
156cm2
Correct answer is D
By Pythagoras, PE\(^2\) = 12\(^2\) - 5\(^2\)
= 144 - 25 = 119
h = PE\(^2\) = √119 = 10.9 ≈ 11cm,
Area of 11gm = b x h
QR = b =(5 + 7)cm = 12cm
area = 12 x 11
= 132cm\(^2\)
\(\frac{5}{9}\)k2 sq. units
\(\frac{1}{3}\)k2 sq. units
\(\frac{8}{9}\)k2 sq. units
\(\frac{7}{9}\)k2 sq. units
\(\frac{2}{3}\)k2 sq. u
Correct answer is A
Area of shaded portion = Area of triangle PQR - Area of inner triangle
Area of triangle given 3 sides a, b, c = \(\sqrt{s(s - a)(s - b)(s - c)}\)
where \(s = \frac{a + b + c}{2} \)
Area of PQR :
\(s = \frac{3 + 5 + 6}{2} = \frac{14}{2} = 7\)
Area = \(\sqrt{7(7 - 3)(7 - 5)(7 - 6)}\)
= \(\sqrt{7(4)(2)(1)} = \sqrt{56}\)
\(\implies K^{2} = \sqrt{56}\)
Area of inner triangle :
\(s = \frac{2 + 4 + \frac{10}{3}}{2} = \frac{14}{3}\)
Area = \(\sqrt{\frac{14}{3} (\frac{14}{3} - 2)(\frac{14}{3} - 4)(\frac{14}{3} - \frac{10}{3})}\)
= \(\sqrt{\frac{14}{3} (\frac{8}{3})(\frac{2}{3})(\frac{4}{3})}\)
= \(\sqrt{\frac{896}{81}}\)
= \(\sqrt{\frac{16}{81}} \times \sqrt{56}\)
= \(\frac{4}{9} K^{2}\)
\(\therefore \text{The area of the shaded portion} = K^{2} - \frac{4}{9}K^{2} = \frac{5}{9}K^{2}\)
45o
35o
40o
30o
42o
Correct answer is D
PSR = 65o x 2
= 130o
PRS = 180o - (130o + 20o)
= 30o
70o
90o
80o
40o
60o
Correct answer is C
Isosceles Triangle PSR:
RSP ≡ RPS → 20º
That is PRQ = 40º
POQ = 2 * 40 = 80º (Angle subtended by chord PQ at centre is twice angle subtended at circumference).
The size of POQ = 80º
If O is the centre of the circle, < POS equls
70o
75o
105o
140o
150o
Correct answer is E
Since O is the centre of the circle < POS = 150o
i.e. < substended at the centre is twice that substended at any part of the circumference
JAMB Subjects
Aptitude Tests