If the ratio of corresponding sides of two similar triangles is 4: 9, then the ratio of areas of these triangles is
- 16:81
- 4:9
- 2:3
- \(\sqrt{2}:\sqrt{3}\)
The Correct Option is A
Solution and Explanation
To solve the problem, we need to find the ratio of areas of two similar triangles when the ratio of their corresponding sides is given.
1. Understanding the Area Ratio for Similar Triangles:
If two triangles are similar, then the ratio of their areas is equal to the square of the ratio of their corresponding sides.
Let the ratio of sides be $ \frac{4}{9} $, then:
$ \frac{\text{Area}_1}{\text{Area}_2} = \left( \frac{4}{9} \right)^2 $
2. Squaring the Side Ratio:
$ \left( \frac{4}{9} \right)^2 = \frac{16}{81} $
Final Answer:
The ratio of the areas of the triangles is $ 16 : 81 $
Top Questions on Triangles
In the adjoining figure, \(PQ \parallel XY \parallel BC\), \(AP=2\ \text{cm}, PX=1.5\ \text{cm}, BX=4\ \text{cm}\). If \(QY=0.75\ \text{cm}\), then \(AQ+CY =\)
- Assertion (A) : \(\triangle ABC \sim \triangle PQR\) such that \(\angle A = 65^\circ\), \(\angle C = 60^\circ\). Hence \(\angle Q = 55^\circ\).
Reason (R) : Sum of all angles of a triangle is \(180^\circ\). - If the orthocentre of the triangle formed by the lines $ y = x + 1 $, $ y = 4x - 8 $, and $ y = mx + c $ is at $ (3, -1) $, then $ m - c $ is:
In the adjoining figure, \( \triangle CAB \) is a right triangle, right angled at A and \( AD \perp BC \). Prove that \( \triangle ADB \sim \triangle CDA \). Further, if \( BC = 10 \text{ cm} \) and \( CD = 2 \text{ cm} \), find the length of } \( AD \).
If a line drawn parallel to one side of a triangle intersecting the other two sides in distinct points divides the two sides in the same ratio, then it is parallel to the third side. State and prove the converse of the above statement.
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