Construction of Triangles — Questions and Answers
Welcome to HSLC Guru. This guide to ASSEB Class 7 New Mathematics Chapter 7, Construction of Triangles, gives step-by-step constructions (equilateral, SSS, SAS, ASA and altitudes) with inline SVG figures, explains triangle inequality, the angle sum property and the exterior angle theorem, and gives fully worked answers to every “Work it Out” box (7.1–7.8), the activities, the “Let us think” prompts and all of Exercise 7.
Summary
A triangle is a closed figure bounded by three line segments; the segments are its sides and each meeting point of two sides is a vertex. Drawing a triangle with only a ruler needs a lot of trial and error, but a compass and a protractor let us construct it accurately. In this chapter we construct a triangle when three sides are given (SSS), when two sides and the included angle are given (SAS), and when two angles and their common side are given (ASA).
The sum of the lengths of any two sides of a triangle is always greater than the third side — this is the triangle inequality, and a triangle can be drawn from three lengths only when it holds. Likewise, two given angles and their common side make a triangle only when the two angles add up to less than 180°.
The three angles of a triangle always add up to 180° (the angle sum property). An exterior angle formed by extending a side equals the sum of the two remote interior angles. The perpendicular drawn from a vertex to the opposite side is an altitude, and every triangle has three altitudes. Triangles are classified by sides as equilateral, isosceles or scalene, and by angles as acute-angled, right-angled or obtuse-angled.
Summary: This ASSEB Class 7 New Mathematics Chapter 7 (Construction of Triangles) guide shows, with clear numbered steps and inline SVG figures, how to construct a triangle by SSS, SAS and ASA, plus the equilateral triangle and the three altitudes. It explains triangle inequality, the angle sum property, the exterior angle theorem and the classification of triangles, and solves every Work it Out box (7.1–7.8), the activities, the Let us think prompts and all of Exercise 7.
Textbook Questions and Answers
Constructing an equilateral triangle (each side 4 cm)
Construct an equilateral triangle each of whose sides measures 4 cm, using a compass.
Answer:
- Step 1: With the ruler, draw the base line segment $\overline{AB}$ of length 4 cm.
- Step 2: With A as centre, draw an arc of radius 4 cm.
- Step 3: With B as centre, draw another arc of radius 4 cm. Mark the point where the two arcs cut as C.
- Step 4: Join $\overline{AC}$ and $\overline{BC}$ with the ruler.
In the triangle so formed, AB = BC = CA = 4 cm, so $\triangle ABC$ is the required equilateral triangle. AC and BC always equal AB because both arcs have radius 4 cm — that is, C lies 4 cm from both A and B.
Construction when three sides are given (SSS)
In $\triangle ABC$, AB = 4 cm, AC = 5 cm and BC = 6 cm. Construct the triangle.
Answer:
- Step 1: Draw the base $\overline{AB}$ of length 4 cm.
- Step 2: With A as centre, draw an arc of radius 5 cm (every point of this arc is 5 cm from A).
- Step 3: With B as centre, draw an arc of radius 6 cm.
- Step 4: Let C be the intersection of the two arcs. Join $\overline{AC}$ and $\overline{BC}$.
Now AC = 5 cm, BC = 6 cm, AB = 4 cm, so $\triangle ABC$ is the required triangle. All three sides are different, so it is a scalene triangle. (Drawing sides 4, 4, 6 cm instead gives two equal sides — an isosceles triangle.)
Work it Out 7.1
1. Write the sides and angles of $\triangle XYZ$.
Answer: Sides — $\overline{XY}$, $\overline{YZ}$ and $\overline{ZX}$. Angles — $\angle X$ (that is $\angle YXZ$), $\angle Y$ ($\angle XYZ$) and $\angle Z$ ($\angle YZX$).
2. Write the name of the side of $\triangle PQR$ opposite to the vertex Q.
Answer: $\overline{PR}$ (the side that does not contain Q).
3. Write the name of the angle of $\triangle LMN$ opposite to the side LN.
Answer: $\angle M$ (that is $\angle LMN$) — the vertex M is not on side LN.
4. Write the name of the vertex of $\triangle RST$ opposite to the side RT.
Answer: Vertex S (the vertex not on side RT).
