Important Questions for Chapter 12 - Heron's Formula

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Important questions based on NCERT syllabus for Chapter 12 - Heron's Formula:

Question 1: A triangular advertisement board has sides 11m,15m and 6m. If the advertisement yields an earning of Rs. 1000/ m^2 per month. So how much the company will earn in 3 and half years?
Solution:
Semi perimeter of a triangle = (a+b+c)/2
⇒ (11+15+6)/2 = 16cm
Area of a triangle = √(s(s-a)(s-b)(s-c))
= √(16(16-11)(16-15)(16-6))
= √800
= 20 √2 cm^2
If the advertisement yields an earning of Rs. 1000 /m^2 per month,
Then the earning in 42 months = 20√2 × 1000 × 42 = Rs. 840000√2

Question 2: Find the area of a quadrilateral ABCD in which AB = 6 cm, BC = 8 cm, CD = 8 cm, DA = 10 cm and AC = 10 cm.
Solution:
Take ΔABC
S = Perimeter/2 = (10+8+6)/2 = 12cm
Area = √(s(s-a)(s-b)(s-c))
= √(12(12-10)(12-8)(12-6))
= √576
= 24 cm^2

Now take ΔCDA
S = Perimeter/2 = (10+8+10)/2 = 14cm
Area = √(s(s-a)(s-b)(s-c) )
= √(14(14-10)(14-8)(14-10))
= √1344 = 6.66 cm^2

Area of Quadrilateral = 24 + 36.66 = 60.66 cm^2

Question 3: A wall is in the shape of a trapezium whose parallel sides are 50 m and 20 m. The non-parallel sides are 28 m and 26 m. Find the area of the wall.
Solution:
herons
Draw a line CE parallel to AD and draw a perpendicular CF on AB.
It can be observed that AECD is a parallelogram.
CE = AD = 26 m
AE = DC = 50 m
BE = 50 − AB = 50m - 20 m = 30m
For ΔBEC,

s = Perimeter/2 = (26+28+30)/2 = 42cm

Area of the ΔBEC = √(s(s-a)(s-b)(s-c))
= √(42(42-26)(42-28)(42-30))
= 336 cm^2

Now, area of ΔBEC = 1/2 × base × height
= 1/2 × BE × CF
= 1/2 × 30 × CF
CF = 672/30
= 22.4cm

Area of AECD = CF × AE
= 22.4 × 50
= 1120 cm^2

Therefore, the area of wall = 1120 - 336 = 784 cm^2.