CLASS 7 · GANITA PRAKASH PART I · CHAPTER 7
Construct triangles and use side and angle rules to understand when they exist, how their angles add up, and how to draw altitudes.
43 worked answers · short steps first, explanations when you need them
Textbook pages 147–151
Question 1
Draw a 4 cm base AB.
Draw radius-4 cm arcs centred at A and B.
Join either arc intersection C to A and B.
AC=AB=BC=4 cm, so ABC is equilateral. The compass guarantees both remaining sides have the required length.
Question 2
Draw AB=4 cm.
Draw an arc of radius 5 cm centred at A and an arc of radius 6 cm centred at B.
Join an intersection point C to A and B.
The intersection C gives AC=5 cm and BC=6 cm, so the required triangle is formed.
Question 3
For (a)–(d), draw the longest side as the base. From its endpoints, draw arcs with radii equal to the two other sides.
Where the arcs meet is the third vertex. Join it to both ends of the base. For (e), use the equal-radius arc construction from question 1.
All five constructions are possible. The arc radii are (a) 4,4 cm; (b) 3,4 cm; (c) 1,5 cm; (d) 4,6 cm. In (e), draw all three sides as 3.5 cm.
Question 4
Choose two points on the circle and use the centre as the third vertex.
Both segments from the centre to the circle have length equal to the radius.
If O is the centre and P,Q lie on the circle, OP=OQ, so ΔOPQ is isosceles.
Question 5
Let C be an intersection. AC and BC are radii, equal to each other and to AB.
Repeat for the other intersection.
ΔABC and the corresponding triangle at the other intersection are equilateral; connecting one centre to points on the other circle also gives isosceles triangles.
Textbook pages 151–159
Question 1
Between two fixed points, a straight segment is the shortest path.
The route through the third vertex is a two-segment path.
Each triangle side is less than the sum of the other two sides.
Question 2
Compare the longest length with the sum of the other two.
No in every case: 8>3+4; 6>2+3; 30>10+15; 7>3+3.
Question 3
Check the largest side against the sum of the two smaller sides.
None exists: 25>20; 20>15; 40>32.
Question 4
Order the lengths x≤y≤z.
The largest z is greater than or equal to each smaller one, so x<y+z and y<x+z.
Yes. The two smaller lengths are each less than the sum of the other two; the largest-side comparison decides whether a triangle exists.
Question 5
Only the longest-side comparison needs checking.
Possible: (b), (d), (e), (g). Impossible: (a), because 5>4; (c), because 8>6; (f), because 35>30.
Question 6
Compare the longest side with the sum of the other two.
The comparison must be strict.
(a) yes, 100<101; (b) no, 9=3+6 gives a straight line, not a triangle; (c) no, 5>2; (d) yes, 12<15.
Question 7
For three equal sides, check s< s+s.
Yes. For every s>0, s<2s, so an equilateral triangle exists.
Question 8
For sides a≤b and third side x, require b−a<x<a+b.
(a) 99.5, 99.4, 100, 100.5, 100.7; range 99<x<101. (b) 1, 3, 4, 4.9, 5; range 0<x<10. (c) 5, 6, 6.4, 7, 8; range 4<x<10.
Textbook pages 160–164
Question 1
Draw AB=5 cm.
Construct a 45° ray at A.
Mark C 4 cm from A on the ray and join BC.
The construction gives the required triangle.
Question 2
At 180°, the two side rays lie on a straight line.
Above 180°, the included angle is not a convex interior angle of a triangle.
No. An included angle must be strictly between 0° and 180°.
Question 3
Each included angle is strictly between 0° and 180°.
Draw the two sides on rays making that angle.
All three are constructible.
Question 4
Draw the 5 cm base.
At its endpoints, draw the two interior angle rays.
Their intersection is the third vertex.
The rays meet because 45°+80°<180°, making the triangle’s third angle 55°.
Question 5
Add the two given angles.
A triangle exists exactly when their sum is less than 180°.
(a) yes, third angle 30°; (b) yes, third angle 95°; (c) yes, third angle 30°.
Question 6
Possible when 0°<B<180°−A.
Impossible when B≥180°−A or B≤0°.
Examples: (a) possible B=50°,90°; impossible 150°,170°. (b) possible 60°,80°; impossible 110°,140°. (c) possible 90°,64°; impossible 126°,134°. (d) possible 20°,35°; impossible 36°,90°.
Question 7
Add the pair; it must be less than 180°.
(a) no, sum 185°; (b) yes, third angle 80°; (c) yes, third angle 5°; (d) no, sum 200°.
Question 8
The third angle is 180°−(A+B).
It must be positive.
A>0°, B>0° and A+B<180°.
Textbook pages 164–172
Question 1
Subtract the two given angles from 180°.
(a) 72°; (b) 15°; (c) 60°; (d) 60°.
Question 2
Three 70° angles total 210°, which is too large.
With two 70° angles, subtract 140° from 180°.
For equal angles, divide 180° by 3.
No; with two angles 70°, the third is 40°; if all angles are equal, each is 60°.
Question 3
The two equal angles share 180°−50°=130°.
Divide by 2.
∠B=∠C=65°.
Question 4
Draw a line through one vertex parallel to the opposite side.
The other two interior angles equal the alternate angles on this line.
These two angles and the vertex angle form a straight angle.