Work it Out 7.2
1. The side lengths (in cm) of triangles are given below. Construct them with ruler and compass — (i) 3.5, 4.5, 5.5 (ii) 5, 5, 6 (iii) 7, 7, 7 (iv) 3, 3, 4 (v) 6, 8, 10.
Answer: Use the SSS method shown above in each case — draw the longest side as the base, then draw arcs of radii equal to the other two sides from its ends and mark their intersection as the third vertex. In every case the sum of any two sides is greater than the third, so all of them can be drawn.
2. Classify the above triangles as equilateral, isosceles or scalene.
| Side lengths | Type |
|---|---|
| (i) 3.5, 4.5, 5.5 | Scalene (all different) |
| (ii) 5, 5, 6 | Isosceles (two equal) |
| (iii) 7, 7, 7 | Equilateral (all equal) |
| (iv) 3, 3, 4 | Isosceles (two equal) |
| (v) 6, 8, 10 | Scalene (also right-angled, since 6² + 8² = 10²) |
3. Each circle has radius 2 cm. Joining the points as shown in Figures 1, 2 and 3, draw equilateral triangles. How many can you draw in each figure? Is there a relationship between the number of triangles and the number of circles? How many for 6 circles?
Answer: Joining the two centres of two overlapping circles to one of their intersection points gives a triangle whose three sides all equal the radius (2 cm), so it is equilateral. Because two circles meet in two points, Figure 1 gives 2 equilateral triangles. Each extra overlapping circle adds more such triangles, so the number of equilateral triangles increases steadily with the number of circles (about two for each adjacent pair in a row). Applying the same relationship lets us find how many equilateral triangles arise with 6 circles.
4. Draw three isosceles triangles inside a circle with one vertex at the centre and the other two on the circle.
Answer: Mark three pairs of points on the circle and join each pair to the centre. In each triangle two sides are radii of the circle, hence equal, so each is an isosceles triangle.
Activity (Triangle inequality)
Groups A, B, C, D and E are given sticks of lengths 3, 4, 5; 2, 3, 5; 3, 4, 8; 4, 6, 8 and 3, 5, 9 cm. Which groups can make a triangle?
Answer: Only groups A (3 + 4 > 5) and D (4 + 6 > 8) can make a triangle. Groups B (2 + 3 = 5), C (3 + 4 = 7 < 8) and E (3 + 5 = 8 < 9) cannot, because the sum of two sticks is not greater than the third. This shows that a triangle can be drawn only when the sum of any two sides is greater than the third side — the triangle inequality.
Work it Out 7.3
1. Give at least three examples of three side lengths with which no triangle can be constructed.
Answer: Any set where two sides do not add up to more than the third — for example (2, 3, 5) [2 + 3 = 5], (1, 2, 4) [1 + 2 < 4] and (3, 4, 9) [3 + 4 < 9].
2. Since 1, 2, 3 cannot form a triangle, is there any other set of three consecutive integers that cannot be used as side lengths?
Answer: No. For three consecutive integers $n$, $n+1$, $n+2$, the two smaller ones sum to $2n+1$; this exceeds $n+2$ when $2n+1 > n+2$, i.e. $n > 1$. So every set of three consecutive positive integers except 1, 2, 3 (where $n = 1$) forms a triangle; only 1, 2, 3 is an exception.
3. Can a triangle be drawn with the following side lengths? Examine.
- (i) 120 mm, 10 cm 20 mm, 0.12 m → all equal 12, 12, 12 cm; 12 + 12 > 12, so yes (equilateral).
- (ii) 12.3, 10.7, 23.5 cm → 12.3 + 10.7 = 23.0 < 23.5, so no.
- (iii) 5, 19, 10 m → 5 + 10 = 15 < 19, so no.
- (iv) 5, 10, 25 m → 5 + 10 = 15 < 25, so no.
4. Pranjal moves from A to C. In the figure AB = 4 m, BC = 3 m and the direct path AC = 5 m. (i) What is the minimum distance he should cover? (ii) The maximum distance?
Answer: (i) Minimum = 5 m — going straight along AC. (ii) Maximum = 4 + 3 = 7 m — going via B (A → B → C). By the triangle inequality the direct side (5) is shorter than the sum of the other two (7).
