In the given figure, PQ RS , RX SQ and PR YS are given. [figure in the original paper] (a) Write the ratio of areas of triangle QRS and parallelogram PRSY. [1] (b) Prove that the areas of…

(a) Ratio of areas

Because

• $PQ\parallel RS$ and $RX\parallel SQ$ ⇒ $PQRS$ is a parallelogram.
• In a parallelogram the diagonal $QR$ divides it into two congruent triangles, therefore

$$\text{Area of }△QRS=\frac{1}{2}\text{Area of }PQRS.$$

The parallelogram $PRSY$ has the same base $RS$ and the same altitude as $PQRS$ (the line through $P$ is parallel to the line through $Y$). Hence

$$\text{Area of }PRSY=\text{Area of }PQRS.$$

Consequently

$$\boxed{\frac{\text{Area of }△QRS}{\text{Area of }PRSY}=\frac{1}{2}}.$$

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(b) Equality of the two parallelograms

$$\text{Parallelogram }PRSY\text{and}\text{Parallelogram }XRSQ$$

Proof

$$\boxed{\text{Area}(PRSY)=\text{Area}(XRSQ)}$$

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(c) Construction

1. Triangle $ABC$

Given:

• $BC=6\text{ cm}$
• $AC=7\text{ cm}$
• $\angle ABC=75^{\circ }$

Construction steps

1. Draw a horizontal line segment $BC$ of length 6 cm.
2. At point $B$ construct an angle of $75^{\circ }$ using a protractor; label the ray $BA$.
3. With centre $C$ and radius 7 cm, draw an arc intersecting the ray $BA$ at point $A$.
4. Join $A$ to $B$ and to $C$.

The required triangle is obtained.

2. Parallelogram $CDEF$ equal in area to $△ABC$

Given: one side $ED=6.5\text{ cm}$.

Step‑wise construction

1. Compute the area of $△ABC$.

   $$\begin{array}{rl}A{B}^2& =B{C}^2+A{C}^2-2(BC)(AC)\cos 75^{\circ }\\ & =6^2+7^2-2(6)(7)(0.258819)\\ & =36+49-21.735\\ & =63.265\\ AB& \approx 7.956\text{ cm}\end{array}$$

   $${\text{Area}}_{△ABC}=\frac{1}{2}BCAB\sin 75^{\circ }=\frac{1}{2}(6)(7.956)(0.965925)\approx 23.07\mathbf{c}{\mathbf{m}}^2.$$

2. Required height of the parallelogram

   $$h=\frac{\text{Area}}{\text{base}}=\frac{23.07}{6.5}\approx 3.55\mathbf{cm}.$$

3. On side $CD$ (to be drawn) mark a segment $ED=6.5\text{ cm}$.

4. Through point $C$ draw a line $CF$ making any convenient angle with $CD$.

5. From point $E$ draw a line parallel to $CF$; let it intersect the extension of $CF$ at $F$.

6. From $F$ draw a line parallel to $CD$ meeting the extension of $CE$ at $D$.

7. The quadrilateral $CDEF$ is a parallelogram; its altitude (distance between $CD$ and $EF$) is set to 3.55 cm by using a ruler and a right‑angle set‑square.

Thus $\text{Area}(CDEF)=6.5\times 3.55=23.07{\text{ cm}}^2$, equal to the area of $△ABC$.

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Constructed figures

Triangle $ABC$

Parallelogram $CDEF$ (shown as a rectangle for clarity; the opposite sides are parallel)

Combined diagram of the original configuration (a & b)

Result summary

• Ratio $\frac{\text{Area }△QRS}{\text{Area }PRSY}=1:2$.
• $\text{Area}(PRSY)=\text{Area}(XRSQ)$.
• Constructed $△ABC$ (BC = 6 cm, AC = 7 cm, ∠B = 75°).
• Constructed parallelogram $CDEF$ with side $ED=6.5$ cm and area 23.07 cm², equal to the triangle’s area.