What is the area of the shaded region?

Mathematics

1 answer:

marshall27 [118]3 years ago

5 0

Area of large triangle
= 1/2 bh
= 1/2 (5)(12)
= 1/2 (60)
= 30

area of un-shaded triangle
= 1/2bh
= 1/2(3)(4)
= 1/2 (12)
= 6

area of shaded = area of large triangle - area of un-shaded triangle
area of shaded = 30 - 6
area of shaded = 24

answer
24

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In ΔJKL, k = 9.6 cm, l = 2.7 cm and ∠J=43°. Find ∠L, to the nearest 10th of a degree.

Bogdan [553]

Answer:

L = 10.64°

Step-by-step explanation:

From the given information:

In triangle JKL;

line k = 9.6 cm

line l = 2.7 cm; &

angle J = 43°

we are to find angle L = ???

We can use the sine rule to determine angle L:

i.e

$\dfrac{j}{SIn \ J} = \dfrac{l}{ SIn \ L}$

Using Pythagoras rule to find j

i,e

j² = k² + l²

j² = 9.6²+ 2.7²

j² = 92.16 + 7.29

j² = 99.45

$j = \sqrt{99.45}$

j = 9.97

∴

$\dfrac{9.97}{Sin \ 43} = \dfrac{2.7}{ Sin \ L}$

${9.97 \times Sin (L ) = (2.7 \times Sin \ 43)$

$= Sin \ L = \dfrac{ (2.7 \times Sin \ 43)}{9.97 } \\ \\ = Sin \ L = \dfrac{ (2.7 \times 0.6819)}{9.97 } \\ \\ = Sin \ L = 0.18466 \\ \\ L = Sin^{-1} (0.18466) \\ \\ L = 10.64 ^0$