Question

Find the area of each shade region
(can you do both for me please)

Answer 1

*FORMULA*

Qu1.

$area \: of \: a \: triangle = \frac{1}{2} base \: \times height$

Height=10

Base=8

$= \frac{1}{2} \times 8 \times 10$

$= 40$

Qu2.
Area of a kite=half×product of digonals
Step by step solution
Summation of 1st diagonal =5+5
=10
Summation of 2nd diagonal =7+5
=12
Product of the diagonals
=10×12
=120
Area=1÷2×120
=60

Answer 2

2a. Answer:   100

Step-by-step explanation:

$Area_{\ shaded\ region}=Area_{\ parallelogram}-Area_{\ triangle}\\\\Area_{\ parallelogram}=base \times height\\\\.\qquad \qquad \qquad \qquad =14\times 10\\\\.\qquad \qquad \qquad \qquad =140\\\\Area_{\ triangle}=\dfrac{base\times height}{2}\\\\\\.\qquad \qquad \quad =\dfrac{8\times 10}{2}\\\\\\.\qquad \qquad \quad =40\\\\Area_{\ shaded\ region}=140-40\\\\.\qquad \qquad \qquad \qquad =100$

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2b. Answer:   114

Step-by-step explanation:

$Area_{\ shaded\ region}=Area_{\ trapezoid}-Area_{\ kite}\\\\Area_{\ trapezoid}=\dfrac{base_1+base_2}{2}\times height\\\\\\.\qquad \qquad \qquad =\dfrac{11+18}{2}\times (5+7)\\\\\\.\qquad \qquad \qquad =\dfrac{29\times 12}{2}\\\\\\.\qquad \qquad \qquad =174\\\\\\Area_{\ kite}=\dfrac{diagonal_1\times diagonal_2}{2}\\\\\\.\qquad \qquad=\dfrac{(5+5)(5+7)}{2}\\\\\\.\qquad \qquad=\dfrac{10\times 12}{2}\\\\\\.\qquad \qquad=60$

$Area_{\ shaded\ region}=174-60\\\\.\qquad \qquad \qquad \qquad =114$