Question

pls tell rhe answer of ques 3 and 4.​

Answer 1

Answer:

Hera are your answers mate↓↓↓↓

Step-by-step explanation:

Q3) The answer is 196 meter square

_____________________________

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!!!