Question 512 marksSection E

OR

Find the area of ΔOPB\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

Area of ΔOPB=12×OP×PB=12×6×23=63 m2≈10.39 m2\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∘\tan 30^\circ.

Answer keywords:right triangle areabase and heightperpendiculartower and groundarea formulaheights and distancesmensuration

Marking scheme

  1. 1.1 mark: identifying that ΔOPB\Delta OPB is right-angled at P, finding PB=23 mPB = 2\sqrt{3} \text{ m}, and correctly stating Area = 12×OP×PB\frac{1}{2} \times OP \times PB.
  2. 2.1 mark: correct substitution and final numeric value 63 m2(≈10.39 m2)6\sqrt{3} \text{ m}^2 (\approx 10.39 \text{ m}^2) with correct unit (m²).

Hint

ΔOPB\Delta OPB is right-angled at P (tower ⟂ ground), so Area = 12×OP×PB\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, ΔOPB\Delta OPB has a right angle at P, so its area is simply half of OP times PB, which comes to 63 m26\sqrt{3} \text{ m}^2.

Common mistakes

  1. 1.Unnecessarily calculating the hypotenuse OB and applying Heron's formula instead of the simple 12×base×height\frac{1}{2} \times \text{base} \times \text{height} for a right triangle.
  2. 2.Using PA instead of PB (confusing this with the AB-only calculation from the OR-alternative part).
  3. 3.Forgetting to square the unit (m²) since this is an area, not a length.

Analysis and explanation

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


Concept

  • OP (horizontal ground) ⊥ PB (vertical tower section), so ∠P=90∘\angle P = 90^\circ; ΔOPB\Delta OPB is right-angled at P.
  • Area of a right triangle = 1/2 × (leg 1) × (leg 2) = 12×OP×PB\frac{1}{2} \times OP \times PB.

Key points

  • PB=6tan⁡30∘=23 mPB = 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 ∠P=90∘\angle P = 90^\circ and instead trying Heron's formula unnecessarily.
  • Using OB (hypotenuse) as a leg by mistake.

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.

Question details

Q51 of 56 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Standard)
Year
2026
Section
Section E
Marks
2
Chapter
Some Applications of Trigonometry
Topic
Heights and Distances — Case Study (Radio Tower/Support-Wire Problem)
Bloom's level
apply
Syllabus topic
Unit 6: Mensuration (Trigonometry application) — area of a right triangle formed by a vertical tower section and its ground base
Author
CBSE
Reviewed by
Padhantu Academic Team

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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 mOP = 6 \text{ m} and PB=23 mPB = 2\sqrt{3} \text{ m}) are needed: Area=12×OP×PB=63 m2Area = \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.