In the given figure, ∠ABC = ∠DCB = 90° AB = 6, DC = 8 then
Problem Set 1 | Q 4 | Page 27
In the given figure, ∠ABC = ∠DCB = 90° AB = 6, DC = 8 then \[\frac{A \left( ∆ ABC \right)}{A \left( ∆ DCB \right)} = ?\]
Given:
∠ABC = ∠DCB = 90°
AB = 6
DC = 8
\[Now, \frac{A \left( ∆ ABC \right)}{A \left( ∆ DCB \right)} = \frac{\frac{1}{2} \times AB \times BC}{\frac{1}{2} \times DC \times BC}\]
\[ = \frac{6}{8}\]
\[ = \frac{3}{4}\]
Answer:_ Given:
- ∠ABC = ∠DCB = 90°
- AB = 6
- DC = 8
To find: Ratio of areas of triangles ∆ABC and ∆DCB
Solution: The area of a triangle is given by the formula A = 1/2 × base × height. Since the two triangles ∆ABC and ∆DCB share a common height (BC), the ratio of their areas can be found by comparing their bases.
Using the given information, we can see that:
- The base of ∆ABC is AB = 6.
- The base of ∆DCB is DC = 8.
Therefore, the ratio of the areas of the two triangles is:
A(∆ABC) / A(∆DCB) = (1/2 × AB × BC) / (1/2 × DC × BC) = AB / DC = 6 / 8 = 3 / 4
Hence, the ratio of the areas of the two triangles ∆ABC and ∆DCB is 3:4.
Problem Set 1 | Q 4 | Page 27
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