Applied Practical Geometry Ex 4.3 Class 8 NCERT

Understanding the Two Adjacent Sides and Three Angles Case

To construct a unique quadrilateral, we require five independent measurements. In Exercise 4.3, these five measurements are provided in a specific combination: two adjacent sides and three angles.

When constructing with these dimensions, you cannot start arbitrarily. You must select one of the given adjacent sides as your base. Once the base is drawn, you construct the known angles at its two endpoints. Using the second known adjacent side, you cut an arc along one of the newly constructed angle rays to establish the third vertex. Finally, you construct the third angle at this new vertex; its ray will intersect the first angle's ray to naturally pinpoint the fourth and final vertex. This method ensures absolute accuracy without guessing where the lines meet.

Step-by-Step Construction of Quadrilateral MORE

1. Step 1: Draw a Rough Sketch — Always draw a quick freehand sketch of the quadrilateral. Label it $MORE$ and write down the given values: $MO = 6\text{ cm}$, $OR = 4.5\text{ cm}$, $\angle M = 60^\circ$, $\angle O = 105^\circ$, and $\angle R = 105^\circ$. This helps you visualize where to start.
2. Step 2: Draw the Base Line Segment — Draw the base line segment $MO = 6\text{ cm}$ using a ruler and pencil.
3. Step 3: Construct Angles at the Base Endpoints — At point $M$, construct an angle of $60^\circ$ using a compass. At point $O$, construct an angle of $105^\circ$ using a compass (or protractor, if allowed by your teacher) and extend the ray.
4. Step 4: Locate Vertex R — Set your compass width to $4.5\text{ cm}$. Place the compass pointer on point $O$ and draw an arc cutting the ray of $105^\circ$. Label the intersection point as $R$ (since $OR = 4.5\text{ cm}$).
5. Step 5: Construct the Third Angle at R — At point $R$, construct an angle of $105^\circ$ pointing inwards towards the ray from $M$.
6. Step 6: Mark the Intersection Point E — Extend the $60^\circ$ ray from $M$ and the $105^\circ$ ray from $R$ until they intersect. Label this intersection point as $E$. $MORE$ is your completed, constructed quadrilateral.

YoLearn Exam Tips & Smart Angle Verification

1. The Angle Sum Property Trap: Sometimes, the question gives you three angles, but one of them is at a vertex you cannot directly reach. For example, in Quadrilateral $PLAN$, you might be given $\angle P$, $\angle A$, and $\angle N$. Since you don't know $\angle L$ (which lies on your base $PL$), you cannot draw the angle at $L$. Always use the Angle Sum Property of a Quadrilateral ($\angle P + \angle L + \angle A + \angle N = 360^\circ$) to calculate the missing base angle before beginning your physical drawing!
2. Sharp Pencils and Firm Compasses: A slight slip or a blunt pencil can throw your final intersection off by several millimeters, causing marks to be deducted. Keep your pencil finely sharpened and ensure your compass hinge is tight.

Practice Questions with Solutions

• Q: Construct Quadrilateral PLAN where PL = 4 cm, LA = 6.5 cm, ∠P = 90°, ∠A = 110°, and ∠N = 85°. A: Step 1: Analyze the given details. We have adjacent sides PL and LA, and angles ∠P, ∠A, and ∠N. We need to find the base angle ∠L to construct the angle at point L. Step 2: Apply the angle sum property of a quadrilateral: ∠P + ∠L + ∠A + ∠N = 360°. 90° + ∠L + 110° + 85° = 360° => 285° + ∠L = 360° => ∠L = 360° - 285° = 75°. Step 3: Draw base PL = 4 cm. Step 4: Construct ∠P = 90° at point P and ∠L = 75° at point L. Step 5: Mark point A on the ray of ∠L at a distance of 6.5 cm from L (LA = 6.5 cm). Step 6: At point A, construct ∠A = 110°. Extend this ray to intersect the 90° ray from P at point N. Final answer: Quadrilateral PLAN is successfully constructed with base PL = 4 cm, LA = 6.5 cm, ∠P = 90°, ∠L = 75°, ∠A = 110°, and ∠N = 85°.
• Q: Construct a Parallelogram HEAR where HE = 5 cm, EA = 6 cm, and ∠R = 85°. A: Step 1: Use the properties of a parallelogram to find all required measurements. In parallelogram HEAR, opposite sides are equal (HE = AR = 5 cm; EA = RH = 6 cm). Step 2: Opposite angles are equal, so ∠E = ∠R = 85°. Step 3: Adjacent angles are supplementary, so ∠H = 180° - 85° = 95°. Also, ∠A = ∠H = 95°. Step 4: Draw base HE = 5 cm. Step 5: Construct ∠H = 95° at H and ∠E = 85° at E. Step 6: Cut an arc of radius 6 cm from point E on its ray to mark point A (since EA = 6 cm). Step 7: Draw an arc of radius 6 cm from point H on its ray to mark point R (since HR = 6 cm). Join R and A. Final answer: Parallelogram HEAR is constructed with HE = 5 cm, EA = 6 cm, and verified angle ∠R = 85°.
• Q: Construct a Rectangle OKAY where OK = 7 cm and KA = 5 cm. A: Step 1: Recall that in a rectangle, opposite sides are equal, and all four interior angles are exactly 90°. Step 2: Therefore, OK = AY = 7 cm, KA = YO = 5 cm, and ∠O = ∠K = ∠A = ∠Y = 90°. Step 3: Draw base OK = 7 cm. Step 4: Construct ∠O = 90° at point O and ∠K = 90° at point K. Step 5: From point K, measure 5 cm along the constructed 90° ray to mark point A (KA = 5 cm). Step 6: From point O, measure 5 cm along the constructed 90° ray to mark point Y (OY = 5 cm). Join Y and A. Final answer: Rectangle OKAY is successfully constructed with adjacent sides of 7 cm and 5 cm, and all angles at 90°.
• Q: Can you construct a quadrilateral ABCD with AB = 5 cm, BC = 8 cm, ∠A = 120°, ∠B = 100°, and ∠C = 150°? Explain your reasoning. A: Step 1: Check the sum of the three given angles: ∠A + ∠B + ∠C = 120° + 100° + 150° = 370°. Step 2: The sum of all four interior angles of any quadrilateral must equal exactly 360°. Step 3: Since the sum of just three angles (370°) already exceeds the maximum permissible sum of all four angles (360°), the remaining angle would have to be negative, which is geometrically impossible. Final answer: No, such a quadrilateral cannot be constructed because the sum of the three given angles exceeds 360°.

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