knowledge-base/class_9/ch10/0149_geometry.md

---
source: ncert_exercises
topic: geometry
class_level: class_9
chapter: ch10
exercise: ex10.1
difficulty: intermediate
---

Class 9 | Ch10: Heron's Formula | Exercise 10.1

Heron's Formula:
Area of triangle = √(s(s-a)(s-b)(s-c))
where s = semi-perimeter = (a+b+c)/2
a, b, c are sides of triangle.

Q1. A traffic signal board is equilateral triangle with side 'a'. Find area using Heron's. If perimeter=180 cm, find area.
Answer: Equilateral triangle: a=b=c. s=(a+a+a)/2=3a/2.
Area=√(3a/2 × a/2 × a/2 × a/2)=√(3a⁴/16)=(√3/4)a².
If perimeter=180 → a=60 cm. Area=(√3/4)×60²=(√3/4)×3600=900√3 cm². ✅

Q2. Triangular side walls of a flyover have sides 122 m, 22 m and 120 m. Advertisements yield ₹5000 per m² per year. Company hired one wall for 3 months. How much rent did it pay?
Answer: a=122, b=22, c=120. Check: 22²+120²=484+14400=14884=122². Right triangle!
s=(122+22+120)/2=264/2=132. Area=½×22×120=1320 m² (easier for right triangle).
Rent = 1320×5000×(3/12) = 1320×1250 = ₹16,50,000. ✅

Q3. There is a slide in a park. One of its side walls has been painted, in some colour, with a message "KEEP THE PARK GREEN AND CLEAN". If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.
Answer: a=15, b=11, c=6. s=(15+11+6)/2=32/2=16.
Area=√(16×(16-15)×(16-11)×(16-6))=√(16×1×5×10)=√800=20√2 m². ✅

Q4. Find area of triangle with two sides 18 cm and 10 cm, and perimeter 42 cm.
Answer: Third side = 42-18-10=14 cm. s=42/2=21.
Area=√(21×(21-18)×(21-10)×(21-14))=√(21×3×11×7)=√4851=21√11 cm². ✅

Q5. Sides of triangle are in ratio 12:17:25 and perimeter is 540 cm. Find area.
Answer: Sides = 12k, 17k, 25k. 12k+17k+25k=540 → 54k=540 → k=10. Sides: 120, 170, 250 cm.
s=270. Area=√(270×150×100×20)=√(270×150×2000)=√81000000=9000 cm². ✅

Q6. An isosceles triangle has perimeter 30 cm and each equal side is 12 cm. Find area.
Answer: Base=30-12-12=6 cm. s=30/2=15.
Area=√(15×(15-12)×(15-12)×(15-6))=√(15×3×3×9)=√1215=9√15 cm². ✅