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3 years ago

13

How do you do this properties and mental math to solve 275+180+120 =

1 answer:

kramer3 years ago

6 0

Make 275 t0 200 and 180 to 100 and 120 to 100

200+100+100=300

now turn 75 to 70 and 80 will stay 80 and 20 will stay 20

70+80+20=170

now add the 5 to 170= 175

now add 300 to 175 to get 475 which is ya answer

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How do I prove exterior angle theorem

soldier1979 [14.2K]

The measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles of the triangle.

m∠4=m∠1+m∠2m∠4=m∠1+m∠2
Proof:
Given: ΔPQRΔPQR
To Prove: m∠4=m∠1+m∠2m∠4=m∠1+m∠2
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5 0

3 years ago

Solve step by step solution then only i can do it plxx

Anuta_ua [19.1K]

Answer:

<h3><u>Let's</u><u> </u><u>understand the concept</u><u>:</u><u>-</u></h3>

Here angle B is 90°

So $\triangle ABC$ and $\triangle ABD$ Are right angled triangle

So we use Pythagoras thereon for solution

<h3><u>Required Answer</u><u>:</u><u>-</u></h3>

First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

We need to find base=b

According to Pythagoras thereon

${\boxed{\sf b^2=h^2-p^2}}$

Substitutethe values

$\longrightarrow$ $\sf b^2=10^2-p^2$

$\longrightarrow$ $\sf b={\sqrt {10^2-8^2}}$

$\longrightarrow$ $\sf b={\sqrt{100-64}}$

$\longrightarrow$ $\bf b={\sqrt {36}}$

$\longrightarrow$ $\sf b=6$

$\therefore$ $\overline{BC}=6cm$

BD=BC+CD

$\longrightarrow$ $BD=9+6$

$\longrightarrow$ $BD=15cm$

Now in $\triangle ABD$

Perpendicular=p=8cm

Base =b=15cm

We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

${\boxed {\sf h^2=p^2+b^2}}$

Substitute the values

$\longrightarrow$ $\sf h^2=8^2+15^2$

$\longrightarrow$ $\sf h={\sqrt {8^2+15^2}}$

$\longrightarrow$ $\sf h={\sqrt {64+225}}$

$\longrightarrow$ $\sf h={\sqrt {289}}$

$\longrightarrow$ $\sf h=17cm$

$\therefore$ ${\underline{\boxed{\bf x=17cm}}}$

3 0

3 years ago

Find x and decide if that side (with the x) is an Altitude

irga5000 [103]

Answer:

b = 15
c = 51
c = 100
b = 25.75

Step-by-step explanation:

Using the Pythagoras theorem equation:

$c^{2}=a^{2}+b^{2}$

Where 'c' is the hypotenuse and 'b' and 'a' the sides of the right triangle.

1) Here c = 17, a = 8 and b = ?

Then, using the above equation we can find the side b.

$17^{2}=8^{2}+b^{2}$

$b=\sqrt{225}$

$b=15$

2) Here a = 24, b = 45 and c = ?

$c=\sqrt{24^{2}+45^{2}}$

$c=51$

3) Here a = 28, b = 96 and c = ?

$c=\sqrt{28^{2}+96^{2}}$

$c=100$

4) Here a = 17, b = ? and c = 28

$b=\sqrt{28^{2}-11^{2}}$

$b=25.75$

Just in the 4 case x is an altitude.

I hope it helps you!

3 0

3 years ago

Find the length of the missing side 15 6

Komok [63]

The answer would be 90

8 0

4 years ago

If a 4 x 16 riangle has the same area as a square, what is the length of a side of the square

m_a_m_a [10]

Answer: $\bold{4\sqrt2}$

<u>Step-by-step explanation:</u>

$A_{triangle}=\bigg(\dfrac{1}{2}\bigg)b\times h$

$A_{triangle}=\bigg(\dfrac{1}{2}\bigg)4\times 16$

$=\bigg(\dfrac{1}{2}\bigg)64$

$=32$

$A_{square}= s^2$

$32 = s^2$

$\sqrt{32}=s$

$\sqrt{\underline{2\cdot 2}\cdot \underline{2\cdot 2}\cdot 2}=s$

$2\cdot 2\sqrt{2}=s$

$4\sqrt{2}=s$

7 0

3 years ago