hi..pls now tell the answer..hope the image is clear
Answers
Answer:
Hera are your answers mate↓↓↓↓
Step-by-step explanation:
Q3) The answer is 196 meter square
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Given:
AB || CD
AB =25 m
CD= 10m
BC=14m
AD= 13m
From point C draw CE || DA . Hence ADCE is a ||grm having CD || AE & AD||CE
AE= CD= 10m
CE= AD= 13m
BE= AB- AE =25- 10=15 m
BE= 15m
In ∆BCE
BC= 14m, CE= 13m, BE= 15m
Semiperimeter (s)= (a+b+c)/2
Semiperimeter(s) =( 14+13+15)/2
s= 42/2= 21m
s= 21m
Area of ∆BCE= √ s(s-a)(s-b)(s-c)
Area of ∆BCE=√ 21(21-14)(21-13)(21-15)
Area of ∆BCE= √ 21×7× 8×6
Area of ∆BCE= √ 7×3× 7× 4×2×2×3
Area of ∆BCE=√7×7×3×3×2×2×4
Area of ∆BCE= 7×3×2×2= 21× 4= 84m²
Area of ∆BCE= 84m²
Area of ∆BCE= 1/2 × base × altitude
Area of ∆BCE= 1/2 × BE ×CL
84= 1/2×15×CL
84×2= 15CL
168= 15CL
CL= 168/15
CL= 56 /5m
Height of trapezium= 56/ 5m
Area of trapezium= 1/2( sum of || sides)( height)
Area of trapezium=1/2(25+10)(56/5)
Area of trapezium= 1/2(35)(56/5)
Area of trapezium= 7×28= 196m²
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Hence the area of field is 196m²
Q4) Given Find the area of the quadrilateral ABCD in which AB=9cm , BC=40cm , CD=28cm, DA =15cm and ∆ABC = 90°
Now ABCD is a quadrilateral.
Triangle ABC = 90
In right angle triangle ABC
AC^2 = AB^2 + BC^2
= 9^2 + 40^2
= 81 + 1600
= 1681
So AC = 41 cm
Now area of triangle = 1/2 x b x h
= 1/2 x 9 x 40 = 180 sq cm
Now to find the other side of triangle using heron's formula we get
s = a + b + c / 2
s = 15 + 41 + 28 / 2
s = 42 cm
Now area of triangle ACD
= √s (s - a)(s - b)(s - c)
= √42(42 - 15)(42 - 41)(42 - 28)
= √42 x 27 x 1 x 14
= 126 sq cm
So Area of quadrilateral will be 126 + 180 = 306 sq cm
Hope this helps you mate!!!