OR

Find the area of $\Delta OPB$ .

Right-angled figure showing tower foot P, midpoint B (top of lower section B) and apex A (top of upper section A) on a v — Fig. for Q51

Some Applications of Trigonometry

Heights and Distances — Case Study (Radio Tower/Support-Wire Problem)

Official Answer

$\text{Area of } \Delta OPB = \frac{1}{2} \times OP \times PB = \frac{1}{2} \times 6 \times 2\sqrt{3} = 6\sqrt{3} \text{ m}^2 \approx 10.39 \text{ m}^2$ , using the right angle at P and PB found from $\tan 30^\circ$ .

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Marking Scheme

11 mark: identifying that $\Delta OPB$ is right-angled at P, finding $PB = 2\sqrt{3} \text{ m}$ , and correctly stating Area = $\frac{1}{2} \times OP \times PB$ .
21 mark: correct substitution and final numeric value $6\sqrt{3} \text{ m}^2 (\approx 10.39 \text{ m}^2)$ with correct unit (m²).

Hint

$\Delta OPB$ is right-angled at P (tower ⟂ ground), so Area = $\frac{1}{2} \times OP \times PB$ — no need to find the hypotenuse OB.

Quick Oral Answer

Since the tower is vertical and the ground is horizontal, $\Delta OPB$ has a right angle at P, so its area is simply half of OP times PB, which comes to $6\sqrt{3} \text{ m}^2$ .

Analysis & Explanation

OR part: find the area of $\Delta OPB$ using the same triangle as Q50, viewed as a mensuration problem.

Concept

OP (horizontal ground) ⊥ PB (vertical tower section), so $\angle P = 90^\circ$ ; $\Delta OPB$ is right-angled at P.
Area of a right triangle = 1/2 × (leg 1) × (leg 2) = $\frac{1}{2} \times OP \times PB$ .

Key points

$PB = 6 \tan 30^\circ = 2\sqrt{3} \text{ m}$ (from the angle of elevation of B).
No need for Heron's formula or the hypotenuse OB once the right angle is spotted.

Common mistakes

Not recognising $\angle P = 90^\circ$ and instead trying Heron's formula unnecessarily.
Using OB (hypotenuse) as a leg by mistake.

Common Mistakes

1Unnecessarily calculating the hypotenuse OB and applying Heron's formula instead of the simple $\frac{1}{2} \times \text{base} \times \text{height}$ for a right triangle.
2Using PA instead of PB (confusing this with the AB-only calculation from the OR-alternative part).
3Forgetting to square the unit (m²) since this is an area, not a length.

Interesting Facts

Any right triangle's area can be found instantly as half the product of the two sides enclosing the right angle — one of the fastest area shortcuts in the CBSE mensuration syllabus.

This 'OR' choice structure (distance vs. area) is a standard CBSE design feature in case studies, letting students pick whichever sub-skill they are more confident with.

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Frequently Asked Questions

Do I need the hypotenuse OB to find this area?

No. Since the triangle is right-angled at P, only the two legs ( $OP = 6 \text{ m}$ and $PB = 2\sqrt{3} \text{ m}$ ) are needed: $Area = \frac{1}{2} \times OP \times PB = 6\sqrt{3} \text{ m}^2$ .

Can I attempt both Q50 and this OR question?

No — in CBSE 'OR' sub-parts, only one of the two alternatives should be attempted; only the first attempted one is usually evaluated.

OR Find the area of ΔOPB\Delta OPBΔOPB.