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melisa1 [442]
3 years ago
8

One side of a right triangle is 2 cm shorter than the hypotenuse and 7 cm longer than the third side. find the lengths of the si

des of the triangle.
Mathematics
1 answer:
astra-53 [7]3 years ago
8 0
The right angles triangle has three sides including the hypotenuse which is opposite to the right angle.
If the hypotenuse - H
Then one side is - H - 2
Third side is 7 cm shorter then H-2
Therefore it’s H -2-7 = H -9
According to Pythagoras theorem
Square of hypotenuse = sum of the squares of the other 2 sides
H² = (H-9)² + (H-2)²
H² = H² -18H + 81 + H² -4H + 4
0 =H² -22H +85
This is a quadratic equation
0 = H² - 17H -5H +85
0 = H (H - 17) -5(H-17)
(H-17) (H-5) =0
H -17=0 or H-5=0
H could be 17 cm or 5 cm
If H =5 cm then
Side 1 -H-2 = 3 cm
And side 2 - H-9 = -4 cm
Since lengths cannot be negative values
H isn’t 5 cm
Therefore H =17 cm
1 side is H-2 = 15 cm
2nd side H -9 = 8 cm
Three sides are 17 cm, 15 cm and 8 cm
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kvv77 [185]
PQ // BC 

then

Angle P = Angle B, A is common angle the two triangles are similar

AP/AB  = AQ/AC

8/18 = 12/ AC

AC = 12 * 18/8 = 27 (your ans: D)

AQ + QC = AC

QC = AC - AQ = 27 - 12 = 15 (if it was required) 

tan60 = y/8

y = s tan60 = 8sqrt(3)  (B)

Next time, post each question separately. 


6 0
2 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

\longrightarrowBD=9+6

\longrightarrowBD=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}}}

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3 years ago
Identify the a b and c values of the following quadratic expression 4x^2+5x=4
yarga [219]

A is the 4

B is the 5

and C is also 4.


A is always the number in front of x^2. B is always the number i front of the normal x, and C is the number that is on the other side of the equal sign!

6 0
3 years ago
Sam is practicing long track speed skating at an skating ring. The distance around the rink is 250 yards . He skated around thr
pentagon [3]
3 miles = 5,280 yards

6(250) = 1,500 yards

5,280 - 1,500 = 3,780 yards
4 0
3 years ago
Evaluate using the order of operations ( 8/9 - 1/3) x 2/3
Korvikt [17]
(8/9-3/9)*2/3=5/9*2/3=10/27
4 0
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