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3 years ago

How would I prove the congruency of these triangles?

Mathematics

2 answers:

Sergeeva-Olga [200]3 years ago

8 0

Answer:

Step-by-step explanation:

In ΔADC & ΔABC,

Statement Reason

AD = BC Given

∠DAC ≅ ∠ACB {AB // BC , AC is transversal, ⇒ alternate interior angles are equal}

AC = CA Reflexive property

ΔADC ≅ ΔABC SAS congruent

AB = CD CPCT -> Corresponding Part of Congruent Triangles

Andre45 [30]3 years ago

3 0

Answer:

SAS

Step-by-step explanation:

line AB = line CD S = SIDE

AB = CD because the shape is a parallelogram and opposite sides of a parallelogram are equal.

Hence, proved AB = CD

You might be interested in

Three expressions that are equal to 5/9 x 8

madam [21]

Answer:

$\frac{13}{9}+3$
$\frac{22}{9}+2$
$\frac{85}{9}-5$

They all equal $\frac{5}{9}*8$ because they all give $\frac{40}{9}$

Step-by-step explanation:

$\frac{13}{9}+3$
$\frac{22}{9}+2$
$\frac{85}{9}-5$

They all equal $\frac{5}{9}*8$ because they all give $\frac{40}{9}$

Done by writing three equations with one random number and the desired result, $\frac{40}{9}$ , returning the answers above.

4 0

3 years ago

Will give brainiest lol

maria [59]

Step-by-step explanation:

1 and 7/8, subtract 4 and 3/8 by 2 and 1/2.

3 0

3 years ago

Read 2 more answers

Show work please \sqrt(x+12)-\sqrt(2x+1)=1

Nesterboy [21]

Answer:

$x=4$

Step-by-step explanation:

Given $\displaystyle\\\sqrt{x+12}-\sqrt{2x+1}=1$ , start by squaring both sides to work towards isolating $x$ :

$\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2$

Recall $(a-b)^2=a^2-2ab+b^2$ and $\sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b}$ :

$\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2\\\implies x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1$

Isolate the radical:

$\displaystyle\\x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1\\\implies -2\sqrt{(x+12)(2x+1)}=-3x-12\\\implies \sqrt{(x+12)(2x+1)}=\frac{-3x-12}{-2}$

Square both sides:

$\displaystyle\\(x+12)(2x+1)=\left(\frac{-3x-12}{-2}\right)^2$

Expand using FOIL and $(a+b)^2=a^2+2ab+b^2$ :

$\displaystyle\\2x^2+25x+12=\frac{9}{4}x^2+18x+36$

Move everything to one side to get a quadratic:

$\displaystyle-\frac{1}{4}x^2+7x-24=0$

Solving using the quadratic formula:

A quadratic in $ax^2+bx+c$ has real solutions $\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ . In $\displaystyle-\frac{1}{4}x^2+7x-24$ , assign values:

$\displaystyle \\a=-\frac{1}{4}\\b=7\\c=-24$

Solving yields:

$\displaystyle\\x=\frac{-7\pm \sqrt{7^2-4\left(-\frac{1}{4}\right)\left(-24\right)}}{2\left(-\frac{1}{4}\right)}\\\\x=\frac{-7\pm \sqrt{25}}{-\frac{1}{2}}\\\\\begin{cases}x=\frac{-7+5}{-0.5}=\frac{-2}{-0.5}=\boxed{4}\\x=\frac{-7-5}{-0.5}=\frac{-12}{-0.5}=24 \:(\text{Extraneous})\end{cases}$

Only $x=4$ works when plugged in the original equation. Therefore, $x=24$ is extraneous and the only solution is $\boxed{x=4}$

4 0

2 years ago

Can anybody help me please

GREYUIT [131]

Question 7: Option 1: x = 33.5°

Question 8: Option 3: x = 14.0°

Step-by-step explanation:

Question 7:

In the given figure, the value of perpendicular and hypotenuse is given, so we have to use any trigonometric ratio to find the value of angle as the given triangle is a right-angled triangle

So,

Perpendicular = P = 32

Hypotenuse = H = 58

So,

$sin\ x = \frac{P}{H}\\sin\ x = \frac{32}{58}\\sin\ x = 0.5517\\x = sin^{-1} ( 0.5517)\\x =33.48$

Rounding off to nearest tenth

x = 33.5°

Question 8:

In the given figure, the value of Base and Perpendicular is given, we will use tangent trigonometric ratio to find the value of x

So,

Perpendicular = P = 5

Base = B = 20

So,

$tan\ x = \frac{P}{B}\\tan\ x = \frac{5}{20}\\tan\ x = 0.25\\x = tan^{-1} (0.25)\\x = 14.036$

Rounding off to nearest tenth

x = 14.0°

Keywords: Right-angled triangle, trigonometric ratios

Learn more about trigonometric ratios at:

#LearnwithBrainly

8 0

3 years ago

Which number is irrational? 6√ 1/3 0.14............ 4√

deff fn [24]

Umm, I think it is the first one.

3 0

3 years ago

Read 2 more answers