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22 Feb 2022, 03:30
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E
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In the diagram above, ∠L = ∠M = 90°, KL = 4, LM = 8, MN = 10, and JN = JK = 13. What is the area of JKLMN?
A. 92
B. 96
C. 100
D. 108
E. 116
Are You Up For the Challenge: 700 Level Questions
Attachment:
pt_img2.png [ 30.51 KiB | Viewed 4461 times ]
22 Feb 2022, 08:46
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This is how I resolved it but if anyone knows a better way to guestimate it please let me know as this method is too long in my opinion.
I will cut the shape in smaller parts.
1) rectangle KLMY (Y = 10 - 6)
Area = 4*8 = 32
2) triangle KZY
Since YK is perpendicular to MZ it is a right triangle
Area = 6*8/2= 24
3) Triangle JKZ
first we find the base KZ
since KZY is a right triangle we can use pythagoras theorem:
KY^2 + ZY^2 = KZ^2
6^2 + 8^2 = 100 -> KZ = 10
To find the area we have to use Hebron's formula:
Heron formula: A = √s(s-a)(s-b)(s-c)
where,
a, b, and c are the sides of the triangle: 13, 13, 10
s is the semi perimeter of the triangle; (13+13+10)/2 = 18
Area of triangle: √18*(5)(5)(8) = 5*2*√18*2 = 10*√36 = 10*6 = 60
Finally we sum every area up...
60 + 32 + 24 = 116
Answer: E
I will cut the shape in smaller parts.
1) rectangle KLMY (Y = 10 - 6)
Area = 4*8 = 32
2) triangle KZY
Since YK is perpendicular to MZ it is a right triangle
Area = 6*8/2= 24
3) Triangle JKZ
first we find the base KZ
since KZY is a right triangle we can use pythagoras theorem:
KY^2 + ZY^2 = KZ^2
6^2 + 8^2 = 100 -> KZ = 10
To find the area we have to use Hebron's formula:
Heron formula: A = √s(s-a)(s-b)(s-c)
where,
a, b, and c are the sides of the triangle: 13, 13, 10
s is the semi perimeter of the triangle; (13+13+10)/2 = 18
Area of triangle: √18*(5)(5)(8) = 5*2*√18*2 = 10*√36 = 10*6 = 60
Finally we sum every area up...
60 + 32 + 24 = 116
Answer: E
05 Mar 2022, 04:00
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Bunuel
MAGOOSH OFFICIAL SOLUTION:
We will subdivide the figure as shown.
Notice, we constructed KR, such that KR is perpendicular to MN. This makes KLMR a rectangle and KRN a right triangle. S is the midpoint of KN, so that JS is the median & altitude of isosceles triangle JKN. (Since it’s easy to find the area of rectangles and right triangles, those make particularly good choices for subdivision units when you have to find area.)
Now, we will figure things out piece-by-piece. First of all, a particularly easy piece is rectangle KLMR: Area = bh = 4*8 = 32.
The problem gives that MN = 10, and we know MR = 4, so NR must equal 6, and KR must equal 8. These are two legs of the Pythagorean triplet (6, 8, 10), which is the (3, 4, 5) triplet times two. This means we know KN = 10. We also know the area of triangle KRN is: Area = 0.5*bh = 0.5*6*8 = 24.
Since we know KN = 10, we know the midpoint divides that in half, so KS = SN = 5. Notice, we now have two right triangles, KJS and NJS, each with a leg of 5 and a hypotenuse of 13. This is another one of the triplets, (5, 12, 13)! Immediately, without further calculation, we know JS = 12. Now we can find the area of the big isosceles triangle JKN: Area = 0.5*bh = 0.5*12*10 = 60.
Total area = (area of rect. KLMR) + (area of triangle KRN) + (area of triangle JKN)
= 32 + 24 + 60 = 116
Answer = E
Attachment:
pt_img6.png [ 88.12 KiB | Viewed 4050 times ]
