What is ratio of area of triangle ADC to area of triangle ADB ?
Consider the following for the next two (02) items that follow :<br>ABC is a triangle right-angled at A. Further, AB=8 cm, BC=10 cm. D is the point on BC such that AD is perpendicular to BC.
- A. 7:15
- B. 9:16 ✓
- C. 2:3
- D. 3:4
Correct Answer: B. 9:16
Explanation
By Pythagoras, AC = 6 cm. The areas of \triangle ADC and \triangle ADB sharing the same height AD are proportional to their bases CD and DB. For a right triangle, AC^2 = CD \times BC and AB^2 = DB \times BC. The ratio \frac{CD}{DB} = \frac{AC^2}{AB^2} = \frac{36}{64} = \frac{9}{16}.
Related questions on Mensuration
- A pendulum swings through an angle of 30° and its end describes an arc of length 55 cm. What is the length of the pendulum? (Take $\pi = 22/...
- A conical tent has an angle of 60° at the vertex. If the curved surface area is 100 m^2, then what is the volume of the tent?
- A right circular cone and a hemisphere have equal base and equal volume. What is the ratio of the height of the cone to the radius of the he...
- A wire is in the form of an equilateral triangle with an area of 36\sqrt{3} cm^2. If it is changed into a semicircle, then what is its r...
- Let the area of the largest possible square inscribed in a circle of unit radius be x. Let the area of the largest possible circle inscribed...