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Marysya12 [62]
3 years ago
11

In ΔIJK, j = 540 inches, ∠I=54° and ∠J=69°. Find the area of ΔIJK, to the nearest square inch.

Mathematics
1 answer:
xenn [34]3 years ago
3 0

Answer:

105963

Step-by-step explanation: delta math

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Answer:

<h3>Option  D) 294 sq cm is correct</h3><h3>∴ the surface area of the rectangular pyramid is 294 sq cm</h3>

Step-by-step explanation:

First we have to split the net into 4 triangles and 1 rectangle

Given a = 12 cm ,b =  6 cm and d = 13 cm

<h3>To find the surface area of the rectangular pyramid:</h3>

Now find the area of the rectangle base

Rectangle base area=b\times a

= 6\times 12

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<h3>∴ Rectangle base area=72 sq cm</h3>

Now to find the area of the triangle on the left

Left triangle=\frac{1}{2}(b)(d)

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<h3>∴ Left triangle=39 sq cm</h3>

Since all the triangles are congruent , you will need to multiply by 2 to get the combined area of the triangle on the left and on the right.

Area of left and right triangles= 2(39)

=78 sq cm

<h3>∴ Area of left and right triangles=78 sq cm</h3>

Find the area of the triangle on the bottom

Bottom triangle area=\frac{1}{2}(a)(a)

=\frac{1}{2}(12)(12)

= 72 sq cm

<h3>∴ Bottom triangle area=72 sq cm</h3>

Since the bottom of the triangle is congruent to the top triangle, multiply that by 2 to get a combined area of the triangle on the bottom and top

Area of top & bottom triangles=2 (72)

= 144 sq cm

<h3>∴ Area of top & bottom triangles= 144 sq cm</h3>

Finally add the area of the 4 triangles to the area of the rectangular base we get

=72 + 78 + 144

= 294 sq cm

<h3>∴ the surface area of the rectangular pyramid is 294 sq cm</h3><h3>∴ option D) 294 sq cm is correct.</h3>
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