In the figure, PQST is a parallelogram and TSR is a straight line. If the area of \(\bigtriangleup\)QRS is 20cm2, find the area of the trapezium PQRT.
35cm2
65cm2
70cm2
140cm2
Correct answer is C
A\(\bigtriangleup\) = \(\frac{1}{2}\) x 8 x h = 20
= \(\frac{1}{2}\) x 8 x h = 4h
h = \(\frac{20}{4}\)
= 5cm
A\(\bigtriangleup\)(PQTS) = L x H
A\(\bigtriangleup\)PQRT = A\(\bigtriangleup\)QSR + A\(\bigtriangleup\)PQTS
20 + 50 = 70cm\(^2\)
ALTERNATIVE METHOD
A\(\bigtriangleup\)PQRT = \(\frac{1}{2}\) x 5 x 28
= 70cm\(^2\)
Similar Questions
Factorize abx2 + 8y - 4bx - 2axy...
Find the sum of the first five terms of the G.P 2,6, 18 .... ...
Let f(x) = 2x + 4 and g(x) = 6x + 7 here g(x) > 0. Solve the inequality \(\frac{f(x)}{g(x)}\) < 1...
If line p = 5x + 3 is parallel to line p = wx + 5. Find the value of w....
Simplify 3\(\frac{1}{3}\) - 1\(\frac{1}{4}\) x \(\frac{2}{3}\) + 1\(\frac{2}{5}\)...
For the set of numbers 2, 3, 5, 6, 7, 7, 8 ...
If (-3, -4) is a point on the line y = mx + 2 find the value of m....
If \(\sin x = \frac{4}{5}\), find \(\frac{1 + \cot^2 x}{\csc^2 x - 1}\)....