Question

Help me find SR, RT and m

Answer 1

Given:-

• A rhombus with

1. ST = 3x + 30
2. SE = 3x
3. ET = 5x + 5
4. EA = 4z - 8
5. SR = 8x - 5
6. angle EAT = 9y + 8°
7. angle ETA = 5y - 2°

To find:-

• The value of SR , RT and angle TAS .

Solution:-

Firstly as we know that all the sides of a rhombus are equal. Hence here , the value of SR and ST will be equal. So that;

`\implies 8x - 5 = 3x +30 \\`

`\implies 8x - 3x = 30 +5 \\`

`\implies 5x = 35 \\`

`\implies x = (35)/(5) \\`

`\implies x = 7\\`

Now substitute this value of x in SR , to find SR as ;

`\implies SR = 8x - 5\\`

`\implies SR = 8(7)-5 \\`

`\implies SR = 56 - 5\\`

`\implies \underline{\underline{ SR = 51 }} \\`

Again, as we know that the diagonals of a rhombus bisect each other at right angles . So the value of SE and EA will be equal . Hence;

`\implies SE = EA \\`

`\implies 3z = 4z - 8 \\`

`\implies 4z - 3z = 8 \\`

`\implies z = 8 \\`

The value of SE will be ,

`\implies 3z = 24 \\`

So , in right angled triangle SER , by Pythagoras theorem ,

`\implies SR^2 = SE^2 + ER^2 \\`

`\implies 51^2 = 24^2 + RE^2 \\`

`\implies RE =√( 2601-576)=√(2025)\\`

`\implies RE = 45 \\`

And ,

`\implies RT = 2RE \\`

`\implies RT = 2(45)\\`

`\implies \underline{\underline{ RT = 90}} \\`

Again , we know that the sum of adjacent angles of a rhombus is 180° . Also we know that the diagonals of a rhombus bisect the angles . So , we have;

`\implies \angle RAT + \angle STA = 180^o \\`

`\implies (1)/(2)(\angle RAT + \angle STA )=(180^o)/(2) \\`

`\implies (1)/(2)\angle RAT +(1)/(2)\angle STA = 90^o \\`

`\implies \angle EAT + \angle ETA = 90^o \\`

`\implies 9y + 8^o + 5y - 2^o = 90^o \\`

`\implies 14y = 90^o - 6^o \\`

`\implies 14y = 84^o\\`

`\implies y =(84)/(14)=\boxed{6^\circ}\\`

Now put this value in the expression of angle EAT

`\implies \angle EAT = 9y + 8^o \\`

`\implies \angle EAT =( 9(6)+8)^o\\`

`\implies \angle EAT = ( 54 + 8)^o\\`

`\implies \angle EAT = 62^o \\`

Also,

`\implies \angle EAT = \angle TAS \\`

Therefore,

`\implies \angle \underline{\underline{TAS = 62^o}} \\`

This is our required answer.

and we are done!

Answer 2

SR = 51

RT = 90

m∠TAS = 62°

Explanation:

All the sides of a rhombus are equal in length, so SR = ST.

Therefore, to find the value of x, equate the expressions for SR and ST and solve for x:

`\begin{aligned}SR&=ST\\\implies 8x-5&=3x+30\\8x-5-3x&=3x+30-3x\\5x-5&=30\\5x-5+5&=30+5\\5x&=35\\(5x)/(5)&=(35)/(5)\\x&=7\end{aligned}`

Substitute the found value of x into the expression for SR:

`\begin{aligned}\implies SR&=8x-5\\&=8(7)-5\\&=56-5\\&=51\end{aligned}`

Therefore, SR = 51.

The diagonals of a rhombus bisect each other, so EA = SE.

Therefore, to find the value of z, equate the expressions for EA and SE and solve for z:

`\begin{aligned}EA&=SE\\\implies 4z-8&=3z\\4z-8-3z&=3z-3z\\z-8&=0\\z-8+8&=0+8\\z&=8\end{aligned}`

As SA bisects RT at point E, RT = 2ET.

Substitute the found value of z to find RT:

`\begin{aligned}\implies RT&=2ET\\&=2(5z+5)\\&=10z+10\\&=10(8)+10\\&=80+10\\&=90\end{aligned}`

Therefore, RT = 90.

As the diagonals of a rhombus bisect each other at 90°, triangle TEA is a right triangle where m∠TEA = 90°.

As the interior angles of a triangle sum to 180°, then the sum of the other two measures of right triangle TEA is 90°:

`\begin{aligned}m \angle TAE + m \angle ETA &= 90^(\circ)\\\implies (9y+8)^(\circ) + (5y-2)^(\circ)&= 90^(\circ)\\9y+8+5y-2&=90\\14y+6&=90\\14y+6-6&=90-6\\14y&=84\\(14y)/(14)&=(84)/(14)\\y&=6\end{aligned}`

As m∠TAS = m∠TAE = (9y + 8)°, substitute the found value of y to find the measure of angle TAS:

`\begin{aligned}\implies m \angle TAS&=(9y + 8)^(\circ)\\&=(9(6) + 8)^(\circ)\\&=(54 + 8)^(\circ)\\&=62^(\circ)\end{aligned}`

Therefore, m∠TAS = 62°.