5. O is any point inside $\triangle PQR$. Which are true? Justify — (i) OP + OQ > PQ (ii) OQ + OR > QR (iii) OR + OP > RP.
Answer: All three are true. O forms the triangles $\triangle OPQ$, $\triangle OQR$ and $\triangle ORP$; in each, by the triangle inequality the sum of two sides is greater than the third. Hence OP + OQ > PQ, OQ + OR > QR and OR + OP > RP.
6. Two sides of Gitanjali’s triangular flag are 8 units and 2 units. Using the triangle inequality, find the least length required for the third side.
Answer: The third side must be greater than the difference of the other two, i.e. more than 8 − 2 = 6 units. So the third side must be a little more than 6 units; taking a whole number, the least length is 7 units.
7. Check which of the following are side lengths (in cm) of a triangle — (i) 3, 6, 8 (ii) 10, 100, 102 (iii) 7, 8, 9 (iv) 10, 11, 12.
Answer: Check whether the two smaller sides add up to more than the third — (i) 3 + 6 = 9 > 8 ✓ (ii) 10 + 100 = 110 > 102 ✓ (iii) 7 + 8 = 15 > 9 ✓ (iv) 10 + 11 = 21 > 12 ✓. So all four represent sides of a triangle.
Construction when two sides and the included angle are given (SAS)
In $\triangle ABC$, AB = 5 cm, BC = 6.5 cm and $\angle B = 60°$. Construct the triangle.
Answer:
- Step 1: Draw the base $\overline{BC}$ of length 6.5 cm.
- Step 2: With a protractor, draw $\angle XBC = 60°$ at B.
- Step 3: With B as centre, draw an arc of radius 5 cm cutting ray BX at A.
- Step 4: Join $\overline{AC}$.
In the resulting $\triangle ABC$, AB = 5 cm, BC = 6.5 cm and $\angle B = 60°$ — the required triangle. Here B is the included (in-between) angle of the two given sides.
Work it Out 7.4
1. Two sides and the included angle are given — draw the triangles: (i) 6 cm, 7 cm, 75° (ii) 8 cm, 8 cm, 60° (iii) 5.5 cm, 7.5 cm, 80°.
Answer: Use the SAS method in each case — draw one side as the base, draw the given angle at its end with the protractor, cut off the second side’s length along that ray and join to the third vertex. Note that in (ii), with two sides 8, 8 cm and included angle 60°, the third side also becomes 8 cm, so it is an equilateral triangle.
2. Draw a right-angled triangle whose two sides forming the right angle are each 4.5 cm.
Answer: Here the included angle is 90°. Draw a 4.5 cm side, draw a $90°$ angle at one end, cut off 4.5 cm along that ray and join the two ends. This gives the required right-angled triangle (which is also isosceles).
3. Two sides of a triangle are 5.5 cm and 6.5 cm and the included angle is 120°. Draw the triangle.
Answer: Draw the 6.5 cm side as base, draw a $120°$ angle at one end, cut off 5.5 cm along that ray and join to the base. This gives the required triangle (obtuse-angled, since 120° is obtuse).
4. For what lengths of two sides and measure of the included angle can a triangle not be constructed? Explain.
Answer: A triangle cannot be constructed if the included angle is $180°$ or more. When the angle is $180°$ the two sides lie along one straight line, so no third vertex is obtained and no triangle is formed. The angle must be between $0°$ and $180°$ for a triangle to be drawn.
Construction when two angles and the common side are given (ASA)
In $\triangle ABC$, $\angle B = 60°$, $\angle C = 70°$ and BC = 5 cm. Construct the triangle.
Answer:
- Step 1: Draw the segment $\overline{BC}$ of length 5 cm.
- Step 2: Draw $\angle CBX = 60°$ at B and $\angle BCY = 70°$ at C.
- Step 3: Let rays BX and CY meet at A.
The resulting $\triangle ABC$ has $\angle B = 60°$, $\angle C = 70°$ and BC = 5 cm — the required triangle. Here BC is the common side of the two given angles.
Work it Out 7.5
1. Two angles and the common side are given — construct the triangles: (i) 65°, 55°, 7 cm (ii) 60°, 50°, 6 cm (iii) 45°, 75°, 7.5 cm.
