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3 years ago

10

Find the missing side of the triangle:

1 answer:

Digiron [165]3 years ago

3 0

Answer:A triangle is a polygon which has three sides and can be categorized into the follow ... sides of the triangle, and s = ½ (a + b + c) is the semi-perimeter of the triangle. ... Example 1: Find the area of a triangle whose base is 14 cm and height is 10 cm. ... a = 5cm and b = 7cm. C = 45o. Solution: Area of a triangle = ½ · a · b · sinC.

Step-by-step explanation:

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What are the next three terms of the pattern 4, -2, -8, -14, ...?

Advocard [28]

I believe the next three terms are -20, -26, -32 if it goes down by 6

6 0

3 years ago

Read 2 more answers

Fill in the blanks and explain the pattern.<br><br> XA, XB, XC, __,__,__

Anit [1.1K]

Answer:

XD,XE,XF

Step-by-step explanation:

XA,XB,XC,XD,XE,XF

IT IS BECAUSE OF THE ALPHABETICAL ORDER AFTER X

8 0

3 years ago

find the value of 10 times the square root of 2 over the square root of 5.. a. 2 b. 2 times the square root of 5. c. 2 the squar

Marizza181 [45]

$\dfrac{10\sqrt2}{\sqrt5}=\dfrac{10\sqrt2\cdot\sqrt5}{\sqrt5\cdot\sqrt5}=\dfrac{10\sqrt{2\cdot5}}{5}=\boxed{2\sqrt{10}}\to\fbox{c.}$

6 0

3 years ago

This problem is lowkey tricky. Can someone help me out

notka56 [123]

Answer:

x = 10

Step-by-step explanation:

Proof: If a line is parallel to one side of the triangle, then it divides the other two sides proportionally.

6/9 = x/15

9x = 90

x = 10

6 0

2 years ago

In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= root 2 cm.

Nata [24]

Given CD is an altitude such that AD=BC , AB=3 cm and CD= √2 cm.

Let AD=x, Since given AB=3

AD+DB=3

x+DB = 3

DB = 3-x

Since ΔBCD is rght angle triangle, let's apply Pythagoras theorem

$BC^{2} = DB^{2} +CD^{2}$

$BC^{2} = (3-x)^{2} +(\sqrt{2} )^{2}$

$BC^{2} =(3-x)^{2} +2$

Since given AD=BC,let us plugin BC=x in above step.

$x^{2} =(3-x)^{2} +2$

$x^{2} =9-6x+x^{2} +2$

6x=11

x= $\frac{11}{6}$

Now we know AD=x= $\frac{11}{6}$ and given CD=√2.

Let us apply Pythagoras theorem for ΔACD

$AC^{2} =AD^{2} +DC^{2}$

$AC^{2} = (\frac{11}{6} )^{2} +(\sqrt{2} )^{2}$

$AC^{2} =\frac{193}{36}$

$AC=\sqrt{\frac{193}{36} }$ = 2.315cm

3 0

3 years ago