Question

Triangles abc and cde have a common vertex c with side ab of triangle abc being parallel to side de of triangle cde. If length of side ab = 4 cm and length of side de = 10 cm and perpendicular distance between sides ab and de is 9.8 cm, then the sum of areas of triangle abc and triangle cde is _________ cm2.

Answer 1

Answer: 94.73 square cm (Approx)

Step-by-step explanation:

Let the height of triangle ABC is x cm

Since, the perpendicular distance between parallel lines AB and DE is 9.8 cm

⇒ The height of triangle CDE = (x+9.8) cm

Since, By Alternative interior angle theorem,

$\angle DEC\cong \angle ABC$

And, $\angle EDC\cong \angle BAC$

Thus, By AA similarity postulate,

$\triangle DEC\sim \triangle ABC$

By the property of similar triangles,

$\frac{x+9.8}{x} = \frac{DE}{AB}$

$\frac{x+9.8}{x} = \frac{10}{4}$

$\frac{x+9.8}{x} = \frac{5}{2}$

$2x+19.6=5x$

$-3x=-19.6$

$x=\frac{19.6}{3} \text{ cm}$

Thus, the sum of area of triangle ABC and triangle DEC

= $\frac{1}{2} \times x\times 4 + \frac{1}{2} \times (x+9.8)\times 10$

= $2x+5(x+9.8)$

= $7x+49$

= $7\times \frac{19.6}{3} + 49$

= $\frac{137.2}{3} + 49$

=$\frac{137.2+147}{3}$

= $\frac{284.2}{3}=94.7333\approx 94.73\text{ square cm}$