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3 years ago

Which statement regarding the diagram is true?

Mathematics

1 answer:

Semmy [17]3 years ago

4 0

Answer:

mKML+mMLK=mJKM

Step-by-step explanation:

A triangle is a polygon with three sides and three angles. Types of triangles are scalene, acute, obtuse, equilateral, isosceles, and right triangle.

As shown in the diagram, polygon MLK is a triangle. The sum of angles in a triangle is equal to 180°, hence:

m∠MLK + m∠KML + m∠MKL = 180° (sum of angles in a triangle).

m∠MKL = 180 - (m∠MLK + m∠KML) (1)

Also along line JL, the sum of angles on one side of a straight line is 180°, therefore:

m∠JKM + m∠MKL = 180° (sum of angles in a straight line)

m∠JKM + m∠MKL = 180° (2)

From equations 1 and 2, equating give:

m∠JKM + [180 - (m∠MLK + m∠KML)) = 180

m∠JKM + 180 - m∠MLK - m∠KML = 180

m∠JKM = m∠MLK + m∠KML

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3 years ago

Bhaskar went for hiking with scouts’ team and there the scouts were given a task to build tents with the help of bamboos, ropes

dimulka [17.4K]

The triangles on Bhaskar's tent are similar triangles.

The length of IJ is 4.5 m
The applicable theorem is the mid-point theorem
The appropriate formula for area is the Heron's formula
The ratio of ABC to DEF is 1 : 9

(a) The length of IJ

The given parameter are:

$\mathbf{EF = 9m}$

I and J are at the midpoint of DE and DF

The above highlight means that

$\mathbf{IJ= \frac 12 \times EF}$ --- midpoint theorem

Substitute 9 for EF

$\mathbf{IJ= \frac 12 \times 9m}$

$\mathbf{IJ= 4.5\ m}$

(b) The property used to find GH and IJ

In (a), the midpoint theorem is used to calculate IJ

GH and IJ are corresponding sides of similar triangles,

So the midpoint theorem can also be used to calculate the length of GH

(c) The area of the triangle

For the given triangles, the lengths of the sides are known.

When side lengths are known, the formula to use for finding the triangle's area is the Heron's formula.

The Heron's formula is:

$\mathbf{Area = \sqrt{s \times (s -a) \times (s - b) \times (s - c)}}$

Where:

$\mathbf{s = a + b + c}\\\mathbf{a,b,c \to sides\ of\ the\ triangle}$

(d): The ratio of the areas:

For the small triangle, we have:

$\mathbf{a= 3.8,\ b = 4,\ c = 3}$

So, we have:

$\mathbf{s = 3.8 + 4 + 3 = 10.8}$

So, the area is:

$\mathbf{Area = \sqrt{s \times (s -a) \times (s - b) \times (s - c)}}$

$\mathbf{A_{small} = \sqrt{10.8 \times (10.8 -3.8) \times (10.8 - 4) \times (10.8 - 3)}}$

$\mathbf{A_{small} = \sqrt{4009.824}}$

For the big triangle, we have:

$\mathbf{a= 11.4,\ b = 12,\ c = 9}$

So, we have:

$\mathbf{s = 11.4 + 12 + 9 = 32.4}$

So, the area is:

$\mathbf{Area = \sqrt{s \times (s -a) \times (s - b) \times (s - c)}}$

$\mathbf{A_{big} = \sqrt{32.4 \times (32.4 -11.4) \times (32.4 - 12) \times (32.4 - 9)}}$

$\mathbf{A_{big} = \sqrt{324795.744}}$

The ratio of the small triangle to the big triangle is:

$\mathbf{Ratio = A_{small} : A_{big}}$

$\mathbf{Ratio = \sqrt{4009.824}:\sqrt{324795.744}}$

Divide by 4009.824

$\mathbf{Ratio = \sqrt{1}:\sqrt{81}}$

Take square roots

$\mathbf{Ratio = 1 : 9}$

Hence, the ratio of ABC to DEF is 1 : 9

Read more about similar triangles at:

brainly.com/question/24874611

7 0

3 years ago

Given the general identity tan X =sin X/cos X , which equation relating the acute angles, A and C, of a right ∆ABC is true?

irakobra [83]

First, note that $m\angle A+m\angle C=90^{\circ}.$ Then

$m\angle A=90^{\circ}-m\angle C \text{ and } m\angle C=90^{\circ}-m\angle A.$

Consider all options:

$\tan A=\dfrac{\sin A}{\sin C}$

By the definition,

$\tan A=\dfrac{BC}{AB},\\ \\\sin A=\dfrac{BC}{AC},\\ \\\sin C=\dfrac{AB}{AC}.$

Now

$\dfrac{\sin A}{\sin C}=\dfrac{\dfrac{BC}{AC}}{\dfrac{AB}{AC}}=\dfrac{BC}{AB}=\tan A.$

Option A is true.

$\cos A=\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}.$

By the definition,

$\cos A=\dfrac{AB}{AC},\\ \\\tan (90^{\circ}-A)=\dfrac{\sin(90^{\circ}-A)}{\cos(90^{\circ}-A)}=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AC}}=\dfrac{AB}{BC},\\ \\\sin (90^{\circ}-C)=\sin A=\dfrac{BC}{AC}.$