ABCD is a square of area $4$, which is divided into four non-overlapping triangles as shown in the figure. Then the sum of the perimeters of the triangles is:
ABCD is a square of area $4$, which is divided into four non-overlapping triangles as shown in the figure. Then the sum of the perimeters of the triangles is:
Show Hint
- $8(2 + \sqrt{2})$
- $8(1 + \sqrt{2})$
- $4(1 + \sqrt{2})$
- $4(2 + \sqrt{2})$
The Correct Option is B
Solution and Explanation
To solve the problem, we begin by understanding the properties of the square ABCD. Given that its area is 4, the side length of the square is \(s\), where \(s^2=4\). Therefore, \(s=2\).
Next, examine the structure of the square divided into four triangles. Each triangle's vertices lie on the vertices and midpoints of the square. We denote the midpoints of sides AB, BC, CD, and DA as points E, F, G, and H, respectively, to form the triangles AEB, BFC, CGD, and DHA.
The side length of each triangle from midpoint to vertex (e.g., AD or DH) is half the side length of the square, so each enclosing triangle has a leg length of 1 unit. The hypotenuse for triangles such as AEB follows the diagonal split from vertex A to midpoint E or B, categorizing them as right-angled isosceles triangles with leg lengths of 1.
Utilize the Pythagorean Theorem to find the hypotenuse of one such triangle:
c=12+12=2
This calculation yields \(\sqrt{2}\).
Now calculate the perimeter for one such triangle:
P=1+1+2=2+2
Therefore, the total perimeter of all four triangles is:
T=4×(2+2)=8(1+2)
Thus, the sum of the perimeters of the triangles is indeed \(8(1+\sqrt{2})\).
Top Questions on Geometry
In the adjoining figure, $\triangle CAB$ is a right triangle, right angled at A and $AD \perp BC$. Prove that $\triangle ADB \sim \triangle CDA$. Further, if $BC = 10$ cm and $CD = 2$ cm, find the length of AD.
In the diagram, the lines QR and ST are parallel to each other. The shortest distance between these two lines is half the shortest distance between the point P and the line QR. What is the ratio of the area of the triangle PST to the area of the trapezium SQRT?
Note: The figure shown is representative
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Questions Asked in CAT exam
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- Para Completion
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- Para Completion