NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Exercise 10.4

Understanding Angle-Side-Angle (ASA) Construction

To construct a unique triangle using the ASA (Angle-Side-Angle) criterion, the side must be the included side—the line segment joining the vertices of the two given angles. Before putting your pencil to paper, you must check the Angle Sum Property: the sum of the two given angles must be strictly less than 180 degrees. If the sum is 180 degrees or more, the lines will never meet to form a third vertex, making the triangle impossible to construct. Sometimes, CBSE questions give you two angles and a non-included side. In such cases, you must use the angle sum property of triangles ($Angle 1 + Angle 2 + Angle 3 = 180^\circ$) to find the correct angle at the end of your given side before starting your construction.

Step-by-Step Guide to Constructing a Triangle (ASA)

1. Draw a Rough Sketch — Always start by drawing a quick freehand sketch of the triangle. Label the given vertices, the known angles, and the given side length to visualize the layout.
2. Draw the Base Line Segment — Using a ruler and a sharp pencil, draw the given line segment (the included side) of the exact specified length. Label the endpoints.
3. Construct the First Angle — Place the center of your protractor on the left endpoint of the base segment. Align the base line of the protractor with your drawn segment, locate the required angle, mark it with a dot, and draw a long ray through it.
4. Construct the Second Angle — Place the protractor on the right endpoint. Align it correctly and mark the second required angle. Draw a ray extending from this endpoint.
5. Identify the Intersection Point — The point where the two rays cross is your third vertex. Label this vertex and erase any excess lines to complete your triangle.

The Angle-Sum Trap in CBSE Exams

Watch out for trick questions! An examiner might ask you to construct a triangle with angles like $110^\circ$ and $80^\circ$. Since $110^\circ + 80^\circ = 190^\circ$ (which is greater than $180^\circ$), a triangle cannot have these angles. In the exam, do not try to draw this. Simply write: 'The sum of the two given angles is greater than $180^\circ$, which violates the Angle Sum Property of a triangle. Therefore, this construction is not possible.'

Practice Questions with Solutions

• Q: Construct $\triangle ABC$ given $m\angle A = 60^\circ$, $m\angle B = 30^\circ$ and $AB = 5.8\text{ cm}$. A: Step 1: Draw a rough sketch of $\triangle ABC$ with $AB = 5.8\text{ cm}$ as the base, $\angle A = 60^\circ$, and $\angle B = 30^\circ$. Step 2: Draw a line segment $AB$ of length $5.8\text{ cm}$ using a ruler. Step 3: At point $A$, place the protractor and draw a ray $AX$ making an angle of $60^\circ$ with $AB$. Step 4: At point $B$, place the protractor and draw a ray $BY$ making an angle of $30^\circ$ with $BA$. Step 5: The rays $AX$ and $BY$ intersect at point $C$. Label the intersection point as $C$. Final answer: $\triangle ABC$ is constructed with the given measurements.
• Q: Construct $\triangle PQR$ if $PQ = 5\text{ cm}$, $m\angle PQR = 105^\circ$ and $m\angle QRP = 40^\circ$. A: Step 1: Analyze the given values. The side given is $PQ$. We are given $\angle Q = 105^\circ$ and $\angle R = 40^\circ$. We do not have $\angle P$ (the other angle on side $PQ$). Step 2: Apply the Angle Sum Property: $\angle P + \angle Q + \angle R = 180^\circ \implies \angle P + 105^\circ + 40^\circ = 180^\circ \implies \angle P = 180^\circ - 145^\circ = 35^\circ$. Step 3: Draw a line segment $PQ = 5\text{ cm}$. Step 4: At point $P$, draw a ray making an angle of $35^\circ$. Step 5: At point $Q$, draw a ray making an angle of $105^\circ$. Step 6: The intersection of these two rays is point $R$. Final answer: $\triangle PQR$ is constructed successfully by finding the third angle first.
• Q: Examine whether you can construct $\triangle DEF$ such that $EF = 7.2\text{ cm}$, $m\angle E = 110^\circ$ and $m\angle F = 80^\circ$. Justify your answer. A: Step 1: Write down the given angles: $m\angle E = 110^\circ$ and $m\angle F = 80^\circ$. Step 2: Calculate the sum of the two angles: $110^\circ + 80^\circ = 190^\circ$. Step 3: Recall the Angle Sum Property of triangles, which states that the sum of all three interior angles of a triangle must equal exactly $180^\circ$. Step 4: Compare: $190^\circ > 180^\circ$. Final answer: Since the sum of the two given angles ($190^\circ$) exceeds the maximum allowable sum of three angles in a triangle ($180^\circ$), the construction of $\triangle DEF$ is not possible.
• Q: Construct an isosceles triangle $XYZ$ where the equal angles are $m\angle X = m\angle Y = 45^\circ$ and the included side $XY = 6\text{ cm}$. A: Step 1: Draw a line segment $XY$ of length $6\text{ cm}$ using a ruler. Step 2: At point $X$, place your protractor and mark a point at $45^\circ$. Draw a ray $XP$ from $X$ through this point. Step 3: At point $Y$, place your protractor and mark a point at $45^\circ$. Draw a ray $YQ$ from $Y$ through this point. Step 4: Label the intersection point of the rays $XP$ and $YQ$ as $Z$. Final answer: $\triangle XYZ$ is the required isosceles triangle where $XZ = YZ$ because the base angles are equal.

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