Find the area of the following
pls...pls help
Answers
Answer:
45.8 cm²
Step-by-step explanation:
In ∆PXT ,
XT = TR - XR
= 12- 9
= 3 cm
Area of ∆PXT
= 1/2 × base × height
= 1/2 × 3 × 4.4
= 3× 2.2
= 6.6 cm²
In ∆ QYR ,
YR = TR - TY
= 12 - 8
= 4 cm
Area of ∆ QYR
= 1/2 × base × height
= 1/2 × 2 × 4
= 4 cm²
For Trapezium (trap.) PQYX ,
XY = 8-(12 -9)= 8-3 = 5 cm
Area
= 1/2 × (sum of parallel sides ) × height
= 1/2 × (4.4 +2) × 5
=1/2× 6.4 × 5
= 3.2 × 5
= 16 cm²
For ∆ RTS,
Area
= 1/2 × base × height
= 1/2 × 3.2 × 12
= 3.2 × 6
= 19.2 cm²
Total area
= 6.6 + 4 + 16 + 19.2
= 25.8 + 20
= 45.8 cm²
In ∆PXT ,
XT = TR - XR
= 12- 9
= 3 cm
Area of ∆PXT
= 1/2 × base × height
= 1/2 × 3 × 4.4
= 3× 2.2
= 6.6 cm²
In ∆ QYR ,
YR = TR - TY
= 12 - 8
= 4 cm
Area of ∆ QYR
= 1/2 × base × height
= 1/2 × 2 × 4
= 4 cm²
For Trapezium (trap.) PQYX ,
XY = 8-(12 -9)= 8-3 = 5 cm
Area
= 1/2 × (sum of parallel sides ) × height
= 1/2 × (4.4 +2) × 5
=1/2× 6.4 × 5
= 3.2 × 5
= 16 cm²
For ∆ RTS,
Area
= 1/2 × base × height
= 1/2 × 3.2 × 12
= 3.2 × 6
= 19.2 cm²
Total area
= 6.6 + 4 + 16 + 19.2
= 25.8 + 20