Question

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Answer 1

Answer:

7) 160 cm2

Step-by-step explanation:

7) Area of triangle ADC= 16 × 12

                                              2

=$96cm^{2}$

Area of ABC = 16 × 8

                          2

= $64cm^{2}$

Total area= 96 + 64 = 160$cm^{2}$

Answer 2

Answer ⤵️

7. Area of the Quadrilateral= Area of ∆ ADC+Area of ∆CBA

Area of the quadrilateral= ½×AC×DM+½×AC×BM

Area of the quadrilateral= ½× AC(DM+BM)

Area of the quadrilateral= ½×16(12+8)

Area of the quadrilateral= 8×20

Area of the quadrilateral= 160 cm².

8. ABCD is a rectangle. PQRD and TBOU is a square

Area of PQRD= (side)²

Area of PQRD= 3²=9 cm²

Area of TBOU= (side)²

Area of TBOU= 3²=9 cm²

Total square area= Area of PQRD+Area of TBOU

Total square area= 9+9=18 cm²

Area of ABCD = L×B

Area of ABCD= 18×12= 216 cm²

Total area= Area of rectangle-Total square area

Total area= 216-18

Total area= 198 cm²

9. Area of Rhombus= ½×AC×BD

120= ½×5× BD

120=5/2×BD

120/5/2=BD

120×2/5=BD

240/5=BD

BD= 48 cm

So, the length of the other diagonal is 48 cm

10. Here,

ABCD is a square

Area of the shaded portion= Area of ABCD - Area of CD( Semi-circle)

Area of ABCD= (side)²

= 20²=400 cm²

Diameter= 14 cm

Radius= 14/2=7 cm

Area of CD= ½×πr²

Area of CD= ½×22/7×7×7

Area of CD= 77 cm²

Step-by-step explanation:

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