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What is the area of quadrilateral \(ABCD\) above?

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Junior | Next Rank: 30 Posts
Posts: 15
Joined: Sat Sep 12, 2020 8:07 pm

Re: What is the area of quadrilateral \(ABCD\) above?

by Nirav Umaretiya » Sat Sep 12, 2020 9:45 pm
There are two ways to solve this one,
You could divide the figure into a rectangle + right angle triangle

or

You may apply the formula of areal of a trapezoid

Let's do the 1st one first.
You could divide the figure into a rectangle + right angle triangle

Area
Area of the rectangle + Area of the right angle triangle which is [hight x base] + [\(\frac{1}{2}\) (hight x base)]
= [4x10] + [\(\frac{1}{2}\) x 4 x (13-10)]
= 40 + 2(6)
= 46

Area
formula of areal of a trapezoid =
= \(\frac{1}{2}\) (sum of paralelle sides) (Distance between them)
= \(\frac{1}{2}\) (10 + 13) (4)
= 20 + 26
= 46

Correct answer is B
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Senior | Next Rank: 100 Posts
Posts: 31
Joined: Thu Sep 17, 2020 9:27 amFollowed by: 1 member

Re: What is the area of quadrilateral \(ABCD\) above?

by DHRJ007 » Mon Sep 21, 2020 8:28 am
area=1/2(sum of parallel sides)x(perpendicular distance between the parallel sides)
=1/2(23x4)
=23x2
=46


OA:B
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