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Alexeev081 [22]
3 years ago
10

Find the area of a triangle with base of 10 inches and altitude to the base of 16 inches. 80 in² 135 in² 160 in²

Mathematics
2 answers:
kotykmax [81]3 years ago
7 0
10 × 16 = 160
160 / 2 = 80

Your answer would be 80<span>in²

Hope that helps!
</span>
Komok [63]3 years ago
3 0

Answer:

Option 1st is correct

area of a triangle is, 80 in²

Step-by-step explanation:

Area of triangle(A) is given by:

A = \frac{1}{2} b \cdot h

where,

b is the base

h is the height or altitude of the triangle respectively.

As per the statement:

Base of a triangle(b) = 10 inches and

Height = altitude(h) = 16 inches

Substitute these in [1] we get:

A = \frac{1}{2}(10) \cdot 16

⇒A = 5 \cdot 16 = 80in²

Therefore, the area of a triangle is, 80 in²

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A parallelogram has sides 10m and 12m and an angle of 45°. Find the distance between the 12m-sides.
stich3 [128]

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The distance between 12m-sides is 5\sqrt{2}m.

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It is given that the parallelogram has sides 10m and 12m and an angle of 45°.

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Since angle C is acute angle therefore it must be 45 degree.

\sin\theta=\frac{perpendicular}{hypotenuse}

\sin C=\frac{BE}{BC}

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\frac{1}{\sqrt{2}}=\frac{d}{10}

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