Question

find perimeter and area of a triangle of which each diagonal has length 25 m and the difference of the two adjacent sides is 5m.

Answer 1

Given the length of the Diagonal of the rectangle as : 25 m

We know that the Diagonal of a Rectangle divides a Rectangle into two Congruent Right angled Triangles with the Diagonal as Hypotenuse and the Length and Breadth of the Rectangle as the Adjacent sides of the Congruent Right angled Triangles.

⇒ From Pythagorean Theorem :

(Diagonal Length)² = (Length)² + (Breadth)²

Given the Length of the Diagonal as 25 m

⇒ 25² = L² + B²

⇒ L² + B² = 625

Given that the Difference of the Two adjacent sides of Rectangle as 5m

We know that Two Adjacent sides of a Rectangle are its Length and Breadth

⇒ L - B = 5

Squaring on both sides we get :

⇒ (L - B)² = 25

⇒ L² + B² - 2.LB = 25

But we know that L² + B² = 625

⇒ 625 - 2 × L × B = 25

⇒ 2 × L × B = 625 - 25

⇒ 2 × L × B = 600

⇒ L × B = 300

we know that L - B = 5

⇒ L = B + 5

substituting the value of L = B + 5 in L × B = 300 we get :

⇒ (B + 5) × B = 300

⇒ B² + 5B - 300 = 0

⇒ B² + 20B - 15B - 300 = 0

⇒ B(B + 20) - 15(B + 20) = 0

⇒ (B + 20)(B - 15) = 0

⇒ B = -20 or B = 15

We know that Breadth cannot be negative, so Breadth of the Rectangle = 15m

if B = 15 then Length = B + 5 = 15 + 5 = 20

⇒ Perimeter of Rectangle =2(L + B) = 2(15 + 20) = 70m

⇒ Area of Rectangle = L × B = 20 × 15 = 300 m²

Answer 2

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