Question

A square of side 10 units lies across two parallel lines, which are 10 units apart, so that two triangular areas of the square lie outside the lines. Find the sum of the perimeters of these two triangles AFE and GHC.​

Answer 1

Step-by-step explanation:

the triplet (3, 5, 4),

32 = 9, 52 = 25, 42 = 16 and 9 + 16 = 25

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ (3, 5, 4) is a pythagorean triplet.

(ii) In the triplet (4, 9, 12),

42 = 16, 92 = 81, 122 = 144 and 16 + 81 = 97 ≠ 144

The square of the largest number is not equal to the sum of the squares of the other two numbers.

∴ (4, 9, 12) is not a pythagorean triplet.

(iii) In the triplet (5, 12, 13),

52 = 25, 122 = 144, 132 = 169 and 25 + 144 = 169

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ (5, 12, 13) is a pythagorean triplet.

(iv) In the triplet (24, 70, 74),

242 = 576, 702 = 4900, 742 = 5476 and 576 + 4900 = 5476

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ (24, 70, 74) is a pythagorean triplet.

(v) In the triplet (10, 24, 27),

102 = 100, 242 = 576, 272 = 729 and 100 + 576 = 676 ≠ 729

The square of the largest number is not equal to the sum of the squares of the other two numbers.

∴ (10, 24, 27) is not a pythagorean triplet.

(vi) In the triplet (11, 60, 61),

112 = 121, 602 = 3600, 612 = 3721 and 121 + 3600 = 3721

The square of the largest number is equal to the sum of the squares of the other two numbers.

∴ (11, 60, 61) is a pythagorean triplet

Answer 2

Given: A square of side 10 units lies across two parallel lines, which are 10 units apart, so that two triangular areas of the square lie outside the lines.

To find: The sum of the perimeters of these two triangles AFE and GHC.​

Solution:

As evident from the information given in the question, the two parallel lines move through the midpoints of the sides of the square. So, E is the mid-point of AB, H is the midpoint of BC, G is the midpoint of CD and F is the midpoint of AD. Thus, AE, AF, CH and CG are equal to half the side of the square.

$\frac{10}{2} = 5 units$

Now, the two triangles formed are right-angled triangles because they are formed at the vertices of a square. Each vertex of a square measures 90°.

In triangle AEF, AE and AF equal 5 units and the value of EF can be calculated using the Pythagoras theorem.

$EF = \sqrt{5^{2} +5^{2} }$

      $= 5\sqrt{2}$

So, the perimeter of the triangle AEF is calculated as

$Perimeter = 5 + 5 + 5\sqrt{2}$

                 $= 10+5\sqrt{2}$

Similarly, the length of GH and the perimeter of the triangle CGH is calculated as

$GH = \sqrt{5^{2} +5^{2} }$

       $= 5\sqrt{2}$

$Perimeter = 5 + 5 + 5\sqrt{2}$

                  $= 10+5\sqrt{2}$

Hence, the sum of their perimeters is

$10+5\sqrt{2} + 10+5\sqrt{2} = 20 + 10\sqrt{2}$

                                  $= 34.14$

Therefore, the sum of the perimeters of these two triangles AFE and GHC is 34.14.