DAV Class 7 Maths Chapter 10 Brain Teasers
Construction of Triangles Brain Teasers
1. A. Tick the correct option.
\( (a) \) A closed figure in a plane having three sides and three angles is known as— \( (i) \) Square \( (ii) \) Rectangle \( (iii) \) Triangle \( (iv) \) Pentagon
Answer \( (iii) \ \color{red} Triangle \)
\( (b) \) What are the possible dimensions for constructing a triangle of perimeter of 5 m? \( (i) \) 1m, 1 m, 3 m \( (ii) \) 1 m, 2 m, 2 m \( (iii) \) 5 m, 1.5 m, 3 m \( (iv) \) 1 m, 1.5 m, 2.5 m
Answer \( (ii) \ \color{red} 1 m, 2 m, 2 m \)
The sum of any two sides of a triangle is greater than the third side. \[ \begin{align*} 1 + 2 = 3 > 2 \\ 2 + 2 = 4 > 2 \\ 1 + 2 = 3 > 2 \\\end{align*} \]
B. Answer the following questions.
(a) How many independent components are required to construct a triangle?
\[ \begin{align*} SSS , SAS , ASA , RHS \end{align*} \]
Answer \( \color{red} Three \)
(b) Mr. Ajay wants to construct a triangular kitchen garden whose perimeter is 42 m. Is it possible to construct a triangle with dimensions 20 m, 12 m, 10 m? Why or why not?
Answer
\[ \begin{align*} 20 + 12 = 32 > 10 \\ 12 + 10 = 22 > 20 \\ 20 + 10 = 30 > 12 \\ \end{align*} \]
Yes it is possible to construct a triangle with dimensions \( 20 m, 12 m, 10 m \), because the sum of any two sides of a triangle is always greater than the third side.
(c) Rohan was asked to construct a triangle with dimensions 20 cm, 13 cm, 7 cm .But he was unable to construct. Give reasons.
Answer
\[ \begin{align*} 20 + 13 = 33 > 7 \\ 20 + 7 = 27 > 13 \\ 13 + 7 = 20 \cancel> 20 \\ \end{align*} \]
Triangle is not possible because the sum of two sides is not greater than the third side \( \color{red} 13 + 7 = 20 \cancel> 20 \).
(d) How wil you proceed to construct a triangle whose two angles and one side (not the included one) is given?
Answer We can find the third angle using the angle sum property and construct a triangle.
(e) How many equilateral triangles are needed to make a regular hexagon?
Answer \( \color{red} 6 \)
2. Construct a triangle ABC in which \( \angle A = 65^\circ , \angle B = 75^\circ \text{ and } BC = 4cm \).
Answer
\[ \begin{align*} In \ \triangle ABC \\ \angle A + \angle B + \angle C & = 180^\circ \\ 65^\circ + 75^\circ + \angle C & = 180^\circ \\ 140^\circ + \angle C & = 180^\circ \\ \angle C & = 180^\circ - 140^\circ \\ \angle C & = 40^\circ \end{align*} \]
3. Construct a triangle ABC with \( \angle C = 90^\circ , \angle A = 45^\circ , BC = 5 cm \).
Answer
\[ \begin{align*} In \ \triangle ABC \\ \angle A + \angle B + \angle C & = 180^\circ \\ 90^\circ + \angle B + 45^\circ & = 180^\circ \\ \angle B + 135^\circ & = 180^\circ \\ \angle B & = 180^\circ - 135^\circ \\ \angle B & = 45^\circ \end{align*} \]
4. Construct a triangle ABC in which AB = 4 cm, AC = 5.2 cm, and \( \angle A = 135^\circ \). Also bisect \( \angle A \text{ and } \angle B \)
Answer
5. Construct a triangle ABC in which AB = 6 cm, BC = 4.5 cm, and AC = 5.9 cm. Also construct perpendicular bisectors of sides AC and AB.
Answer
XY is perpendicular to side AB.
PQ is perpendicular to side AC.
6. Draw a triangle ABC, with AB = 2 cm, BC = 4 cm and CA = 3 cm. Draw a line through A, perpendicular to line BC.
Answer
7. Construct a right angled triangle in which sides containing the right angle are 3 cm and 4 cm. Measure the hypotenuse.
Answer
Hypotenuse = 5cm
8. Construct a right triangle ABC such that BC = 4cm and the hypotenuse makes an angle of \( 30^\circ \) at B. Verify that AB = 2AC.
Answer
\[ \begin{align*} AB &= 4.6cm \\ AC &= 2.3cm \\ 2AC &= 4.6cm \\ \therefore AB &= 2AC \end{align*} \]