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3 years ago

Count box lengths to identify the base and height. Then, find the area of the triangle.

Mathematics

2 answers:

Aleks04 [339]3 years ago

8 0

Answer:

14 units²

Step-by-step explanation:

1/2(7)(4)=14 units²

Luba_88 [7]3 years ago

3 0

Answer:

14 square units

Step-by-step explanation:

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RoseWind [281]

The probability of getting a country CD out of all the CDs is the number of country CDs divided by the total number of CDs in Dante's Collection.

The total number of CDs in Dante's collection is $6+12+8+4+10=40$ .

The total number of country CDs is 12.

So our required probability is $\frac{12}{40}= \frac{3}{10}$ .

In percentage this is 3 divided by 10, and then multiply by 100. We get 30%. Answer choice D is correct.

ANSWER: D

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3 years ago

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ss7ja [257]

Answer:

See below ~

Step-by-step explanation:

$\textsf {Each of the steps has been reasoned below :}$

$\implies \textsf {3x - 2 = 4 (Given)}$

$\implies \textsf {3x = 6 (Addition Property of Equality)}$

$\implies \textsf {x = 2 (Division Property of Equality)}$

$\textsf {In the second step, the property has been applied}\\\textsf {by adding 2 to each side of the equation.}$

$\textsf {In the third step, the property has been applied by dividing 3 on each side.}$

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2 years ago

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find the area of the trapezium whose parallel sides are 25 cm and 13 cm The Other sides of a Trapezium are 15 cm and 15 CM

Snezhnost [94]

$\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}$

Given - A trapezium ABCD with non parallel sides of measure 15 cm each ! along , the parallel sides are of measure 13 cm and 25 cm

To find - Area of trapezium

Refer the figure attached ~

In the given figure ,

AB = 25 cm

BC = AD = 15 cm

CD = 13 cm

Construction -

$draw \: CE \: \parallel \: AD \: \\ and \: CD \: \perp \: AE$

Now , we can clearly see that AECD is a parallelogram !

$\therefore$ AE = CD = 13 cm

Now ,

$AB = AE + BE \\\implies \: BE =AB - AE \\ \implies \: BE = 25 - 13 \\ \implies \: BE = 12 \: cm$

Now , In ∆ BCE ,

$semi \: perimeter \: (s) = \frac{15 + 15 + 12}{2} \\ \\ \implies \: s = \frac{42}{2} = 21 \: cm$

Now , by Heron's formula

$area \: of \: \triangle \: BCE = \sqrt{s(s - a)(s - b)(s - c)} \\ \implies \sqrt{21(21 - 15)(21 - 15)(21 - 12)} \\ \implies \: 21 \times 6 \times 6 \times 9 \\ \implies \: 12 \sqrt{21} \: cm {}^{2}$

Also ,

$area \: of \: \triangle \: = \frac{1}{2} \times base \times height \\ \\\implies 18 \sqrt{21} = \: \frac{1}{\cancel2} \times \cancel12 \times height \\ \\ \implies \: 18 \sqrt{21} = 6 \times height \\ \\ \implies \: height = \frac{\cancel{18} \sqrt{21} }{ \cancel 6} \\ \\ \implies \: height = 3 \sqrt{21} \: cm {}^{2}$

Since we've obtained the height now , we can easily find out the area of trapezium !

$Area \: of \: trapezium = \frac{1}{2} \times(sum \: of \:parallel \: sides) \times height \\ \\ \implies \: \frac{1}{2} \times (25 + 13) \times 3 \sqrt{21} \\ \\ \implies \: \frac{1}{\cancel2} \times \cancel{38 }\times 3 \sqrt{21} \\ \\ \implies \: 19 \times 3 \sqrt{21} \: cm {}^{2} \\ \\ \implies \: 57 \sqrt{21} \: cm {}^{2}$