Question

In the figure below, PQRS and TURV are rectangles. The lengths shown in the figure are in centimetres. If the perimeter of the unshaded rectangle TURV is 32 cm, then the value of x is:​

Answer 1

Answer:

TU = PQ−SV = (8x−6) cm

TV = PS−QU = 4x−x = 3x cm

Perimeter of TURN = 32cm

32cm⇒2(8x−6+3x)=32

⇒11x−16=16

$x = \frac{16 + 6}{11} = \frac{22}{11} = 2cm$

Answer 2

$\large\underline{\sf{Solution-}}$

In the given figure, it is given that,

• PQ = 8x cm

• PS = 4x cm

• SV = 6 cm

• QU = x cm

• Perimeter of rectangle, TURV = 32 cm

Now,

As it is given that, PQRS is a rectangle. We know opposite sides of rectangle are equal.

So, PQ = RS = 8x cm

And, PR = QS = 4x cm

Now, In rectangle TURV

VR = RS - SV = 8x - 6 cm

and

UR = QR - QU = 4x - x = 3x cm

As it is given that,

$\rm \: Perimeter_{(TURV)} = 32$

$\rm \: 2(UR + VR) = 32$

$\rm \: UR + VR = 16$

$\rm \: 8x - 6 + 3x = 16$

$\rm \: 11x - 6 = 16$

$\rm \: 11x = 16 + 6$

$\rm \: 11x = 22$

$\rm\implies \:x = 2 \: cm$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$