∠A+∠B+∠C=180°.
Question 5
Interior ∠C=180°−50°−60°=70°.
The exterior angle forms a linear pair with ∠C.
Exterior angle=110°=∠A+∠B. In general, a triangle’s exterior angle equals the sum of the two opposite interior angles.
Question 6
From a vertex, draw a perpendicular to the line containing the opposite side.
If needed, extend that opposite side.
An altitude is the perpendicular segment from a vertex to the opposite side’s line. For an obtuse triangle, an altitude may meet the extension outside the triangle.
Question 7
Align a ruler with BC.
Slide a set square until its perpendicular edge passes through A.
Draw the perpendicular from A to BC (or its extension).
The resulting segment from A meets BC at 90° and is the altitude.
Question 8
Draw BC=5 cm. Draw an arc of radius 6 cm centred at B and an arc of radius 5 cm centred at C.
Mark an intersection as A, join AB and AC, then draw a perpendicular from A to BC.
The construction is possible. The altitude from A meets BC at 90°. Since AC=BC, the triangle is isosceles at C, but this altitude is not its symmetry altitude.
Question 9
Choose any acute ∠A between 0° and 90°.
Then ∠C=90°−∠A and the hypotenuse remains 5 cm.
Infinitely many: the acute angle can vary continuously.
Question 10
Use equality of side lengths for the first three classes.
Use the largest angle for the angle-based classes.
Equilateral: 3 equal sides; isosceles: 2 equal sides; scalene: no equal sides. Acute: all angles <90°; right: one 90°; obtuse: one >90°. Equilateral cannot be right/obtuse (all angles are 60°). Isosceles can be right (45°,45°,90°) or obtuse (e.g. 40°,40°,100°).
Question 11
For a right isosceles triangle, split the remaining 90° equally.
For an obtuse isosceles triangle, choose equal acute base angles whose sum is below 90°.
Right: 45°,45°,90°. Obtuse: 40°,40°,100°.
Question 12
Unfold adjacent faces into a flat net.
Draw the straight segment between the two corners on the net.
Compare nets that unfold the box in different ways.
The shortest surface path is a straight line on a suitable unfolded net; its length depends on the box dimensions, so the exact route needs those dimensions.
Textbook pages 147–169
Question 4-5-8
Take 8 cm as the base. Draw arcs of radii 4 cm and 5 cm from its endpoints.
The radii add to 9 cm, which is more than the 8 cm distance between the centres, so the arcs cross.
Yes. Their intersection gives the third vertex, so the three lengths form a triangle.
Question circle-cases
Touching: 1,2,3; 2,3,5; 4,5,9 (the two smaller lengths add exactly to the longest).
Separate: 1,2,4; 2,3,6; 3,4,8 (the two smaller lengths add to less than the longest).
If their sum is greater than the longest and their difference is less than it, the circles cross.
Touching examples: (1,2,3), (2,3,5), (4,5,9); separate examples: (1,2,4), (2,3,6), (3,4,8). A triangle exists exactly when the longest side is strictly shorter than the sum of the other two.
Question 40-degree-threshold
At the limiting case, the second arm is parallel to the opposite base line.
Angles on the same side of a transversal between parallel lines add to 180°.
The limiting angle is 180°−40°=140°. At 140° the arms are parallel; above 140° they do not meet on the triangle side. The base length does not change this angle condition.
Question 60-70-check
The angles of a triangle add to 180°.
Subtract the given angles: 180°−60°−70°.
The third angle is 50°. Changing the included side changes the size of the triangle, but not this angle, because the other two angles are fixed.
Question fold-angle-sum
Cut out a triangle and fold its three corners so their tips meet along one straight line.
The three corner angles fit together without a gap or overlap to form a straight angle.
A straight angle is 180°, so the three interior angles total 180°.
Question fold-altitude
Fold the base so that its two ends lie on top of one another.
The crease is the perpendicular bisector of the segment joining those ends, so it meets the base at 90°.
The crease is perpendicular to the base; the part of the crease from the opposite vertex is the altitude.
Textbook pages 170
Question 1
Draw BC=5 cm. With centre B and radius 6 cm, draw an arc; with centre C and radius 5 cm, draw another arc.
Mark an intersection A and join AB and AC. Use a set square to draw from A a perpendicular to BC.
The construction gives AC=BC=5 cm and AB=6 cm. The altitude from A meets BC at 90°. The equal sides meet at C, so this is not the symmetry altitude.
Question 2
Draw RY=4 cm. At R, draw a 140° ray and mark T on it so RT=7 cm. Join T to Y.
Extend RY beyond R. From T, draw a perpendicular to the extended line RY.
The perpendicular from T meets the extension of RY on the side beyond R, so the altitude falls outside the triangle.
Question 3
Choose any angle A strictly between 0° and 90°.
Then angle C=90°−A, while AC remains 5 cm.
Infinitely many triangles are possible because angle A can take infinitely many values between 0° and 90°.
Question 4
In an equilateral triangle, all three angles are 60°.
For an isosceles right triangle, use angles 45°,45°,90°. For an isosceles obtuse triangle, use 40°,40°,100°.
An equilateral triangle cannot be right-angled or obtuse-angled. Isosceles examples are 45°,45°,90° (right) and 40°,40°,100° (obtuse).