Two adjacent sides of a parallelogram are, respectively, 13 cm and 17 cm in length If the length of - Maths - Areas of Parallelograms and Triangles

Question

Two adjacent sides of a parallelogram are, respectively,
13 cm and 17 cm in length If the length of the diagonal
passing through their point of intersection is 20 cm,
then
area of the parallelogram is closest to (take root30 = 5
5)

Lalit Mehra · Accepted Answer

Dear Student!

ABCD is a Parallelogram âˆ´ AB = CD = 17cm (opposite sides of the parallelogram are equal) and AD = BC = 13cm (opposite sides of the parallelogram are equal) Area of Î”ABC = Area of Î”ACD (1) (A diagonal of a parallelogram divides it into two triangles of equal area) Area of Î”ABC can be evaluated using Heron's formula Let a = 17 cm, b = 13 cm and c = 20 cm Semi-perimeter of Î”ABC, S = $\frac{a + b + c}{2} = \frac{17 cm+ 13 cm+ 20 cm}{2} = \frac{50 cm}{2} = 25 cm$ Area of Î”ABC $= \sqrt{s (s - a) (s - b) (s - c)}$ (Heron's formula) $= \sqrt{25 cm×(25 cm - 17 cm) \times (25 cm - 13 cm) \times (25 cm - 20 cm)}$ $= \sqrt{25 cm× 8 cm \times 12 cm \times 5 cm}$ $= \sqrt{\underset{̲}{5 \times 5} \times \underset{̲}{2 \times 2} \times \underset{̲}{2 \times 2} \times 2 \times 3 \times 5} {cm}^{2} = 5 \times 2 \times 2 \times \sqrt{30} {cm}^{2} = 20 \sqrt{30} {cm}^{2}$ Area of parallelogram ABCD = Area of Î”ABC + Area of Î”ACD = 2 Ã— Area of Î”ABC $= 2 \times 20 \sqrt{30} {cm}^{2}$ = 40 Ã— 5 5 cm2 = 220 cm2 Thus, the area of parallelogram is closest to 220 cm2 Cheers!

Two adjacent sides of a parallelogram are, respectively, 13 cm and 17 cm in length. If the length of the diagonal passing through their point of intersection is 20 cm, then area of the parallelogram is closest to (take root30 = 5.5)

Two adjacent sides of a parallelogram are, respectively,

13 cm and 17 cm in length. If the length of the diagonal

passing through their point of intersection is 20 cm, then

area of the parallelogram is closest to (take root30 = 5.5)