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Geometry

Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen

$Math Studyset Vaia Explanations$ Math

Student Edition · 227 pages

ISBN: 9780395977279

Chapter 11: Q22. (page 426)

Consecutive sides of the figures are perpendicular. Find the area of each figure.

Short Answer

Expert verified

Area of the figure $(A) = \frac{49}{2} s q . u n i t s$

Step by step solution

01

Step 1. Given information.

Consecutive sides of the figure is perpendicular.

02

Step 2. Concept Used.

Where, b is the base of the triangles ,i is the height of the triangles and H is the Hypotenuse of the triangles.

$△ A B C$ is- $(45^{0} + 45^{0} + 90^{0})$

Thus, $\begin{array}{l} Cos 45^{0} = \frac{b}{H} \\ \frac{1}{\sqrt{2}} = \frac{b}{H} \\ b = \frac{H}{\sqrt{2}} \end{array}$

The triangle is an isosceles triangle.

Thus

Side opposite to the equal angles are equal.

Therefore, $(l) = (b) = \frac{7}{\sqrt{2}}$

03

Step 3. Let’s calculate the area of the figure.

Now, area of the figure

$\begin{matrix} A = (l \times b) \\ = (\frac{7}{\sqrt{2}}) \times (\frac{7}{\sqrt{2}}) \\ = \frac{49}{2} s q . u n i t s \end{matrix}$

Therefore, the area of the figure $(A) = \frac{49}{2} s q . u n i t s$

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Most popular questions from this chapter

Find the area of each figure:

A rhombus with a $120^{\circ}$ angle and sides 6cm

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Find the area of each figure:

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