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Vesna [10]
4 years ago
11

What is the area of the irregular pentagon

Mathematics
2 answers:
V125BC [204]4 years ago
8 0

Answer:

Answer  is 105 in^2

Step-by-step explanation:   Area of rectangle= length X breadth

= 8X12

=96in^ 2

Area of triangle= 1/2 X  base X height

= 1/2 X 3 X 6

= 9 in^2

Area of irregular pentagon = area of triangle + area of rectangle

= 96+9

=105 in ^2


Ket [755]4 years ago
5 0

Answer:

Answer  is 105 in^2

Step-by-step explanation:  

Area of rectangle= length X breadth = 8X12=96in^ 2Area of triangle= 1/2 X  base X height= 1/2 X 3 X 6= 9 in^2 Area of irregular pentagon = area of triangle + area of rectangle = 96+9=105 in ^2

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In rhombus MOTH, angle HTM=(5x+6) degrees. If angle HMO=(12x-6)degrees, what is angle HMO?
forsale [732]

Answer:

m∠HMO=102°

Step-by-step explanation:

we know that

In a Rhombus the diagonals bisect the angles and opposite angles are equal

In this problem

Angles HMO and HTO are opposite angles

m∠HMO=m∠HTO

m∠HTM=(1/2)m∠HMO ----> because the diagonals bisect the angles and opposite angles are congruent

substitute the values

(5x+6)\°=(1/2)(12x-6)\°

solve for x

10x+12=12x-6\\12x-10x=12+6\\2x=18\\x=9

<em>Find the measure of angle HMO</em>

m∠HMO=(12x-6)°

substitute the value of x

m∠HMO=(12(9)-6)=102°

see the attached figure to better understand the problem

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