Answer: Use the ASA method — draw the common side and draw the two given angles at its ends; the rays meet at the third vertex. In each case the two angles sum to less than 180° (120°, 110°, 120° respectively), so the triangles can be drawn; the third angles are 60°, 70° and 60° respectively.
Role of angles in construction and Let us think
Can a triangle be drawn with two angles 90° and 120° and included side 6 cm? What about 90°, 90° and 110°, 130°?
Answer: No. When 90° and 120° are drawn at the ends of the side, the two rays never meet, so no third vertex is obtained. The same happens for (90°, 90°) and (110°, 130°). The rule is: if both angles are greater than or equal to 90° (or the two angles add up to 180° or more), no triangle can be drawn for any length of the included side. For a triangle, two angles must always add up to less than 180°.
Let us think: In $\triangle ABC$, $\angle A = 30°$, $\angle B = 70°$, AB = 5 cm. Find the third angle $\angle ACB$. Does it change if AB is changed from 5 to 7 cm? What if $\angle A = 50°$, $\angle B = 110°$?
Answer: Draw $\overline{PQ}$ through C parallel to AB. By alternate angles, $\angle PCA = \angle A = 30°$ and $\angle QCB = \angle B = 70°$. Since $\angle PCA + \angle ACB + \angle QCB = 180°$ (straight angle), $\angle ACB = 180° − 30° − 70° = 80°$. Changing AB from 5 to 7 cm does not change the third angle (still 80°), because it depends only on the two given angles. But with $\angle A = 50°$, $\angle B = 110°$, we get $\angle ACB = 180° − 50° − 110° = 20°$ — so changing the angles changes the third angle.
Work it Out 7.6
1. Two angles of a triangle are given; find the third — (i) 70°, 80° (ii) 100°, 60° (iii) 53°, 45° (iv) 90°, 60°.
Answer: Third angle = 180° − (sum of the two given angles). (i) 180° − 150° = 30°; (ii) 180° − 160° = 20°; (iii) 180° − 98° = 82°; (iv) 180° − 150° = 30°.
2. Two angles of a triangle are equal, each 50°. Find the third angle and draw the triangle.
Answer: Third angle = $180° − (50° + 50°) = 80°$. As two angles are equal it is isosceles; draw it by ASA using the two 50° base angles and the 80° top angle.
3. A triangle has two equal base angles and a top angle of 70°. Find each base angle and draw the triangle.
Answer: The two base angles add up to $180° − 70° = 110°$; being equal, each = $110° ÷ 2 = 55°$. So each base angle is 55°.
4. In the figure AB ∥ CD, $\angle ACD = 60°$ and $\angle ABC = 70°$. Find $\angle ACB$.
Answer: AB ∥ CD with AC as transversal, so by alternate angles $\angle BAC = \angle ACD = 60°$. In $\triangle ABC$, $\angle BAC + \angle ABC + \angle ACB = 180°$, i.e. $60° + 70° + \angle ACB = 180°$, giving $\angle ACB = 50°$.
Angle sum property and Activity
Is the sum of the three angles of a triangle always the same? Take three copies (x, y, z) of $\triangle ABC$ and arrange them side by side.
Answer: Draw $\overline{PQ}$ through vertex C parallel to AB. By alternate angles $\angle PCA = \angle CAB$ and $\angle QCB = \angle ABC$. Since $\angle PCA + \angle BCA + \angle QCB = 180°$ (straight angle), $\angle CAB + \angle BCA + \angle ABC = 180°$, i.e. $\angle A + \angle B + \angle C = 180°$. When the three angles $\angle 1$, $\angle 2$, $\angle 3$ of the three copies are placed side by side, they form a straight angle (180°) — proving the angle sum property.
Work it Out 7.7
1. Find the value of the unknown angle $x$ in the following triangles.
Answer: Use the angle sum property (three angles = 180°) in each —
- (a) $\angle B = 50°$, $\angle C = 60°$; so $x = \angle A = 180° − 50° − 60° = 70°$.
- (b) $\angle P = 90°$, $\angle Q = 30°$; so $x = \angle R = 180° − 90° − 30° = 60°$.
- (c) $\angle L = 45°$, $\angle M = 90°$; so $x = \angle N = 180° − 45° − 90° = 45°$.
- (d) $\angle A = 60°$, $\angle B = 60°$; so $x = \angle C = 180° − 60° − 60° = 60°$.
Let us think: Can a triangle have two right angles? Two obtuse angles? All three angles equal to 70°? All three angles less than 60°?
Answer: None is possible. Two right angles already add up to 180°, leaving no room for the third. Two obtuse angles add up to more than 180°. Three 70° angles give 210° ≠ 180°. Three angles each less than 60° give a sum less than 180° — also impossible. So none of these can occur.
Exterior angle
In $\triangle ABC$, AB is extended to D. Find the exterior angle $\angle CBD$ if $\angle A = 50°$, $\angle B = 70°$, $\angle C = 60°$.
Answer: $\angle ABC + \angle CBD = 180°$ (straight angle), so $70° + \angle CBD = 180°$, giving $\angle CBD = 110°$. Notice that the exterior angle $\angle CBD = 110° = \angle A + \angle C = 50° + 60°$ — the exterior angle equals the sum of the two remote interior angles.
Work it Out 7.8
1. The interior angles of a triangle are given; find the three exterior angles — (i) 60°, 60°, 60° (ii) 70°, 80°, 30° (iii) 45°, 45°, 90° (iv) 20°, 40°, 120°.
Answer: Each exterior angle = 180° − the corresponding interior angle (= sum of the two remote interior angles). (i) 120°, 120°, 120°; (ii) 110°, 100°, 150°; (iii) 135°, 135°, 90°; (iv) 160°, 140°, 60°.
2. Find the sum of the three exterior angles in each case above. Can you formulate a rule?
Answer: In every case the sum is 360° — (i) 120+120+120 = 360°; (ii) 110+100+150 = 360°; (iii) 135+135+90 = 360°; (iv) 160+140+60 = 360°. So the rule is: the sum of the three exterior angles of a triangle is always 360°.
3. In the figure, find the measures of the angles $x$, $y$, $z$ (given angles 30°, 60°, 30° and 40°).
Answer: The intersecting lines form several triangles; apply the exterior angle theorem (an exterior angle = sum of the two remote interior angles) together with the angle sum property (180°). Using the exterior-angle relation, $x = 30° + 40° = 70°$ and $y = 60° + 40° = 100°$; and in the bottom triangle formed by the two transversals, $z = 180° − 30° − 40° = 110°$ (the angle next to the 40° angle).
Construction of altitudes
How is the altitude from vertex A to the base BC of $\triangle ABC$ constructed?
- Step 1: Take BC as the base and align the ruler along it; A is the vertex opposite the base.
- Step 2: Place the protractor on the ruler and slide it until its 90° mark touches vertex A.
- Step 3: Draw $\overline{AD}$ from A to the point D where the 90° mark meets the base.
This segment $\overline{AD} \perp BC$ is the altitude of $\triangle ABC$, and its length is the height. Since a triangle has three vertices, it has three altitudes. In an obtuse-angled triangle some altitudes require the base to be extended, so those altitudes lie outside the triangle.
Exercise 7
1. Classify the following triangles as equilateral, isosceles, scalene, right-angled, acute-angled or obtuse-angled.
| Triangle | Given measures | Type |
|---|---|---|
| (i) ABC | AB = 5, BC = 8, CA = 8 cm | Isosceles (and acute-angled) |
| (ii) XYZ | XY = 10, YZ = 6, ZX = 8 cm | Scalene and right-angled (6² + 8² = 10²) |
| (iii) LMN | $\angle L = 20°, \angle M = 40°, \angle N = 120°$ | Scalene and obtuse-angled |
| (iv) ABC | each side 5.2 cm | Equilateral (acute-angled) |
| (v) PQR | PQ = 7, QR = 5, RP = 6 cm | Scalene and acute-angled |
| (vi) RST | $\angle R = 80°, \angle S = 55°, \angle T = 45°$ | Scalene and acute-angled |
2. Will an altitude always lie inside a triangle? If not, draw a simple sketch to establish it.
Answer: No. In an acute-angled triangle all altitudes lie inside, but in an obtuse-angled triangle the altitude from an acute-angle vertex falls outside. In the sketch, drawing the altitude from P requires extending QR to T, so the altitude PT lies outside the triangle.
3. How many altitudes can a triangle have? Draw a triangle with sides 4 cm, 5 cm, 6 cm and draw all its altitudes.
Answer: A triangle has three altitudes (one from each vertex). First draw the triangle by SSS, taking 6 cm as the base and cutting arcs of 4 cm and 5 cm; then, from each vertex, drop a perpendicular to the opposite side with a protractor to get the three altitudes. The three altitudes meet at a common point (the orthocentre).
4. In $\triangle ABC$, $\angle A = 90°$, AB = 6 cm, AC = 8 cm. Draw the triangle. Identify the altitude from vertex C to side AB and draw the other two altitudes as well.
Answer: Since $\angle A = 90°$, AB and AC are perpendicular; draw the triangle by SAS with 90° between the 6 cm and 8 cm sides (now BC = 10 cm). The altitude from C to AB is the side $\overline{CA}$ itself (as CA is already perpendicular to AB). Similarly the altitude from B to AC is $\overline{BA}$; and the third altitude, from A perpendicular to the hypotenuse BC, lies inside the triangle.
5. Match the columns.
Answer: (i) $\angle A = 30°, \angle B = 70°$ → $\angle C = 80°$ → (b) 80°. (ii) $\angle P = 100°, \angle Q = 60°$ → $\angle R = 20°$ → (c) 20°. (iii) $\angle x = 90°, \angle y = 45°$ → $\angle z = 45°$ → (a) 45°.
6. State true or false.
Answer: (i) The altitude always lies inside — False (it lies outside for an obtuse triangle). (ii) Only one altitude can be drawn from a vertex to its opposite side — True. (iii) 40°, 50°, 60° can be the angles of a triangle — False (sum 150° ≠ 180°). (iv) A scalene triangle can have two equal sides — False (a scalene triangle has all sides unequal). (v) Only one unique circle passes through the three vertices — True.
7. Choose the correct option. Assertion (A): The angles of a triangle are 30°, 60°, 90°. Reason (R): The sum of the angles of a triangle is 180°.
Answer: (A) — both A and R are true and R is the correct explanation of A, because $30° + 60° + 90° = 180°$ is exactly why these can be the angles of a triangle.
8. Fill in the gaps.
Answer: (i) The three altitudes of a triangle intersect at one point. (ii) The sum of the three exterior angles is 360°. (iii) The three side lengths always obey the triangle inequality. (iv) The three sides of a scalene triangle are all unequal (different). (v) The exterior angle of a triangle is equal to the sum of the two remote interior angles.
9. (i) In the figure AB ∥ DE; find $x$, $y$, $z$ (given 60° at D and 48° at A).
Answer: AB ∥ DE and the lines cross at C. For transversal AE, alternate angles give $x = \angle A = 48°$; for transversal DB, alternate angles give $y = \angle D = 60°$. Then in $\triangle DCE$, $z = 180° − 60° − 48° = 72°$. So $x = 48°$, $y = 60°$, $z = 72°$.
(ii) In the figure find $x$ and $y$ (at C, $\angle ACB = 30°$; $\angle A = 50°$, $\angle D = 55°$; CB is the inner segment).
Answer: In the left triangle ABC, $\angle A = 50°$, $\angle ACB = 30°$, so $y = \angle ABC = 180° − 50° − 30° = 100°$. As it is a straight line, $\angle DBC = 180° − 100° = 80°$. In the right triangle BCD, $x = \angle BCD = 180° − 80° − 55° = 45°$. So $x = 45°$, $y = 100°$.
(iii) In the figure (a five-pointed star), find $\angle A + \angle B + \angle C + \angle D + \angle E$.
Answer: The sum of the five tip angles of a five-pointed star is always 180°. So $\angle A + \angle B + \angle C + \angle D + \angle E = 180°$.
Additional Questions and Answers
Multiple Choice Questions (MCQ)
1. Each angle of an equilateral triangle is — (a) 45° (b) 50° (c) 60° (d) 90°
Answer: (c) 60°.
2. In the SSS construction with a compass, which side is best taken as base? (a) the shortest (b) the longest (c) the middle one (d) any side
Answer: (b) the longest side.
3. The sum of the three angles of a triangle is — (a) 90° (b) 180° (c) 270° (d) 360°
Answer: (b) 180°.
4. The sum of the three exterior angles of a triangle is — (a) 180° (b) 270° (c) 360° (d) 540°
Answer: (c) 360°.
5. With which set of lengths can no triangle be drawn? (a) 3, 4, 5 (b) 5, 6, 7 (c) 2, 3, 5 (d) 6, 8, 10 (cm)
Answer: (c) 2, 3, 5 (since 2 + 3 = 5, not greater).
6. In an SAS construction the “angle” must be — (a) any angle (b) the included (in-between) angle of the two sides (c) opposite the hypotenuse (d) an exterior angle
Answer: (b) the included (in-between) angle of the two sides.
7. The perpendicular from a vertex to the opposite side is called the — (a) median (b) altitude (c) bisector (d) diagonal
Answer: (b) altitude.
8. In which triangle can an altitude lie outside it? (a) acute-angled (b) right-angled (c) obtuse-angled (d) equilateral
Answer: (c) obtuse-angled triangle.
9. A triangle with one angle of 100° is — (a) acute-angled (b) right-angled (c) obtuse-angled (d) equilateral
Answer: (c) obtuse-angled (100° > 90°).
10. For a triangle to be constructed, the sum of two given angles must be — (a) exactly 180° (b) less than 180° (c) more than 180° (d) exactly 90°
Answer: (b) less than 180°.
Fill in the blanks
1. The sum of any two sides of a triangle is ______ than the third side.
Answer: greater.
2. A triangle with all three sides equal is called an ______ triangle.
Answer: equilateral.
3. A triangle with one angle greater than 90° is called an ______ triangle.
Answer: obtuse-angled.
4. The side opposite the right angle is called the ______.
Answer: hypotenuse.
5. For an SAS construction, along with two sides we must know their ______ angle.
Answer: included (in-between).
True or False
1. A triangle can have two right angles.
Answer: False (two right angles already make 180°).
2. Each angle of an equilateral triangle is 60°.
Answer: True.
3. A triangle has three altitudes.
Answer: True.
4. An exterior angle is always smaller than its adjacent interior angle.
Answer: False (an exterior angle and its adjacent interior angle add up to 180°; the exterior angle equals the sum of the two remote interior angles).
5. A right-angled triangle can be formed with sides 6, 8, 10 cm.
Answer: True (6² + 8² = 10²).
Short Answer Questions
1. What is the triangle inequality? Explain with an example.
Answer: The sum of the lengths of any two sides of a triangle is always greater than the third side. For example, with 3, 4, 5 cm, 3 + 4 = 7 > 5, so a triangle forms; but with 2, 3, 5 cm, 2 + 3 = 5 (not greater), so no triangle can be formed.
2. What is the difference between SAS and ASA construction?
Answer: In SAS two sides and their included angle are given, drawn in the side-angle-side order. In ASA two angles and their common side are given; drawing the two angles at the ends of that side, the rays meet at the third vertex.
3. State the exterior angle property.
Answer: The measure of any exterior angle of a triangle equals the sum of its two remote (opposite) interior angles. For example, if $\angle A = 50°$ and $\angle C = 60°$, the exterior angle at B is $\angle CBD = 50° + 60° = 110°$.
4. Why does the altitude of an obtuse-angled triangle lie outside it?
Answer: An obtuse-angled triangle has one angle greater than 90°. To drop a perpendicular from a vertex next to the obtuse angle onto the opposite side, that side must be extended beyond the triangle, so the altitude falls outside the triangle.
Key Terms
| Term | Meaning |
|---|---|
| Construction of triangle | Drawing a triangle accurately with ruler, compass and protractor |
| Base | The side on which the triangle is taken to stand |
| Arc | Part of a circle drawn with a compass |
| Radius | Distance from the centre to the circle |
| Included angle | The angle between two given sides |
| Common side | The shared side of two given angles |
| Triangle inequality | Sum of any two sides is greater than the third |
| Exterior angle | Angle formed outside when a side is extended |
| Altitude | Perpendicular from a vertex to the opposite side |
| Hypotenuse | The longest side, opposite the right angle |
| Acute-angled triangle | All three angles less than 90° |
| Obtuse-angled triangle | One angle greater than 90° |