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2 years ago

Find the area of quadrilateral ABCD

Mathematics

1 answer:

andreev551 [17]2 years ago

3 0

Answer:

Step-by-step explanation:

a, b, c

P = a + b + c

Semiperimeter s = $\frac{a+b+c}{2}$

A = $\sqrt{s(s-a)(s-b)(s-c)}$

~~~~~~~~~~~~~~~

$P_{ABC}$ = 4.3 + 2.89 + 6.81 = 14

s = 14 ÷ 2 = 7

$A_{ABC}$ = $\sqrt{7(7-4.3)(7-2.89)(7-6.81)}$ = √14.75901 ≈ 3.84

$P_{BCD}$ = 8.59 + 7.58 + 6.81 = 22.98

s = 22.98 ÷ 2 = 11.49

$A_{BCD}$ = $\sqrt{11.49(11.49-8.59)(11.49-7.58)(11.49-6.81)}$ = √609.7343148 ≈ 24.6928

$A_{ABCD}$ = 3.84 + 24.6928 ≈ <em>28.5</em>

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3 years ago

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Given: KL ║ NM , LM = 45, m∠M = 50° KN ⊥ NM , NL ⊥ LM Find: KN and KL

Mice21 [21]

Answer:

$KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08\\ \\KN=45\sin 50^{\circ}\approx 34.47$

Step-by-step explanation:

Given:

KL ║ NM ,

LM = 45

m∠M = 50°

KN ⊥ NM

NL ⊥ LM

Find: KN and KL

1. Consider triangle NLM. This is a right triangle, because NL ⊥ LM. In this triangle,

LM = 45

m∠M = 50°

So,

$\tan \angle M=\dfrac{\text{opposite leg}}{\text{adjacent leg}}=\dfrac{NL}{LM}=\dfrac{NL}{45}\\ \\NL=45\tan 50^{\circ}$

Also

$m\angle LNM=90^{\circ}-50^{\circ}=40^{\circ}$ (angles LNM and M are complementary).

2. Consider triangle NKL. This is a right triangle, because KN ⊥ NM . In this triangle,

$NL=45\tan 50^{\circ}$

$m\angle KLN=m\angle LNM=40^{\circ}$ (alternate interior angles)

$m\angle KNL=90^{\circ}-40^{\circ}=50^{\circ}$ (angles KNL and KLN are complementary).

So,

$\sin \angle KNL=\dfrac{\text{opposite leg}}{\text{hypotenuse}}=\dfrac{KL}{LN}=\dfrac{KL}{45\tan 50^{\circ}}\\ \\KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08$

and

$\cos \angle KNL=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}=\dfrac{KN}{LN}=\dfrac{KN}{45\tan 50^{\circ}}\\ \\KN=45\tan 50^{\circ}\cos 50^{\circ}=45\sin 50^{\circ}\approx 34.47$

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3 years ago

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Daniel [21]

Answer: x = 11 degrees

Step by step explanation:
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2 years ago

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spin [16.1K]

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