Question

as DC JK kg GM ur d DC njgxb RSS s FM nut fj Jennings JK s oh CDs jsshshdhf

Answer 1

For 9th one :
Given: AB || CDAB =25 m CD= 10m BC=14mAD= 13m
From point C draw CE || DA . Hence ADCE is a ||grm having CD || AE & AD||CE
AE= CD= 10mCE= AD= 13mBE= AB- AE =25- 10=15 m BE= 15m
In ∆BCEBC= 14m, CE= 13m, BE= 15m
Semiperimeter (s)= (a+b+c)/2
Semiperimeter(s) =( 14+13+15)/2s= 42/2= 21ms= 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× 4×2×2×3Area of ∆BCE=√7×7×3×3×2×2×4Area of ∆BCE= 7×3×2×2= 21× 4= 84m²
Area of ∆BCE= 84m²
Area of ∆BCE= 1/2 × base × altitudeArea of ∆BCE= 1/2 × BE ×CL
84= 1/2×15×CL
84×2= 15CL168= 15CLCL= 168/15CL= 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

Hence the area of field is 196m²


For 8th one :
semi perimeter of each triangular shaped tile 
s= (35+9+28)/2 = 36 cm
acc to heron's formula,
area of trianlge = √s(s-a)(s-b)(s-c)
⇒√36(36-35)(36-28)(38-9)
⇒36√6 cm² =88.2 cm²
area of 16 tiles = 16*88.2 =1411.2 cm²
∴cost of polishing 1411.2 cm² = 1411.2 *.50= rs 705.60




hope it helps... :)