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In DeltaABC, P is a point on BC such tha...

In `DeltaABC`, P is a point on BC such that BP : PC = 2 : 3 and Q is the midpoint of BP. Then ar `(DeltaABQ): ar(DeltaABC)` is equal to :

A

`1:4`

B

`2:5`

C

`1:5`

D

`2:3`

Text Solution

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To solve the problem, we need to find the ratio of the area of triangle ABQ to the area of triangle ABC given the conditions about points P and Q. ### Step-by-Step Solution: 1. **Understand the Ratios**: We are given that BP : PC = 2 : 3. This means if we let BP = 2x and PC = 3x, then BC = BP + PC = 2x + 3x = 5x. 2. **Assign a Length to BC**: For simplicity, let's assume BC = 10 cm. This gives us: - BP = 2x = 4 cm (since 2/5 of 10 cm is 4 cm) - PC = 3x = 6 cm (since 3/5 of 10 cm is 6 cm) 3. **Identify Point Q**: Q is the midpoint of BP. Therefore, the length of BQ = BP/2 = 4 cm / 2 = 2 cm. 4. **Area of Triangle ABC**: The area of triangle ABC can be calculated using the formula: \[ \text{Area of } \Delta ABC = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times BC \times H = \frac{1}{2} \times 10 \times H = 5H \] 5. **Area of Triangle ABQ**: The area of triangle ABQ can also be calculated using the same formula: \[ \text{Area of } \Delta ABQ = \frac{1}{2} \times BQ \times H = \frac{1}{2} \times 2 \times H = H \] 6. **Calculate the Ratio**: Now we can find the ratio of the areas: \[ \frac{\text{Area of } \Delta ABQ}{\text{Area of } \Delta ABC} = \frac{H}{5H} = \frac{1}{5} \] 7. **Final Result**: Therefore, the ratio of the area of triangle ABQ to the area of triangle ABC is: \[ \text{ar}(\Delta ABQ) : \text{ar}(\Delta ABC) = 1 : 5 \]
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