GATE CH 2023 | GA | Question: 3

Question

In the given figure, PQRS is a parallelogram with $\mathrm{PS}=7 \mathrm{~cm}, \mathrm{PT}=4 \mathrm{~cm}$ and $\text{PV}=5 \mathrm{~cm}$. What is the length of $\mathrm{RS}$ in $\mathrm{cm}?$ (The diagram is representative.)

• A. $\frac{20}{7}$
• B. $\frac{28}{5}$
• C. $\frac{9}{2}$
• D. $\frac{35}{4}$

Comments

S and Q have same angle

sin theta = 5/7 = 4/x

x = 28/5

Answer 1

In the above figure,

Area of a parallelogram = base x height

Taking QR as base and PT as height

Area of parallelogram = QR X PT

                                  = 7 x 4  ------- 1

Taking RS as base and PV as height

Area of a parallelogram = RS X PV

                                      = RS x 5  -------- 2

Equating 1 and 2,

7 x4 = RS x 5

 28 = RS x 5

RS = 28/5

Comments

Really innovative approach @joshika

Answer 2

$RS = ?$

let, $RS = x, \quad SV = a, \quad QT = t$

So, $PQ = x, \quad QR = 7, \quad TR = 7-t, \quad VR = x-a$

Now, In $\triangle$PSV,

$7^{2} = 5^{2} + a^{2}$

$49 - 25 = a^{2}$

$24 = a^{2}$

$a = \sqrt{24}$

$\boxed{SV = a = \sqrt{24}}$

$\boxed{VR = x-a = x-\sqrt{24}}$

Again, In $\triangle$PRV,

$PR^{2} = 5^{2} + (x-a)^{2}$

$PR^{2} = 5^{2} + (x- \sqrt{24})^{2}$         ........................(1)

Again, In $\triangle$PRT,

$PR^{2} = 4^{2} + (7-t)^{2}$         ........................(2)

Again, In $\triangle$PTQ,

$x^{2} = 4^{2} + t^{2}$

$x^{2} - 16 = t^{2}$

$\boxed{t = \sqrt{x^{2} - 16}}$

Putting value of $t$ in eqn (2),

$PR^{2} = 4^{2} + (7-\sqrt{x^{2} - 16})^{2}$         ........................(3)

From eqn (1) and (3),

$5^{2} + (x- \sqrt{24})^{2} = 4^{2} + (7-\sqrt{x^{2} - 16})^{2}$

$25 + x^{2} + 24 - 2 \sqrt{24} x = 16 + 49 + x^{2} - 16 - 14 \sqrt{x^{2} - 16}$

$- 2 \sqrt{24} x = - 14 \sqrt{x^{2} - 16}$

$\sqrt{24} x = 7 \sqrt{x^{2} - 16}$

Squaring both sides,

$24 x^{2} = 49 (x^{2} - 16)$

$24 x^{2} = 49x^{2} - 49 \times 16$

$49x^{2} - 24x^{2} = 49 \times 16$

$25x^{2} = 49 \times 16$

$x^{2} = \dfrac{49 \times 16}{25}$

$x = \dfrac{7 \times 4}{5}$

$\boxed{x = \dfrac{28}{5}}$

Alternate:

Credit: @joshika's answer

$Area(\triangle PQR) = Area(\triangle PQR)$

$\dfrac{1}{2} \times QR \times PT = \dfrac{1}{2} \times RS \times PV$

$\dfrac{1}{2} \times 7 \times 4 = \dfrac{1}{2} \times x \times 5$

$28 = x \times 5$

$\boxed{x = \dfrac{28}{5}}$