Question

Please help me solve 20th question

Answer 1

Answer:

hope it helps u buddy ..!!

don't forget to mark my question as the best ..!!

Answer 2

Answer:

Hey Mate

This is a question from the Perimeter and Area chapter of Mathematics

The trick is to break the figure into several figures and calculate their areas and add them all up to find the total area of the entire figure

Step-by-step explanation:

1. We divide the figure of the question into 3parts:-

a. Triangle ACH

b. Triangle ABC

c. Trapezium DEFG

2. Calculating area of triangles ACH and ABC is done by the formula of Area of triangles

1/2*(Base*Height)

Traingles ACH and ABC are right angled triangles whose hypotenuse (17m) and Perpendicular (8m) are given but the base is not given which is required for calculating their areas

So using Pythagoras theorem

Base² = Hypotenuse² - Perpendicular ²

--> Base² = 17²-8²

Base²= 289-64=225

So Base= ✓225=15

Now using triangle area formula

Area ∆ ACH= 1/2*(15*8) = 1/2*120 = 60m²

Area ∆ABC = Area ∆ACH = 60cm² since both traingles are part of the same rectangular like portion of the figure with the same diagonal and equal perpendiculars portions of those right angled triangles thus being congruent to each other via RHS condition.

3. Area of Trapezium= 1/2*(sum of parallel sides)* height

Parallel sides of DEFG trapezium are EF and GD measuring 9m and 6m respectively so their sum is 15 and height of the trapezium is mentioned inside the trapezium in the perpendicular drawin as 3.6m

Putting these values in the formula

Area = 1/2*15*3.6 = 27m²

4. Total area of figure= Ar. ∆ACH+ Ar.∆ABC + Ar. Trapezium DEFG

---> = 60+60+27= 147m²

Answer: Area of figure= 147m²

Thanks for this wonderful question

Hope the explanation is fruitful and understanding for you

#BeBrainly